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We define chemical equilibrium as the point in a reaction where the rates of the forward and reverse reactions become equal. We also write

$$K_c = \frac{k_f}{k_r}$$

If at equilibrium the rate of the forward reaction and the rate of the reverse reaction become equal then their rate constants should also be same. If rate constants $k_f$ and $k_r$ become the same, then $K_c$ should always be equal to 1. However, the value of $K_c$ is not always 1. Why is it so?

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  • $\begingroup$ Welcome to chemistry.SE! As it looks like you're mixing reaction rate with reaction rate constant. $\endgroup$ – Felipe S. S. Schneider Apr 7 '17 at 12:37
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For an example reaction $\ce{A + B <=> C + D}$

the forward rate is given by

$$r_\mathrm{f} = k_\mathrm{f}[\ce{A}][\ce{B}]$$

and the backward rate is given by

$$r_\mathrm{b} = k_\mathrm{b}[\ce{C}][\ce{D}]$$

For the rates to be equal one simply needs

$$k_\mathrm{f}[\ce{A}][\ce{B}] = r_\mathrm{f} = r_\mathrm{b} = k_\mathrm{b}[\ce{C}][\ce{D}]$$

Note that this does not imply that the two rate constants are the same

$$k_\mathrm{f} \neq k_\mathrm{b} \quad \text{(in general)}$$

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  • $\begingroup$ Although true in principle, in your example the $k_f,~k_b$ have different units so can never be equal, which weakens your argument. $\endgroup$ – porphyrin Apr 8 '17 at 6:31

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