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Introduction to Reversible and Irreversible Reactions

The equilibri...

Question

The equilibrium concentration of $[B]_{e q}$ for the reversible reaction $A ⇌ B$ can be evaluated by the expression:

A

K_{c} [A]_{e q}

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B

\frac{k_{f}}{k_{b}} [A] e_{e q}

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C

k_{f} k_{b}^{- 1} [A]_{e q}

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D

k_{f} k_{b} [A]^{- 1}

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Solution

The correct option is C $k_{f} k_{b}^{- 1} [A]_{e q}$
At equilibrium, rate of forward reaction=rate of backward reaction
$\Rightarrow r_{f} = r_{b}$
$A_{f} B$
$∴ K_{f} [A]_{e q} = K_{b} [B]_{e q}$
$[B]_{e q} = K_{f} K_{b}^{- 1} [A]_{e q}$
Hence, the correct answer is option (c).

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