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Byju's Answer

Standard IX

Chemistry

Ways of Representing the Concentration of a Solution

The equilibri...

Question

The equilibrium $N_{2} (g) + O_{2} (g) ⇌ 2 N O (g)$ is established in a reaction vessel of $2.5 L$ capacity. The amounts of $N_{2} and O_{2}$ taken at the start were $4 and 6 moles$ , respectively, and half a mole of nitrogen was used up at equilibrium. What is the molar concentration of nitric oxide at equilibrium?

0.2

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Solution

The correct option is B $0.4$
Consider the equilibrium, $N_{2} (g) + O_{2} (g) ⇌ 2 N O (g)$

$N_{2} (g)$ $O_{2} (g)$ $N O (g)$

No. of moles initially 4 mol 6 mol 0 mol

No. of moles at equilibrium 4 - 0.5 = 3.5 moles 6 - 0.5 = 5.5 moles $2 \times 0.5 = 1$ mol

Molar concentration of nitric oxide (c) = Number of moles / Volume of the gas
$= \frac{1}{2.5}$
$= 0.4$
(Volume of the container is taken as the volume of the gas)

Hence, option (b) is the correct answer.

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	$N_{2} (g)$	$O_{2} (g)$	$N O (g)$
No. of moles initially	4 mol	6 mol	0 mol
No. of moles at equilibrium	4 - 0.5 = 3.5 moles	6 - 0.5 = 5.5 moles	$2 \times 0.5 = 1$ mol

The equilibrium N2(g)+O2(g)⇌2NO(g) is established in a reaction vessel of 2.5 L capacity. The amounts of N2 and O2 taken at the start were 4 and 6 moles, respectively, and half a mole of nitrogen was used up at equilibrium. What is the molar concentration of nitric oxide at equilibrium?