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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

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The correct Answer is:
`2.0" sec"^(-1)`
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Knowledge Check

  • The activation energy of a reaction is zero . If the rate constant of the reaction at 300K is 3.2xx10^(6)s^(-1) , then at 316K the rate constant of the reaction will be -

    A
    `3.2xx10^(12)s^(-1)`
    B
    `6.4xx10^(6)s^(-1)`
    C
    `3.2xx10^(6)s^(-1)`
    D
    `3.2xx10^(10)s^(-1)`

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