Given:
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g); K_{1}$
$N_{2} (g) + O_{2} (g) ⇌ 2 N O (g); K_{2}$
$H_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ H_{2} O (g); K_{3}$
Calculate the equilibrium constant for the reaction,
$2 N H_{3} (g) + \frac{5}{2} O_{2} (g) ⇌ 2 N O (g) + 3 H_{2} O (g)$

K_{1} K_{2} K_{3}

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\frac{K_{1} K_{2}^{3}}{K_{3}}

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\frac{K_{1} K_{3}^{2}}{K_{2}}

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\frac{K_{2} K_{3}^{3}}{K_{1}}

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Solution

The correct option is D $\frac{K_{2} K_{3}^{3}}{K_{1}}$
For the given Equations:

$N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g); K_{1}$
Reverse the Reaction,
$2 N H_{3} (g) ⇌ N_{2} (g) + 3 H_{2} (g); \frac{1}{K_{1}} . . . . . . e q n (i)$

$N_{2} (g) + O_{2} (g) ⇌ 2 N O (g); K_{2} . . . . e q n (i i)$

$H_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ H_{2} O (g); K_{3} . . . e q n (i i i)$
Multiplying $3$ in eqn(iii) we get,
$3 H_{2} (g) + \frac{3}{2} O_{2} (g) ⇌ 3 H_{2} O (g); K_{3}^{3} . . . e q n (i v)$
Adding $eqn (i) , eqn(ii) and eqn (iv)$ we get,
$2 N H_{3} (g) + \frac{5}{2} O_{2} (g) ⇌ 2 N O (g) + 3 H_{2} O (g) . . . . (v)$
putting values of equlibrium constant we cant get the value of equlibrium constant of eqn(v) ,
$K = K_{2} \times \frac{1}{K_{1}} \times (K_{3})^{3} = \frac{K_{2} \times K_{3}^{3}}{K_{1}}$

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Similar questions

N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g); K_{1}

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

H_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ H_{2} O (g); K_{3}

K_{1}, K_{2}, K_{3}

2 N H_{3} (g) + \frac{5}{2} O_{2} (g) ⇌ 2 N O (g) + 3 H_{2} O (g)

N_{2} + 3 H_{2} ⇌ 2 N H_{3} K_{1}

N_{2} + O_{2} ⇌ 2 N O K_{2}

H_{2} + \frac{1}{2} O_{2} \to H_{2} O K_{3}

2 N H_{3} + \frac{5}{2} O_{2} K ⇌ 2 N O + 3 H_{2} O,

The equilibrium constant $K_{c}$ for the reaction $N_{2}$ (g) + 3 $H_{2}$ (g) $⇋$ 2 $N H_{3}$ (g) is:

3H₂(g) + N₂ (g) ⇌ 2NH₃ (g)

N_{2} + 3 H_{2} \to 2 N H_{3} - K_{1}

N_{2} + O_{2} \to 2 N O - K_{2}

H_{2} + 1 / 2 O_{2} \to H_{2} O - K_{3}

3 H_{2} + 5 / 2 O_{2} \to 2 N O + 3 H_{2} O

K_{1}, K_{2}, K_{3}

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