The Arrhenious equation for the decomposition of methyl nitrite and ethyl nitride are `k_(1)` and `k_(2)` respectively and the activation energy of second is `4 kJ//"mole"` is more than that of first. At what temperature value of rate constant `k_(1)` will be equal to `k_(2)` if
`k_(1) (s) = 10^(13) (e^(-(E_(a_(1)))/(RT)))` and `k_(2)(s^(-1)) = 10^(14) (e^(-(E_(a_(2)))/(RT)))`

Options

A. `650 K`
B. `230 K`
C. `325 K`
D. `425 K`

Answer

B

Explanation

`10^(13)e^(-(Ea_(1))/(RT)) = 10^(14)e^(-(Ea_(2))/(RT)`
`T = 230K`