Home
Class 12
CHEMISTRY
Consider the following reaction. The re...

Consider the following reaction. The reaction is first order in each direction, with an equilibrium constant of `10^(4)`. For the conversion of chair form to boat form, `e^(-Ea//RT)=4.35xx10^(-8)` at 298 K with pre-exponential factor of `10^(12)s^(-1)`. Apparent rate constant `(k_(A))` at 298 K is :

A

`4.35xx10^(4)s^(-1)`

B

`4.35xx10^(8) s^(-1)`

C

`4.35xx10^(-8)s^(-1)`

D

`4.35xx10^(12) s^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
B
`k_(B)=Ae^(-E_(a)//RT)=10^(12) xx4.35xx10^(-8)=4.35 xx10^(4)S^(-1)`
Also, equilibrium constant `=(k_(A))/(k_(B))=10^(4) therefore k_(A) =k_(B) xx10^(4)=4.35 xx10^(8) s^(-1)`
Promotional Banner

Similar Questions

Explore conceptually related problems

The half life of a first order reaction having rate constant k=1.7xx10^(-5) sec^(-1) is :

A first order reaction is found to have a rate constant k= 5.5 xx 10^(-14)s^(-1) . Find half-life of the reaction.

The rate constant for the decompoistion of a certain reaction is described by the equation: log k(s^(-1)) = 14 - (1.25 xx 10^(4) K)/(T) Pre-exponential factor for this reaction is

A first order reaction is found to have a rate constant, k=5.5xx10^(-14)s^(-1) . Find the half-life of the reaction.

A first order reaction is found to have a rate constant k=7.39xx10^(-5)s^(-1) . Find the half life of this reaction.

A first order reaction is found to have a rate constant, k=7.39 xx 10^(-5)s^(-1) . Find the half life of the reaction.

A first order reaction is found to have a rate constant, k = 4.2 xx 10^(-12) s^(-1) . Find the half-life of the reaction.

A first order reaction has a rate constant, k = 5.5 xx 10^(-14) s^(-1) , calculate the half life of reaction.

A reaction has a forward rate constant of 2.3xx10^6 s^(-1) and an equilibrium constant of 4.0xx10^8 .What is the rate constant for the reverse reaction ?