Question

Two reactions R₁ and R₂ have identical pre-exponential factors. Activation energy of R₁ exceeds that of R₂ by 10 kJ mol⁻¹. If k₁ and k₂ are rate constants for reactions R₁ and R₂ respectively at 300 K then ln (k₂/k₁) is equal to ______. (R = 8.314 J mol⁻¹ K⁻¹)

पर्याय

• 6

• 4

• 8

• 12

Answer 1

Two reactions R₁ and R₂ have identical pre-exponential factors. Activation energy of R₁ exceeds that of R₂ by 10 kJ mol⁻¹. If k₁ and k₂ are rate constants for reactions R₁ and R₂ respectively at 300 K then ln (k₂/k₁) is equal to 4.

Explanation:

Given: `E_(a_1) − E_(a_2)` = 10 kJ mol−1=10,000 J mol−1,

T = 300 K, 

R = 8.314

Use Arrhenius: 

\[\ce{k = Ae^{-E_a/RT}}\]

With identical A:

`ln  k_2/k_1 = (-E_(a_2))/(RT) - (-E_(a_1))/(RT)`

= `(E_(a_1) - E_(a_2))/(RT)`

= `10000/(8.314 xx 300)`

= `10000/2194.2`

= 4.01

= 4