Chemical Kinetics – SRMJEEE Study Notes
Goal: Understand how reaction rates are quantified, how to derive rate laws, and how temperature and catalysts influence the speed of a reaction.
1. Simple Explanation
Chemical kinetics studies how fast a reaction proceeds and what influences that speed.
- Rate of reaction: change in concentration of a reactant or product per unit time.
- Rate law: mathematical relationship that links the reaction rate to the concentrations of reactants.
- Rate constant (k): a proportionality factor that depends on temperature, catalysts, etc.
- Activation energy (Ea): energy barrier that must be overcome for the reaction to proceed.
Key Idea: A reaction’s speed is not only dictated by the stoichiometry but also by the microscopic details (energy barriers, collisions, catalysts).
2. Key Points
| Concept | What it Means | How to Write |
|---|---|---|
| Rate of Reaction | (\displaystyle \text{rate} = -\frac{1}{\nu_i}\frac{d[\mathrm{A}_i]}{dt}) (for reactant) or (+\frac{1}{\nu_j}\frac{d[\mathrm{B}_j]}{dt}) (for product) | (\displaystyle \text{rate} = k[\mathrm{A}]^x[\mathrm{B}]^y) |
| Rate Law | Gives the functional dependence of the rate on reactant concentrations. | ( \text{rate} = k[\mathrm{A}]^m[\mathrm{B}]^n ) |
| Reaction Order | Sum of the exponents in the rate law. | ( \text{Overall order} = m+n ) |
| Integrated Rate Laws | Relate concentration directly to time. | • Zero‑order: ([A] = [A]_0 - kt) • First‑order: (\ln [A] = \ln [A]_0 - kt) • Second‑order (A₂): (\frac{1}{[A]} = \frac{1}{[A]_0} + kt) |
| Rate Constant (k) | Temperature‑dependent parameter. | Units vary: zero‑order (M s⁻¹), first‑order (s⁻¹), second‑order (M⁻¹s⁻¹) |
| Activation Energy (Ea) | Energy barrier that must be surpassed. | (\displaystyle k = A e^{-E_a/(RT)}) (Arrhenius) |
| Arrhenius Equation | Relates k to T. | ( \ln k = \ln A - \frac{E_a}{RT} ) |
| Catalyst | Lowers Ea, increases k, no net change in ΔG°. | Usually appears as a surface in heterogeneous catalysis. |
| Reaction Mechanism | Sequence of elementary steps that add up to the overall reaction. | Each elementary step has its own rate law. |
| Heterogeneous Reaction | Reactants in different phases (e.g., gas + solid). | Rate limited by surface area, diffusion, etc. |
3. Worked Example
Problem
A reaction ( \mathrm{2A \rightarrow B} ) is second‑order in A.
Initial concentration ([A]_0 = 0.10,\text{M}).
Rate constant (k = 5.0 \times 10^{2},\text{M}^{-1}\text{s}^{-1}).
Find
- The concentration of A after (t = 2.0,\text{s}).
- The time needed for the concentration of A to drop to (0.025,\text{M}).
Solution
Step 1 – Identify the integrated rate law.
Second‑order (overall) → (\displaystyle \frac{1}{[A]} = \frac{1}{[A]_0} + kt).
1. Concentration after 2 s
[ \begin{aligned} \frac{1}{[A]_{2}} &= \frac{1}{0.10} + (5.0\times10^{2})(2.0) \ &= 10 + 1000 \ &= 1010 \ \Rightarrow [A]_2 &= \frac{1}{1010} \approx 9.90\times10^{-4},\text{M} \end{aligned} ]
2. Time for ([A] = 0.025,\text{M})
[ \begin{aligned} \frac{1}{0.025} &= \frac{1}{0.10} + k t \ 40 &= 10 + (5.0\times10^{2}) t \ 30 &= 500 t \ t &= \frac{30}{500} = 0.060,\text{s} \end{aligned} ]
Answer
- After 2 s, ([A] \approx 9.9 \times 10^{-4},\text{M}).
- To reach 0.025 M, 0.060 s are required.
4. Quick Revision Summary
| Topic | Take‑away |
|---|---|
| Rate of Reaction | ( \text{rate |