1.

In a chemical equilibrium, the rate constant of the backward reaction is `7.5xx10^(-4)` and the equilibrium constant is 1.5. So the rate constant of the forward reaction isA. `5xx10^(-4)`B. `2xx10^(-3)`C. `1.125xx10^(-3)`D. `9.0xx10^(-4)`

Answer» Correct Answer - C
`K_(c)=(K_(f))/(K_(b))`
`K_(f)=K_(c)xxK_(b)=1.5xx7.5xx10^(-4)=1.125xx10^(-3)`


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