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

Answer

A

Explanation

We are given
`k_(1)=10^(16).e^(-2000//T)`
Taking natural logarithm of both sides, we get
`Ink_(1)=In10^(16)-(2000)/(T)`
Conversing to log to the base `10` gives
`2.303logk_(1)=2.303log10^(16)-(2000)/(T)`
or `logk_(1)=16-(2000)/(2.303T)`
Similarly log `k_(2)=15-(1000)/(2.303T)`
If `k_(1)=k_(2)` , then
`16-(2000)/(2.303T)=15-(1000)/(2.303T)`
`1=(1000)/(2.303T)`
or `T=(1000)/(2.303)K`