CHEMISTRY SEMESTER REVIEW

LAST CALL FOR QUESTIONS

Do you remember how to identify the type of reaction

bull 2 H2O rarr 2 H2 + O2

ndash What type of reaction is this

Decomposition Reaction

bull C10H8 + 12 O2 rarr 10 CO2 + 4 H2O

ndash What type of reaction is this

Combustion Reaction

GRAMS to MOLECULES

bull How many molecules are there in 250g of NH3

bull What do we do first

bull First thing to do is find the molar mass of the compound Molar mass is the sum total of every atomrsquos mass in the compound

Whats the molar mass of NH3NITROGEN(N) = 1401gmolHYDROGEN(H) = 1008gmol

Whats the molar mass of NH3

bull N = 1401 g

bull H = 1008 g (times 3 because there are three hydrogen atoms in NH3)

bull The molar mass is 17034 gmole (or 17034 grams per 1 mole of NH3)

Next Step (250g of NH3)

bull Next once you know the total mass of your substance you can convert to moles by dividing the total mass of NH3 (250 g) by the molar mass (17034 gmol)

250 g 1 mol (of NH3)

17034 g1468 mol =

Finally the number of molecules can be calculated

To do this you must convert moles to molecules using Avogadrorsquos which is

602 X 1023

1468 mol602 x 1023

molecules

1 mol8837 x 1024 =

molecules

TYPES OF SYSTEMS

bull open systems can exchange mass and energybull closed systems allow the transfer of energy (he

at) but not massbull isolated systems do not allow transfer of either

mass or energy

HYDROGEN EXPLOSION - combustion

bull Is this reaction taking place in an open closed or isolated system

bull the reacting mixture is the system (hydrogen oxygen and water molecules) everything else is the surroundings

bull This is exothermic because it is a combustion reaction (thermal energy leaves during this and all other exothermic reaction -- a lot of thermal energy is released in this example ）

)(2)(2 222 lOHOgH

open system

Do you remember how to identify the type of reaction

bull 2 H2O rarr 2 H2 + O2

ndash What type of reaction is this

Decomposition Reaction

bull C10H8 + 12 O2 rarr 10 CO2 + 4 H2O

ndash What type of reaction is this

Combustion Reaction

GRAMS to MOLECULES

bull How many molecules are there in 250g of NH3

bull What do we do first

bull First thing to do is find the molar mass of the compound Molar mass is the sum total of every atomrsquos mass in the compound

Whats the molar mass of NH3NITROGEN(N) = 1401gmolHYDROGEN(H) = 1008gmol

Whats the molar mass of NH3

bull N = 1401 g

bull H = 1008 g (times 3 because there are three hydrogen atoms in NH3)

bull The molar mass is 17034 gmole (or 17034 grams per 1 mole of NH3)

Next Step (250g of NH3)

bull Next once you know the total mass of your substance you can convert to moles by dividing the total mass of NH3 (250 g) by the molar mass (17034 gmol)

250 g 1 mol (of NH3)

17034 g1468 mol =

Finally the number of molecules can be calculated

To do this you must convert moles to molecules using Avogadrorsquos which is

602 X 1023

1468 mol602 x 1023

molecules

1 mol8837 x 1024 =

molecules

TYPES OF SYSTEMS

bull open systems can exchange mass and energybull closed systems allow the transfer of energy (he

at) but not massbull isolated systems do not allow transfer of either

mass or energy

HYDROGEN EXPLOSION - combustion

bull Is this reaction taking place in an open closed or isolated system

bull the reacting mixture is the system (hydrogen oxygen and water molecules) everything else is the surroundings

bull This is exothermic because it is a combustion reaction (thermal energy leaves during this and all other exothermic reaction -- a lot of thermal energy is released in this example ）

)(2)(2 222 lOHOgH

open system

GRAMS to MOLECULES

bull How many molecules are there in 250g of NH3

bull What do we do first

bull First thing to do is find the molar mass of the compound Molar mass is the sum total of every atomrsquos mass in the compound

Whats the molar mass of NH3NITROGEN(N) = 1401gmolHYDROGEN(H) = 1008gmol

Whats the molar mass of NH3

bull N = 1401 g

bull H = 1008 g (times 3 because there are three hydrogen atoms in NH3)

bull The molar mass is 17034 gmole (or 17034 grams per 1 mole of NH3)

Next Step (250g of NH3)

bull Next once you know the total mass of your substance you can convert to moles by dividing the total mass of NH3 (250 g) by the molar mass (17034 gmol)

250 g 1 mol (of NH3)

17034 g1468 mol =

Finally the number of molecules can be calculated

To do this you must convert moles to molecules using Avogadrorsquos which is

602 X 1023

1468 mol602 x 1023

molecules

1 mol8837 x 1024 =

molecules

TYPES OF SYSTEMS

bull open systems can exchange mass and energybull closed systems allow the transfer of energy (he

at) but not massbull isolated systems do not allow transfer of either

mass or energy

HYDROGEN EXPLOSION - combustion

bull Is this reaction taking place in an open closed or isolated system

bull the reacting mixture is the system (hydrogen oxygen and water molecules) everything else is the surroundings

bull This is exothermic because it is a combustion reaction (thermal energy leaves during this and all other exothermic reaction -- a lot of thermal energy is released in this example ）

)(2)(2 222 lOHOgH

open system

Whats the molar mass of NH3NITROGEN(N) = 1401gmolHYDROGEN(H) = 1008gmol

Whats the molar mass of NH3

bull N = 1401 g

bull H = 1008 g (times 3 because there are three hydrogen atoms in NH3)

bull The molar mass is 17034 gmole (or 17034 grams per 1 mole of NH3)

Next Step (250g of NH3)

bull Next once you know the total mass of your substance you can convert to moles by dividing the total mass of NH3 (250 g) by the molar mass (17034 gmol)

250 g 1 mol (of NH3)

17034 g1468 mol =

Finally the number of molecules can be calculated

To do this you must convert moles to molecules using Avogadrorsquos which is

602 X 1023

1468 mol602 x 1023

molecules

1 mol8837 x 1024 =

molecules

TYPES OF SYSTEMS

bull open systems can exchange mass and energybull closed systems allow the transfer of energy (he

at) but not massbull isolated systems do not allow transfer of either

mass or energy

HYDROGEN EXPLOSION - combustion

bull Is this reaction taking place in an open closed or isolated system

bull the reacting mixture is the system (hydrogen oxygen and water molecules) everything else is the surroundings

bull This is exothermic because it is a combustion reaction (thermal energy leaves during this and all other exothermic reaction -- a lot of thermal energy is released in this example ）

)(2)(2 222 lOHOgH

open system

Whats the molar mass of NH3

bull N = 1401 g

bull H = 1008 g (times 3 because there are three hydrogen atoms in NH3)

bull The molar mass is 17034 gmole (or 17034 grams per 1 mole of NH3)

Next Step (250g of NH3)

bull Next once you know the total mass of your substance you can convert to moles by dividing the total mass of NH3 (250 g) by the molar mass (17034 gmol)

250 g 1 mol (of NH3)

17034 g1468 mol =

Finally the number of molecules can be calculated

To do this you must convert moles to molecules using Avogadrorsquos which is

602 X 1023

1468 mol602 x 1023

molecules

1 mol8837 x 1024 =

molecules

TYPES OF SYSTEMS

bull open systems can exchange mass and energybull closed systems allow the transfer of energy (he

at) but not massbull isolated systems do not allow transfer of either

mass or energy

HYDROGEN EXPLOSION - combustion

bull Is this reaction taking place in an open closed or isolated system

bull the reacting mixture is the system (hydrogen oxygen and water molecules) everything else is the surroundings

bull This is exothermic because it is a combustion reaction (thermal energy leaves during this and all other exothermic reaction -- a lot of thermal energy is released in this example ）

)(2)(2 222 lOHOgH

open system

Next Step (250g of NH3)

bull Next once you know the total mass of your substance you can convert to moles by dividing the total mass of NH3 (250 g) by the molar mass (17034 gmol)

250 g 1 mol (of NH3)

17034 g1468 mol =

Finally the number of molecules can be calculated

To do this you must convert moles to molecules using Avogadrorsquos which is

602 X 1023

1468 mol602 x 1023

molecules

1 mol8837 x 1024 =

molecules

TYPES OF SYSTEMS

bull open systems can exchange mass and energybull closed systems allow the transfer of energy (he

at) but not massbull isolated systems do not allow transfer of either

mass or energy

HYDROGEN EXPLOSION - combustion

bull Is this reaction taking place in an open closed or isolated system

bull the reacting mixture is the system (hydrogen oxygen and water molecules) everything else is the surroundings

bull This is exothermic because it is a combustion reaction (thermal energy leaves during this and all other exothermic reaction -- a lot of thermal energy is released in this example ）

)(2)(2 222 lOHOgH

open system

Finally the number of molecules can be calculated

To do this you must convert moles to molecules using Avogadrorsquos which is

602 X 1023

1468 mol602 x 1023

molecules

1 mol8837 x 1024 =

molecules

TYPES OF SYSTEMS

bull open systems can exchange mass and energybull closed systems allow the transfer of energy (he

at) but not massbull isolated systems do not allow transfer of either

mass or energy

HYDROGEN EXPLOSION - combustion

bull Is this reaction taking place in an open closed or isolated system

bull the reacting mixture is the system (hydrogen oxygen and water molecules) everything else is the surroundings

bull This is exothermic because it is a combustion reaction (thermal energy leaves during this and all other exothermic reaction -- a lot of thermal energy is released in this example ）

)(2)(2 222 lOHOgH

open system

TYPES OF SYSTEMS

bull open systems can exchange mass and energybull closed systems allow the transfer of energy (he

at) but not massbull isolated systems do not allow transfer of either

mass or energy

HYDROGEN EXPLOSION - combustion

bull Is this reaction taking place in an open closed or isolated system

bull the reacting mixture is the system (hydrogen oxygen and water molecules) everything else is the surroundings

bull This is exothermic because it is a combustion reaction (thermal energy leaves during this and all other exothermic reaction -- a lot of thermal energy is released in this example ）

)(2)(2 222 lOHOgH

open system

bull open systems can exchange mass and energybull closed systems allow the transfer of energy (he

at) but not massbull isolated systems do not allow transfer of either

mass or energy

HYDROGEN EXPLOSION - combustion

bull Is this reaction taking place in an open closed or isolated system

bull the reacting mixture is the system (hydrogen oxygen and water molecules) everything else is the surroundings

bull This is exothermic because it is a combustion reaction (thermal energy leaves during this and all other exothermic reaction -- a lot of thermal energy is released in this example ）

)(2)(2 222 lOHOgH

open system

HYDROGEN EXPLOSION - combustion

bull Is this reaction taking place in an open closed or isolated system

bull the reacting mixture is the system (hydrogen oxygen and water molecules) everything else is the surroundings

bull This is exothermic because it is a combustion reaction (thermal energy leaves during this and all other exothermic reaction -- a lot of thermal energy is released in this example ）

)(2)(2 222 lOHOgH

open system

Whats an example of an isolated system

Calorimetry - the measurement of heat changebull a calorimeter is an insulate

d closed container that creates an ISOLATED SYSTEM

bull a specific quantity of water surrounds a system carrying out a reaction

bull during the reaction heat leaves the inner container and is absorbed by the surrounding water

bull by recording temperature change the heat generated by a reaction can be calculated

bull Heat Capacity ndash heat needed to raise a certain quanity of a su

bstance 1 degree (celsius)

bull Specific Heat ndash the heat needed to raise a 1 gram of a specifi

c substance 1 degree (celsius)

If volume and pressure remain constant then q = Hbull Either heat capacity or specific heat must be a

pplied to find a substances change in heatbull Remember

heat capacity specific heat

Δt x (C) = m x (s) x Δt

tmsqtCq

cc

J

g

J g

degCdegC

Practice Questionbull A quantity of 1435 g naphthalene (C10H8) is bur

ned in a calorimeter The water temp rises from 2028 to 2595 degrees celsius If the heat capacity (C) is 1017 kJcelsius calculate the molar heat combustion of C10H8

bull What is being asked herebull How much heat per mole of napthalene is released into the calorime

ter

To do thisbull You need to figure out how much heat is generated by the

combustionbull You need to convert this total heat to q per mole (kJmol)

solving this problem

kJq

qq

q

qqq

kJq

CCCkJq

tcq

rxn

rxncal

sys

syscalrxn

cal

cal

calcal

6657

0

6657

)28209525)(1710(

We are not done

bull The question asks us to calculate the molar heat of combustion

bull So you have to find the molar mass of naphthalene (C10H8) have to use periodic table

bull molar mass of C10H8 is 1282 g

bull now we can convert from kJ1435 g to kJmol

How to convert

kJmol 101515

mol 1

2128

4351

5766kJ-

3

810

810

810

combustion ofheat molar

x

HC

HCgx

HCg

INTENSIVE and EXTENSIVE PROPERTIES

bull Extensive Properties - are properties that depend on how much matter is being consideredndash for example volume

bull the property of space a substance takes up is a value dependent on how much of the substance there is

ndash Can you think of others

INTENSIVE and EXTENSIVE PROPERTIES

bull Intensive Properties - properties that do NOT depend on how much matter is being consideredndash for example boiling point

bull the boiling point of liquid water into water vapor is the same regardless of how much water there is

ndash can you think of others

MOLE vs MOLARITY

bull mole (mol)bull 1 mol = 602 x 10^23

particles of a particular substance

bull molarity (M)bull molarity or molar co

ncentration is the number of moles of solute per liter of solutionthe dissolved

substanceA solvent is what a solute dissolves in to make a solution

molarity is defined as

molarity = moles of solute liters of solution

So is molarity an intensive or extensive property

intensive property

Change in Enthalpy of a Reaction

rxnH

Standard Enthalpy of Formation

bull DEFINITIONndash the enthalpy of a reaction carried out at 1 atm and at 25 degr

ees Cndash the enthalpy value relates to the heat absorbed or released

during the formation of 1 mole of a specific compound

The symbol for this value is represented as

formation ofenthalpy Standared fH

Once we know these values we can calculate the change in enthalpy of a reaction or

rxnH

Standard Enthalpy of a Reaction rxnH

)reactants()products(H

isequation dgeneralize The

fHfH nnrxn

bull n = the coefficient of the productsbull m = the coefficient of the reactantsbull Σ = the sum ofbull f = formation bull deg = standard state conditions (1 atm and 25 degrees C)

m

The most stable forms of substances will = 0

EXAMPLEGiven the equation 3 O2(g)rarr2 O3(g) ΔH = +2844 kJ

calculate ΔH for the following reaction 32 O2(g) rarr O3(g)REMEMBER

)reactants()products(H

isequation dgeneralize The

fHfH nnrxn

O2 is the most stable form of oxygen So it is equal to 0

Therefore the the change in enthalpy is all about the product

(x products) - (0 reactants) = rxnH = X products = +2844 kJYou can check using the table on P 247

i

P 247bull The ΔHdegf for O3(g) is +1422 bull The ΔHdegf for O2(g) is 0Given the equation 3 O2(g)rarr2 O3(g) ΔH = +2844 kJThere are two O3 in the equation (1422 kJ x 2 = 2844 kJ)

ndash we must remember the coefficientsbull Then the question asks us to calculate ΔH for the

following reaction 32 O2(g) rarr O3(g) bull There is only one O3

bull Since 32 O2(g) rarr O3(g) is frac12 of 3 O2(g)rarr2 O3(g) the enthalpy of the reaction will be frac12 as well

frac12 (+2844kJ)or +1422kJ

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used

in the manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP IWRITE THE BALANCED EQUATIONTo carry any combustion reaction you need oxygen as a reactant Liquid water will always be included as one of the products

__ C4H4S(l) + ___ O2(g) rarr __ CO2(g) + ___ S02(g) + ___ H20(l)1 6 4 1 2

Change of Enthalpy of Reaction

bull The combustion of thiophene C4H4S(l) a compound used in the manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

USE THE COEFFICIENTS FROM YOUR BALANCED EQUATION TO DETERMINE THE TOTAL CHANGE IN ENTHALPY IN THE FORMATION OF EVERY SUBSTANCE USED IN THE REACTION USE THIS INFORMATION TO WRITE YOUR ΔHdegRXN EQUATION

i

)reactants()products(H fHfH nnrxn

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

i

PRODUCTS

4 (________kJmol CO2(g))2 (________kJmol H2O(l))1 (________kJmol SO2(g))

REACTANTS

X kJmol C4H4S(l)

6 (________kJmol O2(g))

-3935

-2858-2968 000

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

i

[4(-3935) + 1(-2968) + 2(-2858)] ndash [1(ΔH) + 6(0)] = -2523

PRODUCTS －

REACTANTS = ΔHRXN

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP III

SOLVE for X

i

-24424 - ΔH(reactants) = -2523

ΔHC4H4S(l) = 813kJmol

CHANGE IN H

0H EXOTHERMIC

0H CENDOTHERMI

Change in Enthalpy in a multi-step reaction

Change in Enthalpy in a multi-step reaction

CHEMICAL EQUILIBRIA

Things to remember

This is how you convert an Equilibrium constant from Kc to Kp

Kp = Kc (00821 T)^Δn

Δn is the sum of all stoichiometric coefficients belonging to products minus the sum of all stoichiometric coefficients belonging to reactants

T is the temperature during the reaction in Kelvin Remember that Kelvin has exactly the same degree of difference between integers as celsius but it does have a different starting point Kelvins zero is 273deg less than celsius For example 2deg C is equal to 275deg K 27deg C is equal to 300deg K

Qc and Kc Questionh

EQUILIBRIUM of 2 or more RXNs

If a reaction can be expressed as the sum of two or more reactions the equilibrium constant for the overall reaction is given by the product of the equilibrium of the individual reactions

Kc = (Kc) (Kc)

Let me elaborate

nn

nn

c BA

DCK

][][

][][ nn

nn

c DC

FEK

][][

][][

c

c

c

K

__________________

K

K

FEBA

FEDC

DCBA

1st Step Reaction 2nd Step Reaction

OverallReaction

To find the overall equilibrium of a multi-step reaction

cnn

nn

nn

nn

cc KDC

FEx

BA

DCKK

][][

][][

][][

][][

2 step reaction1st step reaction

N2(g) + O2(g) harr 2 NO(g) Kc1 = 23 x 10^-19

2nd step reaction

2 NO(g) + O2(g) harr 2 NO2(g) Kc2 = 3 x 10^6

Kc = Kc1 x Kc2 = (23 x 10^-19)(3 x 10^6) Kc = 7 x 10^-13

Write the equilibrium equation for this multi-step reaction

CHEMICAL KINETICS

Reaction rates

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull Suppose that at a particular moment during the reaction molecular hydrogen is reacting at the rate of 0074 Ms

a) At what rate is ammonia being formedb) At what rate is molecular nitrogen reacting

t

NH

t

H

t

N

322

2

1

3

1

1

1

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is ammonia being formed s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

t

NH

t

NH

t

H

sM

3

32

2

10740

3

1

2

1

3

1t

NHsM

30740

3

2

_

-

_

t

NHsM

30490

GIVEN

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is molecular nitrogen reacting s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

-

GIVEN

t

N

t

N

t

H

sM

2

22

1

10740

3

1

1

1

3

1

t

NsM

20740)1(

3

1

t

NsM

20250

-

Describe the difference between Heat of solution and Heat of dilution Give an example of

each one

Heat of solution is the amount of heat absorbed or released during the act of combining a solute with a solvent An example is adding salt to water There are two steps to this process the Lattice Energy (breaking down the molecules) and the heat of hydration (where the ionically charged particles dissolve by attracting to the dipole water molecules

Heat of dilution refers to the amount of heat absorbed or released when water is added to a solution already made In this latter case if the act of making a solution is exothermic adding more water will make it even more exothermic (hotter) For example as you add more and more water to a solution of sulfuric acid and water the more heat it releases (the more it burns)

Neutralization

bull Explain how you would neutralize an acid and a base

Mix an acidic solution with a basic solution wherein the ratio of H+ ions to OH- ions eq

ual to 1

Cabbage Juice as a pH indicator

bull Explain how cabbage juice works as a pH indicator There is anthocyanin in cabbage juice Anthocyanin reacts w

ith hydroxyl ions (OH-) Hydroxyl ions jump off anthocyanin when in an acidic solution thereby changing the chemical composition of the anthocyanin (the pH indicator) This will cause the juice to react with light differently than if these OH- ions had not jumped off the anthocyanin (makes the juice turn a reddish color) Conversely Hydroxyl ions will jump from the solution onto the anthrocyanin molecules when there is an excess of them in the solution (such as when cabbage juice is mixed with a basic solution) This causes the juice

(the pH indicator) to turn a yellowish color

NOT mentioned

bull Balancing redox reactions

bull Kw

bull Ka

bull Kb

bull pH

bull the galvanic cell

bullKNOW THESE CONCEPTS

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Calorimetry - the measurement of heat changebull a calorimeter is an insulate

d closed container that creates an ISOLATED SYSTEM

bull a specific quantity of water surrounds a system carrying out a reaction

bull during the reaction heat leaves the inner container and is absorbed by the surrounding water

bull by recording temperature change the heat generated by a reaction can be calculated

bull Heat Capacity ndash heat needed to raise a certain quanity of a su

bstance 1 degree (celsius)

bull Specific Heat ndash the heat needed to raise a 1 gram of a specifi

c substance 1 degree (celsius)

If volume and pressure remain constant then q = Hbull Either heat capacity or specific heat must be a

pplied to find a substances change in heatbull Remember

heat capacity specific heat

Δt x (C) = m x (s) x Δt

tmsqtCq

cc

J

g

J g

degCdegC

Practice Questionbull A quantity of 1435 g naphthalene (C10H8) is bur

ned in a calorimeter The water temp rises from 2028 to 2595 degrees celsius If the heat capacity (C) is 1017 kJcelsius calculate the molar heat combustion of C10H8

bull What is being asked herebull How much heat per mole of napthalene is released into the calorime

ter

To do thisbull You need to figure out how much heat is generated by the

combustionbull You need to convert this total heat to q per mole (kJmol)

solving this problem

kJq

qq

q

qqq

kJq

CCCkJq

tcq

rxn

rxncal

sys

syscalrxn

cal

cal

calcal

6657

0

6657

)28209525)(1710(

We are not done

bull The question asks us to calculate the molar heat of combustion

bull So you have to find the molar mass of naphthalene (C10H8) have to use periodic table

bull molar mass of C10H8 is 1282 g

bull now we can convert from kJ1435 g to kJmol

How to convert

kJmol 101515

mol 1

2128

4351

5766kJ-

3

810

810

810

combustion ofheat molar

x

HC

HCgx

HCg

INTENSIVE and EXTENSIVE PROPERTIES

bull Extensive Properties - are properties that depend on how much matter is being consideredndash for example volume

bull the property of space a substance takes up is a value dependent on how much of the substance there is

ndash Can you think of others

INTENSIVE and EXTENSIVE PROPERTIES

bull Intensive Properties - properties that do NOT depend on how much matter is being consideredndash for example boiling point

bull the boiling point of liquid water into water vapor is the same regardless of how much water there is

ndash can you think of others

MOLE vs MOLARITY

bull mole (mol)bull 1 mol = 602 x 10^23

particles of a particular substance

bull molarity (M)bull molarity or molar co

ncentration is the number of moles of solute per liter of solutionthe dissolved

substanceA solvent is what a solute dissolves in to make a solution

molarity is defined as

molarity = moles of solute liters of solution

So is molarity an intensive or extensive property

intensive property

Change in Enthalpy of a Reaction

rxnH

Standard Enthalpy of Formation

bull DEFINITIONndash the enthalpy of a reaction carried out at 1 atm and at 25 degr

ees Cndash the enthalpy value relates to the heat absorbed or released

during the formation of 1 mole of a specific compound

The symbol for this value is represented as

formation ofenthalpy Standared fH

Once we know these values we can calculate the change in enthalpy of a reaction or

rxnH

Standard Enthalpy of a Reaction rxnH

)reactants()products(H

isequation dgeneralize The

fHfH nnrxn

bull n = the coefficient of the productsbull m = the coefficient of the reactantsbull Σ = the sum ofbull f = formation bull deg = standard state conditions (1 atm and 25 degrees C)

m

The most stable forms of substances will = 0

EXAMPLEGiven the equation 3 O2(g)rarr2 O3(g) ΔH = +2844 kJ

calculate ΔH for the following reaction 32 O2(g) rarr O3(g)REMEMBER

)reactants()products(H

isequation dgeneralize The

fHfH nnrxn

O2 is the most stable form of oxygen So it is equal to 0

Therefore the the change in enthalpy is all about the product

(x products) - (0 reactants) = rxnH = X products = +2844 kJYou can check using the table on P 247

i

P 247bull The ΔHdegf for O3(g) is +1422 bull The ΔHdegf for O2(g) is 0Given the equation 3 O2(g)rarr2 O3(g) ΔH = +2844 kJThere are two O3 in the equation (1422 kJ x 2 = 2844 kJ)

ndash we must remember the coefficientsbull Then the question asks us to calculate ΔH for the

following reaction 32 O2(g) rarr O3(g) bull There is only one O3

bull Since 32 O2(g) rarr O3(g) is frac12 of 3 O2(g)rarr2 O3(g) the enthalpy of the reaction will be frac12 as well

frac12 (+2844kJ)or +1422kJ

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used

in the manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP IWRITE THE BALANCED EQUATIONTo carry any combustion reaction you need oxygen as a reactant Liquid water will always be included as one of the products

__ C4H4S(l) + ___ O2(g) rarr __ CO2(g) + ___ S02(g) + ___ H20(l)1 6 4 1 2

Change of Enthalpy of Reaction

bull The combustion of thiophene C4H4S(l) a compound used in the manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

USE THE COEFFICIENTS FROM YOUR BALANCED EQUATION TO DETERMINE THE TOTAL CHANGE IN ENTHALPY IN THE FORMATION OF EVERY SUBSTANCE USED IN THE REACTION USE THIS INFORMATION TO WRITE YOUR ΔHdegRXN EQUATION

i

)reactants()products(H fHfH nnrxn

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

i

PRODUCTS

4 (________kJmol CO2(g))2 (________kJmol H2O(l))1 (________kJmol SO2(g))

REACTANTS

X kJmol C4H4S(l)

6 (________kJmol O2(g))

-3935

-2858-2968 000

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

i

[4(-3935) + 1(-2968) + 2(-2858)] ndash [1(ΔH) + 6(0)] = -2523

PRODUCTS －

REACTANTS = ΔHRXN

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP III

SOLVE for X

i

-24424 - ΔH(reactants) = -2523

ΔHC4H4S(l) = 813kJmol

CHANGE IN H

0H EXOTHERMIC

0H CENDOTHERMI

Change in Enthalpy in a multi-step reaction

Change in Enthalpy in a multi-step reaction

CHEMICAL EQUILIBRIA

Things to remember

This is how you convert an Equilibrium constant from Kc to Kp

Kp = Kc (00821 T)^Δn

Δn is the sum of all stoichiometric coefficients belonging to products minus the sum of all stoichiometric coefficients belonging to reactants

T is the temperature during the reaction in Kelvin Remember that Kelvin has exactly the same degree of difference between integers as celsius but it does have a different starting point Kelvins zero is 273deg less than celsius For example 2deg C is equal to 275deg K 27deg C is equal to 300deg K

Qc and Kc Questionh

EQUILIBRIUM of 2 or more RXNs

If a reaction can be expressed as the sum of two or more reactions the equilibrium constant for the overall reaction is given by the product of the equilibrium of the individual reactions

Kc = (Kc) (Kc)

Let me elaborate

nn

nn

c BA

DCK

][][

][][ nn

nn

c DC

FEK

][][

][][

c

c

c

K

__________________

K

K

FEBA

FEDC

DCBA

1st Step Reaction 2nd Step Reaction

OverallReaction

To find the overall equilibrium of a multi-step reaction

cnn

nn

nn

nn

cc KDC

FEx

BA

DCKK

][][

][][

][][

][][

2 step reaction1st step reaction

N2(g) + O2(g) harr 2 NO(g) Kc1 = 23 x 10^-19

2nd step reaction

2 NO(g) + O2(g) harr 2 NO2(g) Kc2 = 3 x 10^6

Kc = Kc1 x Kc2 = (23 x 10^-19)(3 x 10^6) Kc = 7 x 10^-13

Write the equilibrium equation for this multi-step reaction

CHEMICAL KINETICS

Reaction rates

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull Suppose that at a particular moment during the reaction molecular hydrogen is reacting at the rate of 0074 Ms

a) At what rate is ammonia being formedb) At what rate is molecular nitrogen reacting

t

NH

t

H

t

N

322

2

1

3

1

1

1

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is ammonia being formed s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

t

NH

t

NH

t

H

sM

3

32

2

10740

3

1

2

1

3

1t

NHsM

30740

3

2

_

-

_

t

NHsM

30490

GIVEN

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is molecular nitrogen reacting s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

-

GIVEN

t

N

t

N

t

H

sM

2

22

1

10740

3

1

1

1

3

1

t

NsM

20740)1(

3

1

t

NsM

20250

-

Describe the difference between Heat of solution and Heat of dilution Give an example of

each one

Heat of solution is the amount of heat absorbed or released during the act of combining a solute with a solvent An example is adding salt to water There are two steps to this process the Lattice Energy (breaking down the molecules) and the heat of hydration (where the ionically charged particles dissolve by attracting to the dipole water molecules

Heat of dilution refers to the amount of heat absorbed or released when water is added to a solution already made In this latter case if the act of making a solution is exothermic adding more water will make it even more exothermic (hotter) For example as you add more and more water to a solution of sulfuric acid and water the more heat it releases (the more it burns)

Neutralization

bull Explain how you would neutralize an acid and a base

Mix an acidic solution with a basic solution wherein the ratio of H+ ions to OH- ions eq

ual to 1

Cabbage Juice as a pH indicator

bull Explain how cabbage juice works as a pH indicator There is anthocyanin in cabbage juice Anthocyanin reacts w

ith hydroxyl ions (OH-) Hydroxyl ions jump off anthocyanin when in an acidic solution thereby changing the chemical composition of the anthocyanin (the pH indicator) This will cause the juice to react with light differently than if these OH- ions had not jumped off the anthocyanin (makes the juice turn a reddish color) Conversely Hydroxyl ions will jump from the solution onto the anthrocyanin molecules when there is an excess of them in the solution (such as when cabbage juice is mixed with a basic solution) This causes the juice

(the pH indicator) to turn a yellowish color

NOT mentioned

bull Balancing redox reactions

bull Kw

bull Ka

bull Kb

bull pH

bull the galvanic cell

bullKNOW THESE CONCEPTS

• Slide 1
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• Slide 51

bull Heat Capacity ndash heat needed to raise a certain quanity of a su

bstance 1 degree (celsius)

bull Specific Heat ndash the heat needed to raise a 1 gram of a specifi

c substance 1 degree (celsius)

If volume and pressure remain constant then q = Hbull Either heat capacity or specific heat must be a

pplied to find a substances change in heatbull Remember

heat capacity specific heat

Δt x (C) = m x (s) x Δt

tmsqtCq

cc

J

g

J g

degCdegC

Practice Questionbull A quantity of 1435 g naphthalene (C10H8) is bur

ned in a calorimeter The water temp rises from 2028 to 2595 degrees celsius If the heat capacity (C) is 1017 kJcelsius calculate the molar heat combustion of C10H8

bull What is being asked herebull How much heat per mole of napthalene is released into the calorime

ter

To do thisbull You need to figure out how much heat is generated by the

combustionbull You need to convert this total heat to q per mole (kJmol)

solving this problem

kJq

qq

q

qqq

kJq

CCCkJq

tcq

rxn

rxncal

sys

syscalrxn

cal

cal

calcal

6657

0

6657

)28209525)(1710(

We are not done

bull The question asks us to calculate the molar heat of combustion

bull So you have to find the molar mass of naphthalene (C10H8) have to use periodic table

bull molar mass of C10H8 is 1282 g

bull now we can convert from kJ1435 g to kJmol

How to convert

kJmol 101515

mol 1

2128

4351

5766kJ-

3

810

810

810

combustion ofheat molar

x

HC

HCgx

HCg

INTENSIVE and EXTENSIVE PROPERTIES

bull Extensive Properties - are properties that depend on how much matter is being consideredndash for example volume

bull the property of space a substance takes up is a value dependent on how much of the substance there is

ndash Can you think of others

INTENSIVE and EXTENSIVE PROPERTIES

bull Intensive Properties - properties that do NOT depend on how much matter is being consideredndash for example boiling point

bull the boiling point of liquid water into water vapor is the same regardless of how much water there is

ndash can you think of others

MOLE vs MOLARITY

bull mole (mol)bull 1 mol = 602 x 10^23

particles of a particular substance

bull molarity (M)bull molarity or molar co

ncentration is the number of moles of solute per liter of solutionthe dissolved

substanceA solvent is what a solute dissolves in to make a solution

molarity is defined as

molarity = moles of solute liters of solution

So is molarity an intensive or extensive property

intensive property

Change in Enthalpy of a Reaction

rxnH

Standard Enthalpy of Formation

bull DEFINITIONndash the enthalpy of a reaction carried out at 1 atm and at 25 degr

ees Cndash the enthalpy value relates to the heat absorbed or released

during the formation of 1 mole of a specific compound

The symbol for this value is represented as

formation ofenthalpy Standared fH

Once we know these values we can calculate the change in enthalpy of a reaction or

rxnH

Standard Enthalpy of a Reaction rxnH

)reactants()products(H

isequation dgeneralize The

fHfH nnrxn

bull n = the coefficient of the productsbull m = the coefficient of the reactantsbull Σ = the sum ofbull f = formation bull deg = standard state conditions (1 atm and 25 degrees C)

m

The most stable forms of substances will = 0

EXAMPLEGiven the equation 3 O2(g)rarr2 O3(g) ΔH = +2844 kJ

calculate ΔH for the following reaction 32 O2(g) rarr O3(g)REMEMBER

)reactants()products(H

isequation dgeneralize The

fHfH nnrxn

O2 is the most stable form of oxygen So it is equal to 0

Therefore the the change in enthalpy is all about the product

(x products) - (0 reactants) = rxnH = X products = +2844 kJYou can check using the table on P 247

i

P 247bull The ΔHdegf for O3(g) is +1422 bull The ΔHdegf for O2(g) is 0Given the equation 3 O2(g)rarr2 O3(g) ΔH = +2844 kJThere are two O3 in the equation (1422 kJ x 2 = 2844 kJ)

ndash we must remember the coefficientsbull Then the question asks us to calculate ΔH for the

following reaction 32 O2(g) rarr O3(g) bull There is only one O3

bull Since 32 O2(g) rarr O3(g) is frac12 of 3 O2(g)rarr2 O3(g) the enthalpy of the reaction will be frac12 as well

frac12 (+2844kJ)or +1422kJ

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used

in the manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP IWRITE THE BALANCED EQUATIONTo carry any combustion reaction you need oxygen as a reactant Liquid water will always be included as one of the products

__ C4H4S(l) + ___ O2(g) rarr __ CO2(g) + ___ S02(g) + ___ H20(l)1 6 4 1 2

Change of Enthalpy of Reaction

bull The combustion of thiophene C4H4S(l) a compound used in the manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

USE THE COEFFICIENTS FROM YOUR BALANCED EQUATION TO DETERMINE THE TOTAL CHANGE IN ENTHALPY IN THE FORMATION OF EVERY SUBSTANCE USED IN THE REACTION USE THIS INFORMATION TO WRITE YOUR ΔHdegRXN EQUATION

i

)reactants()products(H fHfH nnrxn

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

i

PRODUCTS

4 (________kJmol CO2(g))2 (________kJmol H2O(l))1 (________kJmol SO2(g))

REACTANTS

X kJmol C4H4S(l)

6 (________kJmol O2(g))

-3935

-2858-2968 000

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

i

[4(-3935) + 1(-2968) + 2(-2858)] ndash [1(ΔH) + 6(0)] = -2523

PRODUCTS －

REACTANTS = ΔHRXN

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP III

SOLVE for X

i

-24424 - ΔH(reactants) = -2523

ΔHC4H4S(l) = 813kJmol

CHANGE IN H

0H EXOTHERMIC

0H CENDOTHERMI

Change in Enthalpy in a multi-step reaction

Change in Enthalpy in a multi-step reaction

CHEMICAL EQUILIBRIA

Things to remember

This is how you convert an Equilibrium constant from Kc to Kp

Kp = Kc (00821 T)^Δn

Δn is the sum of all stoichiometric coefficients belonging to products minus the sum of all stoichiometric coefficients belonging to reactants

T is the temperature during the reaction in Kelvin Remember that Kelvin has exactly the same degree of difference between integers as celsius but it does have a different starting point Kelvins zero is 273deg less than celsius For example 2deg C is equal to 275deg K 27deg C is equal to 300deg K

Qc and Kc Questionh

EQUILIBRIUM of 2 or more RXNs

If a reaction can be expressed as the sum of two or more reactions the equilibrium constant for the overall reaction is given by the product of the equilibrium of the individual reactions

Kc = (Kc) (Kc)

Let me elaborate

nn

nn

c BA

DCK

][][

][][ nn

nn

c DC

FEK

][][

][][

c

c

c

K

__________________

K

K

FEBA

FEDC

DCBA

1st Step Reaction 2nd Step Reaction

OverallReaction

To find the overall equilibrium of a multi-step reaction

cnn

nn

nn

nn

cc KDC

FEx

BA

DCKK

][][

][][

][][

][][

2 step reaction1st step reaction

N2(g) + O2(g) harr 2 NO(g) Kc1 = 23 x 10^-19

2nd step reaction

2 NO(g) + O2(g) harr 2 NO2(g) Kc2 = 3 x 10^6

Kc = Kc1 x Kc2 = (23 x 10^-19)(3 x 10^6) Kc = 7 x 10^-13

Write the equilibrium equation for this multi-step reaction

CHEMICAL KINETICS

Reaction rates

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull Suppose that at a particular moment during the reaction molecular hydrogen is reacting at the rate of 0074 Ms

a) At what rate is ammonia being formedb) At what rate is molecular nitrogen reacting

t

NH

t

H

t

N

322

2

1

3

1

1

1

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is ammonia being formed s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

t

NH

t

NH

t

H

sM

3

32

2

10740

3

1

2

1

3

1t

NHsM

30740

3

2

_

-

_

t

NHsM

30490

GIVEN

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is molecular nitrogen reacting s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

-

GIVEN

t

N

t

N

t

H

sM

2

22

1

10740

3

1

1

1

3

1

t

NsM

20740)1(

3

1

t

NsM

20250

-

Describe the difference between Heat of solution and Heat of dilution Give an example of

each one

Heat of solution is the amount of heat absorbed or released during the act of combining a solute with a solvent An example is adding salt to water There are two steps to this process the Lattice Energy (breaking down the molecules) and the heat of hydration (where the ionically charged particles dissolve by attracting to the dipole water molecules

Heat of dilution refers to the amount of heat absorbed or released when water is added to a solution already made In this latter case if the act of making a solution is exothermic adding more water will make it even more exothermic (hotter) For example as you add more and more water to a solution of sulfuric acid and water the more heat it releases (the more it burns)

Neutralization

bull Explain how you would neutralize an acid and a base

Mix an acidic solution with a basic solution wherein the ratio of H+ ions to OH- ions eq

ual to 1

Cabbage Juice as a pH indicator

bull Explain how cabbage juice works as a pH indicator There is anthocyanin in cabbage juice Anthocyanin reacts w

ith hydroxyl ions (OH-) Hydroxyl ions jump off anthocyanin when in an acidic solution thereby changing the chemical composition of the anthocyanin (the pH indicator) This will cause the juice to react with light differently than if these OH- ions had not jumped off the anthocyanin (makes the juice turn a reddish color) Conversely Hydroxyl ions will jump from the solution onto the anthrocyanin molecules when there is an excess of them in the solution (such as when cabbage juice is mixed with a basic solution) This causes the juice

(the pH indicator) to turn a yellowish color

NOT mentioned

bull Balancing redox reactions

bull Kw

bull Ka

bull Kb

bull pH

bull the galvanic cell

bullKNOW THESE CONCEPTS

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
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• Slide 47
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• Slide 49
• Slide 50
• Slide 51

If volume and pressure remain constant then q = Hbull Either heat capacity or specific heat must be a

pplied to find a substances change in heatbull Remember

heat capacity specific heat

Δt x (C) = m x (s) x Δt

tmsqtCq

cc

J

g

J g

degCdegC

Practice Questionbull A quantity of 1435 g naphthalene (C10H8) is bur

ned in a calorimeter The water temp rises from 2028 to 2595 degrees celsius If the heat capacity (C) is 1017 kJcelsius calculate the molar heat combustion of C10H8

bull What is being asked herebull How much heat per mole of napthalene is released into the calorime

ter

To do thisbull You need to figure out how much heat is generated by the

combustionbull You need to convert this total heat to q per mole (kJmol)

solving this problem

kJq

qq

q

qqq

kJq

CCCkJq

tcq

rxn

rxncal

sys

syscalrxn

cal

cal

calcal

6657

0

6657

)28209525)(1710(

We are not done

bull The question asks us to calculate the molar heat of combustion

bull So you have to find the molar mass of naphthalene (C10H8) have to use periodic table

bull molar mass of C10H8 is 1282 g

bull now we can convert from kJ1435 g to kJmol

How to convert

kJmol 101515

mol 1

2128

4351

5766kJ-

3

810

810

810

combustion ofheat molar

x

HC

HCgx

HCg

INTENSIVE and EXTENSIVE PROPERTIES

bull Extensive Properties - are properties that depend on how much matter is being consideredndash for example volume

bull the property of space a substance takes up is a value dependent on how much of the substance there is

ndash Can you think of others

INTENSIVE and EXTENSIVE PROPERTIES

bull Intensive Properties - properties that do NOT depend on how much matter is being consideredndash for example boiling point

bull the boiling point of liquid water into water vapor is the same regardless of how much water there is

ndash can you think of others

MOLE vs MOLARITY

bull mole (mol)bull 1 mol = 602 x 10^23

particles of a particular substance

bull molarity (M)bull molarity or molar co

ncentration is the number of moles of solute per liter of solutionthe dissolved

substanceA solvent is what a solute dissolves in to make a solution

molarity is defined as

molarity = moles of solute liters of solution

So is molarity an intensive or extensive property

intensive property

Change in Enthalpy of a Reaction

rxnH

Standard Enthalpy of Formation

bull DEFINITIONndash the enthalpy of a reaction carried out at 1 atm and at 25 degr

ees Cndash the enthalpy value relates to the heat absorbed or released

during the formation of 1 mole of a specific compound

The symbol for this value is represented as

formation ofenthalpy Standared fH

Once we know these values we can calculate the change in enthalpy of a reaction or

rxnH

Standard Enthalpy of a Reaction rxnH

)reactants()products(H

isequation dgeneralize The

fHfH nnrxn

bull n = the coefficient of the productsbull m = the coefficient of the reactantsbull Σ = the sum ofbull f = formation bull deg = standard state conditions (1 atm and 25 degrees C)

m

The most stable forms of substances will = 0

EXAMPLEGiven the equation 3 O2(g)rarr2 O3(g) ΔH = +2844 kJ

calculate ΔH for the following reaction 32 O2(g) rarr O3(g)REMEMBER

)reactants()products(H

isequation dgeneralize The

fHfH nnrxn

O2 is the most stable form of oxygen So it is equal to 0

Therefore the the change in enthalpy is all about the product

(x products) - (0 reactants) = rxnH = X products = +2844 kJYou can check using the table on P 247

i

P 247bull The ΔHdegf for O3(g) is +1422 bull The ΔHdegf for O2(g) is 0Given the equation 3 O2(g)rarr2 O3(g) ΔH = +2844 kJThere are two O3 in the equation (1422 kJ x 2 = 2844 kJ)

ndash we must remember the coefficientsbull Then the question asks us to calculate ΔH for the

following reaction 32 O2(g) rarr O3(g) bull There is only one O3

bull Since 32 O2(g) rarr O3(g) is frac12 of 3 O2(g)rarr2 O3(g) the enthalpy of the reaction will be frac12 as well

frac12 (+2844kJ)or +1422kJ

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used

in the manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP IWRITE THE BALANCED EQUATIONTo carry any combustion reaction you need oxygen as a reactant Liquid water will always be included as one of the products

__ C4H4S(l) + ___ O2(g) rarr __ CO2(g) + ___ S02(g) + ___ H20(l)1 6 4 1 2

Change of Enthalpy of Reaction

bull The combustion of thiophene C4H4S(l) a compound used in the manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

USE THE COEFFICIENTS FROM YOUR BALANCED EQUATION TO DETERMINE THE TOTAL CHANGE IN ENTHALPY IN THE FORMATION OF EVERY SUBSTANCE USED IN THE REACTION USE THIS INFORMATION TO WRITE YOUR ΔHdegRXN EQUATION

i

)reactants()products(H fHfH nnrxn

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

i

PRODUCTS

4 (________kJmol CO2(g))2 (________kJmol H2O(l))1 (________kJmol SO2(g))

REACTANTS

X kJmol C4H4S(l)

6 (________kJmol O2(g))

-3935

-2858-2968 000

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

i

[4(-3935) + 1(-2968) + 2(-2858)] ndash [1(ΔH) + 6(0)] = -2523

PRODUCTS －

REACTANTS = ΔHRXN

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP III

SOLVE for X

i

-24424 - ΔH(reactants) = -2523

ΔHC4H4S(l) = 813kJmol

CHANGE IN H

0H EXOTHERMIC

0H CENDOTHERMI

Change in Enthalpy in a multi-step reaction

Change in Enthalpy in a multi-step reaction

CHEMICAL EQUILIBRIA

Things to remember

This is how you convert an Equilibrium constant from Kc to Kp

Kp = Kc (00821 T)^Δn

Δn is the sum of all stoichiometric coefficients belonging to products minus the sum of all stoichiometric coefficients belonging to reactants

T is the temperature during the reaction in Kelvin Remember that Kelvin has exactly the same degree of difference between integers as celsius but it does have a different starting point Kelvins zero is 273deg less than celsius For example 2deg C is equal to 275deg K 27deg C is equal to 300deg K

Qc and Kc Questionh

EQUILIBRIUM of 2 or more RXNs

If a reaction can be expressed as the sum of two or more reactions the equilibrium constant for the overall reaction is given by the product of the equilibrium of the individual reactions

Kc = (Kc) (Kc)

Let me elaborate

nn

nn

c BA

DCK

][][

][][ nn

nn

c DC

FEK

][][

][][

c

c

c

K

__________________

K

K

FEBA

FEDC

DCBA

1st Step Reaction 2nd Step Reaction

OverallReaction

To find the overall equilibrium of a multi-step reaction

cnn

nn

nn

nn

cc KDC

FEx

BA

DCKK

][][

][][

][][

][][

2 step reaction1st step reaction

N2(g) + O2(g) harr 2 NO(g) Kc1 = 23 x 10^-19

2nd step reaction

2 NO(g) + O2(g) harr 2 NO2(g) Kc2 = 3 x 10^6

Kc = Kc1 x Kc2 = (23 x 10^-19)(3 x 10^6) Kc = 7 x 10^-13

Write the equilibrium equation for this multi-step reaction

CHEMICAL KINETICS

Reaction rates

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull Suppose that at a particular moment during the reaction molecular hydrogen is reacting at the rate of 0074 Ms

a) At what rate is ammonia being formedb) At what rate is molecular nitrogen reacting

t

NH

t

H

t

N

322

2

1

3

1

1

1

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is ammonia being formed s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

t

NH

t

NH

t

H

sM

3

32

2

10740

3

1

2

1

3

1t

NHsM

30740

3

2

_

-

_

t

NHsM

30490

GIVEN

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is molecular nitrogen reacting s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

-

GIVEN

t

N

t

N

t

H

sM

2

22

1

10740

3

1

1

1

3

1

t

NsM

20740)1(

3

1

t

NsM

20250

-

Describe the difference between Heat of solution and Heat of dilution Give an example of

each one

Heat of solution is the amount of heat absorbed or released during the act of combining a solute with a solvent An example is adding salt to water There are two steps to this process the Lattice Energy (breaking down the molecules) and the heat of hydration (where the ionically charged particles dissolve by attracting to the dipole water molecules

Heat of dilution refers to the amount of heat absorbed or released when water is added to a solution already made In this latter case if the act of making a solution is exothermic adding more water will make it even more exothermic (hotter) For example as you add more and more water to a solution of sulfuric acid and water the more heat it releases (the more it burns)

Neutralization

bull Explain how you would neutralize an acid and a base

Mix an acidic solution with a basic solution wherein the ratio of H+ ions to OH- ions eq

ual to 1

Cabbage Juice as a pH indicator

bull Explain how cabbage juice works as a pH indicator There is anthocyanin in cabbage juice Anthocyanin reacts w

ith hydroxyl ions (OH-) Hydroxyl ions jump off anthocyanin when in an acidic solution thereby changing the chemical composition of the anthocyanin (the pH indicator) This will cause the juice to react with light differently than if these OH- ions had not jumped off the anthocyanin (makes the juice turn a reddish color) Conversely Hydroxyl ions will jump from the solution onto the anthrocyanin molecules when there is an excess of them in the solution (such as when cabbage juice is mixed with a basic solution) This causes the juice

(the pH indicator) to turn a yellowish color

NOT mentioned

bull Balancing redox reactions

bull Kw

bull Ka

bull Kb

bull pH

bull the galvanic cell

bullKNOW THESE CONCEPTS

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Practice Questionbull A quantity of 1435 g naphthalene (C10H8) is bur

ned in a calorimeter The water temp rises from 2028 to 2595 degrees celsius If the heat capacity (C) is 1017 kJcelsius calculate the molar heat combustion of C10H8

bull What is being asked herebull How much heat per mole of napthalene is released into the calorime

ter

To do thisbull You need to figure out how much heat is generated by the

combustionbull You need to convert this total heat to q per mole (kJmol)

solving this problem

kJq

qq

q

qqq

kJq

CCCkJq

tcq

rxn

rxncal

sys

syscalrxn

cal

cal

calcal

6657

0

6657

)28209525)(1710(

We are not done

bull The question asks us to calculate the molar heat of combustion

bull So you have to find the molar mass of naphthalene (C10H8) have to use periodic table

bull molar mass of C10H8 is 1282 g

bull now we can convert from kJ1435 g to kJmol

How to convert

kJmol 101515

mol 1

2128

4351

5766kJ-

3

810

810

810

combustion ofheat molar

x

HC

HCgx

HCg

INTENSIVE and EXTENSIVE PROPERTIES

bull Extensive Properties - are properties that depend on how much matter is being consideredndash for example volume

bull the property of space a substance takes up is a value dependent on how much of the substance there is

ndash Can you think of others

INTENSIVE and EXTENSIVE PROPERTIES

bull Intensive Properties - properties that do NOT depend on how much matter is being consideredndash for example boiling point

bull the boiling point of liquid water into water vapor is the same regardless of how much water there is

ndash can you think of others

MOLE vs MOLARITY

bull mole (mol)bull 1 mol = 602 x 10^23

particles of a particular substance

bull molarity (M)bull molarity or molar co

ncentration is the number of moles of solute per liter of solutionthe dissolved

substanceA solvent is what a solute dissolves in to make a solution

molarity is defined as

molarity = moles of solute liters of solution

So is molarity an intensive or extensive property

intensive property

Change in Enthalpy of a Reaction

rxnH

Standard Enthalpy of Formation

bull DEFINITIONndash the enthalpy of a reaction carried out at 1 atm and at 25 degr

ees Cndash the enthalpy value relates to the heat absorbed or released

during the formation of 1 mole of a specific compound

The symbol for this value is represented as

formation ofenthalpy Standared fH

Once we know these values we can calculate the change in enthalpy of a reaction or

rxnH

Standard Enthalpy of a Reaction rxnH

)reactants()products(H

isequation dgeneralize The

fHfH nnrxn

bull n = the coefficient of the productsbull m = the coefficient of the reactantsbull Σ = the sum ofbull f = formation bull deg = standard state conditions (1 atm and 25 degrees C)

m

The most stable forms of substances will = 0

EXAMPLEGiven the equation 3 O2(g)rarr2 O3(g) ΔH = +2844 kJ

calculate ΔH for the following reaction 32 O2(g) rarr O3(g)REMEMBER

)reactants()products(H

isequation dgeneralize The

fHfH nnrxn

O2 is the most stable form of oxygen So it is equal to 0

Therefore the the change in enthalpy is all about the product

(x products) - (0 reactants) = rxnH = X products = +2844 kJYou can check using the table on P 247

i

P 247bull The ΔHdegf for O3(g) is +1422 bull The ΔHdegf for O2(g) is 0Given the equation 3 O2(g)rarr2 O3(g) ΔH = +2844 kJThere are two O3 in the equation (1422 kJ x 2 = 2844 kJ)

ndash we must remember the coefficientsbull Then the question asks us to calculate ΔH for the

following reaction 32 O2(g) rarr O3(g) bull There is only one O3

bull Since 32 O2(g) rarr O3(g) is frac12 of 3 O2(g)rarr2 O3(g) the enthalpy of the reaction will be frac12 as well

frac12 (+2844kJ)or +1422kJ

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used

in the manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP IWRITE THE BALANCED EQUATIONTo carry any combustion reaction you need oxygen as a reactant Liquid water will always be included as one of the products

__ C4H4S(l) + ___ O2(g) rarr __ CO2(g) + ___ S02(g) + ___ H20(l)1 6 4 1 2

Change of Enthalpy of Reaction

bull The combustion of thiophene C4H4S(l) a compound used in the manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

USE THE COEFFICIENTS FROM YOUR BALANCED EQUATION TO DETERMINE THE TOTAL CHANGE IN ENTHALPY IN THE FORMATION OF EVERY SUBSTANCE USED IN THE REACTION USE THIS INFORMATION TO WRITE YOUR ΔHdegRXN EQUATION

i

)reactants()products(H fHfH nnrxn

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

i

PRODUCTS

4 (________kJmol CO2(g))2 (________kJmol H2O(l))1 (________kJmol SO2(g))

REACTANTS

X kJmol C4H4S(l)

6 (________kJmol O2(g))

-3935

-2858-2968 000

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

i

[4(-3935) + 1(-2968) + 2(-2858)] ndash [1(ΔH) + 6(0)] = -2523

PRODUCTS －

REACTANTS = ΔHRXN

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP III

SOLVE for X

i

-24424 - ΔH(reactants) = -2523

ΔHC4H4S(l) = 813kJmol

CHANGE IN H

0H EXOTHERMIC

0H CENDOTHERMI

Change in Enthalpy in a multi-step reaction

Change in Enthalpy in a multi-step reaction

CHEMICAL EQUILIBRIA

Things to remember

This is how you convert an Equilibrium constant from Kc to Kp

Kp = Kc (00821 T)^Δn

Δn is the sum of all stoichiometric coefficients belonging to products minus the sum of all stoichiometric coefficients belonging to reactants

T is the temperature during the reaction in Kelvin Remember that Kelvin has exactly the same degree of difference between integers as celsius but it does have a different starting point Kelvins zero is 273deg less than celsius For example 2deg C is equal to 275deg K 27deg C is equal to 300deg K

Qc and Kc Questionh

EQUILIBRIUM of 2 or more RXNs

If a reaction can be expressed as the sum of two or more reactions the equilibrium constant for the overall reaction is given by the product of the equilibrium of the individual reactions

Kc = (Kc) (Kc)

Let me elaborate

nn

nn

c BA

DCK

][][

][][ nn

nn

c DC

FEK

][][

][][

c

c

c

K

__________________

K

K

FEBA

FEDC

DCBA

1st Step Reaction 2nd Step Reaction

OverallReaction

To find the overall equilibrium of a multi-step reaction

cnn

nn

nn

nn

cc KDC

FEx

BA

DCKK

][][

][][

][][

][][

2 step reaction1st step reaction

N2(g) + O2(g) harr 2 NO(g) Kc1 = 23 x 10^-19

2nd step reaction

2 NO(g) + O2(g) harr 2 NO2(g) Kc2 = 3 x 10^6

Kc = Kc1 x Kc2 = (23 x 10^-19)(3 x 10^6) Kc = 7 x 10^-13

Write the equilibrium equation for this multi-step reaction

CHEMICAL KINETICS

Reaction rates

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull Suppose that at a particular moment during the reaction molecular hydrogen is reacting at the rate of 0074 Ms

a) At what rate is ammonia being formedb) At what rate is molecular nitrogen reacting

t

NH

t

H

t

N

322

2

1

3

1

1

1

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is ammonia being formed s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

t

NH

t

NH

t

H

sM

3

32

2

10740

3

1

2

1

3

1t

NHsM

30740

3

2

_

-

_

t

NHsM

30490

GIVEN

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is molecular nitrogen reacting s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

-

GIVEN

t

N

t

N

t

H

sM

2

22

1

10740

3

1

1

1

3

1

t

NsM

20740)1(

3

1

t

NsM

20250

-

Describe the difference between Heat of solution and Heat of dilution Give an example of

each one

Heat of solution is the amount of heat absorbed or released during the act of combining a solute with a solvent An example is adding salt to water There are two steps to this process the Lattice Energy (breaking down the molecules) and the heat of hydration (where the ionically charged particles dissolve by attracting to the dipole water molecules

Heat of dilution refers to the amount of heat absorbed or released when water is added to a solution already made In this latter case if the act of making a solution is exothermic adding more water will make it even more exothermic (hotter) For example as you add more and more water to a solution of sulfuric acid and water the more heat it releases (the more it burns)

Neutralization

bull Explain how you would neutralize an acid and a base

Mix an acidic solution with a basic solution wherein the ratio of H+ ions to OH- ions eq

ual to 1

Cabbage Juice as a pH indicator

bull Explain how cabbage juice works as a pH indicator There is anthocyanin in cabbage juice Anthocyanin reacts w

ith hydroxyl ions (OH-) Hydroxyl ions jump off anthocyanin when in an acidic solution thereby changing the chemical composition of the anthocyanin (the pH indicator) This will cause the juice to react with light differently than if these OH- ions had not jumped off the anthocyanin (makes the juice turn a reddish color) Conversely Hydroxyl ions will jump from the solution onto the anthrocyanin molecules when there is an excess of them in the solution (such as when cabbage juice is mixed with a basic solution) This causes the juice

(the pH indicator) to turn a yellowish color

NOT mentioned

bull Balancing redox reactions

bull Kw

bull Ka

bull Kb

bull pH

bull the galvanic cell

bullKNOW THESE CONCEPTS

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
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solving this problem

kJq

qq

q

qqq

kJq

CCCkJq

tcq

rxn

rxncal

sys

syscalrxn

cal

cal

calcal

6657

0

6657

)28209525)(1710(

We are not done

bull The question asks us to calculate the molar heat of combustion

bull So you have to find the molar mass of naphthalene (C10H8) have to use periodic table

bull molar mass of C10H8 is 1282 g

bull now we can convert from kJ1435 g to kJmol

How to convert

kJmol 101515

mol 1

2128

4351

5766kJ-

3

810

810

810

combustion ofheat molar

x

HC

HCgx

HCg

INTENSIVE and EXTENSIVE PROPERTIES

bull Extensive Properties - are properties that depend on how much matter is being consideredndash for example volume

bull the property of space a substance takes up is a value dependent on how much of the substance there is

ndash Can you think of others

INTENSIVE and EXTENSIVE PROPERTIES

bull Intensive Properties - properties that do NOT depend on how much matter is being consideredndash for example boiling point

bull the boiling point of liquid water into water vapor is the same regardless of how much water there is

ndash can you think of others

MOLE vs MOLARITY

bull mole (mol)bull 1 mol = 602 x 10^23

particles of a particular substance

bull molarity (M)bull molarity or molar co

ncentration is the number of moles of solute per liter of solutionthe dissolved

substanceA solvent is what a solute dissolves in to make a solution

molarity is defined as

molarity = moles of solute liters of solution

So is molarity an intensive or extensive property

intensive property

Change in Enthalpy of a Reaction

rxnH

Standard Enthalpy of Formation

bull DEFINITIONndash the enthalpy of a reaction carried out at 1 atm and at 25 degr

ees Cndash the enthalpy value relates to the heat absorbed or released

during the formation of 1 mole of a specific compound

The symbol for this value is represented as

formation ofenthalpy Standared fH

Once we know these values we can calculate the change in enthalpy of a reaction or

rxnH

Standard Enthalpy of a Reaction rxnH

)reactants()products(H

isequation dgeneralize The

fHfH nnrxn

bull n = the coefficient of the productsbull m = the coefficient of the reactantsbull Σ = the sum ofbull f = formation bull deg = standard state conditions (1 atm and 25 degrees C)

m

The most stable forms of substances will = 0

EXAMPLEGiven the equation 3 O2(g)rarr2 O3(g) ΔH = +2844 kJ

calculate ΔH for the following reaction 32 O2(g) rarr O3(g)REMEMBER

)reactants()products(H

isequation dgeneralize The

fHfH nnrxn

O2 is the most stable form of oxygen So it is equal to 0

Therefore the the change in enthalpy is all about the product

(x products) - (0 reactants) = rxnH = X products = +2844 kJYou can check using the table on P 247

i

P 247bull The ΔHdegf for O3(g) is +1422 bull The ΔHdegf for O2(g) is 0Given the equation 3 O2(g)rarr2 O3(g) ΔH = +2844 kJThere are two O3 in the equation (1422 kJ x 2 = 2844 kJ)

ndash we must remember the coefficientsbull Then the question asks us to calculate ΔH for the

following reaction 32 O2(g) rarr O3(g) bull There is only one O3

bull Since 32 O2(g) rarr O3(g) is frac12 of 3 O2(g)rarr2 O3(g) the enthalpy of the reaction will be frac12 as well

frac12 (+2844kJ)or +1422kJ

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used

in the manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP IWRITE THE BALANCED EQUATIONTo carry any combustion reaction you need oxygen as a reactant Liquid water will always be included as one of the products

__ C4H4S(l) + ___ O2(g) rarr __ CO2(g) + ___ S02(g) + ___ H20(l)1 6 4 1 2

Change of Enthalpy of Reaction

bull The combustion of thiophene C4H4S(l) a compound used in the manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

USE THE COEFFICIENTS FROM YOUR BALANCED EQUATION TO DETERMINE THE TOTAL CHANGE IN ENTHALPY IN THE FORMATION OF EVERY SUBSTANCE USED IN THE REACTION USE THIS INFORMATION TO WRITE YOUR ΔHdegRXN EQUATION

i

)reactants()products(H fHfH nnrxn

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

i

PRODUCTS

4 (________kJmol CO2(g))2 (________kJmol H2O(l))1 (________kJmol SO2(g))

REACTANTS

X kJmol C4H4S(l)

6 (________kJmol O2(g))

-3935

-2858-2968 000

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

i

[4(-3935) + 1(-2968) + 2(-2858)] ndash [1(ΔH) + 6(0)] = -2523

PRODUCTS －

REACTANTS = ΔHRXN

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP III

SOLVE for X

i

-24424 - ΔH(reactants) = -2523

ΔHC4H4S(l) = 813kJmol

CHANGE IN H

0H EXOTHERMIC

0H CENDOTHERMI

Change in Enthalpy in a multi-step reaction

Change in Enthalpy in a multi-step reaction

CHEMICAL EQUILIBRIA

Things to remember

This is how you convert an Equilibrium constant from Kc to Kp

Kp = Kc (00821 T)^Δn

Δn is the sum of all stoichiometric coefficients belonging to products minus the sum of all stoichiometric coefficients belonging to reactants

T is the temperature during the reaction in Kelvin Remember that Kelvin has exactly the same degree of difference between integers as celsius but it does have a different starting point Kelvins zero is 273deg less than celsius For example 2deg C is equal to 275deg K 27deg C is equal to 300deg K

Qc and Kc Questionh

EQUILIBRIUM of 2 or more RXNs

If a reaction can be expressed as the sum of two or more reactions the equilibrium constant for the overall reaction is given by the product of the equilibrium of the individual reactions

Kc = (Kc) (Kc)

Let me elaborate

nn

nn

c BA

DCK

][][

][][ nn

nn

c DC

FEK

][][

][][

c

c

c

K

__________________

K

K

FEBA

FEDC

DCBA

1st Step Reaction 2nd Step Reaction

OverallReaction

To find the overall equilibrium of a multi-step reaction

cnn

nn

nn

nn

cc KDC

FEx

BA

DCKK

][][

][][

][][

][][

2 step reaction1st step reaction

N2(g) + O2(g) harr 2 NO(g) Kc1 = 23 x 10^-19

2nd step reaction

2 NO(g) + O2(g) harr 2 NO2(g) Kc2 = 3 x 10^6

Kc = Kc1 x Kc2 = (23 x 10^-19)(3 x 10^6) Kc = 7 x 10^-13

Write the equilibrium equation for this multi-step reaction

CHEMICAL KINETICS

Reaction rates

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull Suppose that at a particular moment during the reaction molecular hydrogen is reacting at the rate of 0074 Ms

a) At what rate is ammonia being formedb) At what rate is molecular nitrogen reacting

t

NH

t

H

t

N

322

2

1

3

1

1

1

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is ammonia being formed s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

t

NH

t

NH

t

H

sM

3

32

2

10740

3

1

2

1

3

1t

NHsM

30740

3

2

_

-

_

t

NHsM

30490

GIVEN

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is molecular nitrogen reacting s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

-

GIVEN

t

N

t

N

t

H

sM

2

22

1

10740

3

1

1

1

3

1

t

NsM

20740)1(

3

1

t

NsM

20250

-

Describe the difference between Heat of solution and Heat of dilution Give an example of

each one

Heat of solution is the amount of heat absorbed or released during the act of combining a solute with a solvent An example is adding salt to water There are two steps to this process the Lattice Energy (breaking down the molecules) and the heat of hydration (where the ionically charged particles dissolve by attracting to the dipole water molecules

Heat of dilution refers to the amount of heat absorbed or released when water is added to a solution already made In this latter case if the act of making a solution is exothermic adding more water will make it even more exothermic (hotter) For example as you add more and more water to a solution of sulfuric acid and water the more heat it releases (the more it burns)

Neutralization

bull Explain how you would neutralize an acid and a base

Mix an acidic solution with a basic solution wherein the ratio of H+ ions to OH- ions eq

ual to 1

Cabbage Juice as a pH indicator

bull Explain how cabbage juice works as a pH indicator There is anthocyanin in cabbage juice Anthocyanin reacts w

ith hydroxyl ions (OH-) Hydroxyl ions jump off anthocyanin when in an acidic solution thereby changing the chemical composition of the anthocyanin (the pH indicator) This will cause the juice to react with light differently than if these OH- ions had not jumped off the anthocyanin (makes the juice turn a reddish color) Conversely Hydroxyl ions will jump from the solution onto the anthrocyanin molecules when there is an excess of them in the solution (such as when cabbage juice is mixed with a basic solution) This causes the juice

(the pH indicator) to turn a yellowish color

NOT mentioned

bull Balancing redox reactions

bull Kw

bull Ka

bull Kb

bull pH

bull the galvanic cell

bullKNOW THESE CONCEPTS

• Slide 1
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• Slide 50
• Slide 51

We are not done

bull The question asks us to calculate the molar heat of combustion

bull So you have to find the molar mass of naphthalene (C10H8) have to use periodic table

bull molar mass of C10H8 is 1282 g

bull now we can convert from kJ1435 g to kJmol

How to convert

kJmol 101515

mol 1

2128

4351

5766kJ-

3

810

810

810

combustion ofheat molar

x

HC

HCgx

HCg

INTENSIVE and EXTENSIVE PROPERTIES

bull Extensive Properties - are properties that depend on how much matter is being consideredndash for example volume

bull the property of space a substance takes up is a value dependent on how much of the substance there is

ndash Can you think of others

INTENSIVE and EXTENSIVE PROPERTIES

bull Intensive Properties - properties that do NOT depend on how much matter is being consideredndash for example boiling point

bull the boiling point of liquid water into water vapor is the same regardless of how much water there is

ndash can you think of others

MOLE vs MOLARITY

bull mole (mol)bull 1 mol = 602 x 10^23

particles of a particular substance

bull molarity (M)bull molarity or molar co

ncentration is the number of moles of solute per liter of solutionthe dissolved

substanceA solvent is what a solute dissolves in to make a solution

molarity is defined as

molarity = moles of solute liters of solution

So is molarity an intensive or extensive property

intensive property

Change in Enthalpy of a Reaction

rxnH

Standard Enthalpy of Formation

bull DEFINITIONndash the enthalpy of a reaction carried out at 1 atm and at 25 degr

ees Cndash the enthalpy value relates to the heat absorbed or released

during the formation of 1 mole of a specific compound

The symbol for this value is represented as

formation ofenthalpy Standared fH

Once we know these values we can calculate the change in enthalpy of a reaction or

rxnH

Standard Enthalpy of a Reaction rxnH

)reactants()products(H

isequation dgeneralize The

fHfH nnrxn

bull n = the coefficient of the productsbull m = the coefficient of the reactantsbull Σ = the sum ofbull f = formation bull deg = standard state conditions (1 atm and 25 degrees C)

m

The most stable forms of substances will = 0

EXAMPLEGiven the equation 3 O2(g)rarr2 O3(g) ΔH = +2844 kJ

calculate ΔH for the following reaction 32 O2(g) rarr O3(g)REMEMBER

)reactants()products(H

isequation dgeneralize The

fHfH nnrxn

O2 is the most stable form of oxygen So it is equal to 0

Therefore the the change in enthalpy is all about the product

(x products) - (0 reactants) = rxnH = X products = +2844 kJYou can check using the table on P 247

i

P 247bull The ΔHdegf for O3(g) is +1422 bull The ΔHdegf for O2(g) is 0Given the equation 3 O2(g)rarr2 O3(g) ΔH = +2844 kJThere are two O3 in the equation (1422 kJ x 2 = 2844 kJ)

ndash we must remember the coefficientsbull Then the question asks us to calculate ΔH for the

following reaction 32 O2(g) rarr O3(g) bull There is only one O3

bull Since 32 O2(g) rarr O3(g) is frac12 of 3 O2(g)rarr2 O3(g) the enthalpy of the reaction will be frac12 as well

frac12 (+2844kJ)or +1422kJ

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used

in the manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP IWRITE THE BALANCED EQUATIONTo carry any combustion reaction you need oxygen as a reactant Liquid water will always be included as one of the products

__ C4H4S(l) + ___ O2(g) rarr __ CO2(g) + ___ S02(g) + ___ H20(l)1 6 4 1 2

Change of Enthalpy of Reaction

bull The combustion of thiophene C4H4S(l) a compound used in the manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

USE THE COEFFICIENTS FROM YOUR BALANCED EQUATION TO DETERMINE THE TOTAL CHANGE IN ENTHALPY IN THE FORMATION OF EVERY SUBSTANCE USED IN THE REACTION USE THIS INFORMATION TO WRITE YOUR ΔHdegRXN EQUATION

i

)reactants()products(H fHfH nnrxn

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

i

PRODUCTS

4 (________kJmol CO2(g))2 (________kJmol H2O(l))1 (________kJmol SO2(g))

REACTANTS

X kJmol C4H4S(l)

6 (________kJmol O2(g))

-3935

-2858-2968 000

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

i

[4(-3935) + 1(-2968) + 2(-2858)] ndash [1(ΔH) + 6(0)] = -2523

PRODUCTS －

REACTANTS = ΔHRXN

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP III

SOLVE for X

i

-24424 - ΔH(reactants) = -2523

ΔHC4H4S(l) = 813kJmol

CHANGE IN H

0H EXOTHERMIC

0H CENDOTHERMI

Change in Enthalpy in a multi-step reaction

Change in Enthalpy in a multi-step reaction

CHEMICAL EQUILIBRIA

Things to remember

This is how you convert an Equilibrium constant from Kc to Kp

Kp = Kc (00821 T)^Δn

Δn is the sum of all stoichiometric coefficients belonging to products minus the sum of all stoichiometric coefficients belonging to reactants

T is the temperature during the reaction in Kelvin Remember that Kelvin has exactly the same degree of difference between integers as celsius but it does have a different starting point Kelvins zero is 273deg less than celsius For example 2deg C is equal to 275deg K 27deg C is equal to 300deg K

Qc and Kc Questionh

EQUILIBRIUM of 2 or more RXNs

If a reaction can be expressed as the sum of two or more reactions the equilibrium constant for the overall reaction is given by the product of the equilibrium of the individual reactions

Kc = (Kc) (Kc)

Let me elaborate

nn

nn

c BA

DCK

][][

][][ nn

nn

c DC

FEK

][][

][][

c

c

c

K

__________________

K

K

FEBA

FEDC

DCBA

1st Step Reaction 2nd Step Reaction

OverallReaction

To find the overall equilibrium of a multi-step reaction

cnn

nn

nn

nn

cc KDC

FEx

BA

DCKK

][][

][][

][][

][][

2 step reaction1st step reaction

N2(g) + O2(g) harr 2 NO(g) Kc1 = 23 x 10^-19

2nd step reaction

2 NO(g) + O2(g) harr 2 NO2(g) Kc2 = 3 x 10^6

Kc = Kc1 x Kc2 = (23 x 10^-19)(3 x 10^6) Kc = 7 x 10^-13

Write the equilibrium equation for this multi-step reaction

CHEMICAL KINETICS

Reaction rates

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull Suppose that at a particular moment during the reaction molecular hydrogen is reacting at the rate of 0074 Ms

a) At what rate is ammonia being formedb) At what rate is molecular nitrogen reacting

t

NH

t

H

t

N

322

2

1

3

1

1

1

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is ammonia being formed s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

t

NH

t

NH

t

H

sM

3

32

2

10740

3

1

2

1

3

1t

NHsM

30740

3

2

_

-

_

t

NHsM

30490

GIVEN

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is molecular nitrogen reacting s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

-

GIVEN

t

N

t

N

t

H

sM

2

22

1

10740

3

1

1

1

3

1

t

NsM

20740)1(

3

1

t

NsM

20250

-

Describe the difference between Heat of solution and Heat of dilution Give an example of

each one

Heat of solution is the amount of heat absorbed or released during the act of combining a solute with a solvent An example is adding salt to water There are two steps to this process the Lattice Energy (breaking down the molecules) and the heat of hydration (where the ionically charged particles dissolve by attracting to the dipole water molecules

Heat of dilution refers to the amount of heat absorbed or released when water is added to a solution already made In this latter case if the act of making a solution is exothermic adding more water will make it even more exothermic (hotter) For example as you add more and more water to a solution of sulfuric acid and water the more heat it releases (the more it burns)

Neutralization

bull Explain how you would neutralize an acid and a base

Mix an acidic solution with a basic solution wherein the ratio of H+ ions to OH- ions eq

ual to 1

Cabbage Juice as a pH indicator

bull Explain how cabbage juice works as a pH indicator There is anthocyanin in cabbage juice Anthocyanin reacts w

ith hydroxyl ions (OH-) Hydroxyl ions jump off anthocyanin when in an acidic solution thereby changing the chemical composition of the anthocyanin (the pH indicator) This will cause the juice to react with light differently than if these OH- ions had not jumped off the anthocyanin (makes the juice turn a reddish color) Conversely Hydroxyl ions will jump from the solution onto the anthrocyanin molecules when there is an excess of them in the solution (such as when cabbage juice is mixed with a basic solution) This causes the juice

(the pH indicator) to turn a yellowish color

NOT mentioned

bull Balancing redox reactions

bull Kw

bull Ka

bull Kb

bull pH

bull the galvanic cell

bullKNOW THESE CONCEPTS

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
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• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51

How to convert

kJmol 101515

mol 1

2128

4351

5766kJ-

3

810

810

810

combustion ofheat molar

x

HC

HCgx

HCg

INTENSIVE and EXTENSIVE PROPERTIES

bull Extensive Properties - are properties that depend on how much matter is being consideredndash for example volume

bull the property of space a substance takes up is a value dependent on how much of the substance there is

ndash Can you think of others

INTENSIVE and EXTENSIVE PROPERTIES

bull Intensive Properties - properties that do NOT depend on how much matter is being consideredndash for example boiling point

bull the boiling point of liquid water into water vapor is the same regardless of how much water there is

ndash can you think of others

MOLE vs MOLARITY

bull mole (mol)bull 1 mol = 602 x 10^23

particles of a particular substance

bull molarity (M)bull molarity or molar co

ncentration is the number of moles of solute per liter of solutionthe dissolved

substanceA solvent is what a solute dissolves in to make a solution

molarity is defined as

molarity = moles of solute liters of solution

So is molarity an intensive or extensive property

intensive property

Change in Enthalpy of a Reaction

rxnH

Standard Enthalpy of Formation

bull DEFINITIONndash the enthalpy of a reaction carried out at 1 atm and at 25 degr

ees Cndash the enthalpy value relates to the heat absorbed or released

during the formation of 1 mole of a specific compound

The symbol for this value is represented as

formation ofenthalpy Standared fH

Once we know these values we can calculate the change in enthalpy of a reaction or

rxnH

Standard Enthalpy of a Reaction rxnH

)reactants()products(H

isequation dgeneralize The

fHfH nnrxn

bull n = the coefficient of the productsbull m = the coefficient of the reactantsbull Σ = the sum ofbull f = formation bull deg = standard state conditions (1 atm and 25 degrees C)

m

The most stable forms of substances will = 0

EXAMPLEGiven the equation 3 O2(g)rarr2 O3(g) ΔH = +2844 kJ

calculate ΔH for the following reaction 32 O2(g) rarr O3(g)REMEMBER

)reactants()products(H

isequation dgeneralize The

fHfH nnrxn

O2 is the most stable form of oxygen So it is equal to 0

Therefore the the change in enthalpy is all about the product

(x products) - (0 reactants) = rxnH = X products = +2844 kJYou can check using the table on P 247

i

P 247bull The ΔHdegf for O3(g) is +1422 bull The ΔHdegf for O2(g) is 0Given the equation 3 O2(g)rarr2 O3(g) ΔH = +2844 kJThere are two O3 in the equation (1422 kJ x 2 = 2844 kJ)

ndash we must remember the coefficientsbull Then the question asks us to calculate ΔH for the

following reaction 32 O2(g) rarr O3(g) bull There is only one O3

bull Since 32 O2(g) rarr O3(g) is frac12 of 3 O2(g)rarr2 O3(g) the enthalpy of the reaction will be frac12 as well

frac12 (+2844kJ)or +1422kJ

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used

in the manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP IWRITE THE BALANCED EQUATIONTo carry any combustion reaction you need oxygen as a reactant Liquid water will always be included as one of the products

__ C4H4S(l) + ___ O2(g) rarr __ CO2(g) + ___ S02(g) + ___ H20(l)1 6 4 1 2

Change of Enthalpy of Reaction

bull The combustion of thiophene C4H4S(l) a compound used in the manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

USE THE COEFFICIENTS FROM YOUR BALANCED EQUATION TO DETERMINE THE TOTAL CHANGE IN ENTHALPY IN THE FORMATION OF EVERY SUBSTANCE USED IN THE REACTION USE THIS INFORMATION TO WRITE YOUR ΔHdegRXN EQUATION

i

)reactants()products(H fHfH nnrxn

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

i

PRODUCTS

4 (________kJmol CO2(g))2 (________kJmol H2O(l))1 (________kJmol SO2(g))

REACTANTS

X kJmol C4H4S(l)

6 (________kJmol O2(g))

-3935

-2858-2968 000

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

i

[4(-3935) + 1(-2968) + 2(-2858)] ndash [1(ΔH) + 6(0)] = -2523

PRODUCTS －

REACTANTS = ΔHRXN

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP III

SOLVE for X

i

-24424 - ΔH(reactants) = -2523

ΔHC4H4S(l) = 813kJmol

CHANGE IN H

0H EXOTHERMIC

0H CENDOTHERMI

Change in Enthalpy in a multi-step reaction

Change in Enthalpy in a multi-step reaction

CHEMICAL EQUILIBRIA

Things to remember

This is how you convert an Equilibrium constant from Kc to Kp

Kp = Kc (00821 T)^Δn

Δn is the sum of all stoichiometric coefficients belonging to products minus the sum of all stoichiometric coefficients belonging to reactants

T is the temperature during the reaction in Kelvin Remember that Kelvin has exactly the same degree of difference between integers as celsius but it does have a different starting point Kelvins zero is 273deg less than celsius For example 2deg C is equal to 275deg K 27deg C is equal to 300deg K

Qc and Kc Questionh

EQUILIBRIUM of 2 or more RXNs

If a reaction can be expressed as the sum of two or more reactions the equilibrium constant for the overall reaction is given by the product of the equilibrium of the individual reactions

Kc = (Kc) (Kc)

Let me elaborate

nn

nn

c BA

DCK

][][

][][ nn

nn

c DC

FEK

][][

][][

c

c

c

K

__________________

K

K

FEBA

FEDC

DCBA

1st Step Reaction 2nd Step Reaction

OverallReaction

To find the overall equilibrium of a multi-step reaction

cnn

nn

nn

nn

cc KDC

FEx

BA

DCKK

][][

][][

][][

][][

2 step reaction1st step reaction

N2(g) + O2(g) harr 2 NO(g) Kc1 = 23 x 10^-19

2nd step reaction

2 NO(g) + O2(g) harr 2 NO2(g) Kc2 = 3 x 10^6

Kc = Kc1 x Kc2 = (23 x 10^-19)(3 x 10^6) Kc = 7 x 10^-13

Write the equilibrium equation for this multi-step reaction

CHEMICAL KINETICS

Reaction rates

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull Suppose that at a particular moment during the reaction molecular hydrogen is reacting at the rate of 0074 Ms

a) At what rate is ammonia being formedb) At what rate is molecular nitrogen reacting

t

NH

t

H

t

N

322

2

1

3

1

1

1

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is ammonia being formed s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

t

NH

t

NH

t

H

sM

3

32

2

10740

3

1

2

1

3

1t

NHsM

30740

3

2

_

-

_

t

NHsM

30490

GIVEN

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is molecular nitrogen reacting s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

-

GIVEN

t

N

t

N

t

H

sM

2

22

1

10740

3

1

1

1

3

1

t

NsM

20740)1(

3

1

t

NsM

20250

-

Describe the difference between Heat of solution and Heat of dilution Give an example of

each one

Heat of solution is the amount of heat absorbed or released during the act of combining a solute with a solvent An example is adding salt to water There are two steps to this process the Lattice Energy (breaking down the molecules) and the heat of hydration (where the ionically charged particles dissolve by attracting to the dipole water molecules

Heat of dilution refers to the amount of heat absorbed or released when water is added to a solution already made In this latter case if the act of making a solution is exothermic adding more water will make it even more exothermic (hotter) For example as you add more and more water to a solution of sulfuric acid and water the more heat it releases (the more it burns)

Neutralization

bull Explain how you would neutralize an acid and a base

Mix an acidic solution with a basic solution wherein the ratio of H+ ions to OH- ions eq

ual to 1

Cabbage Juice as a pH indicator

bull Explain how cabbage juice works as a pH indicator There is anthocyanin in cabbage juice Anthocyanin reacts w

ith hydroxyl ions (OH-) Hydroxyl ions jump off anthocyanin when in an acidic solution thereby changing the chemical composition of the anthocyanin (the pH indicator) This will cause the juice to react with light differently than if these OH- ions had not jumped off the anthocyanin (makes the juice turn a reddish color) Conversely Hydroxyl ions will jump from the solution onto the anthrocyanin molecules when there is an excess of them in the solution (such as when cabbage juice is mixed with a basic solution) This causes the juice

(the pH indicator) to turn a yellowish color

NOT mentioned

bull Balancing redox reactions

bull Kw

bull Ka

bull Kb

bull pH

bull the galvanic cell

bullKNOW THESE CONCEPTS

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INTENSIVE and EXTENSIVE PROPERTIES

bull Extensive Properties - are properties that depend on how much matter is being consideredndash for example volume

bull the property of space a substance takes up is a value dependent on how much of the substance there is

ndash Can you think of others

INTENSIVE and EXTENSIVE PROPERTIES

bull Intensive Properties - properties that do NOT depend on how much matter is being consideredndash for example boiling point

bull the boiling point of liquid water into water vapor is the same regardless of how much water there is

ndash can you think of others

MOLE vs MOLARITY

bull mole (mol)bull 1 mol = 602 x 10^23

particles of a particular substance

bull molarity (M)bull molarity or molar co

ncentration is the number of moles of solute per liter of solutionthe dissolved

substanceA solvent is what a solute dissolves in to make a solution

molarity is defined as

molarity = moles of solute liters of solution

So is molarity an intensive or extensive property

intensive property

Change in Enthalpy of a Reaction

rxnH

Standard Enthalpy of Formation

bull DEFINITIONndash the enthalpy of a reaction carried out at 1 atm and at 25 degr

ees Cndash the enthalpy value relates to the heat absorbed or released

during the formation of 1 mole of a specific compound

The symbol for this value is represented as

formation ofenthalpy Standared fH

Once we know these values we can calculate the change in enthalpy of a reaction or

rxnH

Standard Enthalpy of a Reaction rxnH

)reactants()products(H

isequation dgeneralize The

fHfH nnrxn

bull n = the coefficient of the productsbull m = the coefficient of the reactantsbull Σ = the sum ofbull f = formation bull deg = standard state conditions (1 atm and 25 degrees C)

m

The most stable forms of substances will = 0

EXAMPLEGiven the equation 3 O2(g)rarr2 O3(g) ΔH = +2844 kJ

calculate ΔH for the following reaction 32 O2(g) rarr O3(g)REMEMBER

)reactants()products(H

isequation dgeneralize The

fHfH nnrxn

O2 is the most stable form of oxygen So it is equal to 0

Therefore the the change in enthalpy is all about the product

(x products) - (0 reactants) = rxnH = X products = +2844 kJYou can check using the table on P 247

i

P 247bull The ΔHdegf for O3(g) is +1422 bull The ΔHdegf for O2(g) is 0Given the equation 3 O2(g)rarr2 O3(g) ΔH = +2844 kJThere are two O3 in the equation (1422 kJ x 2 = 2844 kJ)

ndash we must remember the coefficientsbull Then the question asks us to calculate ΔH for the

following reaction 32 O2(g) rarr O3(g) bull There is only one O3

bull Since 32 O2(g) rarr O3(g) is frac12 of 3 O2(g)rarr2 O3(g) the enthalpy of the reaction will be frac12 as well

frac12 (+2844kJ)or +1422kJ

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used

in the manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP IWRITE THE BALANCED EQUATIONTo carry any combustion reaction you need oxygen as a reactant Liquid water will always be included as one of the products

__ C4H4S(l) + ___ O2(g) rarr __ CO2(g) + ___ S02(g) + ___ H20(l)1 6 4 1 2

Change of Enthalpy of Reaction

bull The combustion of thiophene C4H4S(l) a compound used in the manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

USE THE COEFFICIENTS FROM YOUR BALANCED EQUATION TO DETERMINE THE TOTAL CHANGE IN ENTHALPY IN THE FORMATION OF EVERY SUBSTANCE USED IN THE REACTION USE THIS INFORMATION TO WRITE YOUR ΔHdegRXN EQUATION

i

)reactants()products(H fHfH nnrxn

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

i

PRODUCTS

4 (________kJmol CO2(g))2 (________kJmol H2O(l))1 (________kJmol SO2(g))

REACTANTS

X kJmol C4H4S(l)

6 (________kJmol O2(g))

-3935

-2858-2968 000

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

i

[4(-3935) + 1(-2968) + 2(-2858)] ndash [1(ΔH) + 6(0)] = -2523

PRODUCTS －

REACTANTS = ΔHRXN

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP III

SOLVE for X

i

-24424 - ΔH(reactants) = -2523

ΔHC4H4S(l) = 813kJmol

CHANGE IN H

0H EXOTHERMIC

0H CENDOTHERMI

Change in Enthalpy in a multi-step reaction

Change in Enthalpy in a multi-step reaction

CHEMICAL EQUILIBRIA

Things to remember

This is how you convert an Equilibrium constant from Kc to Kp

Kp = Kc (00821 T)^Δn

Δn is the sum of all stoichiometric coefficients belonging to products minus the sum of all stoichiometric coefficients belonging to reactants

T is the temperature during the reaction in Kelvin Remember that Kelvin has exactly the same degree of difference between integers as celsius but it does have a different starting point Kelvins zero is 273deg less than celsius For example 2deg C is equal to 275deg K 27deg C is equal to 300deg K

Qc and Kc Questionh

EQUILIBRIUM of 2 or more RXNs

If a reaction can be expressed as the sum of two or more reactions the equilibrium constant for the overall reaction is given by the product of the equilibrium of the individual reactions

Kc = (Kc) (Kc)

Let me elaborate

nn

nn

c BA

DCK

][][

][][ nn

nn

c DC

FEK

][][

][][

c

c

c

K

__________________

K

K

FEBA

FEDC

DCBA

1st Step Reaction 2nd Step Reaction

OverallReaction

To find the overall equilibrium of a multi-step reaction

cnn

nn

nn

nn

cc KDC

FEx

BA

DCKK

][][

][][

][][

][][

2 step reaction1st step reaction

N2(g) + O2(g) harr 2 NO(g) Kc1 = 23 x 10^-19

2nd step reaction

2 NO(g) + O2(g) harr 2 NO2(g) Kc2 = 3 x 10^6

Kc = Kc1 x Kc2 = (23 x 10^-19)(3 x 10^6) Kc = 7 x 10^-13

Write the equilibrium equation for this multi-step reaction

CHEMICAL KINETICS

Reaction rates

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull Suppose that at a particular moment during the reaction molecular hydrogen is reacting at the rate of 0074 Ms

a) At what rate is ammonia being formedb) At what rate is molecular nitrogen reacting

t

NH

t

H

t

N

322

2

1

3

1

1

1

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is ammonia being formed s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

t

NH

t

NH

t

H

sM

3

32

2

10740

3

1

2

1

3

1t

NHsM

30740

3

2

_

-

_

t

NHsM

30490

GIVEN

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is molecular nitrogen reacting s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

-

GIVEN

t

N

t

N

t

H

sM

2

22

1

10740

3

1

1

1

3

1

t

NsM

20740)1(

3

1

t

NsM

20250

-

Describe the difference between Heat of solution and Heat of dilution Give an example of

each one

Heat of solution is the amount of heat absorbed or released during the act of combining a solute with a solvent An example is adding salt to water There are two steps to this process the Lattice Energy (breaking down the molecules) and the heat of hydration (where the ionically charged particles dissolve by attracting to the dipole water molecules

Heat of dilution refers to the amount of heat absorbed or released when water is added to a solution already made In this latter case if the act of making a solution is exothermic adding more water will make it even more exothermic (hotter) For example as you add more and more water to a solution of sulfuric acid and water the more heat it releases (the more it burns)

Neutralization

bull Explain how you would neutralize an acid and a base

Mix an acidic solution with a basic solution wherein the ratio of H+ ions to OH- ions eq

ual to 1

Cabbage Juice as a pH indicator

bull Explain how cabbage juice works as a pH indicator There is anthocyanin in cabbage juice Anthocyanin reacts w

ith hydroxyl ions (OH-) Hydroxyl ions jump off anthocyanin when in an acidic solution thereby changing the chemical composition of the anthocyanin (the pH indicator) This will cause the juice to react with light differently than if these OH- ions had not jumped off the anthocyanin (makes the juice turn a reddish color) Conversely Hydroxyl ions will jump from the solution onto the anthrocyanin molecules when there is an excess of them in the solution (such as when cabbage juice is mixed with a basic solution) This causes the juice

(the pH indicator) to turn a yellowish color

NOT mentioned

bull Balancing redox reactions

bull Kw

bull Ka

bull Kb

bull pH

bull the galvanic cell

bullKNOW THESE CONCEPTS

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INTENSIVE and EXTENSIVE PROPERTIES

bull Intensive Properties - properties that do NOT depend on how much matter is being consideredndash for example boiling point

bull the boiling point of liquid water into water vapor is the same regardless of how much water there is

ndash can you think of others

MOLE vs MOLARITY

bull mole (mol)bull 1 mol = 602 x 10^23

particles of a particular substance

bull molarity (M)bull molarity or molar co

ncentration is the number of moles of solute per liter of solutionthe dissolved

substanceA solvent is what a solute dissolves in to make a solution

molarity is defined as

molarity = moles of solute liters of solution

So is molarity an intensive or extensive property

intensive property

Change in Enthalpy of a Reaction

rxnH

Standard Enthalpy of Formation

bull DEFINITIONndash the enthalpy of a reaction carried out at 1 atm and at 25 degr

ees Cndash the enthalpy value relates to the heat absorbed or released

during the formation of 1 mole of a specific compound

The symbol for this value is represented as

formation ofenthalpy Standared fH

Once we know these values we can calculate the change in enthalpy of a reaction or

rxnH

Standard Enthalpy of a Reaction rxnH

)reactants()products(H

isequation dgeneralize The

fHfH nnrxn

bull n = the coefficient of the productsbull m = the coefficient of the reactantsbull Σ = the sum ofbull f = formation bull deg = standard state conditions (1 atm and 25 degrees C)

m

The most stable forms of substances will = 0

EXAMPLEGiven the equation 3 O2(g)rarr2 O3(g) ΔH = +2844 kJ

calculate ΔH for the following reaction 32 O2(g) rarr O3(g)REMEMBER

)reactants()products(H

isequation dgeneralize The

fHfH nnrxn

O2 is the most stable form of oxygen So it is equal to 0

Therefore the the change in enthalpy is all about the product

(x products) - (0 reactants) = rxnH = X products = +2844 kJYou can check using the table on P 247

i

P 247bull The ΔHdegf for O3(g) is +1422 bull The ΔHdegf for O2(g) is 0Given the equation 3 O2(g)rarr2 O3(g) ΔH = +2844 kJThere are two O3 in the equation (1422 kJ x 2 = 2844 kJ)

ndash we must remember the coefficientsbull Then the question asks us to calculate ΔH for the

following reaction 32 O2(g) rarr O3(g) bull There is only one O3

bull Since 32 O2(g) rarr O3(g) is frac12 of 3 O2(g)rarr2 O3(g) the enthalpy of the reaction will be frac12 as well

frac12 (+2844kJ)or +1422kJ

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used

in the manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP IWRITE THE BALANCED EQUATIONTo carry any combustion reaction you need oxygen as a reactant Liquid water will always be included as one of the products

__ C4H4S(l) + ___ O2(g) rarr __ CO2(g) + ___ S02(g) + ___ H20(l)1 6 4 1 2

Change of Enthalpy of Reaction

bull The combustion of thiophene C4H4S(l) a compound used in the manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

USE THE COEFFICIENTS FROM YOUR BALANCED EQUATION TO DETERMINE THE TOTAL CHANGE IN ENTHALPY IN THE FORMATION OF EVERY SUBSTANCE USED IN THE REACTION USE THIS INFORMATION TO WRITE YOUR ΔHdegRXN EQUATION

i

)reactants()products(H fHfH nnrxn

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

i

PRODUCTS

4 (________kJmol CO2(g))2 (________kJmol H2O(l))1 (________kJmol SO2(g))

REACTANTS

X kJmol C4H4S(l)

6 (________kJmol O2(g))

-3935

-2858-2968 000

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

i

[4(-3935) + 1(-2968) + 2(-2858)] ndash [1(ΔH) + 6(0)] = -2523

PRODUCTS －

REACTANTS = ΔHRXN

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP III

SOLVE for X

i

-24424 - ΔH(reactants) = -2523

ΔHC4H4S(l) = 813kJmol

CHANGE IN H

0H EXOTHERMIC

0H CENDOTHERMI

Change in Enthalpy in a multi-step reaction

Change in Enthalpy in a multi-step reaction

CHEMICAL EQUILIBRIA

Things to remember

This is how you convert an Equilibrium constant from Kc to Kp

Kp = Kc (00821 T)^Δn

Δn is the sum of all stoichiometric coefficients belonging to products minus the sum of all stoichiometric coefficients belonging to reactants

T is the temperature during the reaction in Kelvin Remember that Kelvin has exactly the same degree of difference between integers as celsius but it does have a different starting point Kelvins zero is 273deg less than celsius For example 2deg C is equal to 275deg K 27deg C is equal to 300deg K

Qc and Kc Questionh

EQUILIBRIUM of 2 or more RXNs

If a reaction can be expressed as the sum of two or more reactions the equilibrium constant for the overall reaction is given by the product of the equilibrium of the individual reactions

Kc = (Kc) (Kc)

Let me elaborate

nn

nn

c BA

DCK

][][

][][ nn

nn

c DC

FEK

][][

][][

c

c

c

K

__________________

K

K

FEBA

FEDC

DCBA

1st Step Reaction 2nd Step Reaction

OverallReaction

To find the overall equilibrium of a multi-step reaction

cnn

nn

nn

nn

cc KDC

FEx

BA

DCKK

][][

][][

][][

][][

2 step reaction1st step reaction

N2(g) + O2(g) harr 2 NO(g) Kc1 = 23 x 10^-19

2nd step reaction

2 NO(g) + O2(g) harr 2 NO2(g) Kc2 = 3 x 10^6

Kc = Kc1 x Kc2 = (23 x 10^-19)(3 x 10^6) Kc = 7 x 10^-13

Write the equilibrium equation for this multi-step reaction

CHEMICAL KINETICS

Reaction rates

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull Suppose that at a particular moment during the reaction molecular hydrogen is reacting at the rate of 0074 Ms

a) At what rate is ammonia being formedb) At what rate is molecular nitrogen reacting

t

NH

t

H

t

N

322

2

1

3

1

1

1

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is ammonia being formed s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

t

NH

t

NH

t

H

sM

3

32

2

10740

3

1

2

1

3

1t

NHsM

30740

3

2

_

-

_

t

NHsM

30490

GIVEN

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is molecular nitrogen reacting s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

-

GIVEN

t

N

t

N

t

H

sM

2

22

1

10740

3

1

1

1

3

1

t

NsM

20740)1(

3

1

t

NsM

20250

-

Describe the difference between Heat of solution and Heat of dilution Give an example of

each one

Heat of solution is the amount of heat absorbed or released during the act of combining a solute with a solvent An example is adding salt to water There are two steps to this process the Lattice Energy (breaking down the molecules) and the heat of hydration (where the ionically charged particles dissolve by attracting to the dipole water molecules

Heat of dilution refers to the amount of heat absorbed or released when water is added to a solution already made In this latter case if the act of making a solution is exothermic adding more water will make it even more exothermic (hotter) For example as you add more and more water to a solution of sulfuric acid and water the more heat it releases (the more it burns)

Neutralization

bull Explain how you would neutralize an acid and a base

Mix an acidic solution with a basic solution wherein the ratio of H+ ions to OH- ions eq

ual to 1

Cabbage Juice as a pH indicator

bull Explain how cabbage juice works as a pH indicator There is anthocyanin in cabbage juice Anthocyanin reacts w

ith hydroxyl ions (OH-) Hydroxyl ions jump off anthocyanin when in an acidic solution thereby changing the chemical composition of the anthocyanin (the pH indicator) This will cause the juice to react with light differently than if these OH- ions had not jumped off the anthocyanin (makes the juice turn a reddish color) Conversely Hydroxyl ions will jump from the solution onto the anthrocyanin molecules when there is an excess of them in the solution (such as when cabbage juice is mixed with a basic solution) This causes the juice

(the pH indicator) to turn a yellowish color

NOT mentioned

bull Balancing redox reactions

bull Kw

bull Ka

bull Kb

bull pH

bull the galvanic cell

bullKNOW THESE CONCEPTS

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
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• Slide 50
• Slide 51

MOLE vs MOLARITY

bull mole (mol)bull 1 mol = 602 x 10^23

particles of a particular substance

bull molarity (M)bull molarity or molar co

ncentration is the number of moles of solute per liter of solutionthe dissolved

substanceA solvent is what a solute dissolves in to make a solution

molarity is defined as

molarity = moles of solute liters of solution

So is molarity an intensive or extensive property

intensive property

Change in Enthalpy of a Reaction

rxnH

Standard Enthalpy of Formation

bull DEFINITIONndash the enthalpy of a reaction carried out at 1 atm and at 25 degr

ees Cndash the enthalpy value relates to the heat absorbed or released

during the formation of 1 mole of a specific compound

The symbol for this value is represented as

formation ofenthalpy Standared fH

Once we know these values we can calculate the change in enthalpy of a reaction or

rxnH

Standard Enthalpy of a Reaction rxnH

)reactants()products(H

isequation dgeneralize The

fHfH nnrxn

bull n = the coefficient of the productsbull m = the coefficient of the reactantsbull Σ = the sum ofbull f = formation bull deg = standard state conditions (1 atm and 25 degrees C)

m

The most stable forms of substances will = 0

EXAMPLEGiven the equation 3 O2(g)rarr2 O3(g) ΔH = +2844 kJ

calculate ΔH for the following reaction 32 O2(g) rarr O3(g)REMEMBER

)reactants()products(H

isequation dgeneralize The

fHfH nnrxn

O2 is the most stable form of oxygen So it is equal to 0

Therefore the the change in enthalpy is all about the product

(x products) - (0 reactants) = rxnH = X products = +2844 kJYou can check using the table on P 247

i

P 247bull The ΔHdegf for O3(g) is +1422 bull The ΔHdegf for O2(g) is 0Given the equation 3 O2(g)rarr2 O3(g) ΔH = +2844 kJThere are two O3 in the equation (1422 kJ x 2 = 2844 kJ)

ndash we must remember the coefficientsbull Then the question asks us to calculate ΔH for the

following reaction 32 O2(g) rarr O3(g) bull There is only one O3

bull Since 32 O2(g) rarr O3(g) is frac12 of 3 O2(g)rarr2 O3(g) the enthalpy of the reaction will be frac12 as well

frac12 (+2844kJ)or +1422kJ

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used

in the manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP IWRITE THE BALANCED EQUATIONTo carry any combustion reaction you need oxygen as a reactant Liquid water will always be included as one of the products

__ C4H4S(l) + ___ O2(g) rarr __ CO2(g) + ___ S02(g) + ___ H20(l)1 6 4 1 2

Change of Enthalpy of Reaction

bull The combustion of thiophene C4H4S(l) a compound used in the manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

USE THE COEFFICIENTS FROM YOUR BALANCED EQUATION TO DETERMINE THE TOTAL CHANGE IN ENTHALPY IN THE FORMATION OF EVERY SUBSTANCE USED IN THE REACTION USE THIS INFORMATION TO WRITE YOUR ΔHdegRXN EQUATION

i

)reactants()products(H fHfH nnrxn

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

i

PRODUCTS

4 (________kJmol CO2(g))2 (________kJmol H2O(l))1 (________kJmol SO2(g))

REACTANTS

X kJmol C4H4S(l)

6 (________kJmol O2(g))

-3935

-2858-2968 000

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

i

[4(-3935) + 1(-2968) + 2(-2858)] ndash [1(ΔH) + 6(0)] = -2523

PRODUCTS －

REACTANTS = ΔHRXN

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP III

SOLVE for X

i

-24424 - ΔH(reactants) = -2523

ΔHC4H4S(l) = 813kJmol

CHANGE IN H

0H EXOTHERMIC

0H CENDOTHERMI

Change in Enthalpy in a multi-step reaction

Change in Enthalpy in a multi-step reaction

CHEMICAL EQUILIBRIA

Things to remember

This is how you convert an Equilibrium constant from Kc to Kp

Kp = Kc (00821 T)^Δn

Δn is the sum of all stoichiometric coefficients belonging to products minus the sum of all stoichiometric coefficients belonging to reactants

T is the temperature during the reaction in Kelvin Remember that Kelvin has exactly the same degree of difference between integers as celsius but it does have a different starting point Kelvins zero is 273deg less than celsius For example 2deg C is equal to 275deg K 27deg C is equal to 300deg K

Qc and Kc Questionh

EQUILIBRIUM of 2 or more RXNs

If a reaction can be expressed as the sum of two or more reactions the equilibrium constant for the overall reaction is given by the product of the equilibrium of the individual reactions

Kc = (Kc) (Kc)

Let me elaborate

nn

nn

c BA

DCK

][][

][][ nn

nn

c DC

FEK

][][

][][

c

c

c

K

__________________

K

K

FEBA

FEDC

DCBA

1st Step Reaction 2nd Step Reaction

OverallReaction

To find the overall equilibrium of a multi-step reaction

cnn

nn

nn

nn

cc KDC

FEx

BA

DCKK

][][

][][

][][

][][

2 step reaction1st step reaction

N2(g) + O2(g) harr 2 NO(g) Kc1 = 23 x 10^-19

2nd step reaction

2 NO(g) + O2(g) harr 2 NO2(g) Kc2 = 3 x 10^6

Kc = Kc1 x Kc2 = (23 x 10^-19)(3 x 10^6) Kc = 7 x 10^-13

Write the equilibrium equation for this multi-step reaction

CHEMICAL KINETICS

Reaction rates

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull Suppose that at a particular moment during the reaction molecular hydrogen is reacting at the rate of 0074 Ms

a) At what rate is ammonia being formedb) At what rate is molecular nitrogen reacting

t

NH

t

H

t

N

322

2

1

3

1

1

1

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is ammonia being formed s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

t

NH

t

NH

t

H

sM

3

32

2

10740

3

1

2

1

3

1t

NHsM

30740

3

2

_

-

_

t

NHsM

30490

GIVEN

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is molecular nitrogen reacting s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

-

GIVEN

t

N

t

N

t

H

sM

2

22

1

10740

3

1

1

1

3

1

t

NsM

20740)1(

3

1

t

NsM

20250

-

Describe the difference between Heat of solution and Heat of dilution Give an example of

each one

Heat of solution is the amount of heat absorbed or released during the act of combining a solute with a solvent An example is adding salt to water There are two steps to this process the Lattice Energy (breaking down the molecules) and the heat of hydration (where the ionically charged particles dissolve by attracting to the dipole water molecules

Heat of dilution refers to the amount of heat absorbed or released when water is added to a solution already made In this latter case if the act of making a solution is exothermic adding more water will make it even more exothermic (hotter) For example as you add more and more water to a solution of sulfuric acid and water the more heat it releases (the more it burns)

Neutralization

bull Explain how you would neutralize an acid and a base

Mix an acidic solution with a basic solution wherein the ratio of H+ ions to OH- ions eq

ual to 1

Cabbage Juice as a pH indicator

bull Explain how cabbage juice works as a pH indicator There is anthocyanin in cabbage juice Anthocyanin reacts w

ith hydroxyl ions (OH-) Hydroxyl ions jump off anthocyanin when in an acidic solution thereby changing the chemical composition of the anthocyanin (the pH indicator) This will cause the juice to react with light differently than if these OH- ions had not jumped off the anthocyanin (makes the juice turn a reddish color) Conversely Hydroxyl ions will jump from the solution onto the anthrocyanin molecules when there is an excess of them in the solution (such as when cabbage juice is mixed with a basic solution) This causes the juice

(the pH indicator) to turn a yellowish color

NOT mentioned

bull Balancing redox reactions

bull Kw

bull Ka

bull Kb

bull pH

bull the galvanic cell

bullKNOW THESE CONCEPTS

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molarity is defined as

molarity = moles of solute liters of solution

So is molarity an intensive or extensive property

intensive property

Change in Enthalpy of a Reaction

rxnH

Standard Enthalpy of Formation

bull DEFINITIONndash the enthalpy of a reaction carried out at 1 atm and at 25 degr

ees Cndash the enthalpy value relates to the heat absorbed or released

during the formation of 1 mole of a specific compound

The symbol for this value is represented as

formation ofenthalpy Standared fH

Once we know these values we can calculate the change in enthalpy of a reaction or

rxnH

Standard Enthalpy of a Reaction rxnH

)reactants()products(H

isequation dgeneralize The

fHfH nnrxn

bull n = the coefficient of the productsbull m = the coefficient of the reactantsbull Σ = the sum ofbull f = formation bull deg = standard state conditions (1 atm and 25 degrees C)

m

The most stable forms of substances will = 0

EXAMPLEGiven the equation 3 O2(g)rarr2 O3(g) ΔH = +2844 kJ

calculate ΔH for the following reaction 32 O2(g) rarr O3(g)REMEMBER

)reactants()products(H

isequation dgeneralize The

fHfH nnrxn

O2 is the most stable form of oxygen So it is equal to 0

Therefore the the change in enthalpy is all about the product

(x products) - (0 reactants) = rxnH = X products = +2844 kJYou can check using the table on P 247

i

P 247bull The ΔHdegf for O3(g) is +1422 bull The ΔHdegf for O2(g) is 0Given the equation 3 O2(g)rarr2 O3(g) ΔH = +2844 kJThere are two O3 in the equation (1422 kJ x 2 = 2844 kJ)

ndash we must remember the coefficientsbull Then the question asks us to calculate ΔH for the

following reaction 32 O2(g) rarr O3(g) bull There is only one O3

bull Since 32 O2(g) rarr O3(g) is frac12 of 3 O2(g)rarr2 O3(g) the enthalpy of the reaction will be frac12 as well

frac12 (+2844kJ)or +1422kJ

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used

in the manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP IWRITE THE BALANCED EQUATIONTo carry any combustion reaction you need oxygen as a reactant Liquid water will always be included as one of the products

__ C4H4S(l) + ___ O2(g) rarr __ CO2(g) + ___ S02(g) + ___ H20(l)1 6 4 1 2

Change of Enthalpy of Reaction

bull The combustion of thiophene C4H4S(l) a compound used in the manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

USE THE COEFFICIENTS FROM YOUR BALANCED EQUATION TO DETERMINE THE TOTAL CHANGE IN ENTHALPY IN THE FORMATION OF EVERY SUBSTANCE USED IN THE REACTION USE THIS INFORMATION TO WRITE YOUR ΔHdegRXN EQUATION

i

)reactants()products(H fHfH nnrxn

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

i

PRODUCTS

4 (________kJmol CO2(g))2 (________kJmol H2O(l))1 (________kJmol SO2(g))

REACTANTS

X kJmol C4H4S(l)

6 (________kJmol O2(g))

-3935

-2858-2968 000

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

i

[4(-3935) + 1(-2968) + 2(-2858)] ndash [1(ΔH) + 6(0)] = -2523

PRODUCTS －

REACTANTS = ΔHRXN

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP III

SOLVE for X

i

-24424 - ΔH(reactants) = -2523

ΔHC4H4S(l) = 813kJmol

CHANGE IN H

0H EXOTHERMIC

0H CENDOTHERMI

Change in Enthalpy in a multi-step reaction

Change in Enthalpy in a multi-step reaction

CHEMICAL EQUILIBRIA

Things to remember

This is how you convert an Equilibrium constant from Kc to Kp

Kp = Kc (00821 T)^Δn

Δn is the sum of all stoichiometric coefficients belonging to products minus the sum of all stoichiometric coefficients belonging to reactants

T is the temperature during the reaction in Kelvin Remember that Kelvin has exactly the same degree of difference between integers as celsius but it does have a different starting point Kelvins zero is 273deg less than celsius For example 2deg C is equal to 275deg K 27deg C is equal to 300deg K

Qc and Kc Questionh

EQUILIBRIUM of 2 or more RXNs

If a reaction can be expressed as the sum of two or more reactions the equilibrium constant for the overall reaction is given by the product of the equilibrium of the individual reactions

Kc = (Kc) (Kc)

Let me elaborate

nn

nn

c BA

DCK

][][

][][ nn

nn

c DC

FEK

][][

][][

c

c

c

K

__________________

K

K

FEBA

FEDC

DCBA

1st Step Reaction 2nd Step Reaction

OverallReaction

To find the overall equilibrium of a multi-step reaction

cnn

nn

nn

nn

cc KDC

FEx

BA

DCKK

][][

][][

][][

][][

2 step reaction1st step reaction

N2(g) + O2(g) harr 2 NO(g) Kc1 = 23 x 10^-19

2nd step reaction

2 NO(g) + O2(g) harr 2 NO2(g) Kc2 = 3 x 10^6

Kc = Kc1 x Kc2 = (23 x 10^-19)(3 x 10^6) Kc = 7 x 10^-13

Write the equilibrium equation for this multi-step reaction

CHEMICAL KINETICS

Reaction rates

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull Suppose that at a particular moment during the reaction molecular hydrogen is reacting at the rate of 0074 Ms

a) At what rate is ammonia being formedb) At what rate is molecular nitrogen reacting

t

NH

t

H

t

N

322

2

1

3

1

1

1

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is ammonia being formed s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

t

NH

t

NH

t

H

sM

3

32

2

10740

3

1

2

1

3

1t

NHsM

30740

3

2

_

-

_

t

NHsM

30490

GIVEN

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is molecular nitrogen reacting s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

-

GIVEN

t

N

t

N

t

H

sM

2

22

1

10740

3

1

1

1

3

1

t

NsM

20740)1(

3

1

t

NsM

20250

-

Describe the difference between Heat of solution and Heat of dilution Give an example of

each one

Heat of solution is the amount of heat absorbed or released during the act of combining a solute with a solvent An example is adding salt to water There are two steps to this process the Lattice Energy (breaking down the molecules) and the heat of hydration (where the ionically charged particles dissolve by attracting to the dipole water molecules

Heat of dilution refers to the amount of heat absorbed or released when water is added to a solution already made In this latter case if the act of making a solution is exothermic adding more water will make it even more exothermic (hotter) For example as you add more and more water to a solution of sulfuric acid and water the more heat it releases (the more it burns)

Neutralization

bull Explain how you would neutralize an acid and a base

Mix an acidic solution with a basic solution wherein the ratio of H+ ions to OH- ions eq

ual to 1

Cabbage Juice as a pH indicator

bull Explain how cabbage juice works as a pH indicator There is anthocyanin in cabbage juice Anthocyanin reacts w

ith hydroxyl ions (OH-) Hydroxyl ions jump off anthocyanin when in an acidic solution thereby changing the chemical composition of the anthocyanin (the pH indicator) This will cause the juice to react with light differently than if these OH- ions had not jumped off the anthocyanin (makes the juice turn a reddish color) Conversely Hydroxyl ions will jump from the solution onto the anthrocyanin molecules when there is an excess of them in the solution (such as when cabbage juice is mixed with a basic solution) This causes the juice

(the pH indicator) to turn a yellowish color

NOT mentioned

bull Balancing redox reactions

bull Kw

bull Ka

bull Kb

bull pH

bull the galvanic cell

bullKNOW THESE CONCEPTS

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
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• Slide 51

Change in Enthalpy of a Reaction

rxnH

Standard Enthalpy of Formation

bull DEFINITIONndash the enthalpy of a reaction carried out at 1 atm and at 25 degr

ees Cndash the enthalpy value relates to the heat absorbed or released

during the formation of 1 mole of a specific compound

The symbol for this value is represented as

formation ofenthalpy Standared fH

Once we know these values we can calculate the change in enthalpy of a reaction or

rxnH

Standard Enthalpy of a Reaction rxnH

)reactants()products(H

isequation dgeneralize The

fHfH nnrxn

bull n = the coefficient of the productsbull m = the coefficient of the reactantsbull Σ = the sum ofbull f = formation bull deg = standard state conditions (1 atm and 25 degrees C)

m

The most stable forms of substances will = 0

EXAMPLEGiven the equation 3 O2(g)rarr2 O3(g) ΔH = +2844 kJ

calculate ΔH for the following reaction 32 O2(g) rarr O3(g)REMEMBER

)reactants()products(H

isequation dgeneralize The

fHfH nnrxn

O2 is the most stable form of oxygen So it is equal to 0

Therefore the the change in enthalpy is all about the product

(x products) - (0 reactants) = rxnH = X products = +2844 kJYou can check using the table on P 247

i

P 247bull The ΔHdegf for O3(g) is +1422 bull The ΔHdegf for O2(g) is 0Given the equation 3 O2(g)rarr2 O3(g) ΔH = +2844 kJThere are two O3 in the equation (1422 kJ x 2 = 2844 kJ)

ndash we must remember the coefficientsbull Then the question asks us to calculate ΔH for the

following reaction 32 O2(g) rarr O3(g) bull There is only one O3

bull Since 32 O2(g) rarr O3(g) is frac12 of 3 O2(g)rarr2 O3(g) the enthalpy of the reaction will be frac12 as well

frac12 (+2844kJ)or +1422kJ

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used

in the manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP IWRITE THE BALANCED EQUATIONTo carry any combustion reaction you need oxygen as a reactant Liquid water will always be included as one of the products

__ C4H4S(l) + ___ O2(g) rarr __ CO2(g) + ___ S02(g) + ___ H20(l)1 6 4 1 2

Change of Enthalpy of Reaction

bull The combustion of thiophene C4H4S(l) a compound used in the manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

USE THE COEFFICIENTS FROM YOUR BALANCED EQUATION TO DETERMINE THE TOTAL CHANGE IN ENTHALPY IN THE FORMATION OF EVERY SUBSTANCE USED IN THE REACTION USE THIS INFORMATION TO WRITE YOUR ΔHdegRXN EQUATION

i

)reactants()products(H fHfH nnrxn

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

i

PRODUCTS

4 (________kJmol CO2(g))2 (________kJmol H2O(l))1 (________kJmol SO2(g))

REACTANTS

X kJmol C4H4S(l)

6 (________kJmol O2(g))

-3935

-2858-2968 000

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

i

[4(-3935) + 1(-2968) + 2(-2858)] ndash [1(ΔH) + 6(0)] = -2523

PRODUCTS －

REACTANTS = ΔHRXN

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP III

SOLVE for X

i

-24424 - ΔH(reactants) = -2523

ΔHC4H4S(l) = 813kJmol

CHANGE IN H

0H EXOTHERMIC

0H CENDOTHERMI

Change in Enthalpy in a multi-step reaction

Change in Enthalpy in a multi-step reaction

CHEMICAL EQUILIBRIA

Things to remember

This is how you convert an Equilibrium constant from Kc to Kp

Kp = Kc (00821 T)^Δn

Δn is the sum of all stoichiometric coefficients belonging to products minus the sum of all stoichiometric coefficients belonging to reactants

T is the temperature during the reaction in Kelvin Remember that Kelvin has exactly the same degree of difference between integers as celsius but it does have a different starting point Kelvins zero is 273deg less than celsius For example 2deg C is equal to 275deg K 27deg C is equal to 300deg K

Qc and Kc Questionh

EQUILIBRIUM of 2 or more RXNs

If a reaction can be expressed as the sum of two or more reactions the equilibrium constant for the overall reaction is given by the product of the equilibrium of the individual reactions

Kc = (Kc) (Kc)

Let me elaborate

nn

nn

c BA

DCK

][][

][][ nn

nn

c DC

FEK

][][

][][

c

c

c

K

__________________

K

K

FEBA

FEDC

DCBA

1st Step Reaction 2nd Step Reaction

OverallReaction

To find the overall equilibrium of a multi-step reaction

cnn

nn

nn

nn

cc KDC

FEx

BA

DCKK

][][

][][

][][

][][

2 step reaction1st step reaction

N2(g) + O2(g) harr 2 NO(g) Kc1 = 23 x 10^-19

2nd step reaction

2 NO(g) + O2(g) harr 2 NO2(g) Kc2 = 3 x 10^6

Kc = Kc1 x Kc2 = (23 x 10^-19)(3 x 10^6) Kc = 7 x 10^-13

Write the equilibrium equation for this multi-step reaction

CHEMICAL KINETICS

Reaction rates

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull Suppose that at a particular moment during the reaction molecular hydrogen is reacting at the rate of 0074 Ms

a) At what rate is ammonia being formedb) At what rate is molecular nitrogen reacting

t

NH

t

H

t

N

322

2

1

3

1

1

1

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is ammonia being formed s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

t

NH

t

NH

t

H

sM

3

32

2

10740

3

1

2

1

3

1t

NHsM

30740

3

2

_

-

_

t

NHsM

30490

GIVEN

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is molecular nitrogen reacting s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

-

GIVEN

t

N

t

N

t

H

sM

2

22

1

10740

3

1

1

1

3

1

t

NsM

20740)1(

3

1

t

NsM

20250

-

Describe the difference between Heat of solution and Heat of dilution Give an example of

each one

Heat of solution is the amount of heat absorbed or released during the act of combining a solute with a solvent An example is adding salt to water There are two steps to this process the Lattice Energy (breaking down the molecules) and the heat of hydration (where the ionically charged particles dissolve by attracting to the dipole water molecules

Heat of dilution refers to the amount of heat absorbed or released when water is added to a solution already made In this latter case if the act of making a solution is exothermic adding more water will make it even more exothermic (hotter) For example as you add more and more water to a solution of sulfuric acid and water the more heat it releases (the more it burns)

Neutralization

bull Explain how you would neutralize an acid and a base

Mix an acidic solution with a basic solution wherein the ratio of H+ ions to OH- ions eq

ual to 1

Cabbage Juice as a pH indicator

bull Explain how cabbage juice works as a pH indicator There is anthocyanin in cabbage juice Anthocyanin reacts w

ith hydroxyl ions (OH-) Hydroxyl ions jump off anthocyanin when in an acidic solution thereby changing the chemical composition of the anthocyanin (the pH indicator) This will cause the juice to react with light differently than if these OH- ions had not jumped off the anthocyanin (makes the juice turn a reddish color) Conversely Hydroxyl ions will jump from the solution onto the anthrocyanin molecules when there is an excess of them in the solution (such as when cabbage juice is mixed with a basic solution) This causes the juice

(the pH indicator) to turn a yellowish color

NOT mentioned

bull Balancing redox reactions

bull Kw

bull Ka

bull Kb

bull pH

bull the galvanic cell

bullKNOW THESE CONCEPTS

• Slide 1
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Standard Enthalpy of Formation

bull DEFINITIONndash the enthalpy of a reaction carried out at 1 atm and at 25 degr

ees Cndash the enthalpy value relates to the heat absorbed or released

during the formation of 1 mole of a specific compound

The symbol for this value is represented as

formation ofenthalpy Standared fH

Once we know these values we can calculate the change in enthalpy of a reaction or

rxnH

Standard Enthalpy of a Reaction rxnH

)reactants()products(H

isequation dgeneralize The

fHfH nnrxn

bull n = the coefficient of the productsbull m = the coefficient of the reactantsbull Σ = the sum ofbull f = formation bull deg = standard state conditions (1 atm and 25 degrees C)

m

The most stable forms of substances will = 0

EXAMPLEGiven the equation 3 O2(g)rarr2 O3(g) ΔH = +2844 kJ

calculate ΔH for the following reaction 32 O2(g) rarr O3(g)REMEMBER

)reactants()products(H

isequation dgeneralize The

fHfH nnrxn

O2 is the most stable form of oxygen So it is equal to 0

Therefore the the change in enthalpy is all about the product

(x products) - (0 reactants) = rxnH = X products = +2844 kJYou can check using the table on P 247

i

P 247bull The ΔHdegf for O3(g) is +1422 bull The ΔHdegf for O2(g) is 0Given the equation 3 O2(g)rarr2 O3(g) ΔH = +2844 kJThere are two O3 in the equation (1422 kJ x 2 = 2844 kJ)

ndash we must remember the coefficientsbull Then the question asks us to calculate ΔH for the

following reaction 32 O2(g) rarr O3(g) bull There is only one O3

bull Since 32 O2(g) rarr O3(g) is frac12 of 3 O2(g)rarr2 O3(g) the enthalpy of the reaction will be frac12 as well

frac12 (+2844kJ)or +1422kJ

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used

in the manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP IWRITE THE BALANCED EQUATIONTo carry any combustion reaction you need oxygen as a reactant Liquid water will always be included as one of the products

__ C4H4S(l) + ___ O2(g) rarr __ CO2(g) + ___ S02(g) + ___ H20(l)1 6 4 1 2

Change of Enthalpy of Reaction

bull The combustion of thiophene C4H4S(l) a compound used in the manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

USE THE COEFFICIENTS FROM YOUR BALANCED EQUATION TO DETERMINE THE TOTAL CHANGE IN ENTHALPY IN THE FORMATION OF EVERY SUBSTANCE USED IN THE REACTION USE THIS INFORMATION TO WRITE YOUR ΔHdegRXN EQUATION

i

)reactants()products(H fHfH nnrxn

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

i

PRODUCTS

4 (________kJmol CO2(g))2 (________kJmol H2O(l))1 (________kJmol SO2(g))

REACTANTS

X kJmol C4H4S(l)

6 (________kJmol O2(g))

-3935

-2858-2968 000

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

i

[4(-3935) + 1(-2968) + 2(-2858)] ndash [1(ΔH) + 6(0)] = -2523

PRODUCTS －

REACTANTS = ΔHRXN

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP III

SOLVE for X

i

-24424 - ΔH(reactants) = -2523

ΔHC4H4S(l) = 813kJmol

CHANGE IN H

0H EXOTHERMIC

0H CENDOTHERMI

Change in Enthalpy in a multi-step reaction

Change in Enthalpy in a multi-step reaction

CHEMICAL EQUILIBRIA

Things to remember

This is how you convert an Equilibrium constant from Kc to Kp

Kp = Kc (00821 T)^Δn

Δn is the sum of all stoichiometric coefficients belonging to products minus the sum of all stoichiometric coefficients belonging to reactants

T is the temperature during the reaction in Kelvin Remember that Kelvin has exactly the same degree of difference between integers as celsius but it does have a different starting point Kelvins zero is 273deg less than celsius For example 2deg C is equal to 275deg K 27deg C is equal to 300deg K

Qc and Kc Questionh

EQUILIBRIUM of 2 or more RXNs

If a reaction can be expressed as the sum of two or more reactions the equilibrium constant for the overall reaction is given by the product of the equilibrium of the individual reactions

Kc = (Kc) (Kc)

Let me elaborate

nn

nn

c BA

DCK

][][

][][ nn

nn

c DC

FEK

][][

][][

c

c

c

K

__________________

K

K

FEBA

FEDC

DCBA

1st Step Reaction 2nd Step Reaction

OverallReaction

To find the overall equilibrium of a multi-step reaction

cnn

nn

nn

nn

cc KDC

FEx

BA

DCKK

][][

][][

][][

][][

2 step reaction1st step reaction

N2(g) + O2(g) harr 2 NO(g) Kc1 = 23 x 10^-19

2nd step reaction

2 NO(g) + O2(g) harr 2 NO2(g) Kc2 = 3 x 10^6

Kc = Kc1 x Kc2 = (23 x 10^-19)(3 x 10^6) Kc = 7 x 10^-13

Write the equilibrium equation for this multi-step reaction

CHEMICAL KINETICS

Reaction rates

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull Suppose that at a particular moment during the reaction molecular hydrogen is reacting at the rate of 0074 Ms

a) At what rate is ammonia being formedb) At what rate is molecular nitrogen reacting

t

NH

t

H

t

N

322

2

1

3

1

1

1

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is ammonia being formed s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

t

NH

t

NH

t

H

sM

3

32

2

10740

3

1

2

1

3

1t

NHsM

30740

3

2

_

-

_

t

NHsM

30490

GIVEN

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is molecular nitrogen reacting s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

-

GIVEN

t

N

t

N

t

H

sM

2

22

1

10740

3

1

1

1

3

1

t

NsM

20740)1(

3

1

t

NsM

20250

-

Describe the difference between Heat of solution and Heat of dilution Give an example of

each one

Heat of solution is the amount of heat absorbed or released during the act of combining a solute with a solvent An example is adding salt to water There are two steps to this process the Lattice Energy (breaking down the molecules) and the heat of hydration (where the ionically charged particles dissolve by attracting to the dipole water molecules

Heat of dilution refers to the amount of heat absorbed or released when water is added to a solution already made In this latter case if the act of making a solution is exothermic adding more water will make it even more exothermic (hotter) For example as you add more and more water to a solution of sulfuric acid and water the more heat it releases (the more it burns)

Neutralization

bull Explain how you would neutralize an acid and a base

Mix an acidic solution with a basic solution wherein the ratio of H+ ions to OH- ions eq

ual to 1

Cabbage Juice as a pH indicator

bull Explain how cabbage juice works as a pH indicator There is anthocyanin in cabbage juice Anthocyanin reacts w

ith hydroxyl ions (OH-) Hydroxyl ions jump off anthocyanin when in an acidic solution thereby changing the chemical composition of the anthocyanin (the pH indicator) This will cause the juice to react with light differently than if these OH- ions had not jumped off the anthocyanin (makes the juice turn a reddish color) Conversely Hydroxyl ions will jump from the solution onto the anthrocyanin molecules when there is an excess of them in the solution (such as when cabbage juice is mixed with a basic solution) This causes the juice

(the pH indicator) to turn a yellowish color

NOT mentioned

bull Balancing redox reactions

bull Kw

bull Ka

bull Kb

bull pH

bull the galvanic cell

bullKNOW THESE CONCEPTS

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
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• Slide 50
• Slide 51

Standard Enthalpy of a Reaction rxnH

)reactants()products(H

isequation dgeneralize The

fHfH nnrxn

bull n = the coefficient of the productsbull m = the coefficient of the reactantsbull Σ = the sum ofbull f = formation bull deg = standard state conditions (1 atm and 25 degrees C)

m

The most stable forms of substances will = 0

EXAMPLEGiven the equation 3 O2(g)rarr2 O3(g) ΔH = +2844 kJ

calculate ΔH for the following reaction 32 O2(g) rarr O3(g)REMEMBER

)reactants()products(H

isequation dgeneralize The

fHfH nnrxn

O2 is the most stable form of oxygen So it is equal to 0

Therefore the the change in enthalpy is all about the product

(x products) - (0 reactants) = rxnH = X products = +2844 kJYou can check using the table on P 247

i

P 247bull The ΔHdegf for O3(g) is +1422 bull The ΔHdegf for O2(g) is 0Given the equation 3 O2(g)rarr2 O3(g) ΔH = +2844 kJThere are two O3 in the equation (1422 kJ x 2 = 2844 kJ)

ndash we must remember the coefficientsbull Then the question asks us to calculate ΔH for the

following reaction 32 O2(g) rarr O3(g) bull There is only one O3

bull Since 32 O2(g) rarr O3(g) is frac12 of 3 O2(g)rarr2 O3(g) the enthalpy of the reaction will be frac12 as well

frac12 (+2844kJ)or +1422kJ

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used

in the manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP IWRITE THE BALANCED EQUATIONTo carry any combustion reaction you need oxygen as a reactant Liquid water will always be included as one of the products

__ C4H4S(l) + ___ O2(g) rarr __ CO2(g) + ___ S02(g) + ___ H20(l)1 6 4 1 2

Change of Enthalpy of Reaction

bull The combustion of thiophene C4H4S(l) a compound used in the manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

USE THE COEFFICIENTS FROM YOUR BALANCED EQUATION TO DETERMINE THE TOTAL CHANGE IN ENTHALPY IN THE FORMATION OF EVERY SUBSTANCE USED IN THE REACTION USE THIS INFORMATION TO WRITE YOUR ΔHdegRXN EQUATION

i

)reactants()products(H fHfH nnrxn

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

i

PRODUCTS

4 (________kJmol CO2(g))2 (________kJmol H2O(l))1 (________kJmol SO2(g))

REACTANTS

X kJmol C4H4S(l)

6 (________kJmol O2(g))

-3935

-2858-2968 000

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

i

[4(-3935) + 1(-2968) + 2(-2858)] ndash [1(ΔH) + 6(0)] = -2523

PRODUCTS －

REACTANTS = ΔHRXN

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP III

SOLVE for X

i

-24424 - ΔH(reactants) = -2523

ΔHC4H4S(l) = 813kJmol

CHANGE IN H

0H EXOTHERMIC

0H CENDOTHERMI

Change in Enthalpy in a multi-step reaction

Change in Enthalpy in a multi-step reaction

CHEMICAL EQUILIBRIA

Things to remember

This is how you convert an Equilibrium constant from Kc to Kp

Kp = Kc (00821 T)^Δn

Δn is the sum of all stoichiometric coefficients belonging to products minus the sum of all stoichiometric coefficients belonging to reactants

T is the temperature during the reaction in Kelvin Remember that Kelvin has exactly the same degree of difference between integers as celsius but it does have a different starting point Kelvins zero is 273deg less than celsius For example 2deg C is equal to 275deg K 27deg C is equal to 300deg K

Qc and Kc Questionh

EQUILIBRIUM of 2 or more RXNs

If a reaction can be expressed as the sum of two or more reactions the equilibrium constant for the overall reaction is given by the product of the equilibrium of the individual reactions

Kc = (Kc) (Kc)

Let me elaborate

nn

nn

c BA

DCK

][][

][][ nn

nn

c DC

FEK

][][

][][

c

c

c

K

__________________

K

K

FEBA

FEDC

DCBA

1st Step Reaction 2nd Step Reaction

OverallReaction

To find the overall equilibrium of a multi-step reaction

cnn

nn

nn

nn

cc KDC

FEx

BA

DCKK

][][

][][

][][

][][

2 step reaction1st step reaction

N2(g) + O2(g) harr 2 NO(g) Kc1 = 23 x 10^-19

2nd step reaction

2 NO(g) + O2(g) harr 2 NO2(g) Kc2 = 3 x 10^6

Kc = Kc1 x Kc2 = (23 x 10^-19)(3 x 10^6) Kc = 7 x 10^-13

Write the equilibrium equation for this multi-step reaction

CHEMICAL KINETICS

Reaction rates

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull Suppose that at a particular moment during the reaction molecular hydrogen is reacting at the rate of 0074 Ms

a) At what rate is ammonia being formedb) At what rate is molecular nitrogen reacting

t

NH

t

H

t

N

322

2

1

3

1

1

1

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is ammonia being formed s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

t

NH

t

NH

t

H

sM

3

32

2

10740

3

1

2

1

3

1t

NHsM

30740

3

2

_

-

_

t

NHsM

30490

GIVEN

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is molecular nitrogen reacting s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

-

GIVEN

t

N

t

N

t

H

sM

2

22

1

10740

3

1

1

1

3

1

t

NsM

20740)1(

3

1

t

NsM

20250

-

Describe the difference between Heat of solution and Heat of dilution Give an example of

each one

Heat of solution is the amount of heat absorbed or released during the act of combining a solute with a solvent An example is adding salt to water There are two steps to this process the Lattice Energy (breaking down the molecules) and the heat of hydration (where the ionically charged particles dissolve by attracting to the dipole water molecules

Heat of dilution refers to the amount of heat absorbed or released when water is added to a solution already made In this latter case if the act of making a solution is exothermic adding more water will make it even more exothermic (hotter) For example as you add more and more water to a solution of sulfuric acid and water the more heat it releases (the more it burns)

Neutralization

bull Explain how you would neutralize an acid and a base

Mix an acidic solution with a basic solution wherein the ratio of H+ ions to OH- ions eq

ual to 1

Cabbage Juice as a pH indicator

bull Explain how cabbage juice works as a pH indicator There is anthocyanin in cabbage juice Anthocyanin reacts w

ith hydroxyl ions (OH-) Hydroxyl ions jump off anthocyanin when in an acidic solution thereby changing the chemical composition of the anthocyanin (the pH indicator) This will cause the juice to react with light differently than if these OH- ions had not jumped off the anthocyanin (makes the juice turn a reddish color) Conversely Hydroxyl ions will jump from the solution onto the anthrocyanin molecules when there is an excess of them in the solution (such as when cabbage juice is mixed with a basic solution) This causes the juice

(the pH indicator) to turn a yellowish color

NOT mentioned

bull Balancing redox reactions

bull Kw

bull Ka

bull Kb

bull pH

bull the galvanic cell

bullKNOW THESE CONCEPTS

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
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• Slide 41
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• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51

The most stable forms of substances will = 0

EXAMPLEGiven the equation 3 O2(g)rarr2 O3(g) ΔH = +2844 kJ

calculate ΔH for the following reaction 32 O2(g) rarr O3(g)REMEMBER

)reactants()products(H

isequation dgeneralize The

fHfH nnrxn

O2 is the most stable form of oxygen So it is equal to 0

Therefore the the change in enthalpy is all about the product

(x products) - (0 reactants) = rxnH = X products = +2844 kJYou can check using the table on P 247

i

P 247bull The ΔHdegf for O3(g) is +1422 bull The ΔHdegf for O2(g) is 0Given the equation 3 O2(g)rarr2 O3(g) ΔH = +2844 kJThere are two O3 in the equation (1422 kJ x 2 = 2844 kJ)

ndash we must remember the coefficientsbull Then the question asks us to calculate ΔH for the

following reaction 32 O2(g) rarr O3(g) bull There is only one O3

bull Since 32 O2(g) rarr O3(g) is frac12 of 3 O2(g)rarr2 O3(g) the enthalpy of the reaction will be frac12 as well

frac12 (+2844kJ)or +1422kJ

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used

in the manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP IWRITE THE BALANCED EQUATIONTo carry any combustion reaction you need oxygen as a reactant Liquid water will always be included as one of the products

__ C4H4S(l) + ___ O2(g) rarr __ CO2(g) + ___ S02(g) + ___ H20(l)1 6 4 1 2

Change of Enthalpy of Reaction

bull The combustion of thiophene C4H4S(l) a compound used in the manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

USE THE COEFFICIENTS FROM YOUR BALANCED EQUATION TO DETERMINE THE TOTAL CHANGE IN ENTHALPY IN THE FORMATION OF EVERY SUBSTANCE USED IN THE REACTION USE THIS INFORMATION TO WRITE YOUR ΔHdegRXN EQUATION

i

)reactants()products(H fHfH nnrxn

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

i

PRODUCTS

4 (________kJmol CO2(g))2 (________kJmol H2O(l))1 (________kJmol SO2(g))

REACTANTS

X kJmol C4H4S(l)

6 (________kJmol O2(g))

-3935

-2858-2968 000

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

i

[4(-3935) + 1(-2968) + 2(-2858)] ndash [1(ΔH) + 6(0)] = -2523

PRODUCTS －

REACTANTS = ΔHRXN

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP III

SOLVE for X

i

-24424 - ΔH(reactants) = -2523

ΔHC4H4S(l) = 813kJmol

CHANGE IN H

0H EXOTHERMIC

0H CENDOTHERMI

Change in Enthalpy in a multi-step reaction

Change in Enthalpy in a multi-step reaction

CHEMICAL EQUILIBRIA

Things to remember

This is how you convert an Equilibrium constant from Kc to Kp

Kp = Kc (00821 T)^Δn

Δn is the sum of all stoichiometric coefficients belonging to products minus the sum of all stoichiometric coefficients belonging to reactants

T is the temperature during the reaction in Kelvin Remember that Kelvin has exactly the same degree of difference between integers as celsius but it does have a different starting point Kelvins zero is 273deg less than celsius For example 2deg C is equal to 275deg K 27deg C is equal to 300deg K

Qc and Kc Questionh

EQUILIBRIUM of 2 or more RXNs

If a reaction can be expressed as the sum of two or more reactions the equilibrium constant for the overall reaction is given by the product of the equilibrium of the individual reactions

Kc = (Kc) (Kc)

Let me elaborate

nn

nn

c BA

DCK

][][

][][ nn

nn

c DC

FEK

][][

][][

c

c

c

K

__________________

K

K

FEBA

FEDC

DCBA

1st Step Reaction 2nd Step Reaction

OverallReaction

To find the overall equilibrium of a multi-step reaction

cnn

nn

nn

nn

cc KDC

FEx

BA

DCKK

][][

][][

][][

][][

2 step reaction1st step reaction

N2(g) + O2(g) harr 2 NO(g) Kc1 = 23 x 10^-19

2nd step reaction

2 NO(g) + O2(g) harr 2 NO2(g) Kc2 = 3 x 10^6

Kc = Kc1 x Kc2 = (23 x 10^-19)(3 x 10^6) Kc = 7 x 10^-13

Write the equilibrium equation for this multi-step reaction

CHEMICAL KINETICS

Reaction rates

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull Suppose that at a particular moment during the reaction molecular hydrogen is reacting at the rate of 0074 Ms

a) At what rate is ammonia being formedb) At what rate is molecular nitrogen reacting

t

NH

t

H

t

N

322

2

1

3

1

1

1

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is ammonia being formed s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

t

NH

t

NH

t

H

sM

3

32

2

10740

3

1

2

1

3

1t

NHsM

30740

3

2

_

-

_

t

NHsM

30490

GIVEN

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is molecular nitrogen reacting s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

-

GIVEN

t

N

t

N

t

H

sM

2

22

1

10740

3

1

1

1

3

1

t

NsM

20740)1(

3

1

t

NsM

20250

-

Describe the difference between Heat of solution and Heat of dilution Give an example of

each one

Heat of solution is the amount of heat absorbed or released during the act of combining a solute with a solvent An example is adding salt to water There are two steps to this process the Lattice Energy (breaking down the molecules) and the heat of hydration (where the ionically charged particles dissolve by attracting to the dipole water molecules

Heat of dilution refers to the amount of heat absorbed or released when water is added to a solution already made In this latter case if the act of making a solution is exothermic adding more water will make it even more exothermic (hotter) For example as you add more and more water to a solution of sulfuric acid and water the more heat it releases (the more it burns)

Neutralization

bull Explain how you would neutralize an acid and a base

Mix an acidic solution with a basic solution wherein the ratio of H+ ions to OH- ions eq

ual to 1

Cabbage Juice as a pH indicator

bull Explain how cabbage juice works as a pH indicator There is anthocyanin in cabbage juice Anthocyanin reacts w

ith hydroxyl ions (OH-) Hydroxyl ions jump off anthocyanin when in an acidic solution thereby changing the chemical composition of the anthocyanin (the pH indicator) This will cause the juice to react with light differently than if these OH- ions had not jumped off the anthocyanin (makes the juice turn a reddish color) Conversely Hydroxyl ions will jump from the solution onto the anthrocyanin molecules when there is an excess of them in the solution (such as when cabbage juice is mixed with a basic solution) This causes the juice

(the pH indicator) to turn a yellowish color

NOT mentioned

bull Balancing redox reactions

bull Kw

bull Ka

bull Kb

bull pH

bull the galvanic cell

bullKNOW THESE CONCEPTS

• Slide 1
• Slide 2
• Slide 3
• Slide 4
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• Slide 50
• Slide 51

P 247bull The ΔHdegf for O3(g) is +1422 bull The ΔHdegf for O2(g) is 0Given the equation 3 O2(g)rarr2 O3(g) ΔH = +2844 kJThere are two O3 in the equation (1422 kJ x 2 = 2844 kJ)

ndash we must remember the coefficientsbull Then the question asks us to calculate ΔH for the

following reaction 32 O2(g) rarr O3(g) bull There is only one O3

bull Since 32 O2(g) rarr O3(g) is frac12 of 3 O2(g)rarr2 O3(g) the enthalpy of the reaction will be frac12 as well

frac12 (+2844kJ)or +1422kJ

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used

in the manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP IWRITE THE BALANCED EQUATIONTo carry any combustion reaction you need oxygen as a reactant Liquid water will always be included as one of the products

__ C4H4S(l) + ___ O2(g) rarr __ CO2(g) + ___ S02(g) + ___ H20(l)1 6 4 1 2

Change of Enthalpy of Reaction

bull The combustion of thiophene C4H4S(l) a compound used in the manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

USE THE COEFFICIENTS FROM YOUR BALANCED EQUATION TO DETERMINE THE TOTAL CHANGE IN ENTHALPY IN THE FORMATION OF EVERY SUBSTANCE USED IN THE REACTION USE THIS INFORMATION TO WRITE YOUR ΔHdegRXN EQUATION

i

)reactants()products(H fHfH nnrxn

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

i

PRODUCTS

4 (________kJmol CO2(g))2 (________kJmol H2O(l))1 (________kJmol SO2(g))

REACTANTS

X kJmol C4H4S(l)

6 (________kJmol O2(g))

-3935

-2858-2968 000

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

i

[4(-3935) + 1(-2968) + 2(-2858)] ndash [1(ΔH) + 6(0)] = -2523

PRODUCTS －

REACTANTS = ΔHRXN

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP III

SOLVE for X

i

-24424 - ΔH(reactants) = -2523

ΔHC4H4S(l) = 813kJmol

CHANGE IN H

0H EXOTHERMIC

0H CENDOTHERMI

Change in Enthalpy in a multi-step reaction

Change in Enthalpy in a multi-step reaction

CHEMICAL EQUILIBRIA

Things to remember

This is how you convert an Equilibrium constant from Kc to Kp

Kp = Kc (00821 T)^Δn

Δn is the sum of all stoichiometric coefficients belonging to products minus the sum of all stoichiometric coefficients belonging to reactants

T is the temperature during the reaction in Kelvin Remember that Kelvin has exactly the same degree of difference between integers as celsius but it does have a different starting point Kelvins zero is 273deg less than celsius For example 2deg C is equal to 275deg K 27deg C is equal to 300deg K

Qc and Kc Questionh

EQUILIBRIUM of 2 or more RXNs

If a reaction can be expressed as the sum of two or more reactions the equilibrium constant for the overall reaction is given by the product of the equilibrium of the individual reactions

Kc = (Kc) (Kc)

Let me elaborate

nn

nn

c BA

DCK

][][

][][ nn

nn

c DC

FEK

][][

][][

c

c

c

K

__________________

K

K

FEBA

FEDC

DCBA

1st Step Reaction 2nd Step Reaction

OverallReaction

To find the overall equilibrium of a multi-step reaction

cnn

nn

nn

nn

cc KDC

FEx

BA

DCKK

][][

][][

][][

][][

2 step reaction1st step reaction

N2(g) + O2(g) harr 2 NO(g) Kc1 = 23 x 10^-19

2nd step reaction

2 NO(g) + O2(g) harr 2 NO2(g) Kc2 = 3 x 10^6

Kc = Kc1 x Kc2 = (23 x 10^-19)(3 x 10^6) Kc = 7 x 10^-13

Write the equilibrium equation for this multi-step reaction

CHEMICAL KINETICS

Reaction rates

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull Suppose that at a particular moment during the reaction molecular hydrogen is reacting at the rate of 0074 Ms

a) At what rate is ammonia being formedb) At what rate is molecular nitrogen reacting

t

NH

t

H

t

N

322

2

1

3

1

1

1

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is ammonia being formed s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

t

NH

t

NH

t

H

sM

3

32

2

10740

3

1

2

1

3

1t

NHsM

30740

3

2

_

-

_

t

NHsM

30490

GIVEN

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is molecular nitrogen reacting s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

-

GIVEN

t

N

t

N

t

H

sM

2

22

1

10740

3

1

1

1

3

1

t

NsM

20740)1(

3

1

t

NsM

20250

-

Describe the difference between Heat of solution and Heat of dilution Give an example of

each one

Heat of solution is the amount of heat absorbed or released during the act of combining a solute with a solvent An example is adding salt to water There are two steps to this process the Lattice Energy (breaking down the molecules) and the heat of hydration (where the ionically charged particles dissolve by attracting to the dipole water molecules

Heat of dilution refers to the amount of heat absorbed or released when water is added to a solution already made In this latter case if the act of making a solution is exothermic adding more water will make it even more exothermic (hotter) For example as you add more and more water to a solution of sulfuric acid and water the more heat it releases (the more it burns)

Neutralization

bull Explain how you would neutralize an acid and a base

Mix an acidic solution with a basic solution wherein the ratio of H+ ions to OH- ions eq

ual to 1

Cabbage Juice as a pH indicator

bull Explain how cabbage juice works as a pH indicator There is anthocyanin in cabbage juice Anthocyanin reacts w

ith hydroxyl ions (OH-) Hydroxyl ions jump off anthocyanin when in an acidic solution thereby changing the chemical composition of the anthocyanin (the pH indicator) This will cause the juice to react with light differently than if these OH- ions had not jumped off the anthocyanin (makes the juice turn a reddish color) Conversely Hydroxyl ions will jump from the solution onto the anthrocyanin molecules when there is an excess of them in the solution (such as when cabbage juice is mixed with a basic solution) This causes the juice

(the pH indicator) to turn a yellowish color

NOT mentioned

bull Balancing redox reactions

bull Kw

bull Ka

bull Kb

bull pH

bull the galvanic cell

bullKNOW THESE CONCEPTS

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
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• Slide 51

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used

in the manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP IWRITE THE BALANCED EQUATIONTo carry any combustion reaction you need oxygen as a reactant Liquid water will always be included as one of the products

__ C4H4S(l) + ___ O2(g) rarr __ CO2(g) + ___ S02(g) + ___ H20(l)1 6 4 1 2

Change of Enthalpy of Reaction

bull The combustion of thiophene C4H4S(l) a compound used in the manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

USE THE COEFFICIENTS FROM YOUR BALANCED EQUATION TO DETERMINE THE TOTAL CHANGE IN ENTHALPY IN THE FORMATION OF EVERY SUBSTANCE USED IN THE REACTION USE THIS INFORMATION TO WRITE YOUR ΔHdegRXN EQUATION

i

)reactants()products(H fHfH nnrxn

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

i

PRODUCTS

4 (________kJmol CO2(g))2 (________kJmol H2O(l))1 (________kJmol SO2(g))

REACTANTS

X kJmol C4H4S(l)

6 (________kJmol O2(g))

-3935

-2858-2968 000

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

i

[4(-3935) + 1(-2968) + 2(-2858)] ndash [1(ΔH) + 6(0)] = -2523

PRODUCTS －

REACTANTS = ΔHRXN

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP III

SOLVE for X

i

-24424 - ΔH(reactants) = -2523

ΔHC4H4S(l) = 813kJmol

CHANGE IN H

0H EXOTHERMIC

0H CENDOTHERMI

Change in Enthalpy in a multi-step reaction

Change in Enthalpy in a multi-step reaction

CHEMICAL EQUILIBRIA

Things to remember

This is how you convert an Equilibrium constant from Kc to Kp

Kp = Kc (00821 T)^Δn

Δn is the sum of all stoichiometric coefficients belonging to products minus the sum of all stoichiometric coefficients belonging to reactants

T is the temperature during the reaction in Kelvin Remember that Kelvin has exactly the same degree of difference between integers as celsius but it does have a different starting point Kelvins zero is 273deg less than celsius For example 2deg C is equal to 275deg K 27deg C is equal to 300deg K

Qc and Kc Questionh

EQUILIBRIUM of 2 or more RXNs

If a reaction can be expressed as the sum of two or more reactions the equilibrium constant for the overall reaction is given by the product of the equilibrium of the individual reactions

Kc = (Kc) (Kc)

Let me elaborate

nn

nn

c BA

DCK

][][

][][ nn

nn

c DC

FEK

][][

][][

c

c

c

K

__________________

K

K

FEBA

FEDC

DCBA

1st Step Reaction 2nd Step Reaction

OverallReaction

To find the overall equilibrium of a multi-step reaction

cnn

nn

nn

nn

cc KDC

FEx

BA

DCKK

][][

][][

][][

][][

2 step reaction1st step reaction

N2(g) + O2(g) harr 2 NO(g) Kc1 = 23 x 10^-19

2nd step reaction

2 NO(g) + O2(g) harr 2 NO2(g) Kc2 = 3 x 10^6

Kc = Kc1 x Kc2 = (23 x 10^-19)(3 x 10^6) Kc = 7 x 10^-13

Write the equilibrium equation for this multi-step reaction

CHEMICAL KINETICS

Reaction rates

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull Suppose that at a particular moment during the reaction molecular hydrogen is reacting at the rate of 0074 Ms

a) At what rate is ammonia being formedb) At what rate is molecular nitrogen reacting

t

NH

t

H

t

N

322

2

1

3

1

1

1

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is ammonia being formed s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

t

NH

t

NH

t

H

sM

3

32

2

10740

3

1

2

1

3

1t

NHsM

30740

3

2

_

-

_

t

NHsM

30490

GIVEN

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is molecular nitrogen reacting s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

-

GIVEN

t

N

t

N

t

H

sM

2

22

1

10740

3

1

1

1

3

1

t

NsM

20740)1(

3

1

t

NsM

20250

-

Describe the difference between Heat of solution and Heat of dilution Give an example of

each one

Heat of solution is the amount of heat absorbed or released during the act of combining a solute with a solvent An example is adding salt to water There are two steps to this process the Lattice Energy (breaking down the molecules) and the heat of hydration (where the ionically charged particles dissolve by attracting to the dipole water molecules

Heat of dilution refers to the amount of heat absorbed or released when water is added to a solution already made In this latter case if the act of making a solution is exothermic adding more water will make it even more exothermic (hotter) For example as you add more and more water to a solution of sulfuric acid and water the more heat it releases (the more it burns)

Neutralization

bull Explain how you would neutralize an acid and a base

Mix an acidic solution with a basic solution wherein the ratio of H+ ions to OH- ions eq

ual to 1

Cabbage Juice as a pH indicator

bull Explain how cabbage juice works as a pH indicator There is anthocyanin in cabbage juice Anthocyanin reacts w

ith hydroxyl ions (OH-) Hydroxyl ions jump off anthocyanin when in an acidic solution thereby changing the chemical composition of the anthocyanin (the pH indicator) This will cause the juice to react with light differently than if these OH- ions had not jumped off the anthocyanin (makes the juice turn a reddish color) Conversely Hydroxyl ions will jump from the solution onto the anthrocyanin molecules when there is an excess of them in the solution (such as when cabbage juice is mixed with a basic solution) This causes the juice

(the pH indicator) to turn a yellowish color

NOT mentioned

bull Balancing redox reactions

bull Kw

bull Ka

bull Kb

bull pH

bull the galvanic cell

bullKNOW THESE CONCEPTS

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
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• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51

Change of Enthalpy of Reaction

bull The combustion of thiophene C4H4S(l) a compound used in the manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

USE THE COEFFICIENTS FROM YOUR BALANCED EQUATION TO DETERMINE THE TOTAL CHANGE IN ENTHALPY IN THE FORMATION OF EVERY SUBSTANCE USED IN THE REACTION USE THIS INFORMATION TO WRITE YOUR ΔHdegRXN EQUATION

i

)reactants()products(H fHfH nnrxn

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

i

PRODUCTS

4 (________kJmol CO2(g))2 (________kJmol H2O(l))1 (________kJmol SO2(g))

REACTANTS

X kJmol C4H4S(l)

6 (________kJmol O2(g))

-3935

-2858-2968 000

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

i

[4(-3935) + 1(-2968) + 2(-2858)] ndash [1(ΔH) + 6(0)] = -2523

PRODUCTS －

REACTANTS = ΔHRXN

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP III

SOLVE for X

i

-24424 - ΔH(reactants) = -2523

ΔHC4H4S(l) = 813kJmol

CHANGE IN H

0H EXOTHERMIC

0H CENDOTHERMI

Change in Enthalpy in a multi-step reaction

Change in Enthalpy in a multi-step reaction

CHEMICAL EQUILIBRIA

Things to remember

This is how you convert an Equilibrium constant from Kc to Kp

Kp = Kc (00821 T)^Δn

Δn is the sum of all stoichiometric coefficients belonging to products minus the sum of all stoichiometric coefficients belonging to reactants

T is the temperature during the reaction in Kelvin Remember that Kelvin has exactly the same degree of difference between integers as celsius but it does have a different starting point Kelvins zero is 273deg less than celsius For example 2deg C is equal to 275deg K 27deg C is equal to 300deg K

Qc and Kc Questionh

EQUILIBRIUM of 2 or more RXNs

If a reaction can be expressed as the sum of two or more reactions the equilibrium constant for the overall reaction is given by the product of the equilibrium of the individual reactions

Kc = (Kc) (Kc)

Let me elaborate

nn

nn

c BA

DCK

][][

][][ nn

nn

c DC

FEK

][][

][][

c

c

c

K

__________________

K

K

FEBA

FEDC

DCBA

1st Step Reaction 2nd Step Reaction

OverallReaction

To find the overall equilibrium of a multi-step reaction

cnn

nn

nn

nn

cc KDC

FEx

BA

DCKK

][][

][][

][][

][][

2 step reaction1st step reaction

N2(g) + O2(g) harr 2 NO(g) Kc1 = 23 x 10^-19

2nd step reaction

2 NO(g) + O2(g) harr 2 NO2(g) Kc2 = 3 x 10^6

Kc = Kc1 x Kc2 = (23 x 10^-19)(3 x 10^6) Kc = 7 x 10^-13

Write the equilibrium equation for this multi-step reaction

CHEMICAL KINETICS

Reaction rates

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull Suppose that at a particular moment during the reaction molecular hydrogen is reacting at the rate of 0074 Ms

a) At what rate is ammonia being formedb) At what rate is molecular nitrogen reacting

t

NH

t

H

t

N

322

2

1

3

1

1

1

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is ammonia being formed s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

t

NH

t

NH

t

H

sM

3

32

2

10740

3

1

2

1

3

1t

NHsM

30740

3

2

_

-

_

t

NHsM

30490

GIVEN

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is molecular nitrogen reacting s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

-

GIVEN

t

N

t

N

t

H

sM

2

22

1

10740

3

1

1

1

3

1

t

NsM

20740)1(

3

1

t

NsM

20250

-

Describe the difference between Heat of solution and Heat of dilution Give an example of

each one

Heat of solution is the amount of heat absorbed or released during the act of combining a solute with a solvent An example is adding salt to water There are two steps to this process the Lattice Energy (breaking down the molecules) and the heat of hydration (where the ionically charged particles dissolve by attracting to the dipole water molecules

Heat of dilution refers to the amount of heat absorbed or released when water is added to a solution already made In this latter case if the act of making a solution is exothermic adding more water will make it even more exothermic (hotter) For example as you add more and more water to a solution of sulfuric acid and water the more heat it releases (the more it burns)

Neutralization

bull Explain how you would neutralize an acid and a base

Mix an acidic solution with a basic solution wherein the ratio of H+ ions to OH- ions eq

ual to 1

Cabbage Juice as a pH indicator

bull Explain how cabbage juice works as a pH indicator There is anthocyanin in cabbage juice Anthocyanin reacts w

ith hydroxyl ions (OH-) Hydroxyl ions jump off anthocyanin when in an acidic solution thereby changing the chemical composition of the anthocyanin (the pH indicator) This will cause the juice to react with light differently than if these OH- ions had not jumped off the anthocyanin (makes the juice turn a reddish color) Conversely Hydroxyl ions will jump from the solution onto the anthrocyanin molecules when there is an excess of them in the solution (such as when cabbage juice is mixed with a basic solution) This causes the juice

(the pH indicator) to turn a yellowish color

NOT mentioned

bull Balancing redox reactions

bull Kw

bull Ka

bull Kb

bull pH

bull the galvanic cell

bullKNOW THESE CONCEPTS

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

i

PRODUCTS

4 (________kJmol CO2(g))2 (________kJmol H2O(l))1 (________kJmol SO2(g))

REACTANTS

X kJmol C4H4S(l)

6 (________kJmol O2(g))

-3935

-2858-2968 000

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

i

[4(-3935) + 1(-2968) + 2(-2858)] ndash [1(ΔH) + 6(0)] = -2523

PRODUCTS －

REACTANTS = ΔHRXN

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP III

SOLVE for X

i

-24424 - ΔH(reactants) = -2523

ΔHC4H4S(l) = 813kJmol

CHANGE IN H

0H EXOTHERMIC

0H CENDOTHERMI

Change in Enthalpy in a multi-step reaction

Change in Enthalpy in a multi-step reaction

CHEMICAL EQUILIBRIA

Things to remember

This is how you convert an Equilibrium constant from Kc to Kp

Kp = Kc (00821 T)^Δn

Δn is the sum of all stoichiometric coefficients belonging to products minus the sum of all stoichiometric coefficients belonging to reactants

T is the temperature during the reaction in Kelvin Remember that Kelvin has exactly the same degree of difference between integers as celsius but it does have a different starting point Kelvins zero is 273deg less than celsius For example 2deg C is equal to 275deg K 27deg C is equal to 300deg K

Qc and Kc Questionh

EQUILIBRIUM of 2 or more RXNs

If a reaction can be expressed as the sum of two or more reactions the equilibrium constant for the overall reaction is given by the product of the equilibrium of the individual reactions

Kc = (Kc) (Kc)

Let me elaborate

nn

nn

c BA

DCK

][][

][][ nn

nn

c DC

FEK

][][

][][

c

c

c

K

__________________

K

K

FEBA

FEDC

DCBA

1st Step Reaction 2nd Step Reaction

OverallReaction

To find the overall equilibrium of a multi-step reaction

cnn

nn

nn

nn

cc KDC

FEx

BA

DCKK

][][

][][

][][

][][

2 step reaction1st step reaction

N2(g) + O2(g) harr 2 NO(g) Kc1 = 23 x 10^-19

2nd step reaction

2 NO(g) + O2(g) harr 2 NO2(g) Kc2 = 3 x 10^6

Kc = Kc1 x Kc2 = (23 x 10^-19)(3 x 10^6) Kc = 7 x 10^-13

Write the equilibrium equation for this multi-step reaction

CHEMICAL KINETICS

Reaction rates

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull Suppose that at a particular moment during the reaction molecular hydrogen is reacting at the rate of 0074 Ms

a) At what rate is ammonia being formedb) At what rate is molecular nitrogen reacting

t

NH

t

H

t

N

322

2

1

3

1

1

1

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is ammonia being formed s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

t

NH

t

NH

t

H

sM

3

32

2

10740

3

1

2

1

3

1t

NHsM

30740

3

2

_

-

_

t

NHsM

30490

GIVEN

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is molecular nitrogen reacting s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

-

GIVEN

t

N

t

N

t

H

sM

2

22

1

10740

3

1

1

1

3

1

t

NsM

20740)1(

3

1

t

NsM

20250

-

Describe the difference between Heat of solution and Heat of dilution Give an example of

each one

Heat of solution is the amount of heat absorbed or released during the act of combining a solute with a solvent An example is adding salt to water There are two steps to this process the Lattice Energy (breaking down the molecules) and the heat of hydration (where the ionically charged particles dissolve by attracting to the dipole water molecules

Heat of dilution refers to the amount of heat absorbed or released when water is added to a solution already made In this latter case if the act of making a solution is exothermic adding more water will make it even more exothermic (hotter) For example as you add more and more water to a solution of sulfuric acid and water the more heat it releases (the more it burns)

Neutralization

bull Explain how you would neutralize an acid and a base

Mix an acidic solution with a basic solution wherein the ratio of H+ ions to OH- ions eq

ual to 1

Cabbage Juice as a pH indicator

bull Explain how cabbage juice works as a pH indicator There is anthocyanin in cabbage juice Anthocyanin reacts w

ith hydroxyl ions (OH-) Hydroxyl ions jump off anthocyanin when in an acidic solution thereby changing the chemical composition of the anthocyanin (the pH indicator) This will cause the juice to react with light differently than if these OH- ions had not jumped off the anthocyanin (makes the juice turn a reddish color) Conversely Hydroxyl ions will jump from the solution onto the anthrocyanin molecules when there is an excess of them in the solution (such as when cabbage juice is mixed with a basic solution) This causes the juice

(the pH indicator) to turn a yellowish color

NOT mentioned

bull Balancing redox reactions

bull Kw

bull Ka

bull Kb

bull pH

bull the galvanic cell

bullKNOW THESE CONCEPTS

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP II

i

[4(-3935) + 1(-2968) + 2(-2858)] ndash [1(ΔH) + 6(0)] = -2523

PRODUCTS －

REACTANTS = ΔHRXN

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP III

SOLVE for X

i

-24424 - ΔH(reactants) = -2523

ΔHC4H4S(l) = 813kJmol

CHANGE IN H

0H EXOTHERMIC

0H CENDOTHERMI

Change in Enthalpy in a multi-step reaction

Change in Enthalpy in a multi-step reaction

CHEMICAL EQUILIBRIA

Things to remember

This is how you convert an Equilibrium constant from Kc to Kp

Kp = Kc (00821 T)^Δn

Δn is the sum of all stoichiometric coefficients belonging to products minus the sum of all stoichiometric coefficients belonging to reactants

T is the temperature during the reaction in Kelvin Remember that Kelvin has exactly the same degree of difference between integers as celsius but it does have a different starting point Kelvins zero is 273deg less than celsius For example 2deg C is equal to 275deg K 27deg C is equal to 300deg K

Qc and Kc Questionh

EQUILIBRIUM of 2 or more RXNs

If a reaction can be expressed as the sum of two or more reactions the equilibrium constant for the overall reaction is given by the product of the equilibrium of the individual reactions

Kc = (Kc) (Kc)

Let me elaborate

nn

nn

c BA

DCK

][][

][][ nn

nn

c DC

FEK

][][

][][

c

c

c

K

__________________

K

K

FEBA

FEDC

DCBA

1st Step Reaction 2nd Step Reaction

OverallReaction

To find the overall equilibrium of a multi-step reaction

cnn

nn

nn

nn

cc KDC

FEx

BA

DCKK

][][

][][

][][

][][

2 step reaction1st step reaction

N2(g) + O2(g) harr 2 NO(g) Kc1 = 23 x 10^-19

2nd step reaction

2 NO(g) + O2(g) harr 2 NO2(g) Kc2 = 3 x 10^6

Kc = Kc1 x Kc2 = (23 x 10^-19)(3 x 10^6) Kc = 7 x 10^-13

Write the equilibrium equation for this multi-step reaction

CHEMICAL KINETICS

Reaction rates

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull Suppose that at a particular moment during the reaction molecular hydrogen is reacting at the rate of 0074 Ms

a) At what rate is ammonia being formedb) At what rate is molecular nitrogen reacting

t

NH

t

H

t

N

322

2

1

3

1

1

1

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is ammonia being formed s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

t

NH

t

NH

t

H

sM

3

32

2

10740

3

1

2

1

3

1t

NHsM

30740

3

2

_

-

_

t

NHsM

30490

GIVEN

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is molecular nitrogen reacting s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

-

GIVEN

t

N

t

N

t

H

sM

2

22

1

10740

3

1

1

1

3

1

t

NsM

20740)1(

3

1

t

NsM

20250

-

Describe the difference between Heat of solution and Heat of dilution Give an example of

each one

Heat of solution is the amount of heat absorbed or released during the act of combining a solute with a solvent An example is adding salt to water There are two steps to this process the Lattice Energy (breaking down the molecules) and the heat of hydration (where the ionically charged particles dissolve by attracting to the dipole water molecules

Heat of dilution refers to the amount of heat absorbed or released when water is added to a solution already made In this latter case if the act of making a solution is exothermic adding more water will make it even more exothermic (hotter) For example as you add more and more water to a solution of sulfuric acid and water the more heat it releases (the more it burns)

Neutralization

bull Explain how you would neutralize an acid and a base

Mix an acidic solution with a basic solution wherein the ratio of H+ ions to OH- ions eq

ual to 1

Cabbage Juice as a pH indicator

bull Explain how cabbage juice works as a pH indicator There is anthocyanin in cabbage juice Anthocyanin reacts w

ith hydroxyl ions (OH-) Hydroxyl ions jump off anthocyanin when in an acidic solution thereby changing the chemical composition of the anthocyanin (the pH indicator) This will cause the juice to react with light differently than if these OH- ions had not jumped off the anthocyanin (makes the juice turn a reddish color) Conversely Hydroxyl ions will jump from the solution onto the anthrocyanin molecules when there is an excess of them in the solution (such as when cabbage juice is mixed with a basic solution) This causes the juice

(the pH indicator) to turn a yellowish color

NOT mentioned

bull Balancing redox reactions

bull Kw

bull Ka

bull Kb

bull pH

bull the galvanic cell

bullKNOW THESE CONCEPTS

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51

Change of Enthalpy of Reaction bull The combustion of thiophene C4H4S(l) a compound used in the

manufacture of pharmaceuticals produces carbon dioxide and sulfur dioxide gases and liquid water The enthalpy change in the combustion of one mole of C4H4S(l) is -2523kJ Use this information and data from Table 64 (p247) to establish ΔHdegf for C4H4S(l)

STEP III

SOLVE for X

i

-24424 - ΔH(reactants) = -2523

ΔHC4H4S(l) = 813kJmol

CHANGE IN H

0H EXOTHERMIC

0H CENDOTHERMI

Change in Enthalpy in a multi-step reaction

Change in Enthalpy in a multi-step reaction

CHEMICAL EQUILIBRIA

Things to remember

This is how you convert an Equilibrium constant from Kc to Kp

Kp = Kc (00821 T)^Δn

Δn is the sum of all stoichiometric coefficients belonging to products minus the sum of all stoichiometric coefficients belonging to reactants

T is the temperature during the reaction in Kelvin Remember that Kelvin has exactly the same degree of difference between integers as celsius but it does have a different starting point Kelvins zero is 273deg less than celsius For example 2deg C is equal to 275deg K 27deg C is equal to 300deg K

Qc and Kc Questionh

EQUILIBRIUM of 2 or more RXNs

If a reaction can be expressed as the sum of two or more reactions the equilibrium constant for the overall reaction is given by the product of the equilibrium of the individual reactions

Kc = (Kc) (Kc)

Let me elaborate

nn

nn

c BA

DCK

][][

][][ nn

nn

c DC

FEK

][][

][][

c

c

c

K

__________________

K

K

FEBA

FEDC

DCBA

1st Step Reaction 2nd Step Reaction

OverallReaction

To find the overall equilibrium of a multi-step reaction

cnn

nn

nn

nn

cc KDC

FEx

BA

DCKK

][][

][][

][][

][][

2 step reaction1st step reaction

N2(g) + O2(g) harr 2 NO(g) Kc1 = 23 x 10^-19

2nd step reaction

2 NO(g) + O2(g) harr 2 NO2(g) Kc2 = 3 x 10^6

Kc = Kc1 x Kc2 = (23 x 10^-19)(3 x 10^6) Kc = 7 x 10^-13

Write the equilibrium equation for this multi-step reaction

CHEMICAL KINETICS

Reaction rates

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull Suppose that at a particular moment during the reaction molecular hydrogen is reacting at the rate of 0074 Ms

a) At what rate is ammonia being formedb) At what rate is molecular nitrogen reacting

t

NH

t

H

t

N

322

2

1

3

1

1

1

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is ammonia being formed s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

t

NH

t

NH

t

H

sM

3

32

2

10740

3

1

2

1

3

1t

NHsM

30740

3

2

_

-

_

t

NHsM

30490

GIVEN

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is molecular nitrogen reacting s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

-

GIVEN

t

N

t

N

t

H

sM

2

22

1

10740

3

1

1

1

3

1

t

NsM

20740)1(

3

1

t

NsM

20250

-

Describe the difference between Heat of solution and Heat of dilution Give an example of

each one

Heat of solution is the amount of heat absorbed or released during the act of combining a solute with a solvent An example is adding salt to water There are two steps to this process the Lattice Energy (breaking down the molecules) and the heat of hydration (where the ionically charged particles dissolve by attracting to the dipole water molecules

Heat of dilution refers to the amount of heat absorbed or released when water is added to a solution already made In this latter case if the act of making a solution is exothermic adding more water will make it even more exothermic (hotter) For example as you add more and more water to a solution of sulfuric acid and water the more heat it releases (the more it burns)

Neutralization

bull Explain how you would neutralize an acid and a base

Mix an acidic solution with a basic solution wherein the ratio of H+ ions to OH- ions eq

ual to 1

Cabbage Juice as a pH indicator

bull Explain how cabbage juice works as a pH indicator There is anthocyanin in cabbage juice Anthocyanin reacts w

ith hydroxyl ions (OH-) Hydroxyl ions jump off anthocyanin when in an acidic solution thereby changing the chemical composition of the anthocyanin (the pH indicator) This will cause the juice to react with light differently than if these OH- ions had not jumped off the anthocyanin (makes the juice turn a reddish color) Conversely Hydroxyl ions will jump from the solution onto the anthrocyanin molecules when there is an excess of them in the solution (such as when cabbage juice is mixed with a basic solution) This causes the juice

(the pH indicator) to turn a yellowish color

NOT mentioned

bull Balancing redox reactions

bull Kw

bull Ka

bull Kb

bull pH

bull the galvanic cell

bullKNOW THESE CONCEPTS

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
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• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51

CHANGE IN H

0H EXOTHERMIC

0H CENDOTHERMI

Change in Enthalpy in a multi-step reaction

Change in Enthalpy in a multi-step reaction

CHEMICAL EQUILIBRIA

Things to remember

This is how you convert an Equilibrium constant from Kc to Kp

Kp = Kc (00821 T)^Δn

Δn is the sum of all stoichiometric coefficients belonging to products minus the sum of all stoichiometric coefficients belonging to reactants

T is the temperature during the reaction in Kelvin Remember that Kelvin has exactly the same degree of difference between integers as celsius but it does have a different starting point Kelvins zero is 273deg less than celsius For example 2deg C is equal to 275deg K 27deg C is equal to 300deg K

Qc and Kc Questionh

EQUILIBRIUM of 2 or more RXNs

If a reaction can be expressed as the sum of two or more reactions the equilibrium constant for the overall reaction is given by the product of the equilibrium of the individual reactions

Kc = (Kc) (Kc)

Let me elaborate

nn

nn

c BA

DCK

][][

][][ nn

nn

c DC

FEK

][][

][][

c

c

c

K

__________________

K

K

FEBA

FEDC

DCBA

1st Step Reaction 2nd Step Reaction

OverallReaction

To find the overall equilibrium of a multi-step reaction

cnn

nn

nn

nn

cc KDC

FEx

BA

DCKK

][][

][][

][][

][][

2 step reaction1st step reaction

N2(g) + O2(g) harr 2 NO(g) Kc1 = 23 x 10^-19

2nd step reaction

2 NO(g) + O2(g) harr 2 NO2(g) Kc2 = 3 x 10^6

Kc = Kc1 x Kc2 = (23 x 10^-19)(3 x 10^6) Kc = 7 x 10^-13

Write the equilibrium equation for this multi-step reaction

CHEMICAL KINETICS

Reaction rates

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull Suppose that at a particular moment during the reaction molecular hydrogen is reacting at the rate of 0074 Ms

a) At what rate is ammonia being formedb) At what rate is molecular nitrogen reacting

t

NH

t

H

t

N

322

2

1

3

1

1

1

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is ammonia being formed s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

t

NH

t

NH

t

H

sM

3

32

2

10740

3

1

2

1

3

1t

NHsM

30740

3

2

_

-

_

t

NHsM

30490

GIVEN

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is molecular nitrogen reacting s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

-

GIVEN

t

N

t

N

t

H

sM

2

22

1

10740

3

1

1

1

3

1

t

NsM

20740)1(

3

1

t

NsM

20250

-

Describe the difference between Heat of solution and Heat of dilution Give an example of

each one

Heat of solution is the amount of heat absorbed or released during the act of combining a solute with a solvent An example is adding salt to water There are two steps to this process the Lattice Energy (breaking down the molecules) and the heat of hydration (where the ionically charged particles dissolve by attracting to the dipole water molecules

Heat of dilution refers to the amount of heat absorbed or released when water is added to a solution already made In this latter case if the act of making a solution is exothermic adding more water will make it even more exothermic (hotter) For example as you add more and more water to a solution of sulfuric acid and water the more heat it releases (the more it burns)

Neutralization

bull Explain how you would neutralize an acid and a base

Mix an acidic solution with a basic solution wherein the ratio of H+ ions to OH- ions eq

ual to 1

Cabbage Juice as a pH indicator

bull Explain how cabbage juice works as a pH indicator There is anthocyanin in cabbage juice Anthocyanin reacts w

ith hydroxyl ions (OH-) Hydroxyl ions jump off anthocyanin when in an acidic solution thereby changing the chemical composition of the anthocyanin (the pH indicator) This will cause the juice to react with light differently than if these OH- ions had not jumped off the anthocyanin (makes the juice turn a reddish color) Conversely Hydroxyl ions will jump from the solution onto the anthrocyanin molecules when there is an excess of them in the solution (such as when cabbage juice is mixed with a basic solution) This causes the juice

(the pH indicator) to turn a yellowish color

NOT mentioned

bull Balancing redox reactions

bull Kw

bull Ka

bull Kb

bull pH

bull the galvanic cell

bullKNOW THESE CONCEPTS

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
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• Slide 31
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• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51

Change in Enthalpy in a multi-step reaction

Change in Enthalpy in a multi-step reaction

CHEMICAL EQUILIBRIA

Things to remember

This is how you convert an Equilibrium constant from Kc to Kp

Kp = Kc (00821 T)^Δn

Δn is the sum of all stoichiometric coefficients belonging to products minus the sum of all stoichiometric coefficients belonging to reactants

T is the temperature during the reaction in Kelvin Remember that Kelvin has exactly the same degree of difference between integers as celsius but it does have a different starting point Kelvins zero is 273deg less than celsius For example 2deg C is equal to 275deg K 27deg C is equal to 300deg K

Qc and Kc Questionh

EQUILIBRIUM of 2 or more RXNs

If a reaction can be expressed as the sum of two or more reactions the equilibrium constant for the overall reaction is given by the product of the equilibrium of the individual reactions

Kc = (Kc) (Kc)

Let me elaborate

nn

nn

c BA

DCK

][][

][][ nn

nn

c DC

FEK

][][

][][

c

c

c

K

__________________

K

K

FEBA

FEDC

DCBA

1st Step Reaction 2nd Step Reaction

OverallReaction

To find the overall equilibrium of a multi-step reaction

cnn

nn

nn

nn

cc KDC

FEx

BA

DCKK

][][

][][

][][

][][

2 step reaction1st step reaction

N2(g) + O2(g) harr 2 NO(g) Kc1 = 23 x 10^-19

2nd step reaction

2 NO(g) + O2(g) harr 2 NO2(g) Kc2 = 3 x 10^6

Kc = Kc1 x Kc2 = (23 x 10^-19)(3 x 10^6) Kc = 7 x 10^-13

Write the equilibrium equation for this multi-step reaction

CHEMICAL KINETICS

Reaction rates

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull Suppose that at a particular moment during the reaction molecular hydrogen is reacting at the rate of 0074 Ms

a) At what rate is ammonia being formedb) At what rate is molecular nitrogen reacting

t

NH

t

H

t

N

322

2

1

3

1

1

1

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is ammonia being formed s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

t

NH

t

NH

t

H

sM

3

32

2

10740

3

1

2

1

3

1t

NHsM

30740

3

2

_

-

_

t

NHsM

30490

GIVEN

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is molecular nitrogen reacting s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

-

GIVEN

t

N

t

N

t

H

sM

2

22

1

10740

3

1

1

1

3

1

t

NsM

20740)1(

3

1

t

NsM

20250

-

Describe the difference between Heat of solution and Heat of dilution Give an example of

each one

Heat of solution is the amount of heat absorbed or released during the act of combining a solute with a solvent An example is adding salt to water There are two steps to this process the Lattice Energy (breaking down the molecules) and the heat of hydration (where the ionically charged particles dissolve by attracting to the dipole water molecules

Heat of dilution refers to the amount of heat absorbed or released when water is added to a solution already made In this latter case if the act of making a solution is exothermic adding more water will make it even more exothermic (hotter) For example as you add more and more water to a solution of sulfuric acid and water the more heat it releases (the more it burns)

Neutralization

bull Explain how you would neutralize an acid and a base

Mix an acidic solution with a basic solution wherein the ratio of H+ ions to OH- ions eq

ual to 1

Cabbage Juice as a pH indicator

bull Explain how cabbage juice works as a pH indicator There is anthocyanin in cabbage juice Anthocyanin reacts w

ith hydroxyl ions (OH-) Hydroxyl ions jump off anthocyanin when in an acidic solution thereby changing the chemical composition of the anthocyanin (the pH indicator) This will cause the juice to react with light differently than if these OH- ions had not jumped off the anthocyanin (makes the juice turn a reddish color) Conversely Hydroxyl ions will jump from the solution onto the anthrocyanin molecules when there is an excess of them in the solution (such as when cabbage juice is mixed with a basic solution) This causes the juice

(the pH indicator) to turn a yellowish color

NOT mentioned

bull Balancing redox reactions

bull Kw

bull Ka

bull Kb

bull pH

bull the galvanic cell

bullKNOW THESE CONCEPTS

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
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• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51

Change in Enthalpy in a multi-step reaction

CHEMICAL EQUILIBRIA

Things to remember

This is how you convert an Equilibrium constant from Kc to Kp

Kp = Kc (00821 T)^Δn

Δn is the sum of all stoichiometric coefficients belonging to products minus the sum of all stoichiometric coefficients belonging to reactants

T is the temperature during the reaction in Kelvin Remember that Kelvin has exactly the same degree of difference between integers as celsius but it does have a different starting point Kelvins zero is 273deg less than celsius For example 2deg C is equal to 275deg K 27deg C is equal to 300deg K

Qc and Kc Questionh

EQUILIBRIUM of 2 or more RXNs

If a reaction can be expressed as the sum of two or more reactions the equilibrium constant for the overall reaction is given by the product of the equilibrium of the individual reactions

Kc = (Kc) (Kc)

Let me elaborate

nn

nn

c BA

DCK

][][

][][ nn

nn

c DC

FEK

][][

][][

c

c

c

K

__________________

K

K

FEBA

FEDC

DCBA

1st Step Reaction 2nd Step Reaction

OverallReaction

To find the overall equilibrium of a multi-step reaction

cnn

nn

nn

nn

cc KDC

FEx

BA

DCKK

][][

][][

][][

][][

2 step reaction1st step reaction

N2(g) + O2(g) harr 2 NO(g) Kc1 = 23 x 10^-19

2nd step reaction

2 NO(g) + O2(g) harr 2 NO2(g) Kc2 = 3 x 10^6

Kc = Kc1 x Kc2 = (23 x 10^-19)(3 x 10^6) Kc = 7 x 10^-13

Write the equilibrium equation for this multi-step reaction

CHEMICAL KINETICS

Reaction rates

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull Suppose that at a particular moment during the reaction molecular hydrogen is reacting at the rate of 0074 Ms

a) At what rate is ammonia being formedb) At what rate is molecular nitrogen reacting

t

NH

t

H

t

N

322

2

1

3

1

1

1

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is ammonia being formed s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

t

NH

t

NH

t

H

sM

3

32

2

10740

3

1

2

1

3

1t

NHsM

30740

3

2

_

-

_

t

NHsM

30490

GIVEN

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is molecular nitrogen reacting s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

-

GIVEN

t

N

t

N

t

H

sM

2

22

1

10740

3

1

1

1

3

1

t

NsM

20740)1(

3

1

t

NsM

20250

-

Describe the difference between Heat of solution and Heat of dilution Give an example of

each one

Heat of solution is the amount of heat absorbed or released during the act of combining a solute with a solvent An example is adding salt to water There are two steps to this process the Lattice Energy (breaking down the molecules) and the heat of hydration (where the ionically charged particles dissolve by attracting to the dipole water molecules

Heat of dilution refers to the amount of heat absorbed or released when water is added to a solution already made In this latter case if the act of making a solution is exothermic adding more water will make it even more exothermic (hotter) For example as you add more and more water to a solution of sulfuric acid and water the more heat it releases (the more it burns)

Neutralization

bull Explain how you would neutralize an acid and a base

Mix an acidic solution with a basic solution wherein the ratio of H+ ions to OH- ions eq

ual to 1

Cabbage Juice as a pH indicator

bull Explain how cabbage juice works as a pH indicator There is anthocyanin in cabbage juice Anthocyanin reacts w

ith hydroxyl ions (OH-) Hydroxyl ions jump off anthocyanin when in an acidic solution thereby changing the chemical composition of the anthocyanin (the pH indicator) This will cause the juice to react with light differently than if these OH- ions had not jumped off the anthocyanin (makes the juice turn a reddish color) Conversely Hydroxyl ions will jump from the solution onto the anthrocyanin molecules when there is an excess of them in the solution (such as when cabbage juice is mixed with a basic solution) This causes the juice

(the pH indicator) to turn a yellowish color

NOT mentioned

bull Balancing redox reactions

bull Kw

bull Ka

bull Kb

bull pH

bull the galvanic cell

bullKNOW THESE CONCEPTS

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51

CHEMICAL EQUILIBRIA

Things to remember

This is how you convert an Equilibrium constant from Kc to Kp

Kp = Kc (00821 T)^Δn

Δn is the sum of all stoichiometric coefficients belonging to products minus the sum of all stoichiometric coefficients belonging to reactants

T is the temperature during the reaction in Kelvin Remember that Kelvin has exactly the same degree of difference between integers as celsius but it does have a different starting point Kelvins zero is 273deg less than celsius For example 2deg C is equal to 275deg K 27deg C is equal to 300deg K

Qc and Kc Questionh

EQUILIBRIUM of 2 or more RXNs

If a reaction can be expressed as the sum of two or more reactions the equilibrium constant for the overall reaction is given by the product of the equilibrium of the individual reactions

Kc = (Kc) (Kc)

Let me elaborate

nn

nn

c BA

DCK

][][

][][ nn

nn

c DC

FEK

][][

][][

c

c

c

K

__________________

K

K

FEBA

FEDC

DCBA

1st Step Reaction 2nd Step Reaction

OverallReaction

To find the overall equilibrium of a multi-step reaction

cnn

nn

nn

nn

cc KDC

FEx

BA

DCKK

][][

][][

][][

][][

2 step reaction1st step reaction

N2(g) + O2(g) harr 2 NO(g) Kc1 = 23 x 10^-19

2nd step reaction

2 NO(g) + O2(g) harr 2 NO2(g) Kc2 = 3 x 10^6

Kc = Kc1 x Kc2 = (23 x 10^-19)(3 x 10^6) Kc = 7 x 10^-13

Write the equilibrium equation for this multi-step reaction

CHEMICAL KINETICS

Reaction rates

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull Suppose that at a particular moment during the reaction molecular hydrogen is reacting at the rate of 0074 Ms

a) At what rate is ammonia being formedb) At what rate is molecular nitrogen reacting

t

NH

t

H

t

N

322

2

1

3

1

1

1

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is ammonia being formed s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

t

NH

t

NH

t

H

sM

3

32

2

10740

3

1

2

1

3

1t

NHsM

30740

3

2

_

-

_

t

NHsM

30490

GIVEN

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is molecular nitrogen reacting s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

-

GIVEN

t

N

t

N

t

H

sM

2

22

1

10740

3

1

1

1

3

1

t

NsM

20740)1(

3

1

t

NsM

20250

-

Describe the difference between Heat of solution and Heat of dilution Give an example of

each one

Heat of solution is the amount of heat absorbed or released during the act of combining a solute with a solvent An example is adding salt to water There are two steps to this process the Lattice Energy (breaking down the molecules) and the heat of hydration (where the ionically charged particles dissolve by attracting to the dipole water molecules

Heat of dilution refers to the amount of heat absorbed or released when water is added to a solution already made In this latter case if the act of making a solution is exothermic adding more water will make it even more exothermic (hotter) For example as you add more and more water to a solution of sulfuric acid and water the more heat it releases (the more it burns)

Neutralization

bull Explain how you would neutralize an acid and a base

Mix an acidic solution with a basic solution wherein the ratio of H+ ions to OH- ions eq

ual to 1

Cabbage Juice as a pH indicator

bull Explain how cabbage juice works as a pH indicator There is anthocyanin in cabbage juice Anthocyanin reacts w

ith hydroxyl ions (OH-) Hydroxyl ions jump off anthocyanin when in an acidic solution thereby changing the chemical composition of the anthocyanin (the pH indicator) This will cause the juice to react with light differently than if these OH- ions had not jumped off the anthocyanin (makes the juice turn a reddish color) Conversely Hydroxyl ions will jump from the solution onto the anthrocyanin molecules when there is an excess of them in the solution (such as when cabbage juice is mixed with a basic solution) This causes the juice

(the pH indicator) to turn a yellowish color

NOT mentioned

bull Balancing redox reactions

bull Kw

bull Ka

bull Kb

bull pH

bull the galvanic cell

bullKNOW THESE CONCEPTS

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51

Things to remember

This is how you convert an Equilibrium constant from Kc to Kp

Kp = Kc (00821 T)^Δn

Δn is the sum of all stoichiometric coefficients belonging to products minus the sum of all stoichiometric coefficients belonging to reactants

T is the temperature during the reaction in Kelvin Remember that Kelvin has exactly the same degree of difference between integers as celsius but it does have a different starting point Kelvins zero is 273deg less than celsius For example 2deg C is equal to 275deg K 27deg C is equal to 300deg K

Qc and Kc Questionh

EQUILIBRIUM of 2 or more RXNs

If a reaction can be expressed as the sum of two or more reactions the equilibrium constant for the overall reaction is given by the product of the equilibrium of the individual reactions

Kc = (Kc) (Kc)

Let me elaborate

nn

nn

c BA

DCK

][][

][][ nn

nn

c DC

FEK

][][

][][

c

c

c

K

__________________

K

K

FEBA

FEDC

DCBA

1st Step Reaction 2nd Step Reaction

OverallReaction

To find the overall equilibrium of a multi-step reaction

cnn

nn

nn

nn

cc KDC

FEx

BA

DCKK

][][

][][

][][

][][

2 step reaction1st step reaction

N2(g) + O2(g) harr 2 NO(g) Kc1 = 23 x 10^-19

2nd step reaction

2 NO(g) + O2(g) harr 2 NO2(g) Kc2 = 3 x 10^6

Kc = Kc1 x Kc2 = (23 x 10^-19)(3 x 10^6) Kc = 7 x 10^-13

Write the equilibrium equation for this multi-step reaction

CHEMICAL KINETICS

Reaction rates

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull Suppose that at a particular moment during the reaction molecular hydrogen is reacting at the rate of 0074 Ms

a) At what rate is ammonia being formedb) At what rate is molecular nitrogen reacting

t

NH

t

H

t

N

322

2

1

3

1

1

1

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is ammonia being formed s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

t

NH

t

NH

t

H

sM

3

32

2

10740

3

1

2

1

3

1t

NHsM

30740

3

2

_

-

_

t

NHsM

30490

GIVEN

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is molecular nitrogen reacting s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

-

GIVEN

t

N

t

N

t

H

sM

2

22

1

10740

3

1

1

1

3

1

t

NsM

20740)1(

3

1

t

NsM

20250

-

Describe the difference between Heat of solution and Heat of dilution Give an example of

each one

Heat of solution is the amount of heat absorbed or released during the act of combining a solute with a solvent An example is adding salt to water There are two steps to this process the Lattice Energy (breaking down the molecules) and the heat of hydration (where the ionically charged particles dissolve by attracting to the dipole water molecules

Heat of dilution refers to the amount of heat absorbed or released when water is added to a solution already made In this latter case if the act of making a solution is exothermic adding more water will make it even more exothermic (hotter) For example as you add more and more water to a solution of sulfuric acid and water the more heat it releases (the more it burns)

Neutralization

bull Explain how you would neutralize an acid and a base

Mix an acidic solution with a basic solution wherein the ratio of H+ ions to OH- ions eq

ual to 1

Cabbage Juice as a pH indicator

bull Explain how cabbage juice works as a pH indicator There is anthocyanin in cabbage juice Anthocyanin reacts w

ith hydroxyl ions (OH-) Hydroxyl ions jump off anthocyanin when in an acidic solution thereby changing the chemical composition of the anthocyanin (the pH indicator) This will cause the juice to react with light differently than if these OH- ions had not jumped off the anthocyanin (makes the juice turn a reddish color) Conversely Hydroxyl ions will jump from the solution onto the anthrocyanin molecules when there is an excess of them in the solution (such as when cabbage juice is mixed with a basic solution) This causes the juice

(the pH indicator) to turn a yellowish color

NOT mentioned

bull Balancing redox reactions

bull Kw

bull Ka

bull Kb

bull pH

bull the galvanic cell

bullKNOW THESE CONCEPTS

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51

Qc and Kc Questionh

EQUILIBRIUM of 2 or more RXNs

If a reaction can be expressed as the sum of two or more reactions the equilibrium constant for the overall reaction is given by the product of the equilibrium of the individual reactions

Kc = (Kc) (Kc)

Let me elaborate

nn

nn

c BA

DCK

][][

][][ nn

nn

c DC

FEK

][][

][][

c

c

c

K

__________________

K

K

FEBA

FEDC

DCBA

1st Step Reaction 2nd Step Reaction

OverallReaction

To find the overall equilibrium of a multi-step reaction

cnn

nn

nn

nn

cc KDC

FEx

BA

DCKK

][][

][][

][][

][][

2 step reaction1st step reaction

N2(g) + O2(g) harr 2 NO(g) Kc1 = 23 x 10^-19

2nd step reaction

2 NO(g) + O2(g) harr 2 NO2(g) Kc2 = 3 x 10^6

Kc = Kc1 x Kc2 = (23 x 10^-19)(3 x 10^6) Kc = 7 x 10^-13

Write the equilibrium equation for this multi-step reaction

CHEMICAL KINETICS

Reaction rates

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull Suppose that at a particular moment during the reaction molecular hydrogen is reacting at the rate of 0074 Ms

a) At what rate is ammonia being formedb) At what rate is molecular nitrogen reacting

t

NH

t

H

t

N

322

2

1

3

1

1

1

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is ammonia being formed s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

t

NH

t

NH

t

H

sM

3

32

2

10740

3

1

2

1

3

1t

NHsM

30740

3

2

_

-

_

t

NHsM

30490

GIVEN

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is molecular nitrogen reacting s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

-

GIVEN

t

N

t

N

t

H

sM

2

22

1

10740

3

1

1

1

3

1

t

NsM

20740)1(

3

1

t

NsM

20250

-

Describe the difference between Heat of solution and Heat of dilution Give an example of

each one

Heat of solution is the amount of heat absorbed or released during the act of combining a solute with a solvent An example is adding salt to water There are two steps to this process the Lattice Energy (breaking down the molecules) and the heat of hydration (where the ionically charged particles dissolve by attracting to the dipole water molecules

Heat of dilution refers to the amount of heat absorbed or released when water is added to a solution already made In this latter case if the act of making a solution is exothermic adding more water will make it even more exothermic (hotter) For example as you add more and more water to a solution of sulfuric acid and water the more heat it releases (the more it burns)

Neutralization

bull Explain how you would neutralize an acid and a base

Mix an acidic solution with a basic solution wherein the ratio of H+ ions to OH- ions eq

ual to 1

Cabbage Juice as a pH indicator

bull Explain how cabbage juice works as a pH indicator There is anthocyanin in cabbage juice Anthocyanin reacts w

ith hydroxyl ions (OH-) Hydroxyl ions jump off anthocyanin when in an acidic solution thereby changing the chemical composition of the anthocyanin (the pH indicator) This will cause the juice to react with light differently than if these OH- ions had not jumped off the anthocyanin (makes the juice turn a reddish color) Conversely Hydroxyl ions will jump from the solution onto the anthrocyanin molecules when there is an excess of them in the solution (such as when cabbage juice is mixed with a basic solution) This causes the juice

(the pH indicator) to turn a yellowish color

NOT mentioned

bull Balancing redox reactions

bull Kw

bull Ka

bull Kb

bull pH

bull the galvanic cell

bullKNOW THESE CONCEPTS

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51

EQUILIBRIUM of 2 or more RXNs

If a reaction can be expressed as the sum of two or more reactions the equilibrium constant for the overall reaction is given by the product of the equilibrium of the individual reactions

Kc = (Kc) (Kc)

Let me elaborate

nn

nn

c BA

DCK

][][

][][ nn

nn

c DC

FEK

][][

][][

c

c

c

K

__________________

K

K

FEBA

FEDC

DCBA

1st Step Reaction 2nd Step Reaction

OverallReaction

To find the overall equilibrium of a multi-step reaction

cnn

nn

nn

nn

cc KDC

FEx

BA

DCKK

][][

][][

][][

][][

2 step reaction1st step reaction

N2(g) + O2(g) harr 2 NO(g) Kc1 = 23 x 10^-19

2nd step reaction

2 NO(g) + O2(g) harr 2 NO2(g) Kc2 = 3 x 10^6

Kc = Kc1 x Kc2 = (23 x 10^-19)(3 x 10^6) Kc = 7 x 10^-13

Write the equilibrium equation for this multi-step reaction

CHEMICAL KINETICS

Reaction rates

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull Suppose that at a particular moment during the reaction molecular hydrogen is reacting at the rate of 0074 Ms

a) At what rate is ammonia being formedb) At what rate is molecular nitrogen reacting

t

NH

t

H

t

N

322

2

1

3

1

1

1

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is ammonia being formed s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

t

NH

t

NH

t

H

sM

3

32

2

10740

3

1

2

1

3

1t

NHsM

30740

3

2

_

-

_

t

NHsM

30490

GIVEN

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is molecular nitrogen reacting s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

-

GIVEN

t

N

t

N

t

H

sM

2

22

1

10740

3

1

1

1

3

1

t

NsM

20740)1(

3

1

t

NsM

20250

-

Describe the difference between Heat of solution and Heat of dilution Give an example of

each one

Heat of solution is the amount of heat absorbed or released during the act of combining a solute with a solvent An example is adding salt to water There are two steps to this process the Lattice Energy (breaking down the molecules) and the heat of hydration (where the ionically charged particles dissolve by attracting to the dipole water molecules

Heat of dilution refers to the amount of heat absorbed or released when water is added to a solution already made In this latter case if the act of making a solution is exothermic adding more water will make it even more exothermic (hotter) For example as you add more and more water to a solution of sulfuric acid and water the more heat it releases (the more it burns)

Neutralization

bull Explain how you would neutralize an acid and a base

Mix an acidic solution with a basic solution wherein the ratio of H+ ions to OH- ions eq

ual to 1

Cabbage Juice as a pH indicator

bull Explain how cabbage juice works as a pH indicator There is anthocyanin in cabbage juice Anthocyanin reacts w

ith hydroxyl ions (OH-) Hydroxyl ions jump off anthocyanin when in an acidic solution thereby changing the chemical composition of the anthocyanin (the pH indicator) This will cause the juice to react with light differently than if these OH- ions had not jumped off the anthocyanin (makes the juice turn a reddish color) Conversely Hydroxyl ions will jump from the solution onto the anthrocyanin molecules when there is an excess of them in the solution (such as when cabbage juice is mixed with a basic solution) This causes the juice

(the pH indicator) to turn a yellowish color

NOT mentioned

bull Balancing redox reactions

bull Kw

bull Ka

bull Kb

bull pH

bull the galvanic cell

bullKNOW THESE CONCEPTS

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51

Let me elaborate

nn

nn

c BA

DCK

][][

][][ nn

nn

c DC

FEK

][][

][][

c

c

c

K

__________________

K

K

FEBA

FEDC

DCBA

1st Step Reaction 2nd Step Reaction

OverallReaction

To find the overall equilibrium of a multi-step reaction

cnn

nn

nn

nn

cc KDC

FEx

BA

DCKK

][][

][][

][][

][][

2 step reaction1st step reaction

N2(g) + O2(g) harr 2 NO(g) Kc1 = 23 x 10^-19

2nd step reaction

2 NO(g) + O2(g) harr 2 NO2(g) Kc2 = 3 x 10^6

Kc = Kc1 x Kc2 = (23 x 10^-19)(3 x 10^6) Kc = 7 x 10^-13

Write the equilibrium equation for this multi-step reaction

CHEMICAL KINETICS

Reaction rates

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull Suppose that at a particular moment during the reaction molecular hydrogen is reacting at the rate of 0074 Ms

a) At what rate is ammonia being formedb) At what rate is molecular nitrogen reacting

t

NH

t

H

t

N

322

2

1

3

1

1

1

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is ammonia being formed s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

t

NH

t

NH

t

H

sM

3

32

2

10740

3

1

2

1

3

1t

NHsM

30740

3

2

_

-

_

t

NHsM

30490

GIVEN

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is molecular nitrogen reacting s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

-

GIVEN

t

N

t

N

t

H

sM

2

22

1

10740

3

1

1

1

3

1

t

NsM

20740)1(

3

1

t

NsM

20250

-

Describe the difference between Heat of solution and Heat of dilution Give an example of

each one

Heat of solution is the amount of heat absorbed or released during the act of combining a solute with a solvent An example is adding salt to water There are two steps to this process the Lattice Energy (breaking down the molecules) and the heat of hydration (where the ionically charged particles dissolve by attracting to the dipole water molecules

Heat of dilution refers to the amount of heat absorbed or released when water is added to a solution already made In this latter case if the act of making a solution is exothermic adding more water will make it even more exothermic (hotter) For example as you add more and more water to a solution of sulfuric acid and water the more heat it releases (the more it burns)

Neutralization

bull Explain how you would neutralize an acid and a base

Mix an acidic solution with a basic solution wherein the ratio of H+ ions to OH- ions eq

ual to 1

Cabbage Juice as a pH indicator

bull Explain how cabbage juice works as a pH indicator There is anthocyanin in cabbage juice Anthocyanin reacts w

ith hydroxyl ions (OH-) Hydroxyl ions jump off anthocyanin when in an acidic solution thereby changing the chemical composition of the anthocyanin (the pH indicator) This will cause the juice to react with light differently than if these OH- ions had not jumped off the anthocyanin (makes the juice turn a reddish color) Conversely Hydroxyl ions will jump from the solution onto the anthrocyanin molecules when there is an excess of them in the solution (such as when cabbage juice is mixed with a basic solution) This causes the juice

(the pH indicator) to turn a yellowish color

NOT mentioned

bull Balancing redox reactions

bull Kw

bull Ka

bull Kb

bull pH

bull the galvanic cell

bullKNOW THESE CONCEPTS

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51

To find the overall equilibrium of a multi-step reaction

cnn

nn

nn

nn

cc KDC

FEx

BA

DCKK

][][

][][

][][

][][

2 step reaction1st step reaction

N2(g) + O2(g) harr 2 NO(g) Kc1 = 23 x 10^-19

2nd step reaction

2 NO(g) + O2(g) harr 2 NO2(g) Kc2 = 3 x 10^6

Kc = Kc1 x Kc2 = (23 x 10^-19)(3 x 10^6) Kc = 7 x 10^-13

Write the equilibrium equation for this multi-step reaction

CHEMICAL KINETICS

Reaction rates

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull Suppose that at a particular moment during the reaction molecular hydrogen is reacting at the rate of 0074 Ms

a) At what rate is ammonia being formedb) At what rate is molecular nitrogen reacting

t

NH

t

H

t

N

322

2

1

3

1

1

1

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is ammonia being formed s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

t

NH

t

NH

t

H

sM

3

32

2

10740

3

1

2

1

3

1t

NHsM

30740

3

2

_

-

_

t

NHsM

30490

GIVEN

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is molecular nitrogen reacting s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

-

GIVEN

t

N

t

N

t

H

sM

2

22

1

10740

3

1

1

1

3

1

t

NsM

20740)1(

3

1

t

NsM

20250

-

Describe the difference between Heat of solution and Heat of dilution Give an example of

each one

Heat of solution is the amount of heat absorbed or released during the act of combining a solute with a solvent An example is adding salt to water There are two steps to this process the Lattice Energy (breaking down the molecules) and the heat of hydration (where the ionically charged particles dissolve by attracting to the dipole water molecules

Heat of dilution refers to the amount of heat absorbed or released when water is added to a solution already made In this latter case if the act of making a solution is exothermic adding more water will make it even more exothermic (hotter) For example as you add more and more water to a solution of sulfuric acid and water the more heat it releases (the more it burns)

Neutralization

bull Explain how you would neutralize an acid and a base

Mix an acidic solution with a basic solution wherein the ratio of H+ ions to OH- ions eq

ual to 1

Cabbage Juice as a pH indicator

bull Explain how cabbage juice works as a pH indicator There is anthocyanin in cabbage juice Anthocyanin reacts w

ith hydroxyl ions (OH-) Hydroxyl ions jump off anthocyanin when in an acidic solution thereby changing the chemical composition of the anthocyanin (the pH indicator) This will cause the juice to react with light differently than if these OH- ions had not jumped off the anthocyanin (makes the juice turn a reddish color) Conversely Hydroxyl ions will jump from the solution onto the anthrocyanin molecules when there is an excess of them in the solution (such as when cabbage juice is mixed with a basic solution) This causes the juice

(the pH indicator) to turn a yellowish color

NOT mentioned

bull Balancing redox reactions

bull Kw

bull Ka

bull Kb

bull pH

bull the galvanic cell

bullKNOW THESE CONCEPTS

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51

2 step reaction1st step reaction

N2(g) + O2(g) harr 2 NO(g) Kc1 = 23 x 10^-19

2nd step reaction

2 NO(g) + O2(g) harr 2 NO2(g) Kc2 = 3 x 10^6

Kc = Kc1 x Kc2 = (23 x 10^-19)(3 x 10^6) Kc = 7 x 10^-13

Write the equilibrium equation for this multi-step reaction

CHEMICAL KINETICS

Reaction rates

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull Suppose that at a particular moment during the reaction molecular hydrogen is reacting at the rate of 0074 Ms

a) At what rate is ammonia being formedb) At what rate is molecular nitrogen reacting

t

NH

t

H

t

N

322

2

1

3

1

1

1

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is ammonia being formed s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

t

NH

t

NH

t

H

sM

3

32

2

10740

3

1

2

1

3

1t

NHsM

30740

3

2

_

-

_

t

NHsM

30490

GIVEN

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is molecular nitrogen reacting s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

-

GIVEN

t

N

t

N

t

H

sM

2

22

1

10740

3

1

1

1

3

1

t

NsM

20740)1(

3

1

t

NsM

20250

-

Describe the difference between Heat of solution and Heat of dilution Give an example of

each one

Heat of solution is the amount of heat absorbed or released during the act of combining a solute with a solvent An example is adding salt to water There are two steps to this process the Lattice Energy (breaking down the molecules) and the heat of hydration (where the ionically charged particles dissolve by attracting to the dipole water molecules

Heat of dilution refers to the amount of heat absorbed or released when water is added to a solution already made In this latter case if the act of making a solution is exothermic adding more water will make it even more exothermic (hotter) For example as you add more and more water to a solution of sulfuric acid and water the more heat it releases (the more it burns)

Neutralization

bull Explain how you would neutralize an acid and a base

Mix an acidic solution with a basic solution wherein the ratio of H+ ions to OH- ions eq

ual to 1

Cabbage Juice as a pH indicator

bull Explain how cabbage juice works as a pH indicator There is anthocyanin in cabbage juice Anthocyanin reacts w

ith hydroxyl ions (OH-) Hydroxyl ions jump off anthocyanin when in an acidic solution thereby changing the chemical composition of the anthocyanin (the pH indicator) This will cause the juice to react with light differently than if these OH- ions had not jumped off the anthocyanin (makes the juice turn a reddish color) Conversely Hydroxyl ions will jump from the solution onto the anthrocyanin molecules when there is an excess of them in the solution (such as when cabbage juice is mixed with a basic solution) This causes the juice

(the pH indicator) to turn a yellowish color

NOT mentioned

bull Balancing redox reactions

bull Kw

bull Ka

bull Kb

bull pH

bull the galvanic cell

bullKNOW THESE CONCEPTS

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
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CHEMICAL KINETICS

Reaction rates

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull Suppose that at a particular moment during the reaction molecular hydrogen is reacting at the rate of 0074 Ms

a) At what rate is ammonia being formedb) At what rate is molecular nitrogen reacting

t

NH

t

H

t

N

322

2

1

3

1

1

1

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is ammonia being formed s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

t

NH

t

NH

t

H

sM

3

32

2

10740

3

1

2

1

3

1t

NHsM

30740

3

2

_

-

_

t

NHsM

30490

GIVEN

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is molecular nitrogen reacting s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

-

GIVEN

t

N

t

N

t

H

sM

2

22

1

10740

3

1

1

1

3

1

t

NsM

20740)1(

3

1

t

NsM

20250

-

Describe the difference between Heat of solution and Heat of dilution Give an example of

each one

Heat of solution is the amount of heat absorbed or released during the act of combining a solute with a solvent An example is adding salt to water There are two steps to this process the Lattice Energy (breaking down the molecules) and the heat of hydration (where the ionically charged particles dissolve by attracting to the dipole water molecules

Heat of dilution refers to the amount of heat absorbed or released when water is added to a solution already made In this latter case if the act of making a solution is exothermic adding more water will make it even more exothermic (hotter) For example as you add more and more water to a solution of sulfuric acid and water the more heat it releases (the more it burns)

Neutralization

bull Explain how you would neutralize an acid and a base

Mix an acidic solution with a basic solution wherein the ratio of H+ ions to OH- ions eq

ual to 1

Cabbage Juice as a pH indicator

bull Explain how cabbage juice works as a pH indicator There is anthocyanin in cabbage juice Anthocyanin reacts w

ith hydroxyl ions (OH-) Hydroxyl ions jump off anthocyanin when in an acidic solution thereby changing the chemical composition of the anthocyanin (the pH indicator) This will cause the juice to react with light differently than if these OH- ions had not jumped off the anthocyanin (makes the juice turn a reddish color) Conversely Hydroxyl ions will jump from the solution onto the anthrocyanin molecules when there is an excess of them in the solution (such as when cabbage juice is mixed with a basic solution) This causes the juice

(the pH indicator) to turn a yellowish color

NOT mentioned

bull Balancing redox reactions

bull Kw

bull Ka

bull Kb

bull pH

bull the galvanic cell

bullKNOW THESE CONCEPTS

• Slide 1
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• Slide 51

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull Suppose that at a particular moment during the reaction molecular hydrogen is reacting at the rate of 0074 Ms

a) At what rate is ammonia being formedb) At what rate is molecular nitrogen reacting

t

NH

t

H

t

N

322

2

1

3

1

1

1

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is ammonia being formed s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

t

NH

t

NH

t

H

sM

3

32

2

10740

3

1

2

1

3

1t

NHsM

30740

3

2

_

-

_

t

NHsM

30490

GIVEN

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is molecular nitrogen reacting s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

-

GIVEN

t

N

t

N

t

H

sM

2

22

1

10740

3

1

1

1

3

1

t

NsM

20740)1(

3

1

t

NsM

20250

-

Describe the difference between Heat of solution and Heat of dilution Give an example of

each one

Heat of solution is the amount of heat absorbed or released during the act of combining a solute with a solvent An example is adding salt to water There are two steps to this process the Lattice Energy (breaking down the molecules) and the heat of hydration (where the ionically charged particles dissolve by attracting to the dipole water molecules

Heat of dilution refers to the amount of heat absorbed or released when water is added to a solution already made In this latter case if the act of making a solution is exothermic adding more water will make it even more exothermic (hotter) For example as you add more and more water to a solution of sulfuric acid and water the more heat it releases (the more it burns)

Neutralization

bull Explain how you would neutralize an acid and a base

Mix an acidic solution with a basic solution wherein the ratio of H+ ions to OH- ions eq

ual to 1

Cabbage Juice as a pH indicator

bull Explain how cabbage juice works as a pH indicator There is anthocyanin in cabbage juice Anthocyanin reacts w

ith hydroxyl ions (OH-) Hydroxyl ions jump off anthocyanin when in an acidic solution thereby changing the chemical composition of the anthocyanin (the pH indicator) This will cause the juice to react with light differently than if these OH- ions had not jumped off the anthocyanin (makes the juice turn a reddish color) Conversely Hydroxyl ions will jump from the solution onto the anthrocyanin molecules when there is an excess of them in the solution (such as when cabbage juice is mixed with a basic solution) This causes the juice

(the pH indicator) to turn a yellowish color

NOT mentioned

bull Balancing redox reactions

bull Kw

bull Ka

bull Kb

bull pH

bull the galvanic cell

bullKNOW THESE CONCEPTS

• Slide 1
• Slide 2
• Slide 3
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• Slide 51

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is ammonia being formed s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

t

NH

t

NH

t

H

sM

3

32

2

10740

3

1

2

1

3

1t

NHsM

30740

3

2

_

-

_

t

NHsM

30490

GIVEN

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is molecular nitrogen reacting s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

-

GIVEN

t

N

t

N

t

H

sM

2

22

1

10740

3

1

1

1

3

1

t

NsM

20740)1(

3

1

t

NsM

20250

-

Describe the difference between Heat of solution and Heat of dilution Give an example of

each one

Heat of solution is the amount of heat absorbed or released during the act of combining a solute with a solvent An example is adding salt to water There are two steps to this process the Lattice Energy (breaking down the molecules) and the heat of hydration (where the ionically charged particles dissolve by attracting to the dipole water molecules

Heat of dilution refers to the amount of heat absorbed or released when water is added to a solution already made In this latter case if the act of making a solution is exothermic adding more water will make it even more exothermic (hotter) For example as you add more and more water to a solution of sulfuric acid and water the more heat it releases (the more it burns)

Neutralization

bull Explain how you would neutralize an acid and a base

Mix an acidic solution with a basic solution wherein the ratio of H+ ions to OH- ions eq

ual to 1

Cabbage Juice as a pH indicator

bull Explain how cabbage juice works as a pH indicator There is anthocyanin in cabbage juice Anthocyanin reacts w

ith hydroxyl ions (OH-) Hydroxyl ions jump off anthocyanin when in an acidic solution thereby changing the chemical composition of the anthocyanin (the pH indicator) This will cause the juice to react with light differently than if these OH- ions had not jumped off the anthocyanin (makes the juice turn a reddish color) Conversely Hydroxyl ions will jump from the solution onto the anthrocyanin molecules when there is an excess of them in the solution (such as when cabbage juice is mixed with a basic solution) This causes the juice

(the pH indicator) to turn a yellowish color

NOT mentioned

bull Balancing redox reactions

bull Kw

bull Ka

bull Kb

bull pH

bull the galvanic cell

bullKNOW THESE CONCEPTS

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
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• Slide 51

N2 + 3H2 rarr 2NH3Molecular hydrogen reacts at a rate of 0074 Ms

bull At what rate is molecular nitrogen reacting s

M

t

H07402

t

NH

t

H

t

N

322

2

1

3

1

1

1

-

GIVEN

t

N

t

N

t

H

sM

2

22

1

10740

3

1

1

1

3

1

t

NsM

20740)1(

3

1

t

NsM

20250

-

Describe the difference between Heat of solution and Heat of dilution Give an example of

each one

Heat of solution is the amount of heat absorbed or released during the act of combining a solute with a solvent An example is adding salt to water There are two steps to this process the Lattice Energy (breaking down the molecules) and the heat of hydration (where the ionically charged particles dissolve by attracting to the dipole water molecules

Heat of dilution refers to the amount of heat absorbed or released when water is added to a solution already made In this latter case if the act of making a solution is exothermic adding more water will make it even more exothermic (hotter) For example as you add more and more water to a solution of sulfuric acid and water the more heat it releases (the more it burns)

Neutralization

bull Explain how you would neutralize an acid and a base

Mix an acidic solution with a basic solution wherein the ratio of H+ ions to OH- ions eq

ual to 1

Cabbage Juice as a pH indicator

bull Explain how cabbage juice works as a pH indicator There is anthocyanin in cabbage juice Anthocyanin reacts w

ith hydroxyl ions (OH-) Hydroxyl ions jump off anthocyanin when in an acidic solution thereby changing the chemical composition of the anthocyanin (the pH indicator) This will cause the juice to react with light differently than if these OH- ions had not jumped off the anthocyanin (makes the juice turn a reddish color) Conversely Hydroxyl ions will jump from the solution onto the anthrocyanin molecules when there is an excess of them in the solution (such as when cabbage juice is mixed with a basic solution) This causes the juice

(the pH indicator) to turn a yellowish color

NOT mentioned

bull Balancing redox reactions

bull Kw

bull Ka

bull Kb

bull pH

bull the galvanic cell

bullKNOW THESE CONCEPTS

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
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• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51

Describe the difference between Heat of solution and Heat of dilution Give an example of

each one

Heat of solution is the amount of heat absorbed or released during the act of combining a solute with a solvent An example is adding salt to water There are two steps to this process the Lattice Energy (breaking down the molecules) and the heat of hydration (where the ionically charged particles dissolve by attracting to the dipole water molecules

Heat of dilution refers to the amount of heat absorbed or released when water is added to a solution already made In this latter case if the act of making a solution is exothermic adding more water will make it even more exothermic (hotter) For example as you add more and more water to a solution of sulfuric acid and water the more heat it releases (the more it burns)

Neutralization

bull Explain how you would neutralize an acid and a base

Mix an acidic solution with a basic solution wherein the ratio of H+ ions to OH- ions eq

ual to 1

Cabbage Juice as a pH indicator

bull Explain how cabbage juice works as a pH indicator There is anthocyanin in cabbage juice Anthocyanin reacts w

ith hydroxyl ions (OH-) Hydroxyl ions jump off anthocyanin when in an acidic solution thereby changing the chemical composition of the anthocyanin (the pH indicator) This will cause the juice to react with light differently than if these OH- ions had not jumped off the anthocyanin (makes the juice turn a reddish color) Conversely Hydroxyl ions will jump from the solution onto the anthrocyanin molecules when there is an excess of them in the solution (such as when cabbage juice is mixed with a basic solution) This causes the juice

(the pH indicator) to turn a yellowish color

NOT mentioned

bull Balancing redox reactions

bull Kw

bull Ka

bull Kb

bull pH

bull the galvanic cell

bullKNOW THESE CONCEPTS

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
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• Slide 49
• Slide 50
• Slide 51

Heat of solution is the amount of heat absorbed or released during the act of combining a solute with a solvent An example is adding salt to water There are two steps to this process the Lattice Energy (breaking down the molecules) and the heat of hydration (where the ionically charged particles dissolve by attracting to the dipole water molecules

Heat of dilution refers to the amount of heat absorbed or released when water is added to a solution already made In this latter case if the act of making a solution is exothermic adding more water will make it even more exothermic (hotter) For example as you add more and more water to a solution of sulfuric acid and water the more heat it releases (the more it burns)

Neutralization

bull Explain how you would neutralize an acid and a base

Mix an acidic solution with a basic solution wherein the ratio of H+ ions to OH- ions eq

ual to 1

Cabbage Juice as a pH indicator

bull Explain how cabbage juice works as a pH indicator There is anthocyanin in cabbage juice Anthocyanin reacts w

ith hydroxyl ions (OH-) Hydroxyl ions jump off anthocyanin when in an acidic solution thereby changing the chemical composition of the anthocyanin (the pH indicator) This will cause the juice to react with light differently than if these OH- ions had not jumped off the anthocyanin (makes the juice turn a reddish color) Conversely Hydroxyl ions will jump from the solution onto the anthrocyanin molecules when there is an excess of them in the solution (such as when cabbage juice is mixed with a basic solution) This causes the juice

(the pH indicator) to turn a yellowish color

NOT mentioned

bull Balancing redox reactions

bull Kw

bull Ka

bull Kb

bull pH

bull the galvanic cell

bullKNOW THESE CONCEPTS

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
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• Slide 10
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• Slide 50
• Slide 51

Neutralization

bull Explain how you would neutralize an acid and a base

Mix an acidic solution with a basic solution wherein the ratio of H+ ions to OH- ions eq

ual to 1

Cabbage Juice as a pH indicator

bull Explain how cabbage juice works as a pH indicator There is anthocyanin in cabbage juice Anthocyanin reacts w

ith hydroxyl ions (OH-) Hydroxyl ions jump off anthocyanin when in an acidic solution thereby changing the chemical composition of the anthocyanin (the pH indicator) This will cause the juice to react with light differently than if these OH- ions had not jumped off the anthocyanin (makes the juice turn a reddish color) Conversely Hydroxyl ions will jump from the solution onto the anthrocyanin molecules when there is an excess of them in the solution (such as when cabbage juice is mixed with a basic solution) This causes the juice

(the pH indicator) to turn a yellowish color

NOT mentioned

bull Balancing redox reactions

bull Kw

bull Ka

bull Kb

bull pH

bull the galvanic cell

bullKNOW THESE CONCEPTS

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
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• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51

Cabbage Juice as a pH indicator

bull Explain how cabbage juice works as a pH indicator There is anthocyanin in cabbage juice Anthocyanin reacts w

ith hydroxyl ions (OH-) Hydroxyl ions jump off anthocyanin when in an acidic solution thereby changing the chemical composition of the anthocyanin (the pH indicator) This will cause the juice to react with light differently than if these OH- ions had not jumped off the anthocyanin (makes the juice turn a reddish color) Conversely Hydroxyl ions will jump from the solution onto the anthrocyanin molecules when there is an excess of them in the solution (such as when cabbage juice is mixed with a basic solution) This causes the juice

(the pH indicator) to turn a yellowish color

NOT mentioned

bull Balancing redox reactions

bull Kw

bull Ka

bull Kb

bull pH

bull the galvanic cell

bullKNOW THESE CONCEPTS

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
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• Slide 49
• Slide 50
• Slide 51

NOT mentioned

bull Balancing redox reactions

bull Kw

bull Ka

bull Kb

bull pH

bull the galvanic cell

bullKNOW THESE CONCEPTS

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
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