RG's E-Learning Centre · Book a free demo class · WhatsApp +91-9379416374

Arrhenius equation: Class 12 practice

k = A e^(−Ea/RT). Read the method, follow the three worked examples, then try the six questions.

What you should know first

  • Logarithms
  • Fractions

The idea behind arrhenius equation

k = A e^(−Ea/RT). Comparing two temperatures: log(k₂/k₁) = (Ea/2.303R)(1/T₁ − 1/T₂), with R = 8.314 J/mol K. A rise in temperature increases k sharply.

Step-by-step method

  1. Write T in kelvin.
  2. Use the two-temperature form.
  3. Solve for Ea (J/mol) or k₂.

Worked example 1 (Basic)

A reaction has Ea = 54 kJ/mol. By what factor does k increase from 298 K to 308 K?

  1. log(k₂/k₁) = (54000/(2.303 × 8.314))(1/298 − 1/308)
  2. k₂/k₁ = 2.03
  3. Answer: 2.03 times

Worked example 2 (Standard)

A reaction has Ea = 43 kJ/mol. By what factor does k increase from 310 K to 320 K?

  1. log(k₂/k₁) = (43000/(2.303 × 8.314))(1/310 − 1/320)
  2. k₂/k₁ = 1.68
  3. Answer: 1.68 times

Worked example 3 (Challenge)

The rate constant of a reaction becomes 7.86 times when T rises from 298 K to 323 K. Find Ea.

  1. Ea = 2.303R log(k₂/k₁)/(1/T₁ − 1/T₂)
  2. = 66.01 kJ/mol
  3. Answer: 66.01 kJ/mol

Common mistakes and misconceptions

  • Using °C
  • Forgetting 2.303 with log₁₀

Practice questions

  1. (Basic) A reaction has Ea = 33 kJ/mol. By what factor does k increase from 310 K to 335 K?
  2. (Basic) A reaction has Ea = 80 kJ/mol. By what factor does k increase from 300 K to 310 K?
  3. (Standard) A reaction has Ea = 91 kJ/mol. By what factor does k increase from 290 K to 315 K?
  4. (Standard) A reaction has Ea = 113 kJ/mol. By what factor does k increase from 310 K to 320 K?
  5. (Challenge) The rate constant of a reaction becomes 3.79 times when T rises from 310 K to 320 K. Find Ea.
  6. (Challenge) The rate constant of a reaction becomes 7.24 times when T rises from 298 K to 318 K. Find Ea.

Answer key with full working

  • Q1: log(k₂/k₁) = (33000/(2.303 × 8.314))(1/310 − 1/335) → k₂/k₁ = 2.6 → Answer: 2.6 times
  • Q2: log(k₂/k₁) = (80000/(2.303 × 8.314))(1/300 − 1/310) → k₂/k₁ = 2.81 → Answer: 2.81 times
  • Q3: log(k₂/k₁) = (91000/(2.303 × 8.314))(1/290 − 1/315) → k₂/k₁ = 19.98 → Answer: 19.98 times
  • Q4: log(k₂/k₁) = (113000/(2.303 × 8.314))(1/310 − 1/320) → k₂/k₁ = 3.93 → Answer: 3.93 times
  • Q5: Ea = 2.303R log(k₂/k₁)/(1/T₁ − 1/T₂) → = 109.91 kJ/mol → Answer: 109.91 kJ/mol
  • Q6: Ea = 2.303R log(k₂/k₁)/(1/T₁ − 1/T₂) → = 78 kJ/mol → Answer: 78 kJ/mol

Syllabus note

This skill is listed under "Chemical Kinetics" for Class 12 in the NCERT/CBSE books. ICSE, IGCSE, IB and state boards may teach it in a different chapter or year.

Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.

Related practice and notes

Frequently asked questions

Which chapter is arrhenius equation in?

"Chemical Kinetics", Class 12 Chemistry (NCERT/CBSE). Other boards may place it differently.

How were the answers checked?

Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.

Book a free demo class