Deduce the equilibrium constant expression, \({K_{\text{c}}}\), for the formation of HI(g).

Determine the equilibrium concentrations, in \({\text{mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\), of hydrogen, iodine and hydrogen iodide.

\(({K_{\text{c}}} = )\frac{{{{{\text{[HI]}}}^2}}}{{{\text{[}}{{\text{H}}_2}{\text{][}}{{\text{I}}_2}{\text{]}}}}/\frac{{{\text{[HI]}}}}{{{{{\text{[}}{{\text{H}}_2}{\text{]}}}^{\frac{1}{2}}}{{{\text{[}}{{\text{I}}_2}{\text{]}}}^{\frac{1}{2}}}}}\);

Do not award mark if brackets are omitted or incorrect.

\(\begin{array}{*{20}{l}} {{{\text{H}}_{\text{2}}}}&{ + {{\text{I}}_{\text{2}}}}&{ \to {\text{2HI}}} \\ {(7.50 \times {{10}^{ - 3}} - x)}&{(7.50 \times {{10}^{ - 3}} - x)}&{{\text{2}}x/} \\ {(1.50 \times {{10}^{ - 2}} - x)}&{(1.50 \times {{10}^{ - 2}} - x)}&{{\text{2}}x/} \\ {{{{\text{[}}{{\text{H}}_{\text{2}}}{\text{]}}}_{{\text{initial}}}} - x}&{{{{\text{[}}{{\text{I}}_{\text{2}}}{\text{]}}}_{{\text{initial}}}} - x}&{{\text{2}}x{\text{;}}} \end{array}\)

Accept [H₂]\(_{initial}\) = [I₂]\(_{initial}\) = 7.50 \( \times \) 10^–3 (mol\(\,\)dm^–3) for M1.

\(53 = \frac{{{{(2x)}^2}}}{{{{(7.50 \times {{10}^{ - 3}} - x)}^2}}}/\sqrt {53}  = \frac{{(2x)}}{{(7.50 \times {{10}^{ - 3}} - x)}}\);

Accept 53 \( = \frac{{{{(2x)}^2}}}{{{{(1.50 \times 1{0^{ - 2}} - x)}^2}}}/\sqrt {53}  = \frac{{(2x)}}{{(1.50 \times 1{0^{ - 2}} - x)}}\)

\([{{\text{H}}_{\text{2}}}] = 1.62 \times {10^{ - 3}}{\text{ }}({\text{mol}}\,{\text{d}}{{\text{m}}^{ - 3}})\) and \([{{\text{I}}_2}] = 1.62 \times {10^{ - 3}}{\text{ }}({\text{mol}}\,{\text{d}}{{\text{m}}^{ - 3}})\);

\({\text{[HI]}} = 1.18 \times {10^{ - 2}}\);

Award [4] for correct final answer for values given in M3 and M4.

Award [2 max] for [H₂] = [I₂] = 7.50 \( \times \) 10^–3 (mol\(\,\)dm^–3) and

[HI] = 5.46 \( \times \) 10\(^{ - 2}\) mol\(\,\)dm^–3.

OR

if \({K_{\text{c}}} = \frac{{{\text{[HI]}}}}{{{{{\text{[}}{{\text{H}}_{\text{2}}}{\text{]}}}^{\frac{1}{2}}}{{{\text{[}}{{\text{I}}_{\text{2}}}{\text{]}}}^{\frac{1}{2}}}}}\) is given in (i).

\(\begin{array}{*{20}{l}} {\frac{1}{2}{{\text{H}}_{\text{2}}}}&{ + \frac{1}{2}{{\text{I}}_{\text{2}}}}&{ \to {\text{HI}}} \\ {(7.50 \times {{10}^{ - 3}} - x)}&{(7.50 \times {{10}^{ - 3}} - x)}&{{\text{2}}x/} \\ {(1.50 \times {{10}^{ - 2}} - x)}&{(1.50 \times {{10}^{ - 2}} - x)}&{{\text{2}}x/} \\ {{{{\text{[}}{{\text{H}}_{\text{2}}}{\text{]}}}_{{\text{initial}}}} - x}&{{{{\text{[}}{{\text{I}}_{\text{2}}}{\text{]}}}_{{\text{initial}}}} - x}&{{\text{2}}x{\text{;}}} \end{array}\)

Accept [H₂]\(_{initial}\) = [I₂]\(_{initial}\) = 7.50 \( \times \) 10^–3 (mol\(\,\)dm^–3) for M1.

\(53 = \frac{{(2{\text{x}})}}{{(7.50 \times {{10}^{ - 3}} - x)}}\);

Accept 53 \( = \frac{{(2x)}}{{(1.50 \times 1{0^{ - 2}}- x)}}\).

\([{{\text{H}}_2}] = 2.73 \times {10^{ - 4}}{\text{ mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\) and \([{{\text{I}}_2}] = 2.73 \times {10^{ - 4}}{\text{ mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\);

\([{\text{HI}}] = 1.45 \times {10^{ - 2}}{\text{ mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\);

Award [4] for correct final answer for values given in M3 and M4.

Award [2 max] for [H₂] = [I₂] = 7.50 \( \times \) 10^–3 (mol\(\,\)dm^–3) and

[HI] = 5.46 \( \times \) 10\(^{ - 2}\) mol\(\,\)dm^–3.

Most candidates scored poor marks in this question because they made mistakes in writing the correct equilibrium concentrations for H₂, I₂ and HI or calculating the value of ‘x’. It was not necessary to solve he quadratic equation to calculate the equilibrium concentrations. Most candidates knew that dimethylamine could form H-bonding with water, but only very few scored the mark for explanation. Reference to electronegativity or the lone pair of electrons on nitrogen was needed for both marks. Few candidates were able to name the amine correctly.

Most candidates scored poor marks in this question because they made mistakes in writing the correct equilibrium concentrations for H₂, I₂ and HI or calculating the value of ‘x’. It was not necessary to solve he quadratic equation to calculate the equilibrium concentrations. Most candidates knew that dimethylamine could form H-bonding with water, but only very few scored the mark for explanation. Reference to electronegativity or the lone pair of electrons on nitrogen was needed for both marks. Few candidates were able to name the amine correctly.