Chemistry - Chemical Equilibrium Question with Solution | TestHub

$\begin{array}{llcc}& {\mathrm{N}}_2{\mathrm{O}}_4(\mathrm{g})\rightleftharpoons 2{\mathrm{NO}}_2(\mathrm{g})& & \\ & \text{ Initial moles }& 1& 0\\ & \text{ Moles of equil. }& (1-\alpha )& 2\alpha \end{array}$ ( $\alpha =$ degree of dissociation $)$ Total number of moles at equil. $$\begin{array}{rl}& =(1-\alpha )+2\alpha \\ & =(1+\alpha )\\ p_{N_2{\mathrm{O}}_4}& =\frac{(1-\alpha )}{(1+\alpha )}\times P\\ p_{N{O}_2}& =\frac{2\alpha }{(1+\alpha )}\times P\\ K_P& =\frac{{\left(p_{N_O}\right)}^2}{p_{N_2{O}_4}}=\frac{{\left(\frac{2\alpha }{(1+\alpha )}\times P\right)}^2}{\left(\frac{1-\alpha }{1+\alpha }\right)\times P}=\frac{4{\alpha }^{2}P}{1-{\alpha }^2}\end{array}$$ $$\begin{array}{rl}& \text{ Given, }{K}_P=2,P=0.5\mathrm{atm}\\ & \therefore K_P=\frac{4{\alpha }^{2}P}{1-{\alpha }^2}\\ & =\frac{4{\alpha }^2\times 0.5}{1-{\alpha }^2}\\ & \alpha =0.707\approx 0.71\\ & \therefore \text{ Percentage dissociation }\\ & =0.71\times 100=71\end{array}$$