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For the reaction, A(g)+2B(g) hArr 2C(g),...

For the reaction, `A(g)+2B(g) hArr 2C(g)`, the rate constant for forward and the reverse reactions are `1xx10^(-4)` and `2.5xx10^(-2)` respectively. The value of equilibrium constant, K for the reaction would be

A

`2xx10^(-4)`

B

`3xx10^(-2)`

C

`4xx10^(-3)`

D

`3xx10^(2)`

Text Solution

Verified by Experts

The correct Answer is:
C
We know, `K=K_(f)/K_(b)=(1xx10^(-4))/(2.5xx10^(-2))=4xx10^(-3)`
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