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The equilibrium $N_{2 (g)} + O_{2 (g)} \to 2 {NO}_{(g)}$ established in a reaction vessel of 2.5 L capacity.
The amount of $N_{2} and O_{2}$ taken at the start were respectively 2 moles and 4 moles. Half a mole of nitrogen has been used up at equilibrium. The molar concentration of nitric oxide is

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Detailed Solution

$N_{2 (g)} + O_{2 (g)} \to 2 {NO}_{(g)}$

Initial : 2 4 0

Conc. at equilibrium

2-X 4-X 2X

According to question on ½ mole of $N_{2}$ is used up at equilibrium,

so $x = \frac{1}{2}$

$n_{0} = 2 X = 2 (1 / 2) = 1$

Molar concentration $= \frac{n}{V} = \frac{1}{2.5} = \frac{10}{25} = 4.0 mol / lit$

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