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. (a) `1.0xx10^(6) s^(-1)` and `9.2 kJ mol^(-1)`B. (b) `6.0 s^(-1)` and `16.6 kJ mol^(-1)`C. (c ) `1.0xx10^(6) s^(-1)` and `16.6 kJ mol^(-1)`D. (d) `1.0xx10^(6) s^(-1)` and `38.3 kJ mol^(-1)` |
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Answer» Correct Answer - d `log k=-(2000)1/T+6.0 …(1)` and, ln k=ln `A-E_(a)/(RT)` `2.303log k=2.303 log A-E_(a)/(RT)` `log k=(-E_(a))/(2.303R)xx1/T+log A …(2)` By eqs. (1) and (2), `(-E_(a))/(2.303R)=-2000` `:. E_(a)=2.303xx8.314xx2000=38.29 kJ` and `log A=6, A=10^(6)` |
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