CHE 112 Kinetics- Practice Questions and Answers
What This Post Covers
Chemical kinetics is the study of reaction rates and the factors that affect them. CHE 112 students typically struggle with three things: rate law calculations, order determination, and half-life problems. This post cuts through the textbook fluff and gives you actual practice with real question types.
Core Concepts You Need to Know First
Before diving into questions, make sure these concepts are solid. If they are not, go back and review. There is no point practicing problems when the foundation is broken.
Rate Law Basics
The general rate law for a reaction aA + bB → products is:
Rate = k[A]m[B]n
- k = rate constant (depends on temperature)
- m = reaction order with respect to A
- n = reaction order with respect to B
- Overall order = m + n
Integrated Rate Laws
The integrated rate laws tell you how concentration changes over time. This is where most students lose points.
| Order | Rate Law | Integrated Form | Half-Life |
|---|---|---|---|
| Zero | Rate = k | [A]t = [A]₀ - kt | t₁/₂ = [A]₀/2k |
| First | Rate = k[A] | ln[A]t = ln[A]₀ - kt | t₁/₂ = 0.693/k |
| Second | Rate = k[A]² | 1/[A]t = 1/[A]₀ + kt | t₁/₂ = 1/k[A]₀ |
The half-life of a first-order reaction is independent of initial concentration. That is the key distinction professors test on.
Arrhenius Equation
k = Ae-Ea/RT
Or the two-point form:
ln(k₂/k₁) = (Ea/R)(1/T₁ - 1/T₂)
Students consistently forget to convert temperature to Kelvin. Every single time. Do not be that student.
Practice Questions
Multiple Choice
Question 1: For the reaction 2NO(g) + Cl₂(g) → 2NOCl(g), the rate law is found to be: Rate = k[NO]²[Cl₂]. What is the overall order of the reaction?
A) 2 B) 3 C) 4 D) Cannot be determined
Question 2: Which statement about catalysts is correct?
A) They increase the activation energy
B) They are consumed in the reaction
C) They provide an alternative pathway with lower activation energy
D) They shift the equilibrium position
Question 3: The rate constant for a first-order reaction is 0.693 min⁻¹. What is the half-life?
A) 0.5 min B) 1 min C) 0.693 min D) 2 min
Question 4: A reaction has a rate law: Rate = k[A][B]². If the concentration of A is doubled and B is kept constant, the rate will:
A) Double B) Triple C) Quadruple D) Stay the same
Question 5: For the reaction A → products, a student plots ln[A] vs. time and gets a straight line with a negative slope. The reaction is:
A) Zero order B) First order C) Second order D) Third order
Free Response / Problem-Solving
Problem 1: The decomposition of N₂O₅ at 45°C is first-order with k = 6.3 × 10⁻⁴ s⁻¹. If the initial concentration is 0.50 M, calculate:
- The concentration after 500 seconds
- The time required for the concentration to drop to 0.10 M
Problem 2: The reaction A + B → products was studied and the following data was collected:
| Experiment | [A]₀ (M) | [B]₀ (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 |
Determine the rate law and calculate the rate constant k.
Problem 3: The rate constant for a reaction increases from 1.5 × 10⁻³ s⁻¹ at 25°C to 4.5 × 10⁻³ s⁻¹ at 45°C. Calculate the activation energy in kJ/mol. (R = 8.314 J/mol·K)
Answers and Explanations
Multiple Choice Answers
Answer 1: B) 3
The overall order equals m + n. From the rate law Rate = k[NO]²[Cl₂], we have m = 2 and n = 1. 2 + 1 = 3. That is it. No tricks.
Answer 2: C) They provide an alternative pathway with lower activation energy
Everything else is wrong. Catalysts lower activation energy, are not consumed, and do not affect equilibrium position. They only speed up the rate at which equilibrium is reached.
Answer 3: B) 1 min
For first-order reactions: t₁/₂ = 0.693/k = 0.693/0.693 = 1 min. The numbers are chosen to work out cleanly. Notice that 0.693 is ln(2). That is not a coincidence.
Answer 4: A) Double
Only A changes. The rate law is Rate = k[A][B]². Doubling [A] means the rate doubles. Changes in B do not matter because B is held constant in this scenario.
Answer 5: B) First order
ln[A] vs. time gives a straight line only for first-order reactions. Zero order gives [A] vs. time. Second order gives 1/[A] vs. time. If the line has negative slope, it confirms the concentration is decreasing.
Free Response Solutions
Problem 1 Solution:
This is a first-order decomposition. Use the integrated first-order rate law:
ln[A]t = ln[A]₀ - kt
Part (a): Finding concentration after 500 seconds
ln[A]t = ln(0.50) - (6.3 × 10⁻⁴)(500)
ln[A]t = -0.693 - 0.315
ln[A]t = -1.008
[A]t = e⁻¹·⁰⁰⁸ = 0.365 M
Part (b): Finding time to reach 0.10 M
ln(0.10/0.50) = -kt
ln(0.20) = -kt
-1.609 = -(6.3 × 10⁻⁴)t
t = 2550 seconds (about 42.5 minutes)
Problem 2 Solution:
Compare experiments to find the order with respect to each reactant.
Between experiments 1 and 2: [A] doubles, [B] stays constant
Rate doubles (2.0 → 8.0)
Therefore, order with respect to A = 1
Between experiments 1 and 3: [B] doubles, [A] stays constant
Rate stays the same (2.0 → 2.0)
Therefore, order with respect to B = 0
The rate law is: Rate = k[A]
To find k, use any experiment. Using experiment 1:
2.0 × 10⁻⁴ = k(0.10)
k = 2.0 × 10⁻³ s⁻¹
Problem 3 Solution:
Use the two-point Arrhenius equation:
ln(k₂/k₁) = (Ea/R)(1/T₁ - 1/T₂)
Convert temperatures to Kelvin:
T₁ = 25 + 273 = 298 K
T₂ = 45 + 273 = 318 K
Substitute values:
ln(4.5 × 10⁻³ / 1.5 × 10⁻³) = (Ea/8.314)(1/298 - 1/318)
ln(3) = (Ea/8.314)(0.003356 - 0.003145)
1.099 = (Ea/8.314)(0.000211)
1.099 = Ea(2.54 × 10⁻⁵)
Ea = 43,300 J/mol = 43.3 kJ/mol
Getting Started: How to Actually Solve Kinetics Problems
Follow this sequence. Every time. No exceptions.
- Identify the question type. Is it asking for rate, order, half-life, or activation energy? Different question types use different equations.
- Write down what you know. List [A]₀, [A]t, k, t, or whatever the problem gives you. Circle what you need to find.
- Choose the right equation. If it involves time and concentration, use the integrated rate law. If it involves two temperatures, use Arrhenius.
- Plug in numbers. Check units. Convert everything to the same units before calculating.
- Check your answer. Does the result make physical sense? A half-life should be positive. Concentration should decrease over time for decomposition reactions.
The most common mistakes are:
- Using the wrong integrated rate law equation
- Forgetting to convert Celsius to Kelvin for Arrhenius problems
- Mixing up first-order and second-order half-life formulas
- Rounding too early in calculations, which compounds errors
Quick Reference: What to Memorize
- First-order half-life: t₁/₂ = 0.693/k
- First-order integrated law: ln[A]t = ln[A]₀ - kt
- Arrhenius: ln(k₂/k₁) = (Ea/R)(1/T₁ - 1/T₂)
- Rate = k[A]m[B]n
- Overall order = m + n
That is the entire memorization list for most CHE 112 kinetics exams. Everything else you can derive. Work through the problems above until you can solve them without checking the answers. Then find additional practice problems in your textbook or course materials.