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

K_{1} = \sqrt{K_{2}}

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

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K_{1} = 2 K_{2}

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

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Solution

The correct option is D $K_{2} = \frac{1}{K_{1}^{2}}$
By reversing first reaction and multiplying it by 2 we will get the equation similar to second equation.
Thus, $K_{1}$ and $K_{2}$ can be related in the above equations as:

$K_{2} = \frac{1}{K_{1}^{2}}$
$\Rightarrow K_{1} = \frac{1}{\sqrt{K_{2}}}$

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Similar questions

K_{1}

K_{2}

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)

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:

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:

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

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