Chemical EquilibriumHard
Question
At a certain temperature, the equilibrium constant of the reaction N2(g) + O2(g) $\rightleftharpoons$2NO(g) is 0.0004. Assuming air to be a mixture of four volumes of nitrogen with one volume of oxygen, the percentage of nitric oxide by volume, in the gas produced by allowing air to reach equilibrium at this temperature is
Options
A.3.0%
B.0.8%
C.0.04%
D.8.0%
Solution
N2(g) + O2(g) $\rightleftharpoons$ 2NO(g)
Initial moles 4a a 0
Equilibrium moles 4a – x a – x 2x
Now, $0.0004 = \frac{\left( 2x^{2} \right)}{(4a - x)(a - x)} \approx \frac{4x^{2}}{4a.a}$
$\Rightarrow \frac{x}{a} = 0.02$
∴ Per cent of NO = $\frac{2x}{5a} \times 100 = 0.8\%$
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