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//2.303K`
C. 1000 K
D. `2000//2.303K`

Answer

B

Explanation

`k_(1)=10^(16)e^(-2000//T)`
`k_(2)=10^(15)e^(-1000//T)`
The temperature at which `k_(1)=k_(2)` will be
`10^(16)e^(-2000//T)=10^(15)e^(-1000//T)`
`(e^(-2000//T))/(e^(-1000//T))=(10^(15))/(10^(16))`
`e^(-1000//T)=10^(-1)`
`log_(e)e^(-1000//T)=log_(e)10^(-1)`
`2.303xxlog_(10)E^(-1000//T)=2.303xxlog_(10)10^(-1)`
`(-1000)/(T)xxlog_(10)e=-1`
On solving, we get
`T=1000//2,303K`