Chapter 6 and 16 notes.notebookrrhsapchem.weebly.com/.../thermochem_notes_2016.pdf · Chapter 6 and...

Chapter 6 and 16 notes.notebook

1

January 19, 2016

Dec 198:52 AM

ThermochemistryChapter 6

Dec 198:53 AM

ThermochemistryEnergy: The capacity to do work or to produce heat

Potential Energy: Energy due to position or composition (distance and strength of bonds)

Kinetic Energy: Energy due to the motion of the object and depends on the mass of the object and its velocity (KE = 1/2mv2)

Law of Conservation of Energy: Energy can be converted from one form to another but can be neither created nor destroyed (Several different forms!)

Temperature: A property that reflects the random motions of the particles in a particular substance (measurement of kinetic energy)

Heat: The transfer of energy between two objects due to a temperature difference (Total Energyboth potential and kinetic)

Introvocabulary


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January 19, 2016

Dec 199:00 AM

ThermochemistryChemical Energy: System: The part of the universe on which we focus on

Surroundings: Include everything else in the universe

Exothermic: Energy flows out of the system as heatA + B C + D + Energy(heat)

Endothermic: Heat flows into the systemA + B + Energy(heat) C + D

Dec 181:07 PM

ThermochemistryEnergy in bonds (but really NOT in bonds)

When atoms are not bonded (or apart from each other) it is said they have ZERO potential energy

The lowest point (the most negative) potential energy is where the bond is formed. ENERGY IS RELEASED when bonds formed (Whether Inter or Intra). This is why bond formation is EXOTHERMIC.


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January 19, 2016

Dec 199:04 AM

ThermochemistryLet's calculate the energy for this reaction using bond energies. List of average bond energies is listed below and you will need to draw the Lewis dot structures to see ALL the bonds present. Show which side of the reaction the energy would be (productexo or reactantendo)

CH4 + 2O2 CO2 + 2H2O

Average Bond Energies (kJ/mol)HC 413HO 467O=O 495C=O 745

Remember:

ΔHrxn = ΣBondsreactants ΣBondsproducts

Dec 199:13 AM

ThermochemistryConsider the following reaction:

CH4 + 2O2 CO2 + 2H2O + Heat

The energy required to break the bonds in the reactants and the energy released when the bonds in the products are formed (means the bonds in the reactants are WEAKER (on average) compared to the bonds in the products)

Potential


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January 19, 2016

Dec 199:19 AM

Thermochemistry

The amount of energy to break the bonds in the reactants is higher than that of the energy released from the forming of the bonds in the products (means the bonds in the products are WEAKER (on average) compared to the bonds in the reactants)

potential

Dec 199:24 AM

ThermochemistryFirst Law of Thermodynamics: The energy of the universe is constant

Enthalpy (H): The energy (both kinetic and potential) flow of the system as heat (at constant pressure). Change in enthalpy is also called Heat of reaction

qp = ΔH


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January 19, 2016

Dec 1910:52 AM

ThermochemistryIn terms of a reaction the enthalpy change is expressed by this equation:

ΔH = Hproducts Hreactants

The reaction is endothermic when the products have a higher enthalpy than the reactantsthis means heat was absorbed by the system

the reaction is exothermic when the products have a lower enthalpy than the reactantsthis means heat was released by the system

Note: this is not the same equation as Bond Energy calculations. This equation looks at it from a different perspective (potential energy (bond energy) versus total energy (enthalpy).

(bonds formed) (bonds broken)

Dec 1910:52 AM

ThermochemistryWhen 1 mole of methane (CH4) is burned at constant pressure, 890 kJ of energy is released as heat. Calculate ΔH for a process in which a 5.8 g sample of methane is burned at constant pressure (Hint: calculate your molar enthalpy (kJ/mol) first). We did this type of math in Heating Curves!

Is enthalpy an intensive or extensive property?


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January 19, 2016

Jan 71:23 PM

ThermochemistryCalorimetry

Heat Capacity (C): the measurement of how much energy must be added to raise the temperature of a substance.

Specific heat capacity: the heat capacity per gram of a substanceJ/C*g or J/K*g

Molar heat capacity: the heat capacity per mole of a substance J/C*mol or J/K*mol

Again, we have already used this!

q = mCΔT

Jan 71:23 PM

ThermochemistryCalorimetryLet's say we mix 50.0 mL of 1.0M HCl at 25.0℃ with 50.0 mL of 1.0M NaOH also at 25.0℃in a calorimeter. After the reactants are mixed by stirring, the temp. is observed to increase to 31.9℃ (CH2O = 4.18 J/℃*g).

First identify the surroundings and system!!!


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January 19, 2016

Jan 710:13 AM

Jan 71:23 PM

ThermochemistryCalorimetryWhen 1.00L of 1.00M Ba(NO3)2 solution at 25.0℃ is mixed with 1.00L of 1.00M Na2SO4 solution at 25.0℃ in a calorimeter, the white solid BaSO4 forms and the temperature of the mixture increases to 28.1℃. (CH2O = 4.18 j/C*g). Calculate the molar enthalpy of BaSO4 formed.

Surroundings? System? Net ionic equationMolar enthalpy


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January 19, 2016

Jan 118:09 AM

ThermochemistryList of possible metals for Lab (J/g*c)

Copper 0.385Lead 0.160Tin 0.210Zinc 0.390Aluminum 0.900Nickel 0.440

Jan 71:23 PM

ThermochemistryBomb CalorimetrySuppose we wish to measure the energy of combustion of octane (C8H18), a component of gasoline. A 0.5269 g sample of octane is placed in a bomb calorimeter known to have a heat capacity of 11.3 kJ/℃. (This means that 11.3 kJ of energy is required to raise the temperature of the water and other parts of the calorimeter by 1℃). The octane is ignited in the presence of excess oxygen, and the temperature increase of the calorimeter is 2.25℃.

q = ΔT x C


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January 19, 2016

Jan 71:23 PM

ThermochemistryCalorimetryIt has been suggested that hydrogen gas might be a substitute for natural gas. To compare the energies of combustion of these fuels, the following experiment was carried out using a bomb calorimeter with a heat capacity of 11.3 kJ/℃.

When a 1.50 g sample of methane gas was burned with excess oxygen in the calorimeter, the temperature increased by 7.3℃.

When a 1.15 g sample of hydrogen gas was burned with excess oxygen, the temperature increase was 14.3℃.

Calculate the energy of combustion (per gram for hydrogen and methane)

Energy released 1.5 g methane = (11.3 kJ/C)(7.3C) = 83 kJ

Energy released 1.15 g hydrogen = (11.3 kJ/C)(14.3C) = 162 kJ

Energy released 1 g Methane = 83 kJ/1.5g = 55 kJ/g

Energy released 1 g hydrogen = 162/1.15g = 141 kJ/g

Jan 71:23 PM

ThermochemistryCharacteristics of Enthalpy Changes

1. If a reaction is reversed, the sign of ΔH is also reversed

2. the magnitude of ΔH is directly proportional to the quantities of reactants and products in a reaction.

2NO2(g) ⇒ N2(g) + 2O2(g) ΔHrxn = 68 kJ

N2(g) + 2O2(g) ⇒ 2NO2(g) ΔHrxn = + 68 kJ

N2(g) + 2O2(g) ⇒ 2NO2(g) ΔHrxn = + 68 kJ

2N2(g) + 4O2(g) ⇒ 4NO2(g) ΔHrxn = + 136 kJ

1/2N2(g) + O2(g) ⇒ NO2(g) ΔHrxn = + 34 kJ


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January 19, 2016

Jan 71:23 PM

ThermochemistryHess's Law: Going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps

N2(g) + 2O2(g) ⇒ 2NO2(g) ΔHrxn = 68 kJ

Let's look at the oxidation of nitrogen to form nitrogen dioxide. It can be written in one step:

However the reaction can be carried out in two distinct steps

N2(g) + O2(g) ⇒ 2NO(g) ΔH1 = 180 kJ

2NO(g) + O2(g) ⇒ 2NO2(g) ΔH2 = 112 kJ

ΔH2 + ΔH3 = 180 kJ + (112 kJ) = 68 kJ = ΔH1

Jan 71:23 PM

ThermochemistryTwo forms of carbon are graphite, the soft, black, slippery material used in "lead" pencils, and diamond, the brilliant, hard gemstone. using the enthalpies of combustion for graphite (394 kJ/mole) and diamond (396 kJ/mole), calculate ΔH for the conversion of graphite to diamond.

Cgraphite(s) + O2(g) ⇒ CO2(g) ΔH = 394 kJ

Cdiamond(s) + O2(g) ⇒ CO2(g) ΔH = 396 kJ

This is one of two ways to calculate using Hess's Law


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January 19, 2016

Jan 71:23 PM

ThermochemistryDiborane (B2H6) is highly reactive boron hydride that was once considered as a possible rocket fuel for the U.S. space program. calculate ΔH for the synthesis of diborane from its elements, according to the equation

2B(s) + 3H2(g) ⇒ B2H6(g)

Use the following data:

(a) 2B(s) + 3/2O2(g) ⇒ B2O3(s) ΔH = 1273 kJ

(b) B2H6(g) + 3O2(g) ⇒ B2O3(s) + 3H2O(g) ΔH = 2035 kJ

(c) H2(g) + 1/2O2(g) ⇒ H2O(l) ΔH = 286 kJ

(d) H2O(l) ⇒ H2O(g) ΔH = 44 kJ

Jan 71:23 PM

ThermochemistryHints for Hess's Law

1. Work backward from the required reaction, using the reactants and products to decide how to manipulate the other given reactions at your disposal

2. reverse any reactions to give the required reactants and products

3. multiply the reactions to give the correct numbers of reactants and products.


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January 19, 2016

Jan 148:19 AM

ThermochemistryN2H4(l) + H2(g) → 2NH3(g)

N2H4(l) + CH4O(l) → CH2O(g) + N2(g) + 3H2 (g) ΔH = 37 kJ

N2(g) + 3H2(g) → 2NH 3(g) ΔH = 46 kJ

CH4O(l) → CH2O(g) + H 2(g) ΔH = 65 kJ

Jan 148:27 AM

Zn(s) + 1/8S8(s) + 2O2(g) → ZnSO4(s)

Zn(s) + 1/8S8(s) → ZnS(s) ΔH = 183.92 kJ

2ZnS(s) + 3O2(g) → 2ZnO(s) + 2SO2(g) ΔH = 927.54 kJ

2SO2(g) + O2(g) → 2SO3(g) ΔH = 196.04 kJ

ZnO(s) + SO3(g) → ZnSO4 (s) ΔH = 230.32 kJ

Thermochemistry


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January 19, 2016

Jan 71:23 PM

ThermochemistryStandard Enthalpies of Formation

The change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states. (ΔHof)

The degree symbol (ΔHo) means that the reaction was carried out under standard conditions

Standard State: A precisely defined reference state

Gaseous substances at exactly 1 atm pressurePure substances in a condensed state (solid or liquid) is the pure liquid or solid Solutions have a concentration of exactly 1 M. Elements is the standard state of the element at 1 atm and 25C (oyxgen is a gas, sodium is a solid, bromine is a liquid, etc.) Elements have a ΔHof of 0 kJ, since they do not need to be formed!

Jan 71:23 PM

Thermochemistry1/2N2(g) + O2(g) ⇒ NO2(g) ΔHof = 34 kJ/mol

Standard Enthalpies of Formation (examples)

Notice that nitrogen and oxygen are both in their standard state and that the reaction is for 1 mole of product.

C(s) + 2H2(g) + 1/2O2(g) ⇒ CH3OH(l) ΔHof = 239 kJ/mol


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January 19, 2016

Jan 71:23 PM

ThermochemistryStandard Enthalpies of Formation We use Hess's Law to determine the heat of formation for a reaction

CH4(g) + 2O2(g) ⇒ CO2(g) + 2H2O(l)

C(s) + 2H2(g) ⇒ CH4(g) ΔHof = 75 kJ/molC(s) + O2(g) ⇒ CO2(g) ΔHof = 394 kJ/mol

H2(g) + 1/2O2(g) ⇒ H2O(l) ΔHof =286 kJ/mol

Jan 71:23 PM


(Second way to calculate using Hess's Law)

4NH3(g) + 7O2(g) ⇒ 4NO2(g) + 6H2O(l)

NH3(g) 46NO2(g) 34H2O(l) 286Al2O3(s) 1676Fe2O3(s) 826CO2(g) 394CH3OH(l) 239C8H18(l) 269

Compound ΔHof (kJ/mol)

ΔHReaction = ΣnpΔHof (Products) ΣnrΔHof (reactants)


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January 19, 2016

Jan 71:23 PM


2Al(s)+ Fe2O3(s) ⇒ Al2O3(s) + 2Fe(s)

NH3(g) 46NO2(g) 34H2O(l) 286Al2O3(s) 1676Fe2O3(s) 826CO2(g) 394CH3OH(l) 239C8H18(l) 269


Jan 71:23 PM

ThermochemistryStandard Enthalpies of Formation Methanol (CH3OH) is often used as a fuel in highperformance engines in race cars. using the data below, compare the standard enthalpy of combustion per gram of methanol with that per gram of gasoline. Gasoline is actually a mixture of compounds but assume for this problem that gasoline is pure liquid octane (C8H18)

2CH3OH(l) + 3O2(g) ⇒ 2CO2(g) + 4H2O(l)NH3(g) 46NO2(g) 34H2O(l) 286Al2O3(s) 1676Fe2O3(s) 826CO2(g) 394CH3OH(l) 239C8H18(l) 269


2C8H18(l) + 25O2(g) ⇒16CO2(g) + 18H2O(l)


16

January 19, 2016

Jan 710:48 AM

Spontaneity, Entropy, and Free Energy

Chapter 16

Jan 2210:17 AM

Spontaneous Processes and Entropy Spontaneous Processes: A process that occurs without outside interventions. This can be fast or slow (Kinetics has nothing to do with Spontaneity)

Thermodynamics will tell us the energy of the reaction and what direction it will occur, but not the speed of the process

I.E.According to thermodynamics, a diamond should change spontaneously to graphite.


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January 19, 2016

Jan 2210:21 AM

Spontaneous Processes and Entropy Spontaneous processes: A ball will roll down a hill spontaneously but will never roll back up the hill

If exposed to air and moisture, steel rusts spontaneously. However, rust does not spontaneously change back to iron metal and oxygen gas

A gas fills its container uniformly. it never spontaneously collects at one end of the container

Heat flow always occurs from a hot object to a cooler one. the reverse process never occurs spontaneously

Wood burns spontaneously in an exothermic reaction to form carbon dioxide and water, but wood is not formed when carbon dioxide and water are heated together

At temperatures below 0C, water spontaneously freezes, and at temperatures above 0C, ice spontaneously melts

Jan 2210:28 AM

Spontaneous Processes and Entropy At one point in time it was thought that reactions are spontaneous because it is exothermic, but the melting of ice is endothermic yet it is spontaneous

Entropy (S): The measurement of molecular randomness or disorder (the driving force for spontaneous reactions)

describes the number of arrangements (positions and/or energy levels) that are available to a system existing in a given state

This is like probability: Nature spontaneously proceeds towards the states that have the highest probabilities of existing.


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January 19, 2016

Mar 191:06 PM

Spontaneous Processes and Entropy Let's look at an ideal gas in a vacuum chamber like below

Arrangement 1 Arrangement 2 Arrangement 3

Arrangement 4 Arrangement 5

1 possible way

1 possible way

4 possible way

4 possible way

6 possible way

?

Mar 191:05 PM

Spontaneous Processes and Entropy Which one has more entropy?

Solid CO2 or gaseous CO2

What is the SIGN for the entropy change in the following (more randomness is positive, less randomness (more order) is negative)?

Solid sugar is added to water to form a solution

Iodine vapor condenses on a cold surface to form crystals


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January 19, 2016

Mar 191:09 PM

Entropy Changes in Chemical Reactions

N2(g) + 3H2(g) ⇒ 2NH3(g)

We start with 4 moles of reactants and end with 2 moles of products. Therefore this lowers the "chaos" in the system leading to a negative entropy value

4NH3(g) + 5O2(g) ⇒ 4NO(g) + 6H2O(g)

We start with 9 moles of reactants and end with 10 moles of products. Therefore this increases the "chaos" in the system leading to a positive entropy value

Mar 191:10 PM

Entropy Changes in Chemical Reactions

CaCO3(s) ⇒ CaO(s) + CO2(g)

2SO2(g) + O2(g) ⇒ 2SO3(g)

Predict the sign of entropy for each of the following reactions


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January 19, 2016

Mar 191:07 PM

Second law of Thermodynamics The Second Law of Thermodynamics: In any spontaneous process there is always an increase in the entropy of the universe

Nature is always moving towards the most probable state available

The entropy of the universe is always increasing

ΔSuniv = ΔSsys + ΔSsurrBut, we are not typically looking at the universe, we are looking at the system, so this method does not work for our purposes.

Mar 191:08 PM

Free EnergyΔG = ΔH TΔS

G = (Gibb's) Free EnergyH = Enthalpy T = Temperature (K)S = Entropy of the system

If ΔG is negative (), then the process is spontaneous (favorable)

If ΔG is also zero, then the reaction is at equilibrium (this will be VERY important here and in later chapters!!!)


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January 19, 2016

Mar 191:07 PM

Free EnergyΔG = ΔH TΔS

Lets look at water melting at 10C, 0C, and 10C

H2O(s) ⇒ H2O(l)

ΔH0 = 6.03 x 103 J/mol ΔS0 = 22.1 J/K*mol

T(K) ΔH0

(J/mol) ΔS0(J/K*mol)

ΔG = ΔH TΔS(J/mol)

263

273

283

6.03x103

6.03x103

6.03x103

22.1

22.1

22.1

+2.2x102

0

2.2x102

This shows that some processes (if ΔS and ΔH are the same sign) that the process will be temperature dependent!

Notice, that at the 0℃, ΔG is 0 (remember phase changes are @ equilibrium

Jan 233:33 PM

Free EnergyΔS positive, ΔH negative

ΔS positive, ΔH positive

ΔS negative, ΔH negative

ΔS negative, ΔH positive

Spontaneous at all temps

Spontaneous at high temperatures (where exothermicity is relatively unimportant

Spontaneous at low temperatures (where exothermicity is dominant)

Process not spontaneous at any temperature (reverse process is spontaneous at all temps)

LOOK AT THE EQUATION!!!


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January 19, 2016

Mar 191:09 PM

Free EnergyAt what temperatures is the following process spontaneous at 1 atm?

Br2(l) ⇒ Br2(g)

ΔH0 = 31.0 kJ/mol and ΔS0 = 93.0 J/K*mol

What is the normal boiling point of liquid bromine?

Jan 245:53 PM

Entropy Changes in Chemical ReactionsThird Law of Thermodynamics: The entropy of a perfect crystal at 0K is zero

This allows for calculations to be done to determine standard entropy (S0) values for common substances at 298K and 1 atm (Hess's Law!)

Since entropy (like enthalpy) is a state function (not pathwaydependent) we can use the equation below to calculate the entropy change in a reaction

ΔS0reaction = ΣnpS0products ΣnrS0reactants


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January 19, 2016

Jan 246:00 PM

Entropy Changes in Chemical ReactionsCalculate the ΔS0 at 25C for the reaction

2NiS(s) + 3O2(g) ⇒ 2SO2(g) + 2NiO(s)with the following standard entropy values S0

Substance S0 (J/K*mol)

SO2(g) 248

NiO(s) 38

O2(g) 205

NiS(s) 53

ΔS0reaction = ΣΔnpS0products ΣnrS0reactants

Mar 191:11 PM

Entropy Changes in Chemical ReactionsCalculate the ΔS0 for the reduction of aluminum oxide by hydrogen

Al2O3(s) + 3H2(g) ⇒ 3H2O(g) + 2Al(s)with the following standard entropy values S0

Substance S0 (J/K*mol)

Al2O3(s) 51

H2(g) 131

Al(s) 28

H2O(s) 189

ΔS0reaction = ΣΔnpS0products ΣnrS0reactants

Also can calculate ΔS using reactions with known ΔS values


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January 19, 2016

Mar 191:11 PM

Free Energy and Chemical Reactions Standard Free Energy Change (ΔG0): The change in free energy that will occur if the reactants in their standard states are converted to the products in their standard states.

N2(g) + 3H2(g) ⇒ 2NH3(g) ΔG0 = 33.33kJ

We cannot measure Free Energy (there is no instrument to measure free energy)

Jan 2812:59 PM

Free Energy and Chemical Reactions There are several ways to calculate Free Energy

ΔG0 = ΔH0 TΔS0

For the reaction:

C(s) + O2(g) ⇒ CO2(g)

The values of ΔH0 + ΔS0 are known to be 393.5 kJ and 3.05 J/K, respectively, calculate ΔG0 at 298K


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January 19, 2016

Jan 281:08 PM

Free Energy and Chemical Reactions Consider the reaction:

2SO2(g) + O2(g) ⇒ 2SO3(g)

Carried out at 250C and 1 atm. calculate ΔH0, ΔS0, and ΔG0 using the following data:

Substance ΔH0f (kJ/mol) S0 (J/K*mol)SO2(g) 297 248SO3(g) 396 257O2(g) 0 205

Mar 191:12 PM

Free Energy and Chemical Reactions Since Free Energy is a State Function we can use this method as well:

2CO(g) + O2(g) ⇒ 2CO2(g) ΔG0 = ?

2CH4(g) + 3O2(g) ⇒ 2CO(g) + 4H2O(g) ΔG0 = 1088 kJ

Use the following data to calculate ΔG0 for the reaction above:

CH4(g) + 2O2(g) ⇒ CO2(g) + 2H2O(g) ΔG0 = 801 kJ


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January 19, 2016

Dec 182:50 PM

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