JEE Main 2012 — Chemistry Physical Chemistry

$\begin{array}{llcc}\text { } & \mathrm{N}_2 \mathrm{O}_4(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_2(\mathrm{~g}) \\ & \text { Initial moles } & 1 & 0 \\ & \text { Moles of equil. } & (1-\alpha) & 2 \alpha\end{array}$ ​N2​O4​( g)⇌2NO2​( g) Initial moles  Moles of equil. ​1(1−α)​02α​ ( $\alpha=$α= degree of dissociation $)$) Total number of moles at equil. $$\begin{aligned} & =(1-\alpha)+2 \alpha \\ & =(1+\alpha) \\ p_{N_2 \mathrm{O}_4} & =\frac{(1-\alpha)}{(1+\alpha)} \times P \\ p_{N O_2} & =\frac{2 \alpha}{(1+\alpha)} \times P \\ K_P & =\frac{\left(p_{N_O}\right)^2}{p_{N_2 O_4}}=\frac{\left(\frac{2 \alpha}{(1+\alpha)} \times P\right)^2}{\left(\frac{1-\alpha}{1+\alpha}\right) \times P}=\frac{4 \alpha^2 P}{1-\alpha^2} \end{aligned}$$pN2​O4​​pNO2​​KP​​=(1−α)+2α=(1+α)=(1+α)(1−α)​×P=(1+α)2α​×P=pN2​O4​​(pNO​​)2​=(1+α1−α​)×P((1+α)2α​×P)2​=1−α24α2P​​ $$\begin{aligned} & \text { Given, } K_P=2, P=0.5 \mathrm{~atm} \\ & \therefore \quad K_P=\frac{4 \alpha^2 P}{1-\alpha^2} \\ & =\frac{4 \alpha^2 \times 0.5}{1-\alpha^2} \\ & \alpha=0.707 \approx 0.71 \\ & \therefore \quad \text { Percentage dissociation } \\ & =0.71 \times 100=71 \end{aligned}$$​ Given, KP​=2,P=0.5 atm∴KP​=1−α24α2P​=1−α24α2×0.5​α=0.707≈0.71∴ Percentage dissociation =0.71×100=71​