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Law of Mass Action

An equilibriu...

Question

An equilibrium mixture in a vessel of capacity $100$ litre contain $1 m o l$ $N_{2}$ , $2 m o l$ $O_{2}$ and $3 m o l$ $N O$ . No. of moles of $O_{2}$ to be added so that an new equiirbium the conc. of $N O$ is found to be $0.04 m o l e / l i t$ is:

(108 / 18)

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(108 / 9)

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(216 / 18)

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none of these

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Solution

The correct option is A $(108 / 18)$
$N_{2} (g) + O_{2} (g) ⇌ 2 N O_{2} (g)$
$K_{c} = \frac{[N o]^{2}}{[N_{2}] . [O_{2}]} .$
$[O_{2}] = \frac{2 m o l e}{100 L} = 0.02 m$
$[N_{2}] = \frac{1}{100} = 0.01 m$
$[N O] = \frac{3}{100} = 0.03 M$
$\Rightarrow K C = \frac{(0.03)^{2}}{(0.01) (0.02)} = \frac{3^{2} . \frac{1}{100^{2}}}{2^{2} . \frac{1}{100} . \frac{1}{100}}$
$K_{c} = 9 / 2$
After adding $O_{2}$ gas [NO] new = 0.04 m
$∵ [N O]$ concentration changes as $0.04 - 0.03$
$= 0.01 m$
So from radius we find that concentration
of $N_{2}$ and $O_{2}$ amust decrease by $1 / 2 . (0.01) m$
So $[N_{2}]_{n e w} = [N_{2}] - \frac{1}{2} .0 .01 M$
$= 0.05 M$
$[O_{2}]_{n e w} = \frac{[N O]_{n e w}^{2}}{k c . [N_{2}]_{n e w}} = \frac{16}{225} M$
$[O_{2}]$ increased $= \frac{16}{225} + \frac{1}{2} .0 .01 = \frac{137}{1800} M$
no. of moles of $O_{2} = \frac{\frac{137}{1800} - 0.02}{1 L} \times 100 L = \frac{108}{18}$

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