The activation energy of a reaction is 145.25 kJ/mol and pre-exponential factor is 5 × 10^13 s^-1. At what temperature (in K) will the reaction have half-life of 1 min? (ln 2 = 0.7, ln 3 = 1.1, ln 7 =

Question

The activation energy of a reaction is 145.25 kJ/mol and pre-exponential factor is 5 × 10^13 s^-1. At what temperature (in K) will the reaction have half-life of 1 min? (ln 2 = 0.7, ln 3 = 1.1, ln 7 = 2.0, R = 8.3 J/K-mol)

Prashant Kotian · Accepted Answer

Correct answer: 0500. Half-life = 60 s, so k = ln 2 / 60 = 0.7 / 60 s^-1. Using Arrhenius equation ln k = ln A - Ea/RT. Ea/RT = ln(A/k) = ln(5 × 10^13 × 60 / 0.7) = ln(3 × 10^15 / 0.7). Resolving the logarithm yields roughly 35.0. 145250 / 8.3T = 35.0 => 17500 / T = 35.0 => T = 500 K. Therefore, correct answer is 0500.

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