Question

Two reactants `R_(1)` and `R_(2)` have identical pre-exponential factors . Activation energy of `R_(1)` exceeds that of `R_(2)` by `10 kJ mol^(-1)` . If `k_(1)` and `k_(2)` are rate constants for reaction `R_(1)` and `R_(2)` respectively at 300 K , then Ln`(k_(2)//k_(1))` is equal to :
( R = `8.314 J mol^(-1) K^(-1)`)
A. 12
B. 6
C. 4
D. 8

Answer 1

Correct Answer - c
`k_(1) = Ae^(-E_(a_(1)) //RT) , k_(2) = Ae^(-(Ea_(1) - 10) //RT)`
Ln `((k_(2))/(k_(1))) = (10)/(RT) = (10)/(8.314 xx 10^(-3) xx 300) = 4`