SATHEE: Equilibrium - Result Question 6

Equilibrium - Result Question 6

####7. For the reaction $N_{2} (g) + O_{2} (g) \to 2 N O (g)$ , the equilibrium constant is $K_{1}$ . The equilibrium constant is $K_{2}$ for the reaction

[2011]

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

What is $K$ for the reaction

$N O_{2} (g) \to \frac{1}{2} N_{2} (g) + O_{2} (g) ?$

(a) $1 / (2 K_{1} K_{2})$

(b) $1 / (4 K_{1} K_{2})$

(d) $1 / (K_{1} K_{2})$

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Solution:

$\begin{aligned} N_{2} (g) + O_{2} (g) \to 2 N O (g) \\ K_{1} = \frac{[N O]^{2}}{[N_{2}] [O_{2}]} \end{aligned}$

For the reaction

$\begin{aligned} 2 N O (g) + O_{2} (g) \to 2 N O_{2} (g) \\ K_{2} = \frac{[N O_{2}]^{2}}{[N O^{2} [O_{2}] .} \end{aligned}$

For the reaction

$\begin{aligned} N O_{2} (g) \to \frac{1}{2} N_{2} (g) + O_{2} (g) \\ K = \frac{[N_{2}]^{\frac{1}{2}} [O_{2}]}{[N O_{2}]} \end{aligned}$

Hence, $K = \sqrt{\frac{1}{K_{1}} \times \frac{1}{K_{2}}}$

$K = \frac{[N_{2}]^{\frac{1}{2}} [O_{2}]^{\frac{1}{2}}}{[N O]} \times \frac{[N O] [O_{2}]^{\frac{1}{2}}}{[N O_{2}]} = \frac{[N_{2}]^{\frac{1}{2}} [O_{2}]}{[N O_{2}]}$

Equilibrium - Result Question 6

Solution:

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Equilibrium - Result Question 6

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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