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^(15)e^(-1000//T)=10^(16)e^(-2000//T)`
`(e^(-2000//T))/(e^(-1000//T))=(10^(15))/(10^(16))`
`e^(-1000//RT)=10^(-1)`
`log_(e)e^(-1000//T)=log_(e)10^(-1)`
`(-1000)/(T)=2.303log_(10)10^(-1)=-2.303`
`T=(1000)/(2.303)K`