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Rate Constant

The equilibri...

Question

The equilibrium constant $K_{c}$ of a reversible reaction is 10. The rate constant for the reverse reaction is 2.8. What is the rate constant for the forward reaction?

0.28

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0.028

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280

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Solution

The correct option is A 28
$K c = \frac{K f}{K b}$ . Now Kc = 10, Kb = 2.8
$∴ k f = 10 \times 2.8$
$∴ K f = 28$

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