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)) implies 10 e^((1)/(T) xx 1000) implies "ln" 10 = (1000)/(T)`
`implies 2.303` log 10 = `(1000)/(T) implies T = (1000)/(2.303) K` .