1.

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

Answer»

`10^(-1)`
`10^(-2)`
`2 XX 10^(-3)`
`2 xx 10^(-2)`

Solution :2.303 log `k= 2.303` log `A - (E_(a))/(RT) , E_(a) = 2.303 RT` (given)
Putting the values of `E_(a)` and dividing by 2.303 on both sides , we get , log k = log A - 1 , log k - log A = -1
log `((k)/(A)) = -1 implies ` TAKING antilog `((k)/(A)) = 10^(-1)` .


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