Calculate the standard enthalpy change of this reaction, using the values of enthalpy of combustion in Table 12 of the Data Booklet.

Calculate the standard entropy change for this reaction, \(\Delta {S^\Theta }\), using Table 11 of the Data Booklet and given:

\({S^\Theta }{\text{(CO)}} = 198{\text{ J}}\,{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}{\text{ and }}{S^\Theta }{\text{(}}{{\text{H}}_{\text{2}}}{\text{)}} = 131{\text{ J}}\,{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}\).

Calculate, stating units, the standard free energy change for this reaction, \(\Delta {G^\Theta }\), at 298 K.

Predict, with a reason, the effect of an increase in temperature on the spontaneity of this reaction.

1.00 mol of \({\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}}\) is placed in a closed container of volume \({\text{1.00 d}}{{\text{m}}^{\text{3}}}\) until equilibrium is reached with CO and \({{\text{H}}_{\text{2}}}\). At equilibrium 0.492 mol of \({\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}}\) are present. Calculate \({K_{\text{c}}}\).

\({\text{C}}{{\text{H}}_3}{\text{OH}} + \frac{3}{2}{{\text{O}}_2} \to {\text{C}}{{\text{O}}_2} + 2{{\text{H}}_2}{\text{O}}\)     \(\Delta H_{\text{c}}^{^\Theta } =  - 726{\text{ (kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}}{\text{)}}\)

\({\text{CO}} + \frac{1}{2}{{\text{O}}_2} \to {\text{C}}{{\text{O}}_2}\)     \(\Delta H_{\text{c}}^{^\Theta } =  - 283{\text{ (kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}}{\text{)}}\)

\({{\text{H}}_2} + \frac{1}{2}{{\text{O}}_2} \to {{\text{H}}_2}{\text{O}}\)     \(\Delta H_{\text{c}}^{^\Theta } =  - 286{\text{ (kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}}{\text{)}}\)

Award [1 max] for three correct values.

Mark can be implicit in calculations.

\((\Delta H_{\text{R}}^{^\Theta } = ){\text{ }}2( - 286) + ( - 283) - ( - 726)\);

\( - {\text{129 (kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}}{\text{)}}\);

Award [3] for correct final answer.

Award [2 max] for +129 (kJ\(\,\)mol^–1).

\((\Delta {S^\Theta } = 240 - 198 - 2 \times 131 = ){\text{ }} - 220{\text{ (J}}\,{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}{\text{)}}\);

\(\left( { - 129 - 298( - 0.220) = } \right){\text{ }} - 63.4{\text{ kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}}\);

Award [1] for correct numerical answer and [1] for correct unit if the conversion has been made from J to kJ for \(\Delta {S^\Theta }\).

not spontaneous at high temperature;

\(T\Delta {S^\Theta } < \Delta {H^\Theta }\) and \(\Delta {G^\Theta }\) positive;

\(n{\text{(CO)}} = 0.508{\text{ (mol)}}\);

\(n({{\text{H}}_2}) = 2 \times 0.508{\text{ (mol)}}\);

\({K_{\text{c}}}{\text{ }}\left( { = \frac{{0.492}}{{0.508 \times {{(2 \times 0.508)}^2}}}} \right) = 0.938\);

Accept answer in range between 0.930 and 0.940.

Award [3] for correct final answer.

Award [2] for K_c = 1.066 if (c)(ii) is correct.

In (i), the most common error was \( + {\text{129 kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}}\) but in (ii) the answer was often correct.

In (i), the most common error was \( + {\text{129 kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}}\) but in (ii) the answer was often correct.

Units tended to get muddled in (iii) and many marks were awarded as “error carried forward”.

Few were able to explain the \(\Delta H\) and \(T\Delta S\) relationship in detail in (iv).

Equilibrium was well understood in general with many candidates gaining one of the two available marks. “Equal rates” was more often given than the constancy of macroscopic properties for the second mark. The \({K_{\text{c}}}\) expression was given correctly by the vast majority of candidates (including the correct brackets and indices) but many had difficulty with the equilibrium concentrations in (iii).

The changes in equilibrium position were well understood for the most part although if a mark were to be lost it was for not mentioning the number of moles of gas.