IB DP Chemistry Reactivity 2.2 How fast? The rate of chemical change HL Paper 2- Exam Style Questions - New Syllabus
Question
A student investigated the kinetics of the reaction between a dye, \(\mathrm{RCl}\), and aqueous sodium hydroxide. The dye has an intense blue colour that fades during the reaction.
The reaction can be represented by: \[ \mathrm{RCl\ (aq) + OH^{-}\ (aq) \rightarrow ROH\ (aq) + Cl^{-}\ (aq)} \]
Blue \(\rightarrow\) Colourless.
In a trial, the student mixed solutions and measured the time for the colour to disappear. The “calculated rate” was defined as \[ \text{Calculated rate}=\frac{[\mathrm{RCl}]_{\text{initial}}}{\text{time for colour to disappear}}. \]
| Initial concentration / \(\mathrm{mol\ dm^{-3}}\) | Time for colour to disappear / s | Calculated rate / \(\mathrm{mol\ dm^{-3}\ s^{-1}}\) |
|---|---|---|
| \([\mathrm{RCl}] = 3.00\times 10^{-6}\), \([\mathrm{OH^-}] = 2.00\times 10^{-2}\) | 110 | \(2.73\times 10^{-8}\) |
(a) The student then monitored the reaction by spectrophotometry and obtained the following initial-rate data:
| Experiment | Initial \([\mathrm{RCl}]\) / \(\mathrm{mol\ dm^{-3}}\) | Initial \([\mathrm{OH^-}]\) / \(\mathrm{mol\ dm^{-3}}\) | Initial rate / \(\mathrm{mol\ dm^{-3}\ s^{-1}}\) |
|---|---|---|---|
| 1 | \(3.00\times 10^{-6}\) | \(2.00\times 10^{-2}\) | \(3.20\times 10^{-8}\) |
| 2 | \(1.50\times 10^{-6}\) | \(1.00\times 10^{-2}\) | \(8.00\times 10^{-9}\) |
| 3 | (from graph) | \(2.00\times 10^{-2}\) | (from graph) |
A graph of \([\mathrm{RCl}]\) versus time for experiment 3 is provided (not shown here). Use the graph to answer (a)(i).
![Graph of [RCl] vs Time](https://sp-ao.shortpixel.ai/client/to_webp,q_glossy,ret_img,w_545,h_286/https://www.iitianacademy.com/wp-content/uploads/2024/10/Screenshot-2024-10-07-194406.png)
(i) Using the graph, determine the missing values for experiment 3 and justify your answers. [3]
(ii) Deduce the order of the reaction with respect to each reactant. [2]
(iii) Calculate the rate constant, with units, at this temperature using the data from experiment 1. [2]
(b) The student repeated the experiment at a higher temperature. Explain why the rate increased. [2]
▶️ Answer/Explanation
Markscheme
(a)(i)
From the \([\mathrm{RCl}]\)-vs-time graph for experiment 3:
• Initial \([\mathrm{RCl}]\) = \(y\)-intercept \(\Rightarrow\) \(\boxed{6.00\times 10^{-6}\ \mathrm{mol\ dm^{-3}}}\).
• Initial rate = magnitude of tangent at \(t=0\): \[ \text{rate}=\left.-\frac{d[\mathrm{RCl}]}{dt}\right|_{t=0} \approx \boxed{6.4\times 10^{-8}\ \mathrm{mol\ dm^{-3}\ s^{-1}}}. \] [3 marks]
From the \([\mathrm{RCl}]\)-vs-time graph for experiment 3:
• Initial \([\mathrm{RCl}]\) = \(y\)-intercept \(\Rightarrow\) \(\boxed{6.00\times 10^{-6}\ \mathrm{mol\ dm^{-3}}}\).
• Initial rate = magnitude of tangent at \(t=0\): \[ \text{rate}=\left.-\frac{d[\mathrm{RCl}]}{dt}\right|_{t=0} \approx \boxed{6.4\times 10^{-8}\ \mathrm{mol\ dm^{-3}\ s^{-1}}}. \] [3 marks]
(a)(ii)
Assume \(\text{rate}=k[\mathrm{RCl}]^{m}[\mathrm{OH^-}]^{n}\).
Compare exp 1 vs 2 (both halved): \[ \frac{\text{rate}_2}{\text{rate}_1}=0.25=\left(\tfrac{1}{2}\right)^{m+n}\Rightarrow m+n=2. \] Compare exp 1 vs 3 (same \([\mathrm{OH^-}]\)): \[ \frac{[\mathrm{RCl}]_3}{[\mathrm{RCl}]_1}=2,\quad \frac{\text{rate}_3}{\text{rate}_1}=2 \Rightarrow 2=2^{\,m}\Rightarrow m=1. \] Hence \(n=1\).
Orders: first order in \(\mathrm{RCl}\), first order in \(\mathrm{OH^-}\) (overall second order). [2 marks]
Assume \(\text{rate}=k[\mathrm{RCl}]^{m}[\mathrm{OH^-}]^{n}\).
Compare exp 1 vs 2 (both halved): \[ \frac{\text{rate}_2}{\text{rate}_1}=0.25=\left(\tfrac{1}{2}\right)^{m+n}\Rightarrow m+n=2. \] Compare exp 1 vs 3 (same \([\mathrm{OH^-}]\)): \[ \frac{[\mathrm{RCl}]_3}{[\mathrm{RCl}]_1}=2,\quad \frac{\text{rate}_3}{\text{rate}_1}=2 \Rightarrow 2=2^{\,m}\Rightarrow m=1. \] Hence \(n=1\).
Orders: first order in \(\mathrm{RCl}\), first order in \(\mathrm{OH^-}\) (overall second order). [2 marks]
(a)(iii)
Using exp 1 and \(\text{rate}=k[\mathrm{RCl}][\mathrm{OH^-}]\): \[ k=\frac{3.20\times 10^{-8}}{(3.00\times 10^{-6})(2.00\times 10^{-2})} =\frac{3.20\times 10^{-8}}{6.00\times 10^{-8}} =0.533\ \mathrm{dm^{3}\ mol^{-1}\ s^{-1}}. \] Units: \(\mathrm{(mol\,dm^{-3}\,s^{-1})/(mol\,dm^{-3})^2=dm^{3}\,mol^{-1}\,s^{-1}}\).
\(\boxed{k=0.533\ \mathrm{dm^{3}\ mol^{-1}\ s^{-1}}}\). [2 marks]
Using exp 1 and \(\text{rate}=k[\mathrm{RCl}][\mathrm{OH^-}]\): \[ k=\frac{3.20\times 10^{-8}}{(3.00\times 10^{-6})(2.00\times 10^{-2})} =\frac{3.20\times 10^{-8}}{6.00\times 10^{-8}} =0.533\ \mathrm{dm^{3}\ mol^{-1}\ s^{-1}}. \] Units: \(\mathrm{(mol\,dm^{-3}\,s^{-1})/(mol\,dm^{-3})^2=dm^{3}\,mol^{-1}\,s^{-1}}\).
\(\boxed{k=0.533\ \mathrm{dm^{3}\ mol^{-1}\ s^{-1}}}\). [2 marks]
(b)
Higher temperature:
• Greater fraction of molecules with \(E \ge E_a\) (Maxwell–Boltzmann shift).
• More frequent/successful collisions \(\Rightarrow\) higher rate. [2 marks]
Higher temperature:
• Greater fraction of molecules with \(E \ge E_a\) (Maxwell–Boltzmann shift).
• More frequent/successful collisions \(\Rightarrow\) higher rate. [2 marks]
Total Marks: 9