A rate constant increases by a factor of 4.00 when temperature rises from 300 K to 320 K. Calculate Ea. Use R = 8.31 J/K/mol.
[4 marks]
Total for this question: 4
All calculations
Activation energy from Arrhenius data
Rate equations
AQA 7405
ORIGINAL
Ea = R ln(k2/k1) / (1/T1 - 1/T2)
Temperatures must be kelvin and the reciprocal-temperature subtraction order must match the logarithm ratio.
Board and spec code confirmed against AQA 7405 · registry checked 2026-07-11How this checking works
Work it, then mark it
Do each calculation on paper first, then reveal the mark scheme and tick the marks you actually earned — the same way you should mark past papers.
Q1
Do the calculation on paper first — then mark it.
Where the marks get lost
- Using Celsius temperatures.
- Leaving Ea in J/mol when kJ/mol is requested.
Exam tip: Keep the logarithm ratio and reciprocal-temperature difference in the same high-to-low order so Ea is positive.
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