Question

Air containing $79$% of nitrogen and $21$% of oxygen by volume is heated at $2200K$ and $1$ atm until equilibrium is established according to the reaction
$N_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2NO\left(g\right)$
If the $K_p$ of the reaction is $1.1\times {10}^{-3}$, calculate the amount of nitric oxide produced in terms of volume percent.

• A. $1.33$
• B. $1.12$
• C. $1.02$
• D. $1.44$

Answer 1

Answer: (A)

$1.33$

$N_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2NO\left(g\right)$

At equilibrium, we have $\left[N_2\right]=0.79(1-\alpha )$; $[O_2=0.21(1-\alpha )$; $\left[NO\right]=2\alpha$

Total number of moles $=0.79(1-\alpha )+0.21(1-\alpha )+2\alpha =1+\alpha$

$P{N}_2=\frac{0.79(1-\alpha )}{1+\alpha }\times 1$;

$P{O}_2=\frac{0.21(1-\alpha )}{1+\alpha }\times 1$;

$P_{NO}=\frac{2\alpha }{1+\alpha }\times 1$

$K_p=\frac{P_{NO}^2}{P_{N_2}.P_{O_2}}$

$1.1\times {10}^{-3}=\frac{4{\alpha }^2}{0.79\times 0.21{(1-\alpha )}^2}$
or $\alpha =0.0067$

$\Rightarrow$ vol% of $NO=2\alpha \times 100$

$=2\times 0.0067\times 100=1.33$%

Hence, option A is correct.