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

Answer

B

Explanation

`k_(1)=10^(16)e^(-) (2000)/(T)`
`k_(2)=10^(15)e^(-) (1000)/(T)`
`k_(1)=k_(2)`
`10=e (2000)/T-(1000)/(T)`
`log_(e) 10=(2000)/T-(1000)/(T)`