1.

The activation energy for a reaction at the temperature T K was found to be `2.303RT J mol^(-1)`. The ratio of the rate constant to Arrhenius factor isA. `10^(-1)`B. `10^(-2)`C. `2 xx 10^(-3)`D. `2 xx 10^(-2)`

Answer» Correct Answer - A
a) According to Arrhenius equation,
log k = log A`-E/(2.303R) xx 1/T`
`2.303 log k = 2.303 log A-(E_(a))/(RT)`
Substituting the value of `E_(a)` as 2.303 RT and dividing by 2.303 on both sides, we get.
log k`-log A=-1`
`log(k/A)=log 10^(-1)`
Taking anti log.
`(k/A)=10^(-1)`


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