Second-Order Kinetics in a Substitution Reaction
The phrase “the following substratituion reaction exhibits second order kinetics” is modeled as a typical bimolecular substitution (SN2-type) process where the rate depends on two reacting species.
Problem
Consider the substitution reaction: CH3Br(aq) + OH−(aq) → CH3OH(aq) + Br−(aq). Initial-rate data are measured at constant temperature.
| Experiment | [CH3Br]0 (M) | [OH−]0 (M) | Initial rate (M·s−1) |
|---|---|---|---|
| 1 | 0.050 | 0.050 | 1.25 × 10−4 |
| 2 | 0.100 | 0.050 | 2.50 × 10−4 |
| 3 | 0.050 | 0.100 | 2.50 × 10−4 |
Tasks: (1) Determine the rate law and overall reaction order. (2) Calculate the rate constant \(k\) with units. (3) Assuming equal starting concentrations \([CH3Br]_0 = [OH^-]_0 = 0.050\,\text{M}\), find the half-life and the time for \([CH3Br]\) to drop to \(0.010\,\text{M}\).
Step 1: Determine the reaction orders (method of initial rates)
Assume a general rate law:
\[ \text{rate} = k\,[\text{CH}_3\text{Br}]^{m}\,[\text{OH}^-]^{n} \]
Compare Experiments 1 and 2 (only [CH3Br] changes):
\[ \frac{\text{rate}_2}{\text{rate}_1} = \frac{2.50 \times 10^{-4}}{1.25 \times 10^{-4}} = 2 = \left(\frac{0.100}{0.050}\right)^{m} = 2^{m} \Rightarrow m = 1 \]
Compare Experiments 1 and 3 (only [OH−] changes):
\[ \frac{\text{rate}_3}{\text{rate}_1} = \frac{2.50 \times 10^{-4}}{1.25 \times 10^{-4}} = 2 = \left(\frac{0.100}{0.050}\right)^{n} = 2^{n} \Rightarrow n = 1 \]
Therefore, the rate law is first order in each reactant and second order overall:
\[ \text{rate} = k\,[\text{CH}_3\text{Br}]\,[\text{OH}^-] \quad\text{(overall order }1 + 1 = 2\text{)} \]
Step 2: Compute the rate constant \(k\)
Use Experiment 1:
\[ k = \frac{\text{rate}}{[\text{CH}_3\text{Br}]\,[\text{OH}^-]} = \frac{1.25 \times 10^{-4}}{(0.050)\times(0.050)} = \frac{1.25 \times 10^{-4}}{2.50 \times 10^{-3}} = 5.0 \times 10^{-2} \]
\[ k = 5.0 \times 10^{-2}\ \text{M}^{-1}\text{s}^{-1} \]
Step 3: Integrated form when \([A]_0=[B]_0\)
With 1:1 stoichiometry and equal initial concentrations, \([\text{CH}_3\text{Br}] = [\text{OH}^-] = [A]\) at all times, so:
\[ \text{rate} = k[A]^2 \]
The integrated second-order law becomes:
\[ \frac{1}{[A]_t} = \frac{1}{[A]_0} + k t \]
Step 4: Half-life for a second-order process
For \(\text{rate} = k[A]^2\), the half-life is:
\[ t_{1/2} = \frac{1}{k[A]_0} \]
\[ t_{1/2} = \frac{1}{(5.0 \times 10^{-2})\times(0.050)} = \frac{1}{2.5 \times 10^{-3}} = 4.00 \times 10^{2}\ \text{s} \]
So \(t_{1/2} = 400\ \text{s}\) (about \(6.7\ \text{min}\)).
Step 5: Time to reach \([A]_t = 0.010\,\text{M}\)
\[ t = \frac{\frac{1}{[A]_t} - \frac{1}{[A]_0}}{k} = \frac{\frac{1}{0.010} - \frac{1}{0.050}}{5.0 \times 10^{-2}} = \frac{100 - 20}{5.0 \times 10^{-2}} = \frac{80}{5.0 \times 10^{-2}} = 1.60 \times 10^{3}\ \text{s} \]
Time required: \(t = 1600\ \text{s}\) (about \(26.7\ \text{min}\)).