Question
The Bombardier beetle sprays a mixture of hydroquinone and hydrogen peroxide to fight off predators. The reaction equation to produce the spray can be written as:
| C6H4(OH)2(aq) + H2O2(aq) | → | C6H4O2(aq) + 2H2O(l) |
| hydroquinone | quinone |
Calculate the enthalpy change, in kJ, for the spray reaction, using the data below.
\(\begin{array}{*{20}{l}} {{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{4}}}{{{\text{(OH)}}}_{\text{2}}}{\text{(aq)}} \to {{\text{C}}_{\text{6}}}{{\text{H}}_{\text{4}}}{{\text{O}}_{\text{2}}}{\text{(aq)}} + {{\text{H}}_{\text{2}}}{\text{(g)}}}&{\Delta {H^\theta } = + {\text{177.0 kJ}}} \\ {{\text{2}}{{\text{H}}_{\text{2}}}{\text{O(l)}} + {{\text{O}}_{\text{2}}}{\text{(g)}} \to {\text{2}}{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}{\text{(aq)}}}&{\Delta {H^\theta } = + {\text{189.2 kJ}}} \\ {{{\text{H}}_{\text{2}}}{\text{O(l)}} \to {{\text{H}}_{\text{2}}}{\text{(g)}} + \frac{{\text{1}}}{{\text{2}}}{{\text{O}}_{\text{2}}}{\text{(g)}}}&{\Delta {H^\theta } = + {\text{285.5 kJ}}} \end{array}\)
The energy released by the reaction of one mole of hydrogen peroxide with hydroquinone is used to heat 850 cm3 of water initially at 21.8°C. Determine the highest temperature reached by the water.
Specific heat capacity of water = 4.18 kJ\(\,\)kg−1\(\,\)K−1.
(If you did not obtain an answer to part (i), use a value of 200.0 kJ for the energy released, although this is not the correct answer.)
Identify the species responsible for the peak at m/z = 110 in the mass spectrum of hydroquinone.
Identify the highest m/z value in the mass spectrum of quinone.
Answer/Explanation
Markscheme
ΔH = 177.0 – \(\frac{{189.2}}{2}\) –285.5 «kJ»
«ΔH =» –203.1 «kJ»
Accept other methods for correct manipulation of the three equations.
Award [2] for correct final answer.
[2 marks]
203.1 «kJ» = 0.850 «kg» x 4.18 «kJ\(\,\)kg–1\(\,\)K–1» x ΔT «K»
OR
«ΔT =» 57.2 «K»
«Tfinal = (57.2 + 21.8) °C =» 79.0 «°C» / 352.0 «K»If 200.0 kJ was used:
200.0 «kJ» = 0.850 «kg» x 4.18 «kJ\(\,\)kg–1\(\,\)K–1» x ΔT «K»
OR
«ΔT =» 56.3 «K»
«Tfinal = (56.3 + 21.8) °C =» 78.1 «°C» / 351.1 «K»
Award [2] for correct final answer.
Units, if specified, must be consistent with the value stated.
[2 marks]
C6H4(OH)2+
Accept “molecular ion”.
Do not accept “C6H4(OH)2” (positive charge missing).
[1 mark]
«highest m/z» 108
Only accept exactly 108, not values close to this.
[1 mark]
Question
To determine the enthalpy change of combustion of methanol, \({\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}}\), 0.230 g of methanol was combusted in a spirit burner. The heat released increased the temperature of \({\text{50.0 c}}{{\text{m}}^{\text{3}}}\) of water from 24.5 °C to 45.8 °C.
Methanol can be produced according to the following equation.
\[{\text{CO(g)}} + {\text{2}}{{\text{H}}_2}{\text{(g)}} \to {\text{C}}{{\text{H}}_3}{\text{OH(l)}}\]
The manufacture of gaseous methanol from CO and \({{\text{H}}_{\text{2}}}\) involves an equilibrium reaction.
\({\text{CO(g)}} + {\text{2}}{{\text{H}}_2}{\text{(g)}} \rightleftharpoons {\text{C}}{{\text{H}}_3}{\text{OH(g)}}\) \(\Delta {H^\Theta } < 0\)
Calculate the standard enthalpy change of this reaction, using the values of enthalpy of combustion in Table 12 of the Data Booklet.[3]
Calculate the standard entropy change for this reaction, \(\Delta {S^\Theta }\), using Table 11 of the Data Booklet and given:
\({S^\Theta }{\text{(CO)}} = 198{\text{ J}}\,{{\text{K}}^{ – 1}}{\text{mo}}{{\text{l}}^{ – 1}}{\text{ and }}{S^\Theta }{\text{(}}{{\text{H}}_{\text{2}}}{\text{)}} = 131{\text{ J}}\,{{\text{K}}^{ – 1}}{\text{mo}}{{\text{l}}^{ – 1}}\).[1]
Calculate, stating units, the standard free energy change for this reaction, \(\Delta {G^\Theta }\), at 298 K.[2]
Predict, with a reason, the effect of an increase in temperature on the spontaneity of this reaction.[3]
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 \({K_{\text{c}}}\).[3]
Answer/Explanation
Markscheme
\({\text{C}}{{\text{H}}_3}{\text{OH}} + \frac{3}{2}{{\text{O}}_2} \to {\text{C}}{{\text{O}}_2} + 2{{\text{H}}_2}{\text{O}}\) \(\Delta H_{\text{c}}^{^\Theta } = – 726{\text{ (kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}{\text{)}}\)
\({\text{CO}} + \frac{1}{2}{{\text{O}}_2} \to {\text{C}}{{\text{O}}_2}\) \(\Delta H_{\text{c}}^{^\Theta } = – 283{\text{ (kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}{\text{)}}\)
\({{\text{H}}_2} + \frac{1}{2}{{\text{O}}_2} \to {{\text{H}}_2}{\text{O}}\) \(\Delta H_{\text{c}}^{^\Theta } = – 286{\text{ (kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}{\text{)}}\)
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 Kc = 1.066 if (c)(ii) is correct.
Examiners report
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.