1. Given the following data:

C(s) + F2(g) → CF4(g); ∆H1 = –680 kJ mol–1

F2(g) → 2F(g); ∆H2 = +158 kJ mol–1

C(s) → C(g); ∆H3 = +715 kJ mol–1

calculate the average bond enthalpy (in kJ mol–1) for the C––F bond.

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2. C(s) + 2F2(g) → CF4(g) ∆H1 = –680 kJ;4F(g) → 2F2(g) ∆H2 = 2(–158) kJ;C(g) → C(s) ∆H3 = –715 kJ;

Accept reverse equations with +∆H values.

C(g) + 4F(g) → CF4(g) ∆H = –1711 kJ,

so average bond enthalpy = 41711–

= –428 kJ mol–1; 4Accept + or – sign.

Lots of ways to do this! The correct answer is very different from the value in the Data Booklet, so award [4] for final answer with/without sign units not needed, but deduct [1] if incorrect units. Accept answer in range of 427 to 428 without penalty for sig figs.

If final answer is not correct use following;

Award [1] for evidence of cycle or enthalpy diagram or adding of equations.Award [1] for 2F2 (g) → 4F(g) 2×158 seen.Award [1] for dividing 1711 or other value by 4.

[4]

3. For the process:

C6H6(l) C6H6(s)

the standard entropy and enthalpy changes are:

∆Hο = –9.83kJ mol–1 and ∆Sο = –35.2J K–1 mol–1.

2

Predict and explain the effect of an increase in temperature on the spontaneity of the process.

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4. (∆Gο = ∆H – T∆Sο)

as T increases, –T∆Sο becomes larger/more positive;∆G increases/becomes more positive/less negative;process becomes less spontaneous/reverse reaction favoured; 3

[3]

5. Consider the following reaction.

N2(g) +3H2(g) → 2NH3(g)

(i) Use values from Table 10 in the Data Booklet to calculate the enthalpy change, Hο, for this reaction.

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(ii) The magnitude of the entropy change, S, at 27°C for the reaction is 62.7 J K–1 mol–1. State, with a reason, the sign of S.

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(iii) Calculate G for the reaction at 27°C and determine whether this reaction is spontaneous at this temperature.

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(Total 8 marks)

6. (i) selection of all the correct bonds or values from Data Booklet;∆H = (N≡≡N) + 3(H—H) – 6(N—H) / 944 + 3(436) – 6(388);= –76 (kJ); 3

Allow ECF for one error (wrong bond energy/wrong coefficient/reverse reaction) but not for two errors(so –611, –857, +76, +1088 all score 2 out of 3).

(ii) negative; decrease in the number of gas molecules/OWTTE; 2

(iii) ∆G = ∆H – T∆S∆G = –76.0 – 300 (–0.0627); 3

Award [1] for 300 K.Award [1] for conversion of units J to kJ or vice versa.Allow ECF from (i) from ∆H.Allow ECF from (ii) for sign of ∆S.

= –57.2 (kJ mol–1) is spontaneous/or non-spontaneous if positive value obtained; [3 max]

[8]

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7. Explain in terms of Gο, why a reaction for which bothHο and Sο are positive is sometimes spontaneous and sometimes not.

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8. a reaction is spontaneous when ∆Gο is negative;

at high T, ∆Gο is negative;

–T∆S ο is larger/greater than ∆Hο;

at low T, ∆Gο is positive because –T∆Sο is smaller than ∆Hο/OWTTE; 4[4]

9. Consider the following reaction.

N2(g) + 3H2(g) → 2NH3(g)

(i) Using the average bond enthalpy values in Table 10 of the Data Booklet, calculate the standard enthalpy change for this reaction.

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(ii) The absolute entropy values, S, at 300 K for N2(g), H3(g) and NH2(g) are 193, 131 and

192 JK–1 mol–1 respectively. Calculate Sο for the reaction and explain the sign of Sο.

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(iii) Calculate Gο for the reaction at 300 K.

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(iv) If the ammonia was produced as a liquid and not as a gas, state and explain the effect this

would have on the value of Hο for the reaction.

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(Total 10 marks)

10. (i) ∆H = (sum of energies of bonds broken) – (sum of energies of bonds formed); Can be implied by working.

Correct substitution of values and numbers of bonds broken; Correct substitution of values and numbers of bonds made;

(∆H = (N≡≡N) + 3(H—H) – 6(N—H) = 944 + 3(436) – 6(388) =) –76 (kJ);4

Allow ECF.Do not penalize for SF or units.

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(ii) ∆Sο = (sum of entropies of products) – (sum of entropies of reactants); 3Can be implied by working.

(= 2×192 – (193 + 3×131) =) –202(J K–1 mol–1);four molecules make two molecules/fewer molecules of gas;

(iii) (∆Gο =∆Hο – T∆Sο = –76.0 – 300(–0.202)) = – 15.4 (kJ mol –1); 1Do not penalize for SF.

(iv) ∆Hο becomes more negative; 2heat released when gas → liquid;

[10]

11. Two reactions occurring in the manufacture of sulfuric acid are shown below:

reaction I S(s) +O2(g) SO2(g) HӨ = –297 kJ

reaction II SO2(g) + 21

O2(g) SO3(g) HӨ = –92 kJ

(i) State the name of the term HӨ. State, with a reason, whether reaction I would be accompanied by a decrease or increase in temperature.

(3)

(ii) At room temperature sulfur trioxide, SO3, is a solid. Deduce, with a reason, whether the

HӨ value would be more negative or less negative if SO3(s) instead of SO3(g) were formed in reaction II.

(2)

(iii) Deduce the HӨ value of this reaction:

S(s) + 211 O2(g) SO3(g)

(1)(Total 6 marks)

12. (a) (i) standard enthalpy (change) of reaction;(temperature) increase;reaction is exothermic/sign of H is negative; 3

(ii) more (negative);heat given out when gas changes to solid/solid has less enthalpy thangas/OWTTE; 2

(iii) –389 kJ; 1[6]

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13. (i) Define the term average bond enthalpy.(3)

(ii) Explain why Br2 is not suitable as an example to illustrate the term average bond enthalpy.

(1)

(iii) Using values from Table 10 of the Data Booklet, calculate the enthalpy change for the following reaction:

CH4(g) + Br2(g) CH3Br(g) + HBr(g)(3)

(iv) Sketch an enthalpy level diagram for the reaction in part (iii).(2)

(v) Without carrying out a calculation, suggest, with a reason, how the enthalpy change for the following reaction compares with that of the reaction in part (iii):

CH3Br(g) + Br2(g) CH2Br2(g) + HBr(g)(2)

(Total 11 marks)

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14. (i) the energy needed to break one bond;(in a molecule in the) gaseous state;value averaged using those from similar compounds; 3

(ii) it is an element/no other species with just a Br-Br bond/OWTTE; 1

(iii) (sum bonds broken =) 412 + 193 = 605;(sum bonds formed =) 276 + 366 = 642;(H =) –37 kJ; 3

Award [3] for correct final answer.

Award [2] for “+ 37”.

Accept answer based on breaking and making extra C-H bonds.

(iv)

Enthalpy CH4 + Br2

CH3Br + HBr ;

2Award [1] for enthalpy label and two horizontal lines, [1] for reactants higher than products.

ECF from sign in (iii), ignore any higher energy level involving atoms.

(v) (about) the same/similar;same (number and type of) bonds being broken and formed; 2

[11]

15. (a) Propyne reacts with hydrogen as follows:

C3H4(g) + 2H2(g) C3H8(g) HӨ = –287 kJ

Calculate the standard entropy change of this reaction, given the following additional information:

SӨ of H2(g) = 131 J K–1 mol–1

(3)

(b) Calculate the standard free energy change at 298 K, GӨ, for the reaction in part (a). Use your answer and relevant information from part (d). If you did not obtain an answer to

part (a), use SӨ = –360 J K–1 (this is not the correct value).(3)

(Total 6 marks)

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16. (a) S = SӨ (products) SӨ (reactants)/suitable cycle;

= 270 248 2131;

= 240 (J K1); 3Units not needed for mark, but penalize incorrect units.

Award [3] for correct final answer.

(b) ΔGӨ = 287 (2980.240);Award [1] for correct substitution of values and [1] for conversion of units.

= 215 kJ; 3Units needed for mark.

Apply ECF from 360 kJ or incorrect answer from (a).[6]

17. (a) The lattice enthalpy of an ionic compound can be calculated using a Born-Haber cycle. Using lithium fluoride as the example, construct a Born-Haber cycle, labelling the cycle with the formulas and state symbols of the species present at each stage.

(6)

(b) Two values of the lattice enthalpies for each of the silver halides are quoted in the Data Booklet. Discuss the bonding in silver fluoride and in silver iodide, with reference to these values.

(2)(Total 8 marks)

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18. (a) 6

L iF (s)

122L i(s ) F (g )+

122L i(g ) F (g )+

L i (g ) F (g)+ -+

L i ( g ) e F ( g )+ -+ +

122L i (g ) e F (g )+ -+ +

Award [6] for completely correct cycle, with endothermic processes in any order.

Deduct [1] for each line in which species symbol and/or state symbol is incorrect or missing.

Penalize missing electrons once only.

(b) bonding in AgF more ionic than in AgI/bonding in AgI more covalent thanin AgF;

Accept AgF is ionic and AgI is covalent.

values closer/in better agreement in AgF/big(ger) difference in values forAgI/OWTTE; 2

[8]

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19. Calculate the enthalpy change, H4 for the reaction

C + 2H2 + 21

O2 CH3OH H4

using Hess’s Law, and the following information.

CH3OH + 211 O2 CO2 + 2H2O H1 = 676 kJ mol1

C + O2 CO2 H2 = 394 kJ mol1

H2 + 21

O2 H2O H3 = 242 kJ mol1

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20. 1×H1/676;

1×H2/ 394;

2×H3/ 484;

H4 = 202 (kJ mol1); 4Accept alternative methods.

Correct answers score [4].

Award [3] for (+)202 or (+)40 (kJ/kJ mol1).[4]

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21. Consider the following reaction:

N2(g) + 3H2(g) 2NH3(g)

(i) Suggest why this reaction is important for humanity.(1)

(ii) Using the average bond enthalpy values in Table 10 of the Data Booklet, calculate the standard enthalpy change for this reaction.

(4)

(iii) The absolute entropy values, S, at 238 K for N2(g), H2(g) and NH3(g) are 192, 131 and

193 J K–1 mol–1 respectively. Calculate ∆Sο for the reaction and explain the sign of ∆Sο.(2)

(iv) Calculate ∆Gο for the reaction at 238 K. State and explain whether the reaction is spontaneous.

(3)

(v) If ammonia was produced as a liquid and not as a gas, state and explain the effect this

would have on the value of ∆Hο for the reaction.(2)

(Total 12 marks)

22. (i) fertilizers/increasing crop yields;

production of explosives for mining; 1 max

(ii) H = (sum of energies of bonds broken) – (sum of energies of bonds formed);Can be implied by working.

correct substitution of values and numbers of bonds broken;correct substitution of values and numbers of bonds made;(H = (NN) + 3(H–H) – 6(N–H) = 944 + 3(436) – 6(388) =) –76.0 (kJ); 4

Allow ECF.Do not penalize for sig. fig. or units.Award [4] for correct final answer.

(iii) (Sο[2×193] – [192 + 3×131]) = –199 (J K–1 mol–1); 2Allow ECF.four gaseous molecules generating two gaseous molecules/fewer molecules of gas;

(iv) (Gο = Hο – TSο = –76.0 – 298(–0.199)) = –16.7 (kJ);

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Spontaneous;G is negative; 3

Do not penalize for SF.

(v) heat released when gas → liquid;

Hο becomes more negative; 2[12]

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