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The activation energy for a reaction at ...

The activation energy for a reaction at temperature T K was found to be `2.303" RT J mol"^(-1)`. The ratio of the rate constant to Arrhenius factor is

A

`10^(-1)`

B

`10^(-2)`

C

`2xx10^(-3)`

D

`2xx10^(-2)`

Text Solution

Verified by Experts

The correct Answer is:
A
`2.303 log k = 2.303log A - E_a/(RT)`
Putting `E_a=2*303 RT` ( Given ) , we get
` 2*303 log k = 2&*303 log A -(2*303RT)/(RT)`
or `log k = log A - 1`
or `log ""k/A=-1` or `k/=10^)-1`
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