| Date | November 2009 | Marks available | 2 | Reference code | 09N.2.hl.TZ0.6 |
| Level | HL | Paper | 2 | Time zone | TZ0 |
| Command term | Deduce | Question number | 6 | Adapted from | N/A |
Question
Consider the following reaction studied at 263 K.
\[{\text{2NO(g)}} + {\text{C}}{{\text{l}}_{\text{2}}}{\text{(g)}} \rightleftharpoons {\text{2NOCl(g)}}\]
It was found that the forward reaction is first order with respect to \({\rm{C}}{{\rm{l}}_2}\) and second order with respect to NO. The reverse reaction is second order with respect to NOCl.
Consider the following equilibrium reaction.
\[\begin{array}{*{20}{c}} {{\text{C}}{{\text{l}}_2}({\text{g)}} + {\text{S}}{{\text{O}}_2}({\text{g)}} \rightleftharpoons {\text{S}}{{\text{O}}_2}{\text{C}}{{\text{l}}_2}({\text{g)}}}&{\Delta {H^\Theta } = - 84.5{\text{ kJ}}} \end{array}\]
In a \({\text{1.00 d}}{{\text{m}}^{\text{3}}}\) closed container, at 375 °C, \({\text{8.60}} \times {\text{1}}{{\text{0}}^{ - 3}}{\text{ mol}}\) of \({\text{S}}{{\text{O}}_{\text{2}}}\) and \({\text{8.60}} \times {\text{1}}{{\text{0}}^{ - 3}}{\text{ mol}}\) of \({\text{C}}{{\text{l}}_{\text{2}}}\) were introduced. At equilibrium, \({\text{7.65}} \times {\text{1}}{{\text{0}}^{ - 4}}{\text{ mol}}\) of \({\text{S}}{{\text{O}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}\) was formed.
State the rate expression for the forward reaction.
Predict the effect on the rate of the forward reaction and on the rate constant if the concentration of NO is halved.
1.0 mol of \({\rm{C}}{{\rm{l}}_2}\) and 1.0 mol of NO are mixed in a closed container at constant temperature. Sketch a graph to show how the concentration of NO and NOCl change with time until after equilibrium has been reached. Identify the point on the graph where equilibrium is established.
Consider the following reaction.
\[{\text{N}}{{\text{O}}_2}{\text{(g)}} + {\text{CO(g)}} \to {\text{NO(g)}} + {\text{C}}{{\text{O}}_2}{\text{(g)}}\]
Possible reaction mechanisms are:
\(\begin{array}{*{20}{l}} {{\text{Above 775 K:}}}&{{\text{N}}{{\text{O}}_2} + {\text{CO}} \to {\text{NO}} + {\text{C}}{{\text{O}}_{\text{2}}}}&{{\text{slow}}} \\ {{\text{Below 775 K:}}}&{{\text{2N}}{{\text{O}}_2} \to {\text{NO}} + {\text{N}}{{\text{O}}_{\text{3}}}}&{{\text{slow}}} \\ {}&{{\text{N}}{{\text{O}}_3} + {\text{CO}} \to {\text{N}}{{\text{O}}_2} + {\text{C}}{{\text{O}}_2}}&{{\text{fast}}} \end{array}\)
Based on the mechanisms, deduce the rate expressions above and below 775 K.
State two situations when the rate of a chemical reaction is equal to the rate constant.
Consider the following graph of \(\ln k\) against \(\frac{1}{T}\) for the first order decomposition of \({{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}\) into \({\text{N}}{{\text{O}}_{\text{2}}}\). Determine the activation energy in \({\text{kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}}\) for this reaction.
Deduce the equilibrium constant expression, \({K_{\text{c}}}\), for the reaction.
Determine the value of the equilibrium constant, \({K_{\text{c}}}\).
If the temperature of the reaction is changed to 300 °C, predict, stating a reason in each case, whether the equilibrium concentration of \({\text{S}}{{\text{O}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}\) and the value of \({K_{\text{c}}}\) will increase or decrease.
If the volume of the container is changed to \({\text{1.50 d}}{{\text{m}}^{\text{3}}}\), predict, stating a reason in each case, how this will affect the equilibrium concentration of \({\text{S}}{{\text{O}}_2}{\text{C}}{{\text{l}}_2}\) and the value of \({K_{\text{c}}}\).
Suggest, stating a reason, how the addition of a catalyst at constant pressure and temperature will affect the equilibrium concentration of \({\text{S}}{{\text{O}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}\).
Markscheme
\({\text{rate}} = k{{\text{[NO]}}^2}{\text{[C}}{{\text{l}}_{\text{2}}}{\text{]}}\);
rate of reaction will decrease by a factor of 4;
no effect on the rate constant;
y axis labelled concentration/\({\text{mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\) and x axis is labelled time/s;
gradient for [NO];
gradient for [NOCl] will be equal and opposite;
equilibrium point identified / two curves level off at same time;
Above 775 K: \({\text{rate}} = k{\text{[N}}{{\text{O}}_2}{\text{][CO]}}\);
Below 775 K: \({\text{rate}} = k{{\text{[N}}{{\text{O}}_2}{\text{]}}^2}\);
zero order reaction;
all concentrations are \({\text{1.0 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\);
\({\text{slope}} = \frac{{9.2 - 8.4}}{{(3.53 - 3.65) \times {{10}^{ - 3}}}} = - 6.67 \times {10^3}\);
\(({E_{\text{a}}} = 6.67 \times {10^3} \times 8.31)\)
\({\text{55.4 (kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}}{\text{)}}\);
Accept in range 55.0 – 56.0
Award [1] if 55454 (J) stated
Award [2] for the correct final answer
\(({K_{\text{c}}}) = \frac{{{\text{[S}}{{\text{O}}_2}{\text{C}}{{\text{l}}_2}{\text{]}}}}{{{\text{[C}}{{\text{l}}_2}{\text{][S}}{{\text{O}}_2}{\text{]}}}}\);
Ignore state symbols.
Square brackets [ ] required for the equilibrium expression.
\({\text{7.84}} \times {\text{1}}{{\text{0}}^{ - 3}}{\text{ mol of S}}{{\text{O}}_2}\) and \({\text{7.84}} \times {\text{1}}{{\text{0}}^{ - 3}}{\text{ mol of C}}{{\text{l}}_2}\);
\({\text{7.84}} \times {\text{1}}{{\text{0}}^{ - 3}}{\text{ mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{ of S}}{{\text{O}}_2}\), \({\text{7.84}} \times {\text{1}}{{\text{0}}^{ - 3}}{\text{ mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{ of C}}{{\text{l}}_2}\) and
\({\text{7.65}} \times {\text{1}}{{\text{0}}^{ - 4}}{\text{ mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{ of S}}{{\text{O}}_2}{\text{C}}{{\text{l}}_2}\);
12.5;
Award [1] for 10.34
Award [3] for the correct final answer
value of \({K_{\text{c}}}\) increases;
\({\text{[S}}{{\text{O}}_2}{\text{C}}{{\text{l}}_2}{\text{]}}\) increases;
decrease in temperature favours (forward) reaction which is exothermic;
Do not allow ECF.
no effect on the value of \({K_{\text{c}}}\) / depends only on temperature;
\({\text{[S}}{{\text{O}}_2}{\text{C}}{{\text{l}}_2}{\text{]}}\) decreases;
increase in volume favours the reverse reaction which has more gaseous moles;
Do not allow ECF.
no effect;
catalyst increases the rate of forward and reverse reactions (equally) / catalyst decreases activation energies (equally);
Examiners report
In part (a) the rate expression was correctly stated although some confused this with an equilibrium constant expression.
Only the better candidates realized that the rate of reaction will decrease by a factor of four and there will be no effect on the rate constant.
Although most candidates were able to correctly sketch the concentration versus time graph many forgot to label the axes or include units.
Part (b) was well answered and candidates demonstrated a good understanding of rate expressions based on reaction mechanism.
The better candidates were able to figure out that the rate of a chemical reaction is equal to the rate constant when all concentrations are \({\text{1.0 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\) or for a zero order reaction.
Most candidates had difficulty in calculating activation energy from the graph in part (d) and some gave the answer in \({\text{J}}\,{\text{mo}}{{\text{l}}^{ - 1}}\) instead of \({\text{kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}}\) which showed that they missed this instruction in the question.
In part (e), the equilibrium constant expression was correctly stated by the majority but calculating the value of\({K_{\text{c}}}\) proved to be difficult.
A large number of candidates obtained the incorrect answer of 10.34 as a result of using the initial concentrations of the reactants instead of equilibrium concentrations.
The application of Le Chatelier’s principle was handled well by the majority with minor omissions such as not using the term gaseous particles in part (iv).
Some candidates stated that the addition of a catalyst does not affect the value of \({K_{\text{c}}}\) or the position of equilibrium, which did not answer the question and scored no marks because they had not commented on the concentration of \({\text{SOC}}{{\text{l}}_{\text{2}}}\). Some candidates correctly stated that a catalyst increases the rate of forward and reverse reactions equally.
