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