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 is

Answer»

`5xx10^(-4)`
`2xx10^(-3)`
`1.125xx10^(-3)`
`9.0xx10^(-4)`

Solution :`K_(c)=(K_(f))/(K_(b))`
`K_(f)=K_(c)xxK_(b)=1.5xx7.5xx10^(-4)=1.125xx10^(-3)`


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