The rate constants `k_(1)` and `k_(2)` for two different reactions are `10^(25)e^(-4000//T)` and `10^(25)e^(-1000//T)`, respectively. The temperature at which `k_(2)=k_(2)` is

Options

A. `2000/2.303K`
B. `2000K`
C. `3000/2.303K`
D. `1000K`

Answer

C

Explanation

`k_(1)=10^(26)e^(-4000//T)`
`k_(2)=10^(25)e^(-1000//T)`
if `k_(1)=k_(2)`, then
`10^(26)e^(-4000//T)=10^(25)e^(-1000//T)`
or `ln_(10)-4000/T=-1000/T`
or `T=3000/2.303K`