Equilibrium constant for the reaction:
$H_{2} O (g) + C O (g) \Leftrightarrow H_{2} (g) + C O_{2} (g)$ is 81.
If the rate constant of the forward reaction is 162 $l i t . m o l^{- 1}, s e c^{- 1}$ , then what is the velocity constant (in $l i t . m o l e^{- 1}$ $s e c^{- 1}$ ) for the backward reaction?

131

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261

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243

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Solution

The correct option is A 2
The expression for the equilibrium constant is $K c = \frac{K f}{K b}$ .
But $K = 81, K f = 162 l i t . m o l e^{- 1}$ .
Substitute values in the above expression.
$81 = \frac{162}{K b}$ .
Thus $K b = 2 l i t . m o l e^{- 1}$ .

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