The equilibrium constant for the following reaction is $64$ . $H_{2} O (l) + C O (g) ⇋ H_{2} (g) + C O_{2} (g)$ . If the rate constant for the forward reaction is $160$ , the rate constant for the backward reaction is:

0.4

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2.5

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6.2

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10.24

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Solution

The correct option is B $2.5$
Given that the equilibrium constant for the reaction, $H_{2} O (l) + C O (g) ⇋ H_{2} (g) + C O_{2} (g)$ is $64$ .

The expression for the equilibrium constant, $K = \frac{k_{f}}{k_{b}}$
GIven, $K = 64, k_{f} = 160$
Substituting values in the above expression, we get
$64 = \frac{160}{k_{b}}$ $⟹ k_{b} = 2.5$

Hence, the rate constant for the backward reaction is $2.5$ .

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