Plot ln k against 1/T.

Six original Cambridge-style questions on the Arrhenius equation.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

Kelvin, ln k and a negative gradient.

01
Calculation
[3 marks]

Two points on an Arrhenius plot are (3.10 × 10−3 K−1, −2.50) and (3.40 × 10−3 K−1, −5.50). Calculate Ea.

  • Gradient = (−5.50 − (−2.50)) / (3.40 × 10−3 − 3.10 × 10−3) = −1.00 × 104 K. ✓
  • Ea = −gradient × R = 1.00 × 104 × 8.31 = 83 100 J mol−1. ✓
  • Ea = 83.1 kJ mol−1. ✓
02
Calculation
[2 marks]

Using question 1, calculate ln A and A.

  • ln A = ln k + Ea/RT = −5.50 + 1.00 × 104 × 3.40 × 10−3 = 28.5. ✓
  • A = e28.5 = 2.38 × 1012 (same units as k). ✓
03
Calculation
[3 marks]

A reaction's rate constant doubles when the temperature rises from 298 K to 308 K. Calculate Ea.

  • ln(k2/k1) = ln 2 = 0.693. ✓
  • 1/298 − 1/308 = 1.09 × 10−4 K−1. ✓
  • Ea = 8.31 × 0.693 / 1.09 × 10−4 = 52 900 J mol−1 = 52.9 kJ mol−1. ✓
04
Explain
[3 marks]

Use the Boltzmann distribution to explain why a small rise in temperature causes a large increase in the rate constant.

  • At higher T the distribution flattens and shifts to higher energies. ✓
  • The area beyond Ea (molecules with E ≥ Ea) increases by a large proportion. ✓
  • So the frequency of successful collisions rises much more than the collision frequency does, and k increases. ✓
05
Calculation
[2 marks]

A = 1.00 × 1010 dm3 mol−1 s−1 and Ea = 50.0 kJ mol−1. Calculate k at 298 K.

  • k = 1.00 × 1010 × e−50 000/(8.31 × 298) = 1.00 × 1010 × 1.70 × 10−9. ✓
  • k = 17.0 dm3 mol−1 s−1. ✓
06
Analysis
[2 marks]

A catalyst lowers Ea from 80 kJ mol−1 to 60 kJ mol−1 at 298 K. Assuming A is unchanged, by what factor does k increase?

  • Factor = e(80 000 − 60 000)/(8.31 × 298) = e8.08. ✓
  • Factor = 3.22 × 103. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.