The rate constants' `k_2 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` is equal to `k_2` is

Options

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

Explanation

(c) Give, `k_1 = 10^(16) e^(-2000 // T)`
`k_2 = 10^(15) e^(-1000 // T)`
As, `k_1 = k_2`
then, `10^(16) e^(-2000 // T) = 10^(15) e^(-1000// T)`
or, `10 e^(-2000// T) = e^(-1000//T)`
Taking log on both sides, we get
`"In" 10- (2000)/(T ) = - (1000)/( T)`
or `2.303- (2000)/( T) = - (1000)/( T)`
or `(1000)/(T ) = 2.303`
or `T= (1000)/( 2.303) K`