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The rate constant for the decompoistion of a certain reaction is described by the equation: `log k(s^(-1)) = 14 - (1.25 xx 10^(4) K)/(T)` Energy of activation (in `kcal`) isA. `57.6 kcal`B. `1.25 xx 10^(4) kcal`C. `14.0 kcal`D. `14 xx 10^(4) kcal` |
Answer» Correct Answer - A `E_(a)//2.303RT = (1.25 xx 10^(4) K)/(T)` `:. E_(a) = 1.25 xx 10^(4) xx 2.303 xx 10^(-3) kcal` `= 57.6 kcal` |
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