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`K_f` and `K_b` are the velocity constants of forward and backward reactions. The equilibrium constant `K_(eq)` of the reversible reaction will beA. `K_b//K_f`B. `K_fxxK_b`C. `K_f//K_b`D. `K_f-K_b` |
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Answer» Correct Answer - C Suppose a reversible reaction occurs by a one-step mechanism: `2A+B=A_2B` The rate of the forward reaction is `Rate_f=K_f[A]^2[B]` The rate of the backward reaction is `Rate_b=K_b[A_2B]` At equilibrium, `Rate_f=Rate_b` `K_f[A]^2[B]=K_b[A_2B]` Dividing both sides of this equation by `K_b` and by `[A]^2[B]` gives `K_(eq)=K_f/K_b=([A_2B])/([A]^2[B])` |
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