(i)     Deduce the equilibrium constant expression for this reaction.

(ii)     Explain the effect on the position of equilibrium and the value of \({K_{\text{c}}}\) when pressure is decreased and temperature is kept constant.

(iii)     2.00 mol of NOCl was placed in a \({\text{1.00 d}}{{\text{m}}^{\text{3}}}\) container and allowed to reach equilibrium at 298 K. At equilibrium, 0.200 mol of NO was present. Determine the equilibrium concentrations of NOCl and \({\text{C}}{{\text{l}}_{\text{2}}}\), and hence calculate the value of \({K_{\text{c}}}\) at this temperature.

(iv)     The value of \({K_{\text{c}}}\) is \(1.60 \times {10^{ - 5}}\) at 318 K. State and explain whether the forward reaction is exothermic or endothermic.

(i)     Compare the two liquids in terms of their boiling points, enthalpies of vaporization and vapour pressures.

(ii)     Explain your answer given for part (b)(i).

Calculate the pH of a \({\text{1.50 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\) solution of ammonia at 298 K to two decimal places, using Table 15 of the Data Booklet.

A buffer solution is made using \({\text{25.0 c}}{{\text{m}}^{\text{3}}}\) of \({\text{0.500 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\) hydrochloric acid, HCl (aq), and \({\text{20.0 c}}{{\text{m}}^{\text{3}}}\) of \({\text{1.50 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\) ammonia solution, \({\text{N}}{{\text{H}}_{\text{3}}}{\text{(aq)}}\).

Describe the meaning of the term buffer solution.

Determine the pH of the buffer solution at 298 K.

A \({\text{1.50 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\) solution of ammonia is added to \({\text{25.0 c}}{{\text{m}}^{\text{3}}}\) of a \({\text{0.500 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\) hydrochloric acid solution in a titration experiment.

Calculate the total volume of the solution at the equivalence point.

Calculate the pH of the solution at the equivalence point, using Table 15 of the Data Booklet.

Identify a suitable indicator for this titration, using Table 16 of the Data Booklet.

(i)     \(({K_{\text{c}}} = )\frac{{{\text{[C}}{{\text{l}}_2}{\text{(g)][NO(g)}}{{\text{]}}^2}}}{{{{{\text{[NOCl(g)]}}}^2}}}\);

Ignore state symbols.

(ii)     equilibrium shifts to right as there are more moles (of gas) on product side;

no change to \({K_{\text{c}}}\) as it is a constant at fixed temperature / OWTTE;

(iii)     \({\text{[NOCl(g)]}} = 1.80{\text{ (mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{)}}\);

\({\text{[C}}{{\text{l}}_2}{\text{(g)]}} = 0.100{\text{ (mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{)}}\);

\({K_{\text{c}}} = \left( {\frac{{0.100 \times {{(0.200)}^2}}}{{{{(1.80)}^2}}}} \right)1.23 \times {10^{ - 3}}{\text{ (mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{)}}\);

Award [3] for correct final answer.

(iv)     exothermic as \({K_{\text{c}}}\) is lower at higher temperature;

(i)     hexane has lower boiling point and enthalpy of vaporization than pentan-1-ol / OWTTE;

hexane has higher vapour pressure than pentan-1-ol / OWTTE;

(ii)     hexane is non-polar / has only van der Waals’/London/dispersion forces / has weaker intermolecular forces than pentan-1-ol;

pentan-1-ol has hydrogen bonding between molecules;

\({\text{[O}}{{\text{H}}^ - }{\text{]}} = \sqrt {1.50 \times 1.78 \times {{10}^{ - 5}}}  = 5.17 \times {10^{ - 3}}{\text{ (mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{)}}\);

\({\text{pH}} = (14 - {\text{pOH}} = 14 - 2.29 = ){\text{ }}11.71\);

Award [2] for correct final answer.

Accept correct answer with more than 2 decimal places.

solution which resists change in pH / changes pH slightly / OWTTE;

when small amounts of acid or base are added;

\({\text{[N}}{{\text{H}}_3}{\text{] = }}\left( {\frac{{(1.50 \times 0.0200) - (0.500 \times 0.0250)}}{{0.0450}} = } \right){\text{ }}0.389{\text{ (mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{)}}\);

\({\text{[NH}}_4^ + {\text{]}} = \left( {\frac{{(0.500 \times 0.0250)}}{{0.0450}} = } \right){\text{ }}0.278{\text{ (mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{)}}\);

\({\text{[O}}{{\text{H}}^ - }{\text{]}} = \left( {\frac{{{K_b}{\text{[N}}{{\text{H}}_3}{\text{]}}}}{{{\text{[NH}}_4^ + {\text{]}}}} = } \right){\text{ }}\frac{{1.78 \times {{10}^{ - 5}} \times 0.389}}{{0.278}} = 2.49 \times {10^{ - 5}}{\text{ (mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{)}}\);

\({\text{pH}} = (14.0 - {\text{pOH}} = 14.0 - 4.60 = ){\text{ }}9.40\);

OR

\({\text{pOH}} = {\text{p}}{K_b} + \log \frac{{[{\text{NH}}_4^ + ]}}{{{\text{[N}}{{\text{H}}_3}]}}{\text{  =  p}}{K_{\text{b}}} + \log \frac{{(12.5/1000)}}{{(17.5/1000)}}\);

\({\text{pOH}} = 4.75 + \log \left( {\frac{{12.5}}{{17.5}}} \right) = 4.75 - 0.146 = 4.604\);

\({\text{pH}} = 14.0 - 4.604 = 9.40\);

Award [4] for the correct final answer.

\(\left( {{\text{V(N}}{{\text{H}}_{\text{3}}}{\text{)}} = \frac{{25.0 \times 0.500}}{{1.50}} = 8.33{\text{ c}}{{\text{m}}^3}} \right)\)

\({\text{V}} = {\text{V(N}}{{\text{H}}_3}{\text{)}} + {\text{V(HCl)}} = 8.33 + 25.0 = 33.3{\text{ c}}{{\text{m}}^3}/0.0333{\text{ d}}{{\text{m}}^3}\);

(\({\text{NH}}_{\text{4}}^ + \) ions are present at equivalence point \({\text{N}}{{\text{H}}_3} + {\text{HCl}} \to {\text{NH}}_4^ +  + {\text{C}}{{\text{l}}^ - }\) at equivalence \({\text{n}}({\text{NH}}_4^ + {\text{ produced}}) = {\text{n}}({\text{N}}{{\text{H}}_3}{\text{ added}}) = {\text{n(HCl)}}\))

\([{\text{NH}}_4^ + ] = \frac{{0.500 \times 0.0250}}{{0.0333}} = 0.375{\text{ (mol}}\,{\text{d}}{{\text{m}}^{ - 3}})\);

\({\text{(NH}}_4^ + {\text{(aq)}} \rightleftharpoons {\text{N}}{{\text{H}}_3}{\text{(aq)}} + {{\text{H}}^ + }{\text{(aq)}}/{\text{NH}}_4^ + {\text{(aq)}} + {{\text{H}}_2}{\text{O(l)}} \rightleftharpoons {\text{N}}{{\text{H}}_3}{\text{(aq)}} + {{\text{H}}_3}{{\text{O}}^ + }{\text{(aq)}}\)

\({\text{p}}{K_{\text{a}}}{\text{(NH}}_4^ + ) = 14 - {\text{p}}{K_{\text{b}}}{\text{(N}}{{\text{H}}_3}) = 14.00 - 4.75 = 9.25)\)

\({K_{\text{a}}} = \frac{{{\text{[N}}{{\text{H}}_3}{\text{(aq)][}}{{\text{H}}^ + }{\text{(aq)]}}}}{{{\text{[NH}}_4^ + {\text{(aq)]}}}} = 5.62 \times {10^{ - 10}}\);

\({\text{[}}{{\text{H}}^ + }{\text{(aq)]}} = \sqrt {5.62 \times {{10}^{ - 10}} \times 0.375}  = 1.45 \times {10^{ - 5}}{\text{ (mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{)}}\);

\({\text{pH}} = 4.84\);

Award [4] for the correct final answer.

bromocresol green / methyl red;

ECF for answer in 7(c)(v) if pH given is below 7.

The construction and use of equilibrium expressions for \({K_{\text{c}}}\) showed good understanding. The prediction of the effect of increasing pressure on the position of equilibria by applying Le Chatelier’s principle was good, but the fact that \({K_{\text{c}}}\) remains constant at fixed temperatures was less well known. 

pH calculations in c(i), c(ii) and c(v) tended to be very good or completely incorrect.

The construction and use of equilibrium expressions for \({K_{\text{c}}}\) showed good understanding. The prediction of the effect of increasing pressure on the position of equilibria by applying Le Chatelier’s principle was good, but the fact that \({K_{\text{c}}}\) remains constant at fixed temperatures was less well known. 

pH calculations in c(i), c(ii) and c(v) tended to be very good or completely incorrect.

The construction and use of equilibrium expressions for \({K_{\text{c}}}\) showed good understanding. The prediction of the effect of increasing pressure on the position of equilibria by applying Le Chatelier’s principle was good, but the fact that \({K_{\text{c}}}\) remains constant at fixed temperatures was less well known. 

pH calculations in c(i), c(ii) and c(v) tended to be very good or completely incorrect.

The construction and use of equilibrium expressions for \({K_{\text{c}}}\) showed good understanding. The prediction of the effect of increasing pressure on the position of equilibria by applying Le Chatelier’s principle was good, but the fact that \({K_{\text{c}}}\) remains constant at fixed temperatures was less well known. 

pH calculations in c(i), c(ii) and c(v) tended to be very good or completely incorrect.

The construction and use of equilibrium expressions for \({K_{\text{c}}}\) showed good understanding. The prediction of the effect of increasing pressure on the position of equilibria by applying Le Chatelier’s principle was good, but the fact that \({K_{\text{c}}}\) remains constant at fixed temperatures was less well known. 

pH calculations in c(i), c(ii) and c(v) tended to be very good or completely incorrect.

The construction and use of equilibrium expressions for \({K_{\text{c}}}\) showed good understanding. The prediction of the effect of increasing pressure on the position of equilibria by applying Le Chatelier’s principle was good, but the fact that \({K_{\text{c}}}\) remains constant at fixed temperatures was less well known. 

pH calculations in c(i), c(ii) and c(v) tended to be very good or completely incorrect.

The construction and use of equilibrium expressions for \({K_{\text{c}}}\) showed good understanding. The prediction of the effect of increasing pressure on the position of equilibria by applying Le Chatelier’s principle was good, but the fact that \({K_{\text{c}}}\) remains constant at fixed temperatures was less well known. 

pH calculations in c(i), c(ii) and c(v) tended to be very good or completely incorrect.

The construction and use of equilibrium expressions for \({K_{\text{c}}}\) showed good understanding. The prediction of the effect of increasing pressure on the position of equilibria by applying Le Chatelier’s principle was good, but the fact that \({K_{\text{c}}}\) remains constant at fixed temperatures was less well known. 

pH calculations in c(i), c(ii) and c(v) tended to be very good or completely incorrect.