SATHEE: Equilibrium - Result Question 4

Equilibrium - Result Question 4

####4. If the equilibrium constant for

$N_{2} (g) + O_{2} (g) \to 2 N O (g)$ is $K$ , the equilibrium constant for

[2015 RS] $\frac{1}{2} N_{2} (g) + \frac{1}{2} O_{2} (g) \to N O (g)$ will be:

(a)

(b) $\frac{1}{2} K$

(c) $K$

(d) $K^{2}$

Show Answer

Solution:

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

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

$\begin{aligned} \frac{1}{2} N_{2} + \frac{1}{2} O_{2} \to N O \\ K^{'} = \frac{[N O]}{[N_{2}]^{1 / 2} [O_{2}]^{1 / 2}} = K^{1 / 2} \end{aligned}$

5 .

$(c) 2 S O_{2} + O_{2} \to 2 S O_{3} K = 278 (given)$

$S O_{3} \to S O_{2} + \frac{1}{2} O_{2} K^{'} = (\sqrt{\frac{1}{K}})$

$= \sqrt{\frac{1}{278}} = 6 \times 10^{- 2}$

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