For the hypothetical reactions, the equilibrium constant (K) values are given:
$A ⇌ B; K_{1} = 2; B ⇌ C; K_{2} = 4; C ⇌ D; K_{3} = 3$
The equilibrium constant (K) for the reaction $A ⇌ D$ is:

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Solution

The correct option is C 24
Given that three hypothetical reactions are-

$A ⇌ B; K_{1} = 2$

$B ⇌ C; K_{2} = 4$

$C ⇌ D; K_{3} = 3$

By adding these equilibrium constant values, $A ⇌ D, K$ is obtained.
$K = K_{1} \times K_{2} \times K_{3} = 2 \times 4 \times 3 = 24.$

So, the correct option is $C$

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Similar questions

A ⇌ B; K_{1} = 2; B ⇌ C; K_{2} = 4; C ⇌ D; K_{3} = 3

A ⇌ D

(K)

A ⇌ B; K_{1} = 2 B ⇌ C; K_{2} = 4 C ⇌ D; K_{3} = 3

A ⇌ D

A \to B (k = k_{1}) B \to C (k = k_{2}) C \to D (k = k_{3})

A \to D

A ⇌ B, B ⇌ C, C ⇌ D

K_{1}, K_{2}, K_{3}

A ⇌ D

(K)

A ⇌ B; K_{1} = 2 B ⇌ C; K_{2} = 4 C ⇌ D; K_{3} = 3

A ⇌ D

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