Home
Class 12
CHEMISTRY
The rate constant for forward reaction A...

The rate constant for forward reaction `A(g) hArr 2B(g)` is `1.5 xx 10^(-3) s^(-1)` at `100 K`. If `10^(-5)` moles of `A` and `100 "moles"` of `B` are present in a `10-L` vessel at equilibrium, then the rate constant for the backward reaction at this temperature is

A

`A. 1.50 xx 10^(-4) L mol s^(-1)`

B

`B. 1.50 xx 10^(-11) L mol^(-1) s^(-1)`

C

`C. 1.50 xx 10^(-10) L mol^(-1) s^(-1)`

D

D. None of these

Text Solution

Verified by Experts

The correct Answer is:
B
`k_(eq) = ([B]^(2))/([A]) = ([100//10]^(2))/([10^(-5)//10]) = 10^(8)`
`k_(eq) = (k_(f))/(k_(b)) = rArr k_(b) = (1.5 xx 10^(-3))/(10^(8)) = 1.5 xx 10^(-11) L mol^(-1) s^(-1)`
Promotional Banner

Topper's Solved these Questions

  • CHEMICAL KINETICS

    CENGAGE CHEMISTRY|Exercise Ex 4.3 More Than One Correct|5 Videos

  • CHEMICAL KINETICS

    CENGAGE CHEMISTRY|Exercise Ex4.4 Objective|10 Videos

  • CHEMICAL KINETICS

    CENGAGE CHEMISTRY|Exercise Ex 4.2 (Objective)|10 Videos

  • CARBOXYLIC ACIDS AND THEIR DERIVATIVES

    CENGAGE CHEMISTRY|Exercise Exercises Archives (Analytical And Descriptive)|34 Videos

  • COORDINATION COMPOUNDS

    CENGAGE CHEMISTRY|Exercise Archives Subjective|18 Videos

Similar Questions

Explore conceptually related problems

For the reaction, A(g)+2B(g)hArr2C(g) one mole of A and 1.5 mol of B are taken in a 2.0 L vessel. At equilibrium, the concentration of C was found to be 0.35 M. The equilibrium constant (K_(c)) of the reaction would be

In a vessel of 5L,26 moles of A and 4 moles of B were placed. At equilibrium 1 mole of C was present. The K_(C) for the reaction : A+2bhArrC is

The equilibrium constant of a reaction is 20.0 . At equilibrium, the reate constant of forward reaction is 10.0 . The rate constant for backward reaction is :

The rate constans of forward and backward reactions are 8.5 xx 10^(-5) and 2.38 xx 10^(-4) respectively. The equilibrium constant if

In a chemical equilibrium, the rate constant for the backward reaction is 7.5xx10^(-4) and the equilibrium constant is 1.5 the rate constant for the forward reaction is: