Question
Chlorine gas reacts with water to produce hypochlorous acid and hydrochloric acid.
Cl2 (g) + H2O (l) HClO (aq) + HCl (aq)
Ethoxyethane (diethyl ether) can be used as a solvent for this conversion. Draw the structural formula of ethoxyethane. [1]
Answer/Explanation
Ans
/ CH3CH2OCH2CH3
Question
The \({\text{p}}{K_{\text{a}}}\) value for propanoic acid is given in Table 15 of the Data Booklet.
State the equation for the reaction of propanoic acid with water.[1]
Calculate the hydrogen ion concentration (in \({\text{mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\)) of an aqueous solution of \({\text{0.100 mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) propanoic acid.[2]
The graph below shows a computer simulation of a titration of \({\text{25.0 c}}{{\text{m}}^{\text{3}}}\) of \({\text{0.100 mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) hydrochloric acid with \({\text{0.100 mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) sodium hydroxide and the pH range of phenol red indicator.
Sketch the graph that would be obtained for the titration of \({\text{25.0 c}}{{\text{m}}^{\text{3}}}\) of \({\text{0.100 mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) propanoic acid with \({\text{0.100 mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) potassium hydroxide using bromophenol blue as an indicator. (The pH range of bromophenol blue can be found in Table 16 of the Data Booklet).
[3]
Answer/Explanation
Markscheme
\({\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{COOH}} + {{\text{H}}_2}{\text{O}} \rightleftharpoons {\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{CO}}{{\text{O}}^ – } + {{\text{H}}_3}{{\text{O}}^ + }\) /
\({\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{COOH}} \rightleftharpoons {\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{COO}}{{\text{H}}^ – } + {{\text{H}}^ + }\);
\( \rightleftharpoons \) required for mark.
(\({\text{p}}{K_{\text{a}}}\) for propanoic acid = 4.87)
\({{\text{[}}{{\text{H}}^ + }{\text{]}}^2} = 0.100 \times {K_{\text{a}}}\);
\({\text{[}}{{\text{H}}^ + }{\text{]}} = 1.16 \times {10^{ – 3}}{\text{ (mol}}\,{\text{d}}{{\text{m}}^{ – 3}})\);
sketch to show:
indicator range between pH 3.0 and pH 4.6 (with “yellow” at pH 3.0 and “blue” at pH 4.6);
initial pH of acid at 2.9 ± 1.0 (when no KOH has been added);
half-equivalence point (does not need to be named) at pH 4.9 when \({\text{12.5 c}}{{\text{m}}^{\text{3}}}\) of KOH have been added;
equivalence point at approx pH 8.5 – 9.0 when \({\text{25.0 c}}{{\text{m}}^{\text{3}}}\) of KOH(aq) added;
upper part of curve from 25.0 – 50.0 \({\text{c}}{{\text{m}}^{\text{3}}}\) added identical to original curve;
Award [1] each for any three points.
Examiners report
The equation of propanoic acid with water was problematic for many candidates who omitted the equilibrium arrow \(( \rightleftharpoons )\) in part (a)(i). Although candidates were referred to the Data Booklet, some candidates did not know the formula of propanoic acid.
Part (a)(ii) was answered well by about half the candidates.
Part (b) also caused difficulties, with many candidates scoring only the mark for showing the pH range of bromophenol blue. Some candidates were thrown by the choice of indicator and selected a more appropriate indicator for these reagents. It is important to answer the question on the paper as the indicator was deliberately chosen to be different to the indicator used in the example. Graphs were generally badly and roughly drawn. Even candidates who had correctly calculated \({\text{[}}{{\text{H}}^ + }{\text{]}}\) in part (a) often did not start the graph at the correct pH. Most graphs finished too low at a pH of 10 or less, and the vertical part of the graph was frequently at a volume less than \({\text{25 c}}{{\text{m}}^{\text{3}}}\). Rarely did a candidate get the half-equivalence value correct.
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.