Question

The rate constants k₁ and k₂ for two different
reactions are 10¹⁶ . e ⁻²⁰⁰⁰∕т and 10¹⁵ . e⁻¹⁰⁰⁰∕т ,
respectively. The temperature at which k₁ = k₂ is :

(a) 1000 K (b)2000/2.303 K
(c) 2000 K (d) 1000/2.303 K

Answer 1

(d) 1000/2.303 K

Explanation:

Given: k1 = 10^16*e^(-2000/T)

k2 = 10^15*e^(-2000/T)

If we take log on both sides, for k1 = k2, we get:

log k1 = long 10^16 - 2000/2.303t

log k2 = long 10^15 - 1000/2.303t

So equilibrium will be T = 1000/2.303k

Option D is the answer.