Consider the following reaction in a sealed vessel at equilibrium with concentrations of $\mathrm{N}

Question

Consider the following reaction in a sealed vessel at equilibrium with concentrations of $\mathrm{N}_2=3.0 	imes 10^{-3} \mathrm{M}, \mathrm{O}_2=4.2 	imes 10^{-3} \mathrm{M}$ and $ \begin{aligned} & \mathrm{NO}=2.8 	imes 10^{-3} \mathrm{M} . \ & 2 \mathrm{NO}_{(\mathrm{g})} ightleftharpoons \mathrm{N}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})} \end{aligned} $ If $0.1 \mathrm{~mol} \mathrm{~L}^{-1}$ of $\mathrm{NO}_{(g)}$ \is taken in a closed vessel, what will be degree of dissociation ( $\alpha$ ) of $\mathrm{NO}_{(g)}$ at equilibrium?

Professor PC Thomas Classes · Answer

(d) :  $ K_c=\frac{\frac{x}{2} 	imes \frac{x}{2}}{(0.1-x)^2} $\ Again $K_c==\frac{\left[\mathrm{N}_2ight]\left[\mathrm{O}_2ight]}{[\mathrm{NO}]^2}=\frac{3 	imes 10^{-3} 	imes 4.2 	imes 10^{-3}}{\left(2.8 	imes 10^{-3}ight)^2}=1.607$ $ \begin{aligned} & \Rightarrow \frac{x^2}{4(0.1-x)^2}=1.607 \Rightarrow \frac{x}{2(0.1-x)}=\sqrt{1.607} \ & \Rightarrow \frac{x}{0.2-2 x}=1.26 \Rightarrow x=0.252-2.52 x \ & \Rightarrow 3.52 x=0.252 \Rightarrow x=\frac{0.252}{3.52}=0.0716 \end{aligned} $ \ Degree of dissociation, $\alpha=\frac{0.0716}{0.1}=0.716$