Home
Class 12
CHEMISTRY
The equilibrium constant Kc for A((g))hA...

The equilibrium constant Kc for `A_((g))hArrB_((g))` is `2.5xx10^(-2)`. The rate constant of the forward reaction is `0.05" sec"^(-1)`. Calculate the rate constant of the reverse reaction.

Text Solution

Verified by Experts

The correct Answer is:
`2.0" sec"^(-1)`
Promotional Banner

Topper's Solved these Questions

  • CHEMICAL EQUILIBRIUM - II

    NCERT TAMIL|Exercise SELF EVALUATION (C. Answer not exceeding 60 words)|3 Videos
  • CHEMICAL KINETICS

    NCERT TAMIL|Exercise EXAMPLE|10 Videos
  • CHEMICAL KINETICS-II

    NCERT TAMIL|Exercise SELF EVALUATION ((D) Solve the problems :)|7 Videos

Similar Questions

Explore conceptually related problems

Rate constant of a first order reaction is 0.45 sec^(-1) , calculate its half life.

The rate constant for a first order reaction is 1.54xx10^(-3)s^(-1) . Calculate its half life time.

The rate constant of a reaction is 5.8xx10^(-2)s^(-1) . The order of the reaction is

The rate constant of a reaction is 5.8xx10^(-2)s^(-1) . The order of the reaction is

In a chemical equilibrium, the rate constant for the forward reaction is 2.5xx10^(2) and the equilibrium constant, is 50. The rate constant for the reverse reaction is