Equilibrium constants $K_{1}$ and $K_{2}$ for the following equilibria are related as:
$N O (g) + \frac{1}{2} O_{2}_{1} {N O}_{2} (g)$
$2 {N O}_{2} (g)_{2} 2 N O (g) + O_{2} (g)$

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

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

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

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

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Solution

The correct option is D $K_{2} = \frac{1}{K_{1}^{2}}$
(i) $N O (g) + \frac{1}{2} O_{2}_{1} {N O}_{2} (g)$

So, $K_{1} = \frac{[N O_{2}]}{[N O] {[O_{2}]}^{1 / 2}} . . . (i)$

(ii) $2 {N O}_{2} (g)_{2} 2 N O (g) + O_{2} (g)$

So, $K_{2} = \frac{{[N O]}^{2} [O_{2}]}{{[N O_{2}]}^{2}} . . . (i i)$

From eq. (i) when we square both side:

$K_{1}^{2} = \frac{{[N O_{2}]}^{2}}{{[N O]}^{2} [O_{2}]}$

or $\frac{1}{K_{1}^{2}} = \frac{{[N O]}^{2} [O_{2}]}{{[N O_{2}]}^{2}} . . . (i i i)$

From Eqs (ii) and (iii)
$K_{2} = \frac{1}{K_{1}^{2}}$

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Similar questions

K_{1}

K_{2}

N O (g) + \frac{1}{2} O_{2} (g) K_{1} ⇌ N O_{2} (g)

2 N O_{2} (g) K_{2} ⇌ 2 N O (g) + O_{2} (g)

K_{1}

K_{2}

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

2 N O_{2 (g)} ⇌ 2 N O_{(g)} + O_{2 (g)}

Equilibrium constants $K_{1}$ and $K_{2}$ for the following equilibria are

$N O (g) + 0.5 O_{2} (g) \to K_{1} N O_{2} (g)$ and
$2 N O_{2} (g) \to K_{2} 2 N O (g) + O_{2} (g)$

Equilibrium constants $K_{1}$ and $K_{2}$ for the following equilibria are

$N O (g) + 0.5 O_{2} (g) \to K_{1} N O_{2} (g)$ and
$2 N O_{2} (g) \to K_{2} 2 N O (g) + O_{2} (g)$

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

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

(ii) $2 N O_{2} (g) ⇌ 2 N O (g) + O_{2} (g)$ ,

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

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