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

Answer

B

Explanation

`10^(16)e^((-2000)/(T))=10^(15)e^((-1000)/(T))`
`10=(e^((-1000)/(T)))/(e^((-2000)/(T)))implies10=e^((1)/(T)xx1000)impliesln10=(1000)/(T)`
`implies2.303 log 10=(1000)/(T)impliesT=(1000)/(2.303)K`