Consider the two gaseous equilibrium involving ${S O}_{2}$ and the corresponding equilibrium constants at 299K
${S O}_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ {S O}_{3} (g); K_{1}$
$4 {S O}_{3} (g) ⇌ 4 S O_{2} (g) + {2 O}_{2} (g); K_{2}$
The value of the equilibrium constant are related by:

K_{2} = \frac{1}{{(K_{1})}^{4}}

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K_{2} = K_{1}^{4}

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K_{2} = {(\frac{1}{K_{1}})}^{\frac{1}{4}}

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

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Solution

The correct option is A $K_{2} = \frac{1}{{(K_{1})}^{4}}$
$S O_{2} + \frac{1}{2} O_{2} ⇌ S O_{3} K_{1}$ ..... $(i)$
$4 S O_{3} ⇌ 4 S O_{2} + 2 O_{2} K_{2}$ ....... $(i i)$
$R x 2$ is nothing but $(- R_{p m} (1)) \times 4$
$K_{1} = \frac{[S O_{3}]}{[S O_{2}] [O_{2}] \frac{1}{2}}$
$K_{2} = \frac{[O_{2}]^{2} [S O_{2}]^{4}}{[S O_{3}]^{4}}$
${⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ \frac{[S O_{3}]}{[O_{2}]^{\frac{1}{2}} [S O_{2}]} ⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭}^{- 4} = K_{1}^{- 4}$

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Similar questions

S O_{2}

299 K S O_{2} (g) +^{1 / 2} O_{2} (g) ⇌ S O_{3} (g); K_{1}

4 S O_{3} (g) ⇌ 4 S O_{2} (g) + 2 O_{2} (g); K_{2}

S O_{2}

S O_{2} (g) + 1 / 2 O_{2} (g) ⇌ S O_{3} (g); K_{1}

4 S O_{3} (g) ⇌ 4 S O_{2} (g) + 2 O_{2} (g); K_{2}

K_{1}

K_{2}

S O_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ S O_{3} (g) 2 S O_{3} (g) ⇌ 2 S O_{2} (g) + O_{2} (g)

S O_{2}

S O_{2} (g) + \frac{1}{2} O_{2} (g) \Leftrightarrow S O_{3} (g); K_{1}

2 S O_{3} (g) \Leftrightarrow 2 S O_{2} (g) + O_{2} (g); K_{2}

For the two gaseous reactions (i) and (ii), at $25^{0} C$

$(i) N O (g) + \frac{1}{2} O_{2} (g) ⟵ N O_{2} (g)$

$(i i) 2 N O_{2} (g) ⟵ 2 N O (g) + O_{2} (g)$ ,

The equilibrium constants are $K_{1}$ and $K_{2}$ respectively. these two constants are related as:

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