Question

At a certain temperature, equilibrium constant $K_c=4\times {10}^{-2}$ for the reaction $N_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2NO\left(g\right)$
If we take $1.5mol$ of NO and $2mol$ each of $N_{2}and{O}_2$ in $5L$ vessel, what would be the equilibrium concentration of $NO$ (in $mol/L$)?

Answer 1

$Q_c=\frac{(1.5/5{)}^2}{(2/5{)}^2}=0.5625\because Q_c>K_c$
Therefore, the reaction will proceed in backward direction $\begin{array}{cccccc}& N_2\left(g\right)& +& O_2\left(g\right)& \rightleftharpoons & 2NO\left(g\right)\\ Initialmoles& 2& & 2& & 1.5\\ Conc.ate{q}^m& \frac{2+x}{5}& & \frac{2+x}{5}& & \frac{1.5-2x}{5}\end{array}$
$K_c=\frac{[NO{]}^2}{\left[N_2\right]\left[O_2\right]}=\frac{{\left(\frac{1.5-2x}{5}\right)}^2}{{\left(\frac{2+x}{5}\right)}^2}$
$\frac{1.5-2x}{2+x}=\sqrt{0.04}=0.2$
$1.5-2x=0.4+0.2x$
$x=0.5$
$\therefore \text{Equilibrium concentration of}NO=\frac{1.5-2x}{5}=0.1mol/L$