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Q.

For the reaction, $N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$ equilibrium concentrations of $N_{2} O_{4} (g)$ and $N O_{2} (g)$ respectively are 0.2 and 0.1 mol/lit. Then, KC for the process is

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a

$0.05 M$

b

$0.05 M^{2}$

c

$0.005 M$

d

$20 M$

answer is B.

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Detailed Solution

$N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$

at eq 0.2 0.1

$K_{c} = \frac{{[N O_{2}]}^{2}}{[N_{2} O_{4}]}$

$K_{c} = \frac{{[0.1]}^{2}}{[0.2]}$ = $0.05 M$

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