1.

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 beA. `2xx10^(-4)`B. `3xx10^(-2)`C. `4xx10^(-3)`D. `3xx10^(2)`

Answer» Correct Answer - C
We know, `K=K_(f)/K_(b)=(1xx10^(-4))/(2.5xx10^(-2))=4xx10^(-3)`


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