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Equilibrium Constant and Standard Free Energy Change

If K1 and ...

Question

If $K_{1}$ and $K_{2}$ are the equilibrium constants of the equilibriums $(A)$ and $(B)$ respectively, then what is the relationship between the two constants?
$(A)$ $S O_{2} (g) + 1 / 2 O_{2} (g) ⇋ S O_{3} (g), K_{1}$
$(B)$ $2 S O_{3} (g) ⇋ 2 S O_{2} (g) + O_{2} (g), K_{2}$

K_{1} = K_{2}

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

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

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

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Solution

The correct option is D $K_{1}^{2} = \frac{1}{K_{2}}$
When the equilibrium reaction $(A)$ $S O_{2} (g) + \frac{1}{2} O_{2} (g) ⇋ S O_{3} (g), K_{1}$ is reversed, a new equilibrium $(C)$ $S O_{3} (g) ⇋ S O_{2} (g) + \frac{1}{2} O_{2} (g), K_{3}$ is obtained.
The relationship between $K_{1}$ and $K_{3}$ is $K_{1} = \frac{1}{K_{3}}$ .
When equilibrium $(C)$ $S O_{3} (g) ⇋ S O_{2} (g) + \frac{1}{2} O_{2} (g), K_{3}$ is multiplied by $2$ , equilibrium ( $B$ ) $2 S O_{3} (g) ⇋ 2 S O_{2} (g) + O_{2} (g), K_{2}$ is obtained.
The relationship between $K_{2}$ and $K_{3}$ is $\frac{1}{K_{2}} = {(\frac{1}{K_{3}})}^{2}$ .
Thus, $\frac{1}{K_{2}} = {(\frac{1}{K_{3}})}^{2} = {K_{1}}^{2}$ .

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$(B)$ $2 S O_{3} (g) ⇋ 2 S O_{2} (g) + O_{2} (g), K_{2}$