Chemical Kinetics
1. Introduction
Chemical kinetics studies the rates of chemical reactions and the factors affecting them. It helps predict how quickly reactions proceed and the mechanisms by which they occur.
2. Rate of Reaction
Rate = -d[R]/dt = +d[P]/dt (for a reaction R → P).
Instantaneous rate: slope of concentration vs time curve at a given instant.
2.1 Factors Affecting Rate
- Concentration of reactants
- Temperature
- Presence of catalyst
- Surface area (for solids)
- Nature of reactants
3. Rate Law and Order
For a reaction aA + bB → products: Rate = k[A]ˣ[B]ʸ
Order = x + y (determined experimentally) Molecularity = number of molecules colliding in the elementary step (theoretical)
'Order can be zero, fractional, or negative. Molecularity is always a positive integer. Order is experimental, molecularity is theoretical.'
4. Integrated Rate Equations
4.1 Zero Order
Rate = k. [R] = [R]₀ - kt. t_{1/2} = [R]₀/2k.
4.2 First Order
Rate = k[R]. [R] = [R]₀ e^{-kt}. k = (2.303/t) log([R]₀/[R]). t_{1/2} = 0.693/k (independent of initial concentration).
4.3 Second Order
t_{1/2} ∝ 1/[R]₀.
5. Temperature Dependence of Reaction Rates
5.1 Arrhenius Equation
k = Ae^{-E_a/RT}, where A is the pre-exponential factor and E_a is activation energy.
In logarithmic form: log k = log A - E_a/(2.303 RT).
For two temperatures: log(k₂/k₁) = (E_a/2.303R)(1/T₁ - 1/T₂).
6. Catalysis
A catalyst increases the rate of reaction by providing an alternative path with lower activation energy. It does not affect ΔH or equilibrium constant.
7. Worked Problems
Problem 1: The half-life of a first-order reaction is 100 s. How long will it take for 90% completion? Solution: k = 0.693/100 = 0.00693 s^{-1}. For 90%: [R]₀/[R] = 100/10 = 10. t = (2.303/0.00693) log(10) = 332.2 × 1 = 332.2 s.
Problem 2: A first-order reaction is 50% complete in 20 min. Calculate the time for 75% completion. Solution: t_{1/2} = 20 min. For 75%, two half-lives are needed: t = 40 min.
Problem 3: The rate constant doubles when temperature increases from 300K to 310K. Find E_a. Solution: log(2) = (E_a/2.303×8.314)(1/300 - 1/310). 0.3010 = (E_a/19.147)(10/93000). E_a = 0.3010×19.147×9300 = 53.6 kJ/mol.
8. Common Mistakes
'Students often confuse the order of a reaction with molecularity. Order is determined experimentally and can be fractional. Molecularity is theoretical and must be a whole number.'
9. ISC Exam Focus
| Topic | Theory Marks | Practical Marks |
|---|---|---|
| Rate law and order | 3 | 2 |
| Integrated rate equations | 4 | 2 |
| Half-life | 3 | 2 |
| Arrhenius equation | 3 | 2 |
10. Self-Test Questions
- The rate of a reaction doubles when concentration is doubled. What is the order?
- Derive the integrated rate equation for a first-order reaction.
- The half-life of a first-order reaction is 50 min. How much of a 10 g sample remains after 150 min?
- Define activation energy. How does a catalyst affect it?
- The rate constant at 400K is 1.2×10^{-3} and at 450K it is 4.8×10^{-3}. Calculate E_a.
11. Additional Worked Problems for ISC Practice
Problem A: For a first-order reaction, 10% of the reactant decomposes in 30 minutes. Find the half-life period.
Solution: k = (2.303/t) log([R]₀/[R]) = (2.303/30) log(100/90) = (2.303/30) × 0.0458 = 0.00351 min^{-1}. t_{1/2} = 0.693/k = 0.693/0.00351 = 197.4 min.
Problem B: The rate constant for a reaction is 2.5×10^{-3} s^{-1} at 300K and 1.0×10^{-2} s^{-1} at 320K. Calculate E_a.
Solution: log(k₂/k₁) = (E_a/2.303R)(1/T₁ - 1/T₂). log(1.0×10^{-2}/2.5×10^{-3}) = log(4) = 0.6021. 0.6021 = (E_a/2.303×8.314)(1/300 - 1/320) = (E_a/19.147)(20/96000). E_a = 0.6021 × 19.147 × 4800 = 55.3 kJ/mol.
Problem C: For a zero-order reaction, the concentration decreases from 0.5 M to 0.3 M in 20 minutes. Find the rate constant and the half-life.
Solution: For zero order, k = ([R]₀ - [R])/t = (0.5 - 0.3)/20 = 0.01 M min^{-1}. t_{1/2} = [R]₀/2k = 0.5/(2×0.01) = 25 min.
Problem D: Show that for a first-order reaction, the time required for 99.9% completion is about 10 times the half-life.
Solution: For 99.9% completion, [R]₀/[R] = 1000. t = (2.303/k) log(1000) = (2.303/k) × 3 = 6.909/k. t_{1/2} = 0.693/k. Ratio = 6.909/0.693 = 9.97 ≈ 10. Hence t(99.9%) ≈ 10 × t_{1/2}.
12. Key Equations Summary
| Reaction Order | Rate Law | Integrated Equation | Half-life | Units of k |
|---|---|---|---|---|
| Zero | Rate = k | [R] = [R]₀ - kt | [R]₀/2k | mol L^{-1} s^{-1} |
| First | Rate = k[R] | [R] = [R]₀e^{-kt} | 0.693/k | s^{-1} |
| Second | Rate = k[R]² | 1/[R] = 1/[R]₀ + kt | 1/k[R]₀ | L mol^{-1} s^{-1} |
| Arrhenius | - | k = Ae^{-E_a/RT} | - | - |
13. ISC Examination Strategy
- For numerical problems, first identify the order of the reaction from given data.
- Use the appropriate integrated rate equation and solve for the unknown.
- Pay attention to units — time must be consistent throughout.
- For Arrhenius problems, remember to convert E_a to J/mol (not kJ) when using R = 8.314 J/mol·K.
- When using the half-life method, remember that for a first-order reaction, t_{1/2} is independent of initial concentration.
- Common ISC questions ask to calculate the order from experimental data using the initial rate method or half-life variation.
