A reversible reaction $A ⇌ C$ takes place in two steps $A ⇌ B; B ⇌ C$ , which are also equilibrium steps. If the equilibrium constants of the two steps are $K_{1}$ and $K_{2}$ respectively, the overall equilibrium constant $K$ is equal to:

\frac{K_{1}}{K_{2}}

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\frac{K_{2}}{K_{1}}

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K_{1} \times K_{2}

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K_{1} - K_{2}

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Solution

The correct option is D $K_{1} \times K_{2}$
The reaction is $A ⇌ C$ ......(1)
It is obtained by adding the following two reactions,
$A ⇌ B$ ......(2)
$B ⇌ C$ ......(3)
The value of the equilibrium constant of reaction (1) is obtained by multiplying the values of the equilibrium constants of the reactions (2) and (3).
Thus, $K = K_{1} \times K_{2}$ .

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Q. A reaction takes place in three steps having rate constants

K_{1}

K_{2}

K_{3}

respectively. The overall rate constant

K = \frac{K_{1} K_{3}}{K_{2}}

. If energies of activation for the three steps 40, 30, 20 kJ respectively, the overall energy of activation is:

A reaction takes place in three steps. The rate constant of the three steps are $k_{1}, k_{2}$ and $k_{3}$ respectively. The overall rate constant $k = \frac{k_{1} k_{3}}{k_{2}} .$ The energy of activation for the three steps are 40, 30 and 20 kJ respectively. Therefore: