Question
(a) Determine the change in enthalpy, ΔH, for the combustion of but-2-ene, using data booklet. [3]
CH3CH=CHCH3 (g) + 6O2 (g) → 4CO2 (g) + 4H2O (g)
Answer/Explanation
Ans:
a
Bonds broken: 2(C–C) + 1(C=C) + 8(C–H) + 6O=O / 2(346) + 1(614) + 8(414) + 6(498) / 7606 «kJ»
Bonds formed: 8(C=O) + 8(O–H) / 8(804) + 8(463) / 10 136 «kJ»
Enthalpy change:
«Bonds broken – Bonds formed = 7606 kJ – 10 136 kJ =» –2530 «kJ»
Award [3] for correct final answer.
Award [2 max] for «+» 2530 «kJ».
Question
Two students were asked to use information from the Data Booklet to calculate a value for the enthalpy of hydrogenation of ethene to form ethane.
\[{{\text{C}}_2}{{\text{H}}_4}{\text{(g)}} + {{\text{H}}_2}{\text{(g)}} \to {{\text{C}}_2}{{\text{H}}_6}{\text{(g)}}\]
John used the average bond enthalpies from Table 10. Marit used the values of enthalpies of combustion from Table 12.
Determine the value for the enthalpy of hydrogenation of ethene using the values for the enthalpies of combustion of ethene, hydrogen and ethane given in Table 12.[2]
Suggest one reason why John’s answer is slightly less accurate than Marit’s answer and calculate the percentage difference.[2]
Answer/Explanation
Markscheme
\(\Delta H = -1411 + ( – 286) – ( – 1560)\) / correct energy cycle drawn;
\( = – 137{\text{ kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}\);
Award [1 max] for incorrect or missing sign.
the actual values for the specific bonds may be different to the average values / the combustion values referred to the specific compounds / OWTTE;
(percentage difference) \( = \frac{{(137 – 125)}}{{137}} \times 100 = 8.76\% \);
Accept \(\frac{{(137 – 125)}}{{125}} \times 100 = 9.60\% \).
Examiners report
In part (b) the formula involving enthalpies of formation was often used instead of a correct enthalpy cycle for the combustion. This caused the majority of candidates to score half marks for these questions.
A few candidates could suggest a reason why one answer was slightly less accurate than the other in part (c). Most could correctly calculate the percentage difference. Surprisingly, several candidates calculated part (a) correctly and part (b) incorrectly, and then determined a percentage difference of more than 200% without seeming to notice that this does not reflect two slightly different answers.
Question
Consider the following reaction.
\[{\text{2C}}{{\text{H}}_{\text{3}}}{\text{OH(g)}} + {{\text{H}}_{\text{2}}}{\text{(g)}} \to {{\text{C}}_{\text{2}}}{{\text{H}}_{\text{6}}}{\text{(g)}} + {\text{2}}{{\text{H}}_{\text{2}}}{\text{O(g)}}\]
The standard enthalpy change of formation for \({\text{C}}{{\text{H}}_{\text{3}}}{\text{OH(g)}}\) at 298 K is \( – 201{\text{ kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}\) and for \({{\text{H}}_{\text{2}}}{\text{O(g)}}\) is \( – 242{\text{ kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}\). Using information from Table 11 of the Data Booklet, determine the enthalpy change for this reaction.[2]
The standard entropy for \({\text{C}}{{\text{H}}_{\text{3}}}{\text{OH(g)}}\) at 298 K is \({\text{238 J}}\,{{\text{K}}^{ – 1}}{\text{mo}}{{\text{l}}^{ – 1}}\), for \({{\text{H}}_{\text{2}}}{\text{(g)}}\) is \({\text{131 J}}\,{{\text{K}}^{ – 1}}{\text{mo}}{{\text{l}}^{ – 1}}\) and for \({{\text{H}}_{\text{2}}}{\text{O(g)}}\) is \({\text{189 J}}\,{{\text{K}}^{ – 1}}{\text{mo}}{{\text{l}}^{ – 1}}\). Using information from Table 11 of the Data Booklet, determine the entropy change for this reaction.[2]
Calculate the standard change in free energy, at 298 K, for the reaction and deduce whether the reaction is spontaneous or non-spontaneous.[3]
Answer/Explanation
Markscheme
\(\Delta H_{{\text{reaction}}}^\Theta = \Sigma \Delta H_{\text{f}}^\Theta {\text{(products)}} – \Sigma \Delta H_{\text{f}}^\Theta {\text{(reactants)}}\)
\( = [(1)( – 85) + (2)( – 242)] – [(2)( – 201)]\);
\( = – 167{\text{ (kJ/kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}{\text{)}}\);
Award [1] for (+)167.
\(\Delta S_{{\text{reaction}}}^\Theta = \Sigma {S^\Theta }{\text{(products)}} – \Sigma {S^\Theta }{\text{(reactants)}}\)
\( = [(1)(230) + (2)(189)] – [(2)(238) + (1)(131)]\);
\( = 1{\text{ (J}}\,{{\text{K}}^{ – 1}}{\text{/J}}\,{{\text{K}}^{ – 1}}{\text{mo}}{{\text{l}}^{ – 1}}{\text{)}}\);
\(\Delta G_{{\text{reaction}}}^\Theta = (\Delta {H^\Theta } – T\Delta {S^\Theta }) = ( – 167) – (298)(0.001)\);
Award [1] for correct substitution of values.
\( = – 167{\text{ kJ/}} – 167000{\text{ J}}\);
Units needed for mark in (c) only.
Accept –167 kJ\(\,\)mol–1 or –167000 J\(\,\)mol–1.
spontaneous;
Award marks for final correct answers throughout in each of (a), (b) and (c).
Examiners report
In (a) the most common mistakes included: failure to consider the correct amount of moles of products/reactants, incorrect identification of values or wrong use of convention. It also should be noted that the correct units of \(\Delta {H^\Theta }\) here in the answer will be kJ, since \(n\) is used in the equation, as explained in previous subject reports.
Part (b) was another question where the vast majority of candidates scored full marks.
Free energy calculations (c) continues to prove problematic for many candidates. Candidates very often lost the first mark due to wrong use of units. ECF allowed them to score the second. In contrast most candidates showed a clear understanding of the relationship between the sign of \(\Delta {G^\Theta }\) and spontaneity.