The equilibrium constant for the reaction is . The value of for the reaction given below is

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Solution

Q.
The equilibrium constant for the reaction ${SO}_{3} (g) ⇌ {SO}_{2} (g) + \frac{1}{2} O_{2} (g)$ is $K_{C} = 4.9 \times 10^{- 2}$ . The value of $K_{C}$ for the reaction given below is

$2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$ is : [2024]

1 4.9

2 49

3 41.6

4 416

Ans.
(4)

${SO}_{3} (g) ⇌ {SO}_{2} (g) + \frac{1}{2} O_{2} (g), K_{c} = 4.9 \times 10^{- 2}$ - I

Reverse the above equation. On reversing a chemical equation, equilibrium constant gets inversed.

${SO}_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ {SO}_{3} (g), K_{c}^{1} = \frac{1}{4.9 \times 10^{- 2}}$ - II

Multiply II by 2. On multiplying an equation by $n$ , equilibrium constant gets raised by a power $n$ .

$2 {SO}_{2} (g) + O_{2} (g) ⇌ 2 {SO}_{3} (g), K_{c}^{11} = {(\frac{1}{4.9 \times 10^{- 2}})}^{2} = 416.49$

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