Question

An equilibrium mixture in a vessel of capacity 100 litre contain 1mol $N_2$, 2mol $O_2$ and 3mol $NO$. Number of moles of $O_2$ to be added so that at new equilibrium the conc. of $NO$ is found to be 0.04 mol/lit is:

• A. $101/18$
• B. $101/9$
• C. $202/9$
• D. none of these

Answer 1

The correct option is A $101/18$

$H_2\left(g\right)+O_2\left(g\right)\leftrightharpoons 2NO\left(g\right)$

$MolesatEquilibrium123K_C=\frac{{\left[3\right]}^2}{\left[2\right]\left[1\right]}=\frac{9}{2}DistributedEquilibrium12+x3Q_C=\frac{9}{[2+x]}NewEquilibrium1-y2+x-y3+2y$

as $Q_C<K_C$ thus Reaction will go in forward direction.

Given $V=100L$

Concentrating of $NO$ at new equilibrium $=0.04mol/l$ Thus $\frac{3+2y}{100}=0.04\Rightarrow y=\frac{1}{2}$
Thus moles of $B{N}_2$ at new equilibrium $1-y=-\frac{1}{2}=\frac{1}{2}moles$

$K_C=\frac{[NO{]}^2}{\left[N_2\right]\left[O_2\right]}=\frac{|0.04{|}^2}{\left[\frac{1/2}{100}\frac{[}{\frac{2+x-\frac{1}{2}}{100}}\right]]}=\frac{16}{\left[1/2\right][\frac{3}{2}+x]}{K}_C=\frac{9}{2}=\frac{32}{[\frac{3}{2}+x]}\Rightarrow \frac{3}{2}+x=\frac{64}{9}x=5.6=\frac{101}{18}$

Thus $5.6$ moles of $O_2$ was added to make concentration of $NO=0.04mol/l$