Question

The degree of dissociation of dinitrogen tetraoxide$N_2{O}_4\left(g\right)⟶{2NO}_2\left(g\right)$ at temperature $T$ and total pressure $p$ is$\alpha$.Which one of the following is the correct expression for the equilibrium constant $\left(K_p\right)$ at this temperature ?

• A. $2\alpha /(1-{\alpha }^2)$
• B. ${\alpha }^{2}P/(1-\alpha )$
• C. $4{\alpha }^2/(1-{\alpha }^2)$
• D. $4{\alpha }^{2}P/(1-{\alpha }^2)$

Answer 1

Answer: (A)

$2\alpha /(1-{\alpha }^2)$

$N_2{O}_4\rightleftharpoons 2{NO}_2$

$1$ $0$ initial

$1-\alpha$ $2\alpha$ equilibrium

$P_{N_2{O}_4}=\frac{P(1-\alpha )}{P(1+\alpha )}{P}_{{NO}_2}=\frac{P\left(2\alpha \right)}{(1+\alpha )}$

$K_P=\frac{{\left(P_{{NO}_2}\right)}^2}{\left(P_{N_2{O}_4}\right)}=\frac{{\left(\frac{P\left(2\alpha \right)}{(1+\alpha )}\right)}^2}{\frac{P(1-\alpha )}{P(1+\alpha )}}=\frac{P^2{4\alpha }^2}{{\left(P(1-\alpha )\right)}^2}=\frac{{4\alpha }^2}{(1-{\alpha }^2)}$