The rate constant `k_(1)` and `k_(2)` for two different reactions are `10^(16) e^(-2000//T)` and `10^(15) e^(-1000//T)`, respectively. The temperature at which `k_(1) = k_(2)` is

Options

A. `1000 K`
B. `(2000)/(2.303)K`
C. `2000 K`
D. `(1000)/(2.303)K`

Answer

D

Explanation

`k_1=10^16eF^(-2000//T), k_2=10^(15)e^(-1000//T`
When `k_1=k_2, 10^16e^(-2000//T)=10^15e^(-1000//T)`
or `10e^(-2000//T)=e^(=-1000//T)`
Taking natural logarithm of both sides, we get
In `10-2000/T=(-1000)/T" or "2*303-2000/T=(-1000)/T`
or `1000/T=2*303` or `T=1000/(2*303)K`