Consider the equilibria 1 and 2 with equilibrium constants K 1 and K 2 respectively- S O 2 g -12 O 2 g - S O 3 g - 1 2S O 3 g - 2S O 2 g - O 2 g - 2 K 1 and K 2 are related as-

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Standard XII

Chemistry

Characteristics of Equilibrium Constant

Consider the ...

Question

Consider the equilibria (1) and (2) with equilibrium constants $K_{1}$ and $K_{2}$ respectively.
$S O_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ S O_{3} (g)$ ...... (1)
$2 S O_{3} (g) ⇌ 2 S O_{2} (g) + O_{2} (g)$ ...... (2)
$K_{1}$ and $K_{2}$ are related as:

2 K_{1} = K_{2}^{2}

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K_{1}^{2} = \frac{1}{K_{2}}

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K_{2}^{2} = \frac{1}{K_{1}}

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K_{2} = \frac{2}{K_{1}^{2}}

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Solution

The correct option is B $K_{1}^{2} = \frac{1}{K_{2}}$
Reaction (2) is twice the reverse reaction of reaction (1)

Therefore equilibrium constants are related as:

$K_{2} = \frac{1}{K_{1}^{2}}$ or $K_{1}^{2} = \frac{1}{K_{2}}$

Because on doubling by changing the coefficient the equilibrium constant term gets squared and on reversing the reaction, it becomes reciprocal.

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Similar questions

Q. Two gaseous equilibria

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and

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Consider the equilibria (1) and (2) with equilibrium constants K1 and K2 respectively.SO2(g)+12O2(g)⇌SO3(g) ...... (1)2SO3(g)⇌2SO2(g)+O2(g) ...... (2)K1 and K2 are related as:

The correct option is B K21=1K2Reaction (2) is twice the reverse reaction of reaction (1)Therefore equilibrium constants are related as:K2=1K21 or K21=1K2Because on doubling by changing the coefficient the equilibrium constant term gets squared and on reversing the reaction, it becomes reciprocal.

Consider the equilibria (1) and (2) with equilibrium constants $K_{1}$ and $K_{2}$ respectively.
$S O_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ S O_{3} (g)$ ...... (1)
$2 S O_{3} (g) ⇌ 2 S O_{2} (g) + O_{2} (g)$ ...... (2)
$K_{1}$ and $K_{2}$ are related as: