Question

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)`

Answer 1

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