New Way Chemistry for Hong Kong A-Level Book 11 Energetics 6.1What is Energetics? 6.2Enthalpy...

New Way Chemistry for Hong Kong A-Level Book 11

EnergeticsEnergetics

6.16.1 What is Energetics?What is Energetics?

6.26.2 Enthalpy Changes Related to Breaking and Enthalpy Changes Related to Breaking and Forming of BondsForming of Bonds

6.36.3 Standard Enthalpy ChangesStandard Enthalpy Changes

6.46.4 Experimental Determination of Enthalpy Experimental Determination of Enthalpy ChaChanges by Calorimetrynges by Calorimetry

6.56.5 Hess’s LawHess’s Law

6.66.6 Calculations involving Standard Enthalpy Calculations involving Standard Enthalpy Changes of ReactionsChanges of Reactions

66


What is energetics?What is energetics?Energetics is the study of energy changes associated with chemical reactions.Energetics is the study of energy changes associated with chemical reactions.

Thermochemistry is the study of heat changes associated with chemical reactions.Thermochemistry is the study of heat changes associated with chemical reactions.

Some terms

Enthalpy(H) = heat content in a substance

Enthalpy change(H) = heat content of products - heat content of reactants= Hp - Hr

6.1 What is energetics? (SB p.136)


Law of conservation of energyLaw of conservation of energy


The law of conservation of energy states that energy can neither be created nor destroyed.The law of conservation of energy states that energy can neither be created nor destroyed.


Internal energy and enthalpyInternal energy and enthalpy

e.g. Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)



Internal energy and enthalpyInternal energy and enthalpy6.1 What is energetics? (SB p.138)

(Heat change at constant volume)

Enthalpy change

Heat change atconstant pressure

=Change in internal energy

Work done on the surroundings

+


Exothermic and endothermic Exothermic and endothermic reactionsreactions

An exothermic reaction is a reaction that releases heat energy to the surroundings. (H = -ve)

An exothermic reaction is a reaction that releases heat energy to the surroundings. (H = -ve)



An endothermic reaction is a reaction that absorbs heat energy from the surroundings. (H = +ve)

An endothermic reaction is a reaction that absorbs heat energy from the surroundings. (H = +ve)


Exothermic and endothermic Exothermic and endothermic reactionsreactions

Check Point 6-1Check Point 6-1


6.6.22Energy Changes Energy Changes

Related to Related to Breaking and Breaking and

Forming of BondsForming of Bonds


Enthalpy changes related to breaking Enthalpy changes related to breaking and forming of bondsand forming of bonds

e.g. CH4 + 2O2 CO2 + 2H2O

6.2 Enthalpy changes related to breaking and forming of bonds (SB p.140)




In an exothermic reaction, the energy required in

breaking the bonds in the reactants is less than

the energy released in forming the bonds in the

products (products contain stronger bonds).

In an exothermic reaction, the energy required in

breaking the bonds in the reactants is less than

the energy released in forming the bonds in the

products (products contain stronger bonds).



In an endothermic reaction, the energy required in breaking the bonds in the reactants is more than the energy released in forming the bonds in the products (reactants contain stronger bonds).

In an endothermic reaction, the energy required in breaking the bonds in the reactants is more than the energy released in forming the bonds in the products (reactants contain stronger bonds).




6.6.33 Standard Standard

Enthalpy Enthalpy ChangesChanges


Standard enthalpy changesStandard enthalpy changes

CH4(g) + 2O2(g) CO2(g) + 2H2O(g)

H = -802 kJ mol-1

CH4(g) + 2O2(g) CO2(g) + 2H2O(l)

H = -890 kJ mol-1

6.3 Standard enthalpy changes (SB p.141)


Standard enthalpy changesStandard enthalpy changes

As enthalpy changes depend on temperature and pressure, it is necessary to define standard states and conditions:

1. elements or compounds in their normal physical states;2. a pressure of 1 atm (101325 Nm-2); and3. a temperature of 25oC (298 K)

1. elements or compounds in their normal physical states;2. a pressure of 1 atm (101325 Nm-2); and3. a temperature of 25oC (298 K)

Enthalpy change under standard conditions denoted by symbol: H

ø



Standard enthalpy changes of Standard enthalpy changes of neutralizationneutralization

Standard enthalpy change of neutralization (Hn

eut) is the enthalpy change when one mole of wat

er is formed from the neutralization of an acid by an alkali under standard conditions.

Standard enthalpy change of neutralization (Hn

eut) is the enthalpy change when one mole of wat

er is formed from the neutralization of an acid by an alkali under standard conditions.

ø

e.g. H+(aq) + OH-(aq) H2O(l)

Hneut = -57.3 kJ mol-1

ø



H+(aq) + OH-(aq) H2O(l)

-57.1

-57.2

-52.2

-68.6

NaOH

KOH

NH3

NaOH

HCl

HCl

HCl

HF

Hneu AlkaliAcid ø






Enthalpy level diagram for the neutralization of a strong acid and a strong alkali


Standard enthalpy change of solution (Hsoln) is the enthalpy change when one mole of a solute is completely dissolved in a sufficiently large volume of solvent to form an infinitely dilute solution under standard conditions.

Standard enthalpy change of solution (Hsoln) is the enthalpy change when one mole of a solute is completely dissolved in a sufficiently large volume of solvent to form an infinitely dilute solution under standard conditions.

ø


Standard enthalpy change of Standard enthalpy change of solutionsolution


e.g. NaCl(s) + water Na+(aq)+Cl-(aq)

Hsoln= +3.9 kJ mol-1

ø



Enthalpy level diagram for the dissolution of Na

Cl



Standard enthalpy change of Standard enthalpy change of solutionsolutione.g. LiCl(s) + water Li+(aq) + Cl-(aq)

Hsoln= -37.2 kJ mol-1

ø

Enthalpy level diagram for the dissolution of Li

Cl in water


-44.7

+3.9

-57.8

+20.0

NaOH

NaCl

KOH

KBr

Hsoln(kJ mol-1)Salt ø




Standard enthalpy change of formation (Hf ) i

s the enthalpy change of the reaction when one mole of the compound in its standard state is formed from its constituent elements under standard conditions.

Standard enthalpy change of formation (Hf ) i

s the enthalpy change of the reaction when one mole of the compound in its standard state is formed from its constituent elements under standard conditions.

ø


Standard enthalpy change of Standard enthalpy change of formationformation


Standard enthalpy change of formation of NaCl is -411 kJ mol-1.

e.g. 2Na(s) + Cl2(g) 2NaCl(s)

H = -822 kJ mol-1ø

Na(s) + ½Cl2(g) NaCl(s)

Hf = -411 kJ mol-1

ø




The enthalpy change of formation of an element is always zero.The enthalpy change of formation of an element is always zero.

N2(g) N2(g)

What is Hf [N2(g)] ?

ø

Hf [N2(g)] = 0

ø




e.g. C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)

H1 = -2220 kJ

2C3H8(g) + 10O2(g) 6CO2(g) + 8H2O(l)

H2 = ?H2 = -4440 kJ

It is more convenient to report enthalpy changes per mole of the main reactant reacted/product formed.


Standard enthalpy change of Standard enthalpy change of combustioncombustion


Standard enthalpy change of combustion (Hc ) of a substance is the enthalpy change when one mole of the substance burns completely under standard conditions.

Standard enthalpy change of combustion (Hc ) of a substance is the enthalpy change when one mole of the substance burns completely under standard conditions.

ø

e.g. C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)

1 mole

Hc = -2220 kJ mol-1

ø




-285.8

-395.4

-393.5

-283.0

-890.4

H2(g)

C (diamond)

C (graphite)

CO(g)

CH4(g)

Hc (kJ mol-1)Substance

ø





6.6.44Experimental DeteExperimental Determination of Enthrmination of Enthalpy Changes by Calpy Changes by C

alorimetryalorimetry


Experimental determination of enthalpy chaExperimental determination of enthalpy changes by calorimetrynges by calorimetry

Calorimeter = a container used for measuring the temperature change of solution

6.4 Experimental determination of enthalpy changes by calorimetry (SB p.148)


Determination of enthalpy change of Determination of enthalpy change of neutralizationneutralization



Heat evolved = (m1c1 + m2c2) ΔT

where m1 is the mass of the solution,

m2 is the mass of calorimeter,

c1 is the specific heat capacity of the solution,

c2 is the specific heat capacity of calorimeter,

ΔT is the temperature change of the reaction


Example 6-4AExample 6-4A



Determination of enthalpy change Determination of enthalpy change of combustionof combustion

The Philip Harris calorimeter used for

determining the enthalpy change of

combustion of a liquid fuel


6.4 Experimental determination of enthalpy changes by calorimetry (SB p.151) Determination of enthalpy change Determination of enthalpy change

of combustionof combustion

A simple apparatus used to determine the enthalpy change of combustion of ethanol


Heat evolved = (m1c1 + m2c2) ΔT

Where m1 is the mass of water in the calorimeter,

m2 is the mass of the calorimeter,

c1 is the specific heat capacity of the water,

c2 is the specific heat capacity of calorimeter,

ΔT is the temperature change of the reaction


Example 6-4BExample 6-4B



Determination of enthalpy change Determination of enthalpy change of solutionof solution

• By measuring the temperature change when a known mass of solute is added to a known volume of solvent in a calorimeter

• Heat change = (m1c1 + m2c2) T

Example 6-4CExample 6-4C



Determination of enthalpy change Determination of enthalpy change of formationof formation

• The enthalpy change of formation of a substance can be quite high

• Found out by applying Hess’s law of constant heat summation



6.6.55 Hess’s LawHess’s Law


Hess’s Hess’s LawLaw

A + B C + DRoute 1

H1

E

H2 H3

Route 2 H1 = H2 + H3 H1 = H2 + H3

Hess’s law of constant heat summation states that the total enthalpy change accompanying a chemical reaction is independent of the route by which the chemical reaction takes place.

Hess’s law of constant heat summation states that the total enthalpy change accompanying a chemical reaction is independent of the route by which the chemical reaction takes place.

6.5 Hess’s law (SB p.153)



Enthalpy level Enthalpy level diagramdiagram• Relate substances together in terms of

enthalpy changes of reactions

Enthalpy level diagram for the

oxidation of C(graphite) to

CO2(g)



Enthalpy cycle (Born-Haber cycle)Enthalpy cycle (Born-Haber cycle)• Relate the various equations involved in a

reaction

Enthalpy cycle for the oxidation of

C(graphite) to CO2(g)


Importance of Hess’s lawImportance of Hess’s lawThe enthalpy change of some chemical reactions cannot be determined directly because:

• the reactions cannot be performed in the laboratory• the reaction rates are too slow• the reactions may involve the formation of side

products

But the enthalpy change of such reactions can be determined indirectly by applying Hess’s Law.



Enthalpy change of formation of CO(g)Enthalpy change of formation of CO(g)

= -393.5 - (-283.0 )

= -110.5 kJ mol-1

Given: Hf [CO2(g)] = -393.5 kJ mol-1;

Hc [CO(g)] = -283.0 kJ mol-1

øø

H2

+ ½O2

(g)CO2(g)

H1

+ ½O2

(g)

C(graphite) + ½O2(g) CO(g)Hf [CO(g)]ø

Hf [CO(g)] + H2 = H1Hf [CO(g)] + H2 = H1

ø

Hf [CO(g)] = H1 - H2

ø



6.5 Hess’s law (SB p.153)Enthalpy change of formation of Enthalpy change of formation of CaCOCaCO33(s)(s)

Ca(s) + C(graphite) + O223 CaCO3(s)

CaO(s) + CO2

(g)

H1H2

Hf [CaCO3(s)]

øHf [CaCO3(s)] = H1 + H2

= -1028.5 kJ mol-1 + (-178.0) kJ mol-1

= -1206.5 kJ mol-1

ø


Enthalpy change of hydration of Enthalpy change of hydration of MgSOMgSO44(s)(s)

aq

MgSO4(s) + 7H2O(l) MgSO4·7H2O(s)

Mg2+(aq) + SO42-(aq) + 7H2O(l)

ΔH

ΔH2 aq

ΔH1

ø

ΔH = enthalpy of hydration of MgSO4(s)

ΔH1 = molar enthalpy change of solution of anhydrous magnesium sulphate(VI)

ΔH2 = molar enthalpy change of solution of magnesium

sulphate(VI)-7-water

ΔH = ΔH1 - ΔH2

ø

ø




6.6.66Calculations Calculations

involving involving Standard Enthalpy Standard Enthalpy

Changes of Changes of ReactionsReactions


Calculation of standard enthalpy Calculation of standard enthalpy change of reaction from standard change of reaction from standard enthalpy changes of formationenthalpy changes of formation

Hreaction = Hf [products] - Hf [reactants]Hreaction = Hf [products] - Hf [reactants]ø ø

6.6 Calculations involving standard enthalpy changes of reactions (SB p.159)



Example 6-6AExample 6-6A Example 6-6BExample 6-6B

Example 6-6DExample 6-6DExample 6-6CExample 6-6C


6.6 Calculations involving standard enthalpy changes of reactions (SB p.162) Calculation of standard enthalpy Calculation of standard enthalpy

change of formation from standard change of formation from standard enthalpy changes of combustionenthalpy changes of combustion

Hf = Hc [products] - Hc [reactants]Hf = Hc [products] - Hc [reactants]ø øø



Example 6-6EExample 6-6E Example 6-6FExample 6-6F



6.6.77 Entropy Entropy

ChangeChange

6.7 Entropy change (SB p.164)



EntropyEntropy

A process is said to be spontaneous

• If no external forces are required to keep the process going

• The process may be physical change or a chemical change

• Example of spontaneous physical change: cooling of hot water

• Example of spontaneous chemcial change: burning of wood once the fire is started



EntropyEntropy

• Exothermicity is the reason for the spontaneity of a process

• Some spontaneous changes are endothermic

• Examples: Melting of ice, dissolution of ammonium nitrate in water



Melting of ice

Dissolution of ammonium nitrate in

water



EntropyEntropy• Entropy is a measure of the randomness or

the degree of disorder of a system

Solid Liquid Gas

Entropy increases



Entropy Entropy changechange

• Entropy change means the change in the degree of disorder of a system

S = Sfinal - Sinitial


6.7 Entropy change (SB p.166) Positive Positive

entropyentropy

• Increase in entropy

• Final state has a larger entropy that the initial state

• Example:

Ice (less entropy) Water (more entropy)

S = Swater – Sice = +ve


6.7 Entropy change (SB p.166) Negative Negative

entropyentropy

• Decrease in entropy

• Initial state has a larger entropy that the final state

• Example:

Water (more entropy) Ice (less entropy)

S = Sice – Swater = -ve



6.6.88 Free Free

Energy Energy ChangeChange


6.8 Free energy change (SB p.168)

Free energy Free energy changechange• Entropy is temperature dependent

• At a higher temp, the entropy of the system is higher

• At a lower temp, the entropy of the system is lower



Free Free energyenergy• Another driving force for a process

• Called free energy (G)

G = H – TS

where H is the enthalpy

T is Kelvin temperature

S is the entropy



Free energy changeFree energy change

G = H – TS

• At a given temp, there are two driving forces for a process to occur

• Overall enthalpy of the system tends to be low

• Overall entropy of the system tends to be high



Significance of the equationSignificance of the equation

• Process favoured by:

H = -ve; S = +ve

• Process not favoured by:

H = +ve; S = -ve



Significance of the equationSignificance of the equation

• A process is spontaneous or favourable when G is negative

• A process is not spontaneous or favourable as indicated when G is positive, but is spontaneous in the opposite direction



How How H and H and S affect the S affect the spontaneity of a processspontaneity of a process

H S G = H - TS Result

-ve +ve -ve Process is spontansous at all temepratures

+ve -ve +ve Process is not spontaneous at any

temperature (reverse process is spontaneous

at all temperatures)



Effects of relative magnitudes of Effects of relative magnitudes of H and H and S on the spontaneity of a processS on the spontaneity of a process

H S Condition G = H - TS

Result

+ve +ve At high temp, TS > H

G = -ve Process is spontaneous at

high temp

+ve +ve At low temp, H > TS

G = +ve Process is not spontaneous

-ve -ve At high temp, TS > H

G = +ve Process is not spontaneous

-ve -ve At low temp, H > TS

G = -ve Process is spontaneous at

low temp


The END


State whether the following processes are exothermic or endothermic.

(a) Melting of ice.

(b) Dissolution of table salt.

(c) Condensation of steam.

Back

Answer


(a) Endothermic

(b) Endothermic

(c) Exothermic


(a)State the difference between exothermic and endothermic reactions with respect to

(i) the sign of H;

(ii) the heat change with the surroundings;

(iii) the total enthalpy of reactants and products.

Answer

(a) (i) Exothermic reactions: H = -ve; endothermic reactions: H = +ve

(ii) Heat is given out to the surroundings in exothermic reactions whereas heat is taken in from the surroundings in endothermic reactions.

(iii) In exothermic reactions, the total enthalpy of products is less than that of the reactants. In endothermic reactions, the total enthalpy is greater than that of the reactants.



(b)Draw an enthalpy level diagram for a reaction which is

(i) endothermic, having a large activation energy.

(ii) exothermic, having a small activation energy.

Answer




(b) (i)



(ii)

Back


(a)Why must the condition “burnt completely in oxygen” be emphasized in the definition of standard enthalpy change of combustion?

Answer


(a) If the substance is not completely burnt in excess oxygen, other products such as C(s) and CO(g) may be formed. The enthalpy change of combustion measured will not be accurate.


(b) The enthalpy change of the following reaction under standard conditions is –566.0 kJ.

2CO(g) + O2(g) 2CO2(g)

What is the standard enthalpy change of combustion of carbon monoxide?

Answer


(b) Standard enthalpy change of combustion of CO

= (-566.0) kJ

= -283.0 kJ21


(c)What terms may be given for the enthalpy change of the following reaction?

N2(g) + O2(g) NO2(g)21

Answer


(c) Enthalpy change of combustion of nitrogen or enthalpy change of formation of nitrogen dioxide.

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Determine the enthalpy change of neutralization of 25 cm3 of 1.25 M hydrochloric acid and 25 cm3 of 1.25 M sodium hydroxide solution using the following data:

Mass of calorimeter = 100 g

Initial temperature of acid = 15.5 oC (288.5 K)

Initial temperature of alkali = 15.5 oC (288.5 K)

Final temperature of the reaction mixture = 21.6 oC (294.6 K)

The specific heat capacities of water and calorimeter are 4200 J kg-1 K-1 and 800 J kg-1 K-1 respectively.

Answer



Assume that the density of the reaction mixture is the same as that of water, i.e. 1 g cm-3.

Mass of the reaction mixture = (25 + 25) cm3 1 g cm-3 = 50 g = 0.05 kg

Heat given out = (m1c1 + m2c2) T

= (0.05 kg 4200 J kg-1 K-1 + 0.1 kg 800 J kg-1 K-1) (294.6 – 288.5) K

= 1769 J

H+(aq) + OH-(aq) H2O(l)

Number of moles of HCl = 1.25 mol dm-3 25 10-3 dm3 = 0.03125 mol

Number of moles of NaOH = 1.25 mol dm-3 25 10-3 dm3 = 0.03125 mol

Number of moles of H2O formed = 0.03125 mol


Back


Heat given out per mole of H2O formed

=

= 56608 J mol-1

The enthalpy change of neutralization is –56.6 kJ mol-1.

mol 0.03125J 1769



Determine the enthalpy change of combustion of ethanol using the following data:

Mass of spirit lamp before experiment = 45.24 g

Mass of spirit lamp after experiment = 44.46 g

Mass of water in copper calorimeter = 50 g

Mass of copper calorimeter without water = 380 g

Initial temperature of water = 18.5 oC (291.5 K)

Final temperature of water = 39.4 oC (312.4 K)

The specific heat capacities of water and copper calorimeter are 4200 J kg-1 K-1 and 2100 J kg-1 K-1 respectively. Answer



Heat evolved by the combustion of ethanol

= Heat absorbed by the copper calorimeter

= (m1c1 + m2c2) T

= (0.05 kg 4200 J kg-1 K-1 + 0.38 kg 2100 J kg-1 K-1) (312.4 – 291.5)K

= 21067 J

C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l)

Mass of ethanol burnt = (45.24 – 44.46) g = 0.78 g

Number of moles of ethanol burnt = = 0.017 mol 1mol g 46.0g 0.78



Heat given out per mole of ethanol

=

= 1239235 J mol-1

= 1239 kJ mol-1

The enthalpy change of combustion of ethanol is –1239 kJ mol-1.

There was heat loss by the system to the surroundings, and incomplete combustion of ethanol might occur. Also, the experiment was not carried out under standard conditions. Therefore, the experimentally determined value (-1239 kJ mol-1) is less than the theoretical value of the standard enthalpy change of combustion of ethanol (-1371 kJ mol-1).

mol 0.017J 21067

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0.02 mol of anhydrous ammonium chloride was added to

45 g of water in a polystyrene cup to determine the enthalpy change of solution of anhydrous ammonium chloride. It is found that there was a temperature drop from 24.5 oC to 23.0 oC in the solution.

Given that the specific heat capacity of water is 4200 J kg-1 K-1 and

NH4Cl(s) + aq NH4Cl(aq)

Calculate the enthalpy change of solution of anhydrous ammonium chloride.

(Neglect the specific heat capacity of the polystyrene cup.)

Answer



Heat absorbed = m1c1T ( c2 0)

= 0.045 kg 4200 J kg-1 K-1 (297.5 – 296) K

= 283.5 J (0.284 kJ)

Heat absorbed per mole of ammonium chloride =

= 14.2 kJ mol-1

The enthalpy change of solution of anhydrous ammonium chloride is +14.2 kJ mol-1.

mol 0.02g 0.284

Back


(a) A student tried to determine the enthalpy change of neutralization by putting 25.0 cm3 of 1.0 M HNO3 in a polystyrene cup and adding 25.0 cm3 of 1.0 M NH3 into it. The temperature rise recorded was 3.11 oC. Given that the mass of the polystyrene cup is 250 g, the specific heat capacities of water and the polystyrene cup are

4200 J kg-1 K-1 and 800 J kg-1 K-1 respectively. Determine the enthalpy change of neutralization of nitric acid and aqueous ammonia. (Density of water = 1 g cm-3)Answer




(a) Heat evolved = m1c1T + m2c2 T

= 0.050 kg 4200 J kg-1 K-1 3.11 K + 0.25 kg 800 J kg-1 K-1 3.11 K

= (653.1 + 622) J

= 1275.1 J

No. of moles of HNO3 used = 1.0 M 25 10-3 m3

= 0.025 mol



(a) No. of moles of NH3 used = 1.0 M 25 10-3 dm3

= 0.025 mol

No. of moles of H2O formed = 0.025 mol

Heat evolved per mole of H2O formed

=

= 51.004 kJ mol-1

The enthalpy change of neutralization of nitric acid and aqueous ammonia is –51.004 kJ mol-1.

mol 0.025J 1275.1


(b)When 0.05 mol of silver nitrate was added to 50 g of water in a polystyrene cup, a temperature drop of 5.2 o

C was recorded. Assuming that there was no heat absorption by the polystyrene cup, calculate the enthalpy change of solution of silver nitrate.

(Specific heat capacity of water = 4200 J kg-1 K-1)

Answer




(b) Energy absorbed = mcT

= 0.05 kg 4200 J kg-1 K-1 5.2 K

= 1092 J

No. of moles of AgNO3 used = 0.05 mol

Energy absorbed per mole of AgNO3 used =

= 21.84 kJ mol-1

The enthalpy change of solution of silver nitrate is +21.84 kJ mol-1.

mol 0.05J 1092


(c) A student used a calorimeter as shown in Fig. 6-15 to determine the enthalpy change of combustion of methanol. In the experiment, 1.60 g of methanol was used and 50 g of water was heated up, raising the temperature by 33.2 oC. Given that the specific heat capacities of water and copper calorimeter are 4200 J kg-1 K-1 and 2100 J kg-1 K-1 respectively and the mass of the calorimeter is 400 g, calculate the enthalpy change of combustion of methanol.

Answer




(c) Heat evolved = m1c1T + m2c2 T

= 0.05 kg 4200 J kg-1 K-1 33.2 K + 0.4 kg 2100 J kg-1 K-1 33.2 K

= (6972 + 27888) J

= 34860 J

No. of moles of methanol used =

= 0.05 mol

Heat evolved per mole of methanol used =

= 697.2 kJ mol-1

The enthalpy change of combustion of methanol is –697.2 kJ mol-1.

1-mol g 32.0g 1.60

mol 0.05J 34860

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(a)Given the following thermochemical equation:

2H2(g) + O2(g) 2H2O(l)

(i) Is the reaction endothermic or exothermic?

(ii) What is the enthalpy change for the following reactions?

(1) 2H2O(l) 2H2(g) + O2(g)

(2) H2(g) + O2(g) H2O(l)

(iii) If the enthalpy change for the reaction H2O(l) H2O(g) is +41.1 kJ mol-1, calculate the H for 2H2

(g) + O2(g) 2H2O(g).

21

Answer



(a) (i) Exothermic

(ii) (1) +571.6 kJ mol-1

(2) –285.8 kJ mol-1

(iii)

H = [-571.6 + 2 (+41.1)] kJ mol-1 = -489.4 kJ mol-1



(b)Given the following information about the enthalpy change of combustion of allotropes of carbon:

Hc [C(graphite)] = -393.5 kJ mol-1

Hc [C(diamond)] = -395.4 kJ mol-1

(i) Which allotrope of carbon is more stable?

(ii) What is the enthalpy change for the following process?

C(graphite) C(diamond)

ø

ø

Answer



(b) (i) Graphite

(ii)

H = [-393.5 – (-395.4)] kJ mol-1 = +1.9 kJ mol-1



(c)The formation of ethyne (C2H2(g) can be represented by the following equation:

2C(graphite) + H2(g) C2H2(g)

(i) Draw an enthalpy level diagram relating the above equation to carbon dioxide and water.

(ii) Calculate the standard enthalpy change of formation of ethyne.

(Given: Hc [C(graphite)] = -393.5 kJ mol-1;

Hc [H2(g)] = -285.8 kJ mol-1;

Hc [C2H2(g)] = -1299 kJ mol-1)ø

ø

ø

Answer



(c) (i)

(ii) Hf [C2H2(g)] = Hc [C(graphite)] 2 + Hc [H2(g)] – Hc [C2H2

(g)]

= [(-393.5) 2 + (-285.8) – (-1299)] kJ mol-1

= +226.2 kJ mol-1

øø øø

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Given the following information, find the standard enthalpy change of the reaction:

C2H4(g) + H2(g) C2H6(g)

Hf [C2H4(g)] = +52.3 kJ mol-1

Hf [C2H6(g)] = -84.6 kJ mol-1

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Answer



Note: H1 = [Hf (reactants)] = Hf [C2H4(g)] + Hf [H2(g)]

H2 = [Hf (products)] = Hf [C2H6(g)]

Applying Hess’s law,

H1 + H = H2

H = H2 - H1

= Hf [C2H6(g)] – (Hf [C2H4(g)] + Hf [H2(g)])

= [-84.6 – (+52.3 + 0)] kJ mol-1 =-136.9 kJ mol-1

The standard enthalpy change of the reaction is –136.9 kJ mol-1.

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6PbO(s) + O2(g) 2Pb3O4(s)

Hf [PbO(g)] = -220.0 kJ mol-1

Hf [Pb3O4(g)] = -737.5 kJ mol-1

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Answer



Note: H1 = [Hf (reactants)] = 6 Hf [PbO(s)] + Hf [O2(g)]

H2 = [Hf (products)] = 2 Hf [Pb3O4(s)]


H1 + H = H2

H = H2 - H1

= 2 Hf [Pb3O4(s)] – (6 Hf [PbO(s)] + Hf [O2(g)])

= [2 (-737.5) – 6 (-222.0) – 0] kJ mol-1 =-155.0 kJ mol-1


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Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g)

Hf [Fe2O3(s)] = -822.0 kJ mol-1

Hf [CO(g)] = -110.5 kJ mol-1

Hf [CO2(g)] = -393.5 kJ mol-1

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Answer



Note: H1 = [Hf (reactants)] = Hf [Fe2O3(s)] + 2 Hf [CO(g)]

H2 = [Hf (products)] = 2 Hf [Fe(s)] + 3 Hf [CO2(g)]


H1 + H = H2

H = H2 - H1

= 2 Hf [Fe(s)] + 3 Hf [CO2(g)] - Hf [Fe2O3(s)] - 3 Hf [CO(g)]

= [2 (0) + 3 (-393.5) –(-822.0) – 3 (-110.5)] kJ mol-1

=-27.0 kJ mol-1


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4CH3 · NH · NH2(l) + 5N2O4(l)

4CO2(g) + 12H2O(l) + 9N2(g)

Hf [CH3 · NH · NH2(l)] = +53 kJ mol-1

Hf [N2O4(l)] = -20 kJ mol-1

Hf [CO2(g)] = -393.5 kJ mol-1

Hf [H2O(l)] = -285.8 kJ mol-1

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Answer



Note:H1 = [Hf (reactants)] = 4 Hf [CH3·NH ·NH2(l)] + 5 Hf [N2O4(l)]

H2 = [Hf (products)] = 4 Hf [CO2(g)] + 12 Hf [H2O(l)] + 9 Hf [N2(g)]


H1 + H = H2

H = H2 - H1

= (4 Hf [CO2(g)] + 12 Hf [H2O(l)] + 9 Hf [N2(g)] – (3 Hf [CH3·NH ·NH2(l)] + 5 Hf [N2O4(l)]

= [4 (-393.5) + 12 (-285.8) + 9 (0) – 4 (+53) – 5 (-20)] kJ mol-1

=- 5115.6 kJ mol-1


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Given the following information, find the standard enthalpy change of formation of methane gas.

C(graphite) + O2(g) CO2(g)


H2(g) + O2(g) H2O(l) Hc [H2(g)] = -285.8 kJ mol-1

CH4(g) + 2O2(g) CO2(g) + 2H2O(l)

Hc [CH4(g)] = -20 kJ mol-1

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Answer



Direct measurement of ΔHf [CH4(g)] is impossible because carbon(graphite) and hydrogen do not combine directly, and methane does not decompose directly to form carbon(graphite) and hydrogen. Since methane contain carbon and hydrogen only, they can be related to carbon dioxide and water by the combustion of methane and its constituent elements as shown in the diagram below.

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Note: H1 = Hc [C(graphite)]

H2 = 2 Hc [H2(g)]

H3 = Hc [CH4(g)]


Hf [CH4(g)] + H3 = H1 + H2

Hf [CH4(g)] = H1 + H2 - H3

= Hc [C(graphite)] + 2 Hc [H2(g)] - Hc [CH4(g)]

= [-393.5 + 2 (-285.8) –(-890.4)] kJ mol-1

= -74.7 kJ mol-1

The standard enthalpy change of formation of methane gas is –74.7 kJ mol-1.

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Given the following information, find the standard enthalpy change of formation of methanol.

C(graphite) + O2(g) CO2(g)


H2(g) + O2(g) H2O(l) Hc [H2(g)] = -285.8 kJ mol-1

C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l)

Hc [C2H5OH(l)] = -1371 kJ mol-1

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Answer



Note: H1 = 2 Hc [C(graphite)]

H2 = 3 Hc [H2(g)]

H3 = Hc [C2H5OH(l)]


Hf [C2H5OH(l)] + H3 = H1 + H2

Hf [C2H5OH(l)] = H1 + H2 - H3

= 2 Hc [C(graphite)] + 3 Hc [H2(g)] - Hc [C2H5OH(l)]

= [2 (-393.5) + 3 (-285.8) –(-1371)] kJ mol-1

= -273.4 kJ mol-1

The standard enthalpy change of formation of ethanol is –273.4 kJ mol-1.

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(a) Find the standard enthalpy change of formation of butane gas (C4H10(g)).

Given: Hc [C(graphite)] = -393.5 kJ mol-1

Hc [H2(g)] = -285.8 kJ mol-1

Hc [C4H10(g)] = -2877 kJ mol-1

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Hf [C4H10(g)]

= Hc [C(graphite)] 4 + Hc [H2(g)] 5 - Hc [C4H10(g)]

= [(-393.5) 4 + (-285.8) 5 – (-2877)] kJ mol-1

= -126 kJ mol-1

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(b) Find the standard enthalpy change of the reaction:

Br2(l) + C2H4(g) C2H4Br2(l)

Given: Hf [C2H4(g)] = +52.3 kJ mol-1

Hf [C2H4Br2(l)] = -80.7 kJ mol-1

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H

= [Hf (products)] - [Hf (reactants)]

= [-80.7 – (+52.3) – 0)] kJ mol-1

= -133 kJ mol-1

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Predict whether the following changes or reactions involve an increase or a decrease in entropy.• Dissolving salt in water to form salt solution• Condensation of steam on a cold mirror• Complete combustion of carbon• Complete combustion of carbon monoxide• Oxidation of sulphur dioxide to sulphur trioxide

Answer


(a) Increase

(b) Decrease

(c) Increase

(d) Decrease

(e) Decrease

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In the process of changing of ice to water, at what temperature do you think G equals 0?

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G equals 0 means that neither the forward nor the reverse process is spontaneous. The system is therefore in equilibrium. Melting point of ice is 0 oC (273 K) at which the process of changing ice to water and the process of water turning to ice are at equilibrium. At 0 oC, G of the processes equals 0.

Answer


(a)At what temperatures is the following process spontaneous at 1 atmosphere?

Water Steam

(b)What are the two driving forces that determine the spontaneity of a process?

Answer


(a) 100 oC

(b) Enthalpy and entropy


(c)State whether each of the following cases is spontaneous at all temperatures, not spontaneous at any temperature, spontaneous at high temperatures or spontaneous at low temperatures.

(i) positive S and positive H

(ii) positive S and negative H

(iii) negative S and positive H

(iv) negative S and negative H

Answer

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(i) Spontaneous at high temperatures

(ii) Spontaneous at all temperatures

(iii) Not spontaneous at any temperature

(iv) Spontaneous at low temperatures


New Way Chemistry for Hong Kong A-Level Book 11 Energetics 6.1What is Energetics? 6.2Enthalpy...

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Transcript of New Way Chemistry for Hong Kong A-Level Book 11 Energetics 6.1What is Energetics? 6.2Enthalpy...