Calculate the enthalpy change of combustion of methanol.

Using the theoretical value in Table 12 of the Data Booklet, discuss the experimental results, including one improvement that could be made.

Methanol can be produced according to the following equation.

\[{\text{CO(g)}} + {\text{2}}{{\text{H}}_2}{\text{(g)}} \to {\text{C}}{{\text{H}}_3}{\text{OH(l)}}\]

Calculate the standard enthalpy change of this reaction using the following data:

\[\begin{array}{*{20}{l}} {{\text{I:     2C}}{{\text{H}}_3}{\text{OH(l)}} + {\text{3}}{{\text{O}}_2}{\text{(g)}} \to {\text{2C}}{{\text{O}}_2}{\text{(g)}} + {\text{4}}{{\text{H}}_{\text{2}}}{\text{O(l)}}}&{\Delta {H^\Theta } =  - 1452{\text{ kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}}} \\ {{\text{II:     2CO(g)}} + {{\text{O}}_2}{\text{(g)}} \to {\text{2C}}{{\text{O}}_2}{\text{(g)}}}&{\Delta {H^\Theta } =  - 566{\text{ kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}}} \\ {{\text{III:     2}}{{\text{H}}_2}{\text{(g)}} + {{\text{O}}_2}{\text{(g)}} \to {\text{2}}{{\text{H}}_2}{\text{O(l)}}}&{\Delta {H^\Theta } =  - 572{\text{ kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}}} \end{array}\]

Outline the characteristics of a chemical equilibrium.

Deduce the equilibrium constant expression, \({K_{\text{c}}}\), for this reaction.

Increase in temperature.

Increase in pressure.

Addition of a catalyst.

\((q = mc\Delta T = ){\text{ }}0.0500 \times 4.18 \times 21.3 = 4.45{\text{ (kJ)}}\);

Do not accept m = 0.05023 kg.

\({\text{(}}n{\text{ }}methanol = {\text{) }}\frac{{0.230}}{{{\text{32.05}}}} = 7.18 \times {10^{ - 3}}{\text{ (mol)}}\);

\(\Delta H = \frac{{4.45}}{{7.18 \times {{10}^{ - 3}}}}\);

\(\Delta H =  - 6.20 \times {10^2}{\text{ kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}}\);

Accept integer values of molar mass.

Final answer must have negative sign and correct units.

Award [4] for correct final answer with correct units.

less heat is liberated than theoretically/\( - 726{\text{ kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}}\);

probably due to heat loss/incomplete combustion;

determine heat capacity of calorimeter and take heat absorbed by calorimeter into account / any suitable insulation method / measure temperature with time and extrapolation of graph to compensate heat loss / OWTTE;

If the value calculated in (a) (i) is more exothermic than theoretically, allow ECF for M1 and for improvement if consistent.

\(\Delta {H^\Theta } = \frac{1}{2}{\text{II}} + {\text{III}} - \frac{1}{2}{\text{I}}\) / correct diagram/energy cycle;

\( - 283 - 572 - ( - 726)\);

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

Award [3] for correct final answer.

rate of forward reaction equals rate of backward reaction;

concentrations of reactants and products do not change / constant macroscopic properties;

\({K_{\text{c}}} = \frac{{{\text{[C}}{{\text{H}}_3}{\text{OH]}}}}{{{\text{[CO][}}{{\text{H}}_2}{{\text{]}}^2}}}\);

Do not award mark if incorrect brackets are used or brackets omitted.

shifts to left/reactants;

to endothermic side / (forward) reaction is exothermic;

shifts to the right/products;

to the side with fewer gas molecules/moles of gas;

no effect on equilibrium;

rate of forward and backward reaction increase equally / activation energy of forward and backward reaction lowered equally;

Many candidates used the mass of methanol in their calculation and most did not convert the mass of methanol to moles.

Students did not make a comparison between their calculated value and the theoretical value, often just stating the numbers. Most candidates were aware that heat was lost but improvements were generally simplistic.

The energy cycle was fairly well done, though working out could be shown better.

Many students had no problem with the characteristics of a chemical equilibrium.

The expression for\({K_{\text{c}}}\) was done quite well.

The effect of changes on the equilibrium position was answered quite well, though candidates had difficulty in explaining the rationale, omitting often gas molecules (ii) and increasing equally in (iii).

The effect of changes on the equilibrium position was answered quite well, though candidates had difficulty in explaining the rationale, omitting often gas molecules (ii) and increasing equally in (iii).

The effect of changes on the equilibrium position was answered quite well, though candidates had difficulty in explaining the rationale, omitting often gas molecules (ii) and increasing equally in (iii).