Question

For which of the following reaction, the degree of dissociation ($\alpha$) and equilibrium constant ($K_p$) are related as $K_p=\frac{4{\alpha }^{2}P}{(1-{\alpha }^2)}$?

• A. $N_2{O}_4\left(g\right)\rightleftharpoons 2{NO}_2\left(g\right)$
• B. $H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)$
• C. $N_2\left(g\right)+3H_2\left(g\right)⟶2{NH}_3\left(g\right)$
• D. $P{Cl}_3\left(g\right)+{Cl}_2\left(g\right)\rightleftharpoons P{Cl}_5\left(g\right)$

Answer 1

The correct option is A $N_2{O}_4\left(g\right)\rightleftharpoons 2{NO}_2\left(g\right)$

$\underset{1}{{N_2{O}_4}_{\left(g\right)}}\rightleftharpoons \underset{0}{{2N{O}_2}_{\left(g\right)}}$

Finally:

conc. of $N_2{O}_4=1-\alpha$

Conc. of $N{O}_2=\alpha$

$n_T=1-\alpha +2\alpha =1+\alpha$

Mole fraction $\left(X\right)=\frac{\text{no. of moles}}{\text{total moles}}$

$P=\text{Total pressure}$

$K_p=\frac{{\left(p_{N{O}_2}\right)}^2}{\left(p_{N_2{O}_4}\right)}=\frac{{(X_{N{O}_2}.P)}^2}{(X_{N_2{O}_4}.P)}=\frac{{\left(\frac{2\alpha }{1+\alpha }P\right)}^2}{\left(\frac{1-\alpha }{1+\alpha }P\right)}=\frac{\frac{4{\alpha }^2}{{(1+\alpha )}^2}{P}^2}{\frac{1-\alpha }{1+\alpha }P}$

$K_p=\frac{4{\alpha }^{2}P}{(1-\alpha )(1+\alpha )}=\frac{4{\alpha }^{2}P}{1-{\alpha }^2}$