At a temperature just above 700 K it is found that when 1.60 mol of hydrogen and 1.00 mol of iodine are allowed to reach equilibrium in a \({\text{4.00 d}}{{\text{m}}^{\text{3}}}\) flask, the amount of hydrogen iodide formed in the equilibrium mixture is 1.80 mol. Determine the value of the equilibrium constant at this temperature.

Outline why the value for the standard enthalpy change of formation of hydrogen is zero.

Calculate the standard enthalpy change for the hydrogenation of propene.

Calculate the standard entropy change for the hydrogenation of propene.

Determine the value of \(\Delta {G^\Theta }\) for the hydrogenation of propene at 298 K.

At 298 K the hydrogenation of propene is a spontaneous process. Determine the temperature above which propane will spontaneously decompose into propene and hydrogen.

amount of \({{\text{H}}_2}\) remaining at equilibrium \( = 1.60 - \frac{{1.80}}{2} = 0.70{\text{ mol}}\);

amount of \({{\text{I}}_2}\) remaining at equilibrium \( = 1.0 - \frac{{1.80}}{2} = 0.10{\text{ mol}}\);

\({K_{\text{c}}} = \frac{{{{(1.80/4.0)}^2}}}{{(0.70/4.00) \times (0.10/4.00)}}/\frac{{{{1.80}^2}}}{{0.70 \times 0.10}}\);

\({K_{\text{c}}} = \frac{{{{(1.80)}^2}}}{{0.70 \times 0.10}} = 46.3\);

Award [4] for correct final answer.

by definition \(\Delta H_{\text{f}}^\Theta \) of elements (in their standard states) is zero / no reaction involved / OWTTE;

\(\Delta H =  - 104 - ( + 20.4)\);

\( =  - 124.4{\text{ (kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}}{\text{)}}\);

Award [1 max] for 124.4 (kJ\(\,\)mol⁻¹).

Award [2] for correct final answer.

\(\Delta S = 270 - (267 + 131)\);

\( =  - 128{\text{ (J}}\,{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}{\text{)}}\);

Award [1 max] for +128 ( J\(\,\)K⁻¹mol⁻¹).

Award [2] for correct final answer.

\(\Delta G = \Delta H - {\text{T}}\Delta S =  - 124.4 - \frac{{( - 128 \times 298)}}{{1000}}\);

\( =  - 86.3{\text{ kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}}\);

Units needed for the mark.

Award [2] for correct final answer.

Allow ECF if only one error in first marking point.

\(\Delta G = \Delta H - {\text{T}}\Delta S = 0/\Delta H = {\text{T}}\Delta S\);

\({\text{T}} = \frac{{ - 124.4}}{{ - 128/1000}} = 972{\text{ K}}/699{\text{ }}^\circ {\text{C}}\);

Only penalize incorrect units for T and inconsistent ΔS value once in (iv) and (v).

This was the most popularly answered question. Most candidates were able to give a good description of the characteristics of homogenous equilibrium, and apply Le Chatelier‟s Principle to explain the effect of catalysts and changes of temperature and pressure on the position of equilibrium and the equilibrium constant. A good majority were able to calculate the value of \({K_{\text{c}}}\) although a significant number of candidates incorrectly used the initial rather than the equilibrium concentrations.

Although most candidates clearly understood the concept of standard enthalpy change of formation many were unable to explain why the value for hydrogen is zero. Many responses neglected to mention that \({{\text{H}}_{\text{2}}}\) is an element in its standard state.

Most candidate were able to calculate \(\Delta H\) and \(\Delta S\) although some inverted the equation and gave a positive value instead of negative answer or confused the values for propane and propene.

There were some inconsistencies in the use of units and significant figures when calculating \(\Delta G\) from \(\Delta H\) and \(\Delta S\) values although there was a significant improvement in this area compared to previous.

There were some inconsistencies in the use of units and significant figures when calculating \(\Delta G\) from \(\Delta H\) and \(\Delta S\) values although there was a significant improvement in this area compared to previous.

There were some inconsistencies in the use of units and significant figures when calculating \(\Delta G\) from \(\Delta H\) and \(\Delta S\) values although there was a significant improvement in this area compared to previous. This error resulted in some very strange temperatures for the thermal decomposition of propane to propene.