In Arrhenius equation for a certain reaction, the value of A and E_a (activation energy) are 4 x `10^(13) s^(-1)` and 98.6 kJ `mol^(-1)` respectively If the reaction is of first order, then at what temperature will its half life period be 10 minutes? [log 2 88 = 0.4594]

Explanation

We have , B `k = Ae^(-E//Rt)`
In `k - In A - E/(RT)`
` 2.303 log k = 2.303 log A - E/(RT)`
or `log k = log A - E/(2.303 RT)`
Given that `A = 4xx106(13) s^(-1) , E = 98.6 " kJ mol"^(-1)` .
`t_(1/2) = 10 xx 60 s `
For first - order reaction k ` = (0.6932)/(t_(1/2)) = (0.6932)/(600) s^(-1)`
Thus (1) becomes
`log. (0.6932)/(600) = log (4xx10^(13)) - (98.6)/(2.303xx8.314 xx10^(-3) xxT) `
`[ R = 8.314 xx10^(-3) "kJ/K/mol" ] `
` :. T = 311.2 K `