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The rate constants `k_(1)` and `k_(2)` for two different reactions are `10^(16) * e^(-2000//T)` and `10^(15) * e^(-1000//T)` , respectively . The temperature at which `k_(1) = k_(2)` isA. 2000 KB. `(1000)/(2.303) K`C. 1000 KD. `(2000)/(2.303)K` |
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Answer» Correct Answer - b `10^(16) e^((-2000)/(T)) = 10^(15) e^((-1000)/(T))` `10 = e^((-1000)/(T))/(e^(-2000)/(T)) implies 10 e^((1)/(T) xx 1000) implies "ln" 10 = (1000)/(T)` `implies 2.303` log 10 = `(1000)/(T) implies T = (1000)/(2.303) K` . |
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