Given below are two reversible reactions :
$A + B ⇌ Z$ ; $K_{c_{1}} = 24$
$2 B + C ⇌ 2 Y$ ; $K_{c_{2}} = 16$
The equilibrium constant $K_{c}$ for the reaction :
$A + Y ⇌ Z + \frac{C}{2}$ is

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Solution

$A + B ⇌ Z$ ; $K_{c_{1}} = 24$ ......(1)
$2 B + C ⇌ 2 Y$ ; $K_{c_{2}} = 16$ ......(2)
The reaction 2 is reversed and divided by 2 to obtain
$Y ⇌ B + \frac{1}{2} C$ ; $K_{c_{3}} = \frac{1}{\sqrt{K_{c_{2}}}} = \frac{1}{\sqrt{16}} = \frac{1}{4}$ ......(3).
The reactions (1) and (3) are added to obtain the reaction (4).
$A + Y ⇌ Z + \frac{C}{2}$ ......(4)
The equilibrium constant $K_{c}$ for the reaction
(4) is
$K_{c_{4}} = K_{c_{1}} \times K_{c_{3}} = 24 \times \frac{1}{4} = 6$ .

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