SATHEE: Equilibrium - Result Question 13

Equilibrium - Result Question 13

####14. The following equilibrium constants are given:

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

[2003, 2007]

$N_{2} + O_{2} \to 2 N O; K_{2}$

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

The equilibrium constant for the oxidation of 2 moles $N H_{3}$ by oxygen to give $N O$ is

(a) $\frac{K_{2} K_{3}^{2}}{K_{1}}$

(b) $\frac{K_{2}^{2} K_{3}}{K_{1}}$

(d) $\frac{K_{2} K_{3}^{3}}{K_{1}}$

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Solution:

(d) Given,

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

$N_{2} + O_{2} \to 2 N O; K_{2}$

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

We have to calculate

$4 N H_{3} + 5 O_{2} ⟶ 4 N O + 6 H_{2} O; K =$ ?

or $2 N H_{3} + \frac{5}{2} O_{2} ⟶ 2 N O + 3 H_{2} O$

For this equation, $K = \frac{[N O]^{2} [H_{2} O]^{3}}{[N H_{3}]^{2} [O_{2}]^{5 / 2}}$

but $K_{1} = \frac{[N H_{3}]^{2}}{[N_{2}] [H_{2}]^{3}}, K_{2} = \frac{[N O]^{2}}{[N_{2}] [O_{2}]}$

$Misplaced &$ or $K_{3}^{3} = \frac{[H_{2} O]^{3}}{[H_{2}]^{3} [O_{2}]^{3 / 2}}$

Now operate, $\frac{K_{2} \cdot K_{3}^{3}}{K_{1}}$

$= \frac{[N O]^{2}}{[N_{2}] [O_{2}]} \times \frac{[H_{2} O]^{3}}{[H_{2}]^{3} [O_{2}]^{3 / 2}} \cdot \frac{[N_{2}] [H_{2}]^{3}}{[N H_{3}]^{2}}$

$= \frac{[N O^{2} [H_{2} O]^{3} .}{[N H_{3}]^{2} [O_{2}]^{5 / 2}} = K$

$∴ K = \frac{K_{2} \cdot K_{3}^{3}}{K_{1}}$

Equilibrium - Result Question 13

Solution:

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Equilibrium - Result Question 13

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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