Question

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

• A. $N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$
• B. $H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)$
• C. $N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$
• D. $PC{l}_3\left(g\right)+C{l}_2\left(g\right)\rightleftharpoons PC{l}_5\left(g\right)$

Answer 1

Answer: (A)

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

$N_2{O}_4\left(g\right)\rightleftharpoons 2{NO}_2\left(g\right)$

$\begin{array}{c}\text{Initial:}10\\ \text{Final}:1-\alpha 2\alpha \end{array}$

$\begin{array}{c}{X}_A=\text{Mole}\text{fraction}=\frac{no.\text{of moles of A}}{\text{total moles}}\\ P=\text{Total bressure}\end{array}$

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

$\begin{array}{c}{K}_p=\frac{{\left(P_{N{O}_2}\right)}^2}{P_{N_2{O}_4}}=\frac{{\left(X_{N{0}_2}P\right)}^2}{X_{N_2{O}_4}P}=\frac{{\left(\frac{2\alpha }{1+\alpha }P\right)}^2}{\left(\frac{1-\alpha }{1+\alpha }\right)P}\end{array}$

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

$\therefore K_p=\frac{4{\alpha }^{2}p}{1-{\alpha }^2}$