(i) Calculate ΔH^θ, in kJ, for this similar reaction below using \(\Delta H_{\rm{f}}^\theta \) data from section 12 of the data booklet. \(\Delta H_{\rm{f}}^\theta \) of HOCH₂CH₂OH(l) is –454.8kJmol^-1.

2CO (g) + 3H₂ (g) \( \rightleftharpoons \) HOCH₂CH₂OH (l)

(ii) Deduce why the answers to (a)(iii) and (b)(i) differ.

(iii) ΔS^θ for the reaction in (b)(i) is –620.1JK^-1. Comment on the decrease in entropy.

(iv) Calculate the value of ΔG^θ, in kJ, for this reaction at 298 K using your answer to (b)(i). (If you did not obtain an answer to (b)(i), use –244.0 kJ, but this is not the correct value.)

(v) Comment on the statement that the reaction becomes less spontaneous as temperature is increased.

Predict the ¹HNMR data for ethanedioic acid and ethane-1,2-diol by completing the table.

i
«ΔH = Σ ΔH_f products – ΣΔH_f reactants = –454.8 kJ mol^-1 – 2(–110.5 kJ mol^-1) =» –233.8 «kJ»

ii
in (a)(iii) gas is formed and in (b)(i) liquid is formed
OR
products are in different states
OR
conversion of gas to liquid is exothermic
OR
conversion of liquid to gas is endothermic
OR
enthalpy of vapourisation needs to be taken into account

Accept product is «now» a liquid.
Accept answers referring to bond enthalpies being means/averages.

iii
«ΔS is negative because five mols of» gases becomes «one mol of» liquid
OR
increase in complexity of product «compared to reactants»
OR
product more ordered «than reactants»

Accept “fewer moles of gas” but not “fewer molecules”.

iv
ΔS = \(\left( {\frac{{ - 620.1}}{{1000}}} \right)\)«kJ K^-1»
ΔG = –233.8 kJ – (298 K \(\left( {\frac{{ - 620.1}}{{1000}}} \right)\) kJ K^-1) = –49.0 «kJ»

Award [2] for correct final answer.
Award [1 max] for «+»185 × 10³.

If –244.0 kJ used, answer is:
ΔG = –244.0 kJ – (298 K \(\left( {\frac{{ - 620.1}}{{1000}}} \right)\)kJ K^-1) = –59.2 «kJ»
Award [2] for correct final answer.

v
increasing T makes ΔG larger/more positive/less negative
OR
–TΔS will increase

Accept “none/no splitting” for singlet.