InterviewSolution
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InterviewSolution
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For a first order reaction `A rarr P`, the temperature `(T)` dependent rate constant `(k)` was found to follow the equation `log k = -2000(1//T) + 6.0`. The pre-exponential factor `A` and the activation energy `E_(a)`, respective, areA. `1.0 xx 10^(6) s^(-1)` and `9.2 kJ mol^(-1)`B. `6.0 s^(-1)` and `16.6 kJ mol^(-1)`C. `1.0 xx 10^(6) s^(-1)` and `16.6 kJ mol^(-1)`D. `1.0 xx 10^(6) s^(-1)` and `38.3 kJ mol^(-1)` |
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Answer» Correct Answer - D `k = Ae^(E_(a)//RT)` or `log k = logA - (E_(a))/(2.303 RT)` `log k = -2000((1)/(T)) + 6.0` On comparison, we get `log A = 6` or `A = 10^(6)s^(-1)` `:. (E_(a))/(2.303R) = 2000` or `E_(a) = 2.303 xx 8.314 xx 2000` `~~ 38.3 kJ mol^(-1)`Correct Answer - D `k = Ae^(E_(a)//RT)` or `log k = logA - (E_(a))/(2.303 RT)` `log k = -2000((1)/(T)) + 6.0` On comparison, we get `log A = 6` or `A = 10^(6)s^(-1)` `:. (E_(a))/(2.303R) = 2000` or `E_(a) = 2.303 xx 8.314 xx 2000` `~~ 38.3 kJ mol^(-1)` |
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