Chemical Kinetics Practice Problems with Answers
Chemical Kinetics Practice Problems with Answers
If you're studying reaction rates and need to actually understand chemical kinetics—not just memorize formulas—this is for you. I've put together real practice problems with working solutions. No fluff.
What You Need to Know First
Before touching these problems, make sure you're solid on these concepts:
- Rate laws – how rate depends on concentration
- Order of reaction – what happens when you double a reactant
- Half-life equations – especially for first-order reactions
- Activation energy – Arrhenius equation
- Reaction mechanisms – rate-determining step
If these terms don't ring bells, go back and review. These problems assume you have the basics down.
Rate Law Problems
Problem 1: Determining Rate Law from Experimental Data
Question: The reaction A + B → Products was studied. Initial rates were measured:
| Experiment | [A] (M) | [B] (M) | Initial Rate (M/s) |
|---|---|---|---|
| 1 | 0.10 | 0.10 | 2.0 × 10⁻⁴ |
| 2 | 0.20 | 0.10 | 8.0 × 10⁻⁴ |
| 3 | 0.10 | 0.20 | 2.0 × 10⁻⁴ |
Find: The rate law and calculate the rate constant.
Solution:
Compare experiments 1 and 2: [A] doubles, [B] stays same, rate goes from 2.0 to 8.0 (4× faster).
That means the reaction is second order in A. The exponent is 2.
Compare experiments 1 and 3: [B] doubles, [A] stays same, rate stays the same.
That means zero order in B. The exponent is 0.
The rate law is:
Rate = k[A]²[B]⁰ = k[A]²
Now solve for k using experiment 1:
2.0 × 10⁻⁴ = k(0.10)²
k = 2.0 × 10⁻⁴ / 0.01 = 0.020 M⁻¹s⁻¹
Problem 2: Using the Rate Law
Question: For the reaction 2NO + Cl₂ → 2NOCl, the rate law is Rate = k[NO]²[Cl₂].
If [NO] decreases from 0.050 M to 0.025 M, by what factor does the rate change?
Solution:
[NO] is squared in the rate law. Halving [NO] means the rate changes by (1/2)² = 1/4.
The rate becomes 1/4 of the original.
Half-Life Problems
Problem 3: First-Order Half-Life Calculation
Question: The decomposition of N₂O₅ has a half-life of 4.0 minutes at 25°C.
a) What is the rate constant?
b) How much N₂O₅ remains after 12 minutes if you start with 2.0 g?
Solution:
Part a:
For first-order reactions:
t½ = 0.693/k
4.0 min = 0.693/k
k = 0.693/4.0 = 0.173 min⁻¹
Part b:
12 minutes = 3 half-lives
After 1st half-life: 2.0 g → 1.0 g
After 2nd half-life: 1.0 g → 0.50 g
After 3rd half-life: 0.50 g → 0.25 g
Or using the first-order equation: [A] = [A]₀ × (1/2)ⁿ where n = 12/4 = 3
[A] = 2.0 × (1/2)³ = 2.0 × 0.125 = 0.25 g
Problem 4: Zero-Order Half-Life
Question: A zero-order reaction has k = 1.5 × 10⁻³ M/s and an initial concentration of 0.50 M.
Calculate the half-life.
Solution:
For zero-order reactions:
t½ = [A]₀ / 2k
t½ = 0.50 / (2 × 1.5 × 10⁻³)
t½ = 0.50 / 0.0030 = 167 seconds
Zero-order half-life depends on initial concentration. That's the key difference from first-order reactions.
Arrhenius Equation Problems
Problem 5: Calculating Activation Energy
Question: A reaction has rate constants of 3.0 × 10⁻² s⁻¹ at 300 K and 9.0 × 10⁻² s⁻¹ at 320 K.
Calculate the activation energy (Ea).
Solution:
Use the two-point form of Arrhenius:
ln(k₂/k₁) = Ea/R × (1/T₁ - 1/T₂)
ln(9.0 × 10⁻² / 3.0 × 10⁻²) = Ea/8.314 × (1/300 - 1/320)
ln(3.0) = Ea/8.314 × (0.00333 - 0.00313)
1.099 = Ea/8.314 × 0.00020
1.099 = Ea × 2.41 × 10⁻⁵
Ea = 1.099 / 2.41 × 10⁻⁵
Ea = 45,600 J/mol = 45.6 kJ/mol
Reaction Mechanism Problems
Problem 6: Deriving Rate Law from Mechanism
Question: The proposed mechanism for a reaction is:
Step 1: 2NO ⇌ N₂O₂ (fast equilibrium)
Step 2: N₂O₂ + H₂ → N₂O + H₂O (slow)
Find: The rate law for the overall reaction.
Solution:
The slow step is the rate-determining step.
Rate = k₂[N₂O₂][H₂]
But N₂O₂ is an intermediate. Eliminate it using the fast equilibrium from Step 1.
For the equilibrium: k₁[NO]² = k₋₁[N₂O₂]
N₂O₂ = k₁[NO]² / k₋₁ = K[NO]²
Substitute into the rate law:
Rate = k₂ × K[NO]² × [H₂]
Rate = k[NO]²[H₂]
Common Mistakes to Avoid
- Confusing order with stoichiometry – they're not the same. You determine order experimentally.
- Using first-order half-life formula for zero-order reactions – or vice versa. Know which equation applies.
- Forgetting units – rate constants have units. Zero-order: M/s, first-order: s⁻¹, second-order: M⁻¹s⁻¹.
- Temperature in Kelvin only – don't mix Celsius and Kelvin in Arrhenius calculations.
- Solving for the wrong variable – read the question twice before plugging in numbers.
Quick Reference: Half-Life Equations
| Reaction Order | Half-Life Equation | Half-Life Depends on [A]₀? |
|---|---|---|
| Zero-order | t½ = [A]₀ / 2k | Yes |
| First-order | t½ = 0.693 / k | No |
| Second-order (1:1) | t½ = 1 / k[A]₀ | Yes |
How to Approach Any Kinetics Problem
- Identify what information is given – concentrations, rate constants, time, temperature.
- Determine the reaction order if not given – look for patterns when concentrations change.
- Pick the right equation – rate law, half-life, or Arrhenius. Don't force a formula that doesn't fit.
- Check your units – make sure everything's consistent before calculating.
- Ask yourself if the answer makes sense – doubling concentration should increase rate (or not), half-life should be positive.
Final Thoughts
These problems cover the core concepts you'll encounter. Work through each one until the method clicks, not just the answer. Kinetics problems follow patterns—once you see the pattern, you can solve any variation.
If you're still struggling, focus on rate law determination first. That's where most students lose points, and it's the foundation for everything else.