Reaction kinetics: practice problems with step-by-step solutions

Free Reaction kinetics practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: first order, conversion and second order, Arrhenius.

Free practice problemEasy
t (min)C/C₀t½​ = 5 min0.5

A first-order reaction has a half-life of 5 min. Find its rate constant k.

t½ = 5 min
min⁻¹
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What you practise

  1. Easyfirst order
  2. Mediumconversion and second order
  3. HardArrhenius

Key formulas

First order
ln(CA/CA0) = −kt
Half-life (first order)
t½ = ln 2/k
Second order
1/CA = 1/CA0 + kt
Arrhenius
k = A·e−Ea/(RT)
Two temperatures
ln(k2/k1) = (Ea/R)(1/T1 − 1/T2)

Learn Reaction kinetics from scratch

1What reaction kinetics is

Chemical reactions take time. Kinetics is about how fast a reaction goes, not how far. It tells you how big a reactor must be and how long a batch must run.

For a reaction A → products, the rate is how fast the concentration of A drops, in mol/(L·min). The rate usually depends on how much A is left: fewer molecules, fewer collisions, slower reaction.

The simplest case is a first-order reaction, where the rate is proportional to the concentration CA:

−dCA/dt = k · CA

The rate constant k measures how fast the reaction is. For first order its unit is 1/time, for example min⁻¹. A larger k means a faster reaction.

Self-check The concentration of A in a first-order reaction is doubled. What happens to the rate?
  1. It becomes four times larger
  2. It doubles
  3. It halves
  4. It stays the same

Answer: It doubles. First order means rate = k·C. Double C and the rate doubles. Four times would be second order, where the rate goes with C².

2First-order reactions

Solving the rate equation gives an exponential decay. The concentration after a time t is:

CA = CA0 · e−k·t or ln(CA/CA0) = −k·t
t (min)C (mol/L)C₀C₀/2t½2t½

The half-life t½ is the time to lose half of A. For first order it does not depend on where you start: every half-life halves what is left.

t½ = ln 2 / k = 0.693 / k

The conversion X is the fraction of A that has reacted, so CA = CA0·(1 − X). The time to reach a conversion X is:

t = −ln(1 − X) / k
Self-check A first-order reaction has k = 0.1 min⁻¹. What is its half-life?
  1. 0.0693 min
  2. 10 min
  3. 5 min
  4. 6.93 min

Answer: 6.93 min. t½ = ln 2/k = 0.693/0.1 = 6.93 min. 10 min is 1/k, the time to fall to e⁻¹ = 37%, not to 50%.

3Second order and temperature

In a second-order reaction the rate goes with the square of the concentration: −dCA/dt = k·CA². Then k has units like L/(mol·min), and:

1/CA = 1/CA0 + k·t

Reactions speed up when it gets warmer. The Arrhenius equation describes how k depends on the absolute temperature T:

k = A · e−Ea/(R·T)

Ea is the activation energy in J/mol and R = 8.314 J/(mol·K). With k known at one temperature, divide the equations for two temperatures:

ln(k2/k1) = (Ea/R) · (1/T1 − 1/T2)

Use kelvin, and Ea in J/mol (kJ/mol × 1000).

Self-check A second-order reaction has k = 0.5 L/(mol·min) and C₀ = 1 mol/L. What is C after 2 min?
  1. 0 mol/L
  2. 0.368 mol/L
  3. 0.5 mol/L
  4. 2 mol/L

Answer: 0.5 mol/L. 1/C = 1/1 + 0.5·2 = 2 L/mol, so C = 0.5 mol/L. Using the first-order formula gives e⁻¹ = 0.368, and forgetting to take 1/2 at the end gives 2.

4The method step by step

  1. Find the order from the question: the units of k or the rate law (k·C or k·C²) tell you.
  2. Write down k, CA0 and the time or conversion you know. Check that k and t use the same time unit.
  3. First order: CA = CA0·e−kt, t½ = ln 2/k, or t = −ln(1 − X)/k.
  4. Second order: 1/CA = 1/CA0 + kt, then take the reciprocal.
  5. Temperature: convert both to kelvin and use the two-temperature Arrhenius equation.
  6. Check: the concentration must fall with time and stay above zero.

5A worked example

Follow every step. Try to predict the next one before you open it.

Worked example

A first-order reaction A → products has k = 0.05 min⁻¹. The initial concentration of A is 1.2 mol/L. Find the concentration CA after 10 min.

1

Picture the decay

t (min)C (mol/L)1.2C = ?10

First order, so the concentration falls exponentially from 1.2 mol/L.

2

The exponent

k and t are both in minutes, so they can be multiplied directly.

k · t = 0.05 · 10 = 0.5
3

The decay factor

e−0.5 = 0.6065

So 60.65% of A is still there after 10 min.

4

Concentration

CA = 1.2 · 0.6065 = 0.728 mol/L
5

Check the answer

The half-life is 0.693/0.05 = 13.86 min. After 10 min, less than one half-life has passed, so C must lie between 0.6 and 1.2 mol/L. 0.728 mol/L fits.

Answer

CA ≈ 0.728 mol/L after 10 min.

6Common mistakes

  • Using ln(X) or ln(0.9) for 90% conversion. What is left is 1 − X = 0.1, so t = −ln(0.1)/k.
  • Using log₁₀ instead of the natural logarithm ln.
  • Using the first-order formula for a second-order reaction.
  • Arrhenius with °C, or with Ea in kJ/mol next to R = 8.314 J/(mol·K).
  • Mixing time units: k in min⁻¹ with t in seconds.
Self-check A first-order batch reaction has k = 0.1 min⁻¹. How long does it take to reach 90% conversion?
  1. 1.05 min
  2. 10 min
  3. 23.0 min
  4. 9 min

Answer: 23.0 min. 90% converted means 10% is left: t = −ln(0.1)/0.1 = 23.0 min. Using ln(0.9) gives 1.05 min, which is the time for only 10% conversion.

7How exam questions look

  • Easy: first order. Find the concentration after a time, the half-life, or k from a half-life.
  • Medium: the time for a given conversion, or a second-order reaction with 1/C = 1/C0 + kt.
  • Hard: Arrhenius. Find k at a new temperature, or the activation energy from two rate constants.

Arrhenius example: k1 = 0.02 min⁻¹ at 40 °C, Ea = 40 kJ/mol. At 70 °C: ln(k2/k1) = (40 000/8.314)·(1/313.15 − 1/343.15) = 1.343.

k2 = 0.02 · e1.343 = 0.0766 min⁻¹

Learn it step by step in the appLearn it step by step in the app

Worked examples

Example 1Easy
t (min)C (mol/L)60 minC = ?

A zero-order reaction A → products has a rate constant of 0.01 mol/(L·min). The initial concentration of A is 0.8 mol/L. Find the concentration CA after 60 min.

k = 0.01 mol/(L·min)CA0 = 0.8 mol/Lt = 60 min

Find CA (mol/L)

Show hints
  1. Zero order: the rate −dC/dt = k does not depend on the concentration.
  2. Integrating gives C = C₀ − kt (valid until A runs out).
  3. kt = 0.6 mol/L.
Show the answer and the step-by-step solution

Integrated rate laws: zero order: C = C₀ − kt (straight line, t½ = C₀/(2k)); first order: ln(C₀/C) = kt (t½ = ln 2/k); second order: 1/C − 1/C₀ = kt (t½ = 1/(kC₀)). Conversion X means C = C₀(1 − X).

Integrated zero-order law

CA = CA0 − kt = 0.8 − 0.01·60 = 0.2 mol/L

Answer

CA = 0.2 mol/L
Example 2Medium
t (min)X (%)t = ?99%

A first-order reaction in a batch reactor has k = 0.02 min⁻¹. How long does it take to reach 99% conversion? Find t.

k = 0.02 min⁻¹X = 99 %

Find t (min)

Show hints
  1. Conversion X means C = C₀(1 − X).
  2. Substitute into C = C₀·e^(−kt): 1 − X = e^(−kt), so t = −ln(1 − X)/k.
  3. 1 − X = 0.01.
Show the answer and the step-by-step solution

Reaction rates: for a first-order reaction −dC/dt = kC, so C = C₀·e−kt and the half-life ln 2/k does not depend on concentration. For a second-order reaction 1/C = 1/C₀ + kt. The rate constant rises with temperature following Arrhenius: k = A·e−Ea/(RT).

Conversion and concentration

C/C0 = 1 − X = 0.01

Solve for the time

t = −ln(1 − X)/k = −ln(0.01)/0.02 = 230.3 min

Answer

t = 230.3 min
Example 3Hard
Tk30 °C0.170 °C?

A reaction has a rate constant of 0.1 min⁻¹ at 30 °C and an activation energy of 120 kJ/mol. Find the rate constant k2 at 70 °C.

k1 = 0.1 min⁻¹T1 = 30 °CT2 = 70 °CEa = 120 kJ/molR = 8.314 J/(mol·K)

Find k2 (min⁻¹)

Show hints
  1. Write Arrhenius at both temperatures and divide: ln(k₂/k₁) = −(Ea/R)·(1/T₂ − 1/T₁).
  2. Use kelvin, and Ea in J/mol so it matches R = 8.314 J/(mol·K).
  3. 1/T2 − 1/T1 = -3.845e-4 K⁻¹.
Show the answer and the step-by-step solution

Reaction rates: for a first-order reaction −dC/dt = kC, so C = C₀·e−kt and the half-life ln 2/k does not depend on concentration. For a second-order reaction 1/C = 1/C₀ + kt. The rate constant rises with temperature following Arrhenius: k = A·e−Ea/(RT).

Absolute temperatures

T1 = 303.1 K, T2 = 343.1 K

Arrhenius ratio

ln(k2/k1) = −(120000/8.314)·(1/343.1 − 1/303.1) = 5.55
k2 = 0.1·e5.55 = 25.72 min⁻¹

Answer

k2 = 25.72 min⁻¹

Questions and answers

What is Reaction kinetics?

Chemical reactions take time. Kinetics is about how fast a reaction goes, not how far. It tells you how big a reactor must be and how long a batch must run.

For a reaction A → products, the rate is how fast the concentration of A drops, in mol/(L·min). The rate usually depends on how much A is left: fewer molecules, fewer collisions, slower reaction.

How do I solve Reaction kinetics problems step by step?

Integrated rate laws: zero order: C = C₀ − kt (straight line, t½ = C₀/(2k)); first order: ln(C₀/C) = kt (t½ = ln 2/k); second order: 1/C − 1/C₀ = kt (t½ = 1/(kC₀)). Conversion X means C = C₀(1 − X).

What are common mistakes in Reaction kinetics?
  • That is the first-order law. Zero order means the rate is constant, so C falls linearly: C = C₀ − kt.
  • That is the amount that has reacted. Subtract it from the starting concentration.
What do the Reaction kinetics exercises cover?
  • Easy: first order
  • Medium: conversion and second order
  • Hard: Arrhenius

Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.

How practising works

  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Your progress, mistakes and saved problems are kept, so you can review before the exam.

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Printable worksheet

Reaction kinetics worksheet with answers
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