| Date | May 2014 | Marks available | 2 | Reference code | 14M.2.hl.TZ1.4 |
| Level | HL | Paper | 2 | Time zone | TZ1 |
| Command term | State | Question number | 4 | Adapted from | N/A |
Question
Buta-1,3-diene can be hydrogenated to produce butane, according to the reaction below.
\[{{\text{C}}_{\text{4}}}{{\text{H}}_{\text{6}}}{\text{(g)}} + {\text{2}}{{\text{H}}_{\text{2}}}{\text{(g)}} \to {{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}{\text{(g)}}\]
State the conditions necessary for this reaction.
Determine the standard enthalpy change of reaction, \(\Delta {H^\Theta }\), in \({\text{kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}}\), at 298 K for the hydrogenation reaction, using Table 11 of the Data Booklet.
Calculate the standard free energy change, \(\Delta {G^\Theta }\), in \({\text{kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}}\), at 298 K for the hydrogenation reaction, using Table 11 of the Data Booklet.
(i) Determine the standard entropy change of the reaction, \(\Delta {S^\Theta }\), at 298 K, in \({\text{kJ}}\,{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}\), using your answers from (b) and (c).
(ii) Explain why the standard entropy change for the hydrogenation of buta-1,3-diene has a negative sign.
(iii) Predict whether the hydrogenation reaction becomes more or less spontaneous as the temperature increases.
(iv) Determine the temperature, in K, at which the spontaneity changes.
(v) Determine the standard entropy, \({S^\Theta }\), for hydrogen in \({\text{J}}\,{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}\), using Table 11 of the Data Booklet and your answer for (d)(i).
Markscheme
heat /warm / 140–225 °C;
Do not accept high temperature.
(finely divided) catalyst / Zn/Cu/Ni/Pd/Pt;
\(\Delta {H^\Theta } = \left( {\Sigma \Delta H_f^\Theta {\text{(products)}} - \Sigma \Delta H_f^\Theta {\text{(reactants)}} = - 127 - (110 + 0) = } \right) - 327{\text{ (kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}});\)
\(\Delta {G^\Theta } = \left( {\Sigma \Delta G_f^\Theta {\text{(products)}} - \Sigma \Delta G_f^\Theta {\text{(reactants)}} = - 16 - (152 + 0) = } \right) - 168{\text{ (kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}});\)
(i) \(\Delta {S^\Theta } = \left( {\frac{{\Delta {H^\Theta } - \Delta {G^\Theta }}}{T} = } \right)\frac{{ - 237 - ( - 168)}}{{209}}\);
\( = - 0.232{\text{ (kJ}}\,{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}{\text{)}}\);
Award [2] for correct final answer.
Award [2] for –232 J K–1 mol–1 (units must be given).
(ii) 3 mol of gaseous reactants and 1 mol of gaseous products / fewer moles of gas in products;
(iii) spontaneity decreases (as temperature increases because \(T\Delta {S^\Theta }\) becomes a larger negative value/\(\Delta {G^\Theta }\) becomes positive at higher temperatures);
(iv) \(\Delta {G^\Theta } = \Delta {H^\Theta } - T\Delta {S^\Theta } = 0/ - 237 - T( - 0.232) = 0\);
\(T = 1020{\text{ }}(K)\);
Remember to allow ECF from 4(d)(i).
(v) \(\Delta {S^\Theta } = \Sigma {S^\Theta }{\text{(products)}} - \Sigma {S^\Theta }{\text{(reactants)}}/ - 232 = 310 - (279 + 2{S^\Theta }({H_2}))\);
\({S^\Theta }({H_2}) = \frac{1}{2}(310 - 279 + 232) = 132{\text{ J}}\,{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}})\);
Award [2] for correct final answer.
Remember to allow ECF from 4(d)(i).
Examiners report
Conditions for the reaction, even though the mark scheme accepted fairly vague answers, were not well known. The calculations of enthalpy, entropy and Gibbs free energy scored well; it was pleasing to note that many realised the importance of conversion of units in part d(v). The link between the changes in temperature and the effect on spontaneity was understood, but many lost credit on part d(ii) for failing to mention the change in the number of gaseous moles. In d(v) most candidates missed the fact that 2 moles of hydrogen were present in the equation.
Conditions for the reaction, even though the mark scheme accepted fairly vague answers, were not well known. The calculations of enthalpy, entropy and Gibbs free energy scored well; it was pleasing to note that many realised the importance of conversion of units in part d(v). The link between the changes in temperature and the effect on spontaneity was understood, but many lost credit on part d(ii) for failing to mention the change in the number of gaseous moles. In d(v) most candidates missed the fact that 2 moles of hydrogen were present in the equation.
Conditions for the reaction, even though the mark scheme accepted fairly vague answers, were not well known. The calculations of enthalpy, entropy and Gibbs free energy scored well; it was pleasing to note that many realised the importance of conversion of units in part d(v). The link between the changes in temperature and the effect on spontaneity was understood, but many lost credit on part d(ii) for failing to mention the change in the number of gaseous moles. In d(v) most candidates missed the fact that 2 moles of hydrogen were present in the equation.
Conditions for the reaction, even though the mark scheme accepted fairly vague answers, were not well known. The calculations of enthalpy, entropy and Gibbs free energy scored well; it was pleasing to note that many realised the importance of conversion of units in part d(v). The link between the changes in temperature and the effect on spontaneity was understood, but many lost credit on part d(ii) for failing to mention the change in the number of gaseous moles. In d(v) most candidates missed the fact that 2 moles of hydrogen were present in the equation.
