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Two reactions R1 and R2 have the same pre-exponential (Arrhenius) factor. The activation energy of R1 is greater than that of R2 by 10 kJ/mol. If k1 and k2 are their rate constants at 300 K, find ln(k2/k1). (R = 8.314 J mol⁻¹ K⁻¹)
- 8
- 12
- 6
- 4
Correct answer: 4
Solution
With equal A, ln(k2/k1) = (Ea1 - Ea2)/(RT) = 10000/(8.314 x 300) = 10000/2494 approx 4.0.
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