The activation energy for a given reaction is 50.3 kJ/mol. If the rate constant for the reaction is 0.471 hr-1 at 245 °C , at what temperature, in °C, will the reaction have a rate constant of 0.0107 hr-1 ?
ln (K2/K1) = (Ea/R) [1/T1 - 1/T2]
ln (0.0107 / 0.471) = (50.3 x 1000 / 8.314) [1/518 - 1/T2]
- 3.78 = 50300 / 8.314 [0.0019 - 1/T2]
- 0.00062 = 0.0019 - 1/T2
1/T2 = 0.00252
T2 = 396.8 K
T2 = 123.80C
answer = 124 0C
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