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Arrhenius Acid

For the react...

Question

For the reaction, $N_{2} (g) + O_{2} (g) ⇌ 2 N O (g)$
Equilibrium constant $K_{c} = 2$
Degree of association is?

\frac{1}{1 - \sqrt{2}}

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\frac{1}{1 + \sqrt{2}}

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\frac{2}{1 + \sqrt{2}}

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\frac{2}{1 - \sqrt{2}}

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Solution

The correct option is
C
$\frac{1}{1 - \sqrt{2}}$

$N_{2} + O_{2} ⇌ 2 N O$
$K_{c} = 2$
degree of association is option (A) $= \frac{1}{1 - \sqrt{2}}$
$N_{2} = (1 - a)$
$O_{2} = (1 - a)$
$N O = 2 a$
degree of association $= \frac{{(2 a)}^{2}}{{(1 - a)}^{2}}$ = 2
$\frac{2 a}{1 - a} = \sqrt{2}$
$2 a = \sqrt{2} - \sqrt{2 a}$
$2 a + \sqrt{2 a} = \sqrt{2}$
$a (2 + \sqrt{2}) = \sqrt{2}$
$a = \frac{\sqrt{2}}{\sqrt{2} + 2} = \frac{1}{1 + \sqrt{2}}$

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Similar questions

For the reaction, $N_{2} (g) + O_{2} (g) ⇌ 2 NO (g)$ , equilibrium constant is $K_{c} = 2$ .

Degree of association is:

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N_{2} (g) + O_{2} (g) ⇌ 2 N O (g) . . . . . . (1)

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

N_{2} (g) + O_{2} (g) ⇌ 2 N O (g)

K_{1}

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2 N O (g) + O_{2} (g) ⇌ 2 N O_{2} (g)

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

K_{1}

K_{2}

N_{2} (g) + O_{2} (g) ⇌ 2 N O (g)

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

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x \times 10^{- 16} {l i t r e}^{1 / 2} {m o l}^{- 1 / 2}

\frac{1}{2} N_{2} (g) + \frac{1}{2} O_{2} (g) + \frac{1}{2} B r_{2} (g) ⇌ N O B r (g)

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N O (g) + \frac{1}{2} B r_{2} (g) ⇌ N O B r (g)

K_{c_{2}} = 1.4 {l i t r e}^{1 / 2} {m o l}^{- 1 / 2}

100 x

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