Question

At temperature $T$, a compound $A{B}_2\left(g\right)$ dissociates according to the reaction $2A{B}_2\left(g\right)\rightleftharpoons 2AB\left(g\right)+B_2\left(g\right)$ with degree of dissociation $\alpha$ (which is very small). The expression for $K_P$ in terms of $\alpha$ and the total pressure, $P_T$ is:

• A. $\frac{P_T{\alpha }^3}{2}$
• B. $\frac{P_T{\alpha }^2}{3}$
• C. $\frac{P_T{\alpha }^3}{3}$
• D. $\frac{P_T{\alpha }^2}{4}$

Answer 1

The correct option is A $\frac{P_T{\alpha }^3}{2}$

For the given equilbrium,
Let the initial moles of reactant be 1,
so the equilibrium concentration are
$\begin{array}{cccccc}& 2A{B}_2\left(g\right)& \rightleftharpoons & 2AB\left(g\right)& +& B_2\left(g\right)\\ \text{Initial moles}& 1& & 0& & 0\\ \text{Equi. moles}& (1-\alpha )& & \alpha & & \frac{\alpha }{2}\end{array}$
Total moles = $1+\frac{\alpha }{2}$

Finding Partial pressures,

$P_{A{B}_2}=\frac{1-\alpha }{1+\frac{\alpha }{2}}{P}_T$

$P_{AB}=\frac{\alpha }{1+\frac{\alpha }{2}}{P}_T$

$P_{B_2}=\frac{\frac{\alpha }{2}}{1+\frac{\alpha }{2}}{P}_T$

putting all the values,
$\therefore K_P=\frac{\left(P_{B_2}\right)(P_{AB}{)}^2}{(P_{A{B}_2}{)}^2}=\frac{\frac{\alpha }{2}\times (\alpha {)}^2\times P_T}{\left[1\right(1-\alpha ){]}^2\left[1(1+\frac{\alpha }{2})\right]}$
$K_P=\frac{a^3\times P_T}{2(1-\alpha {)}^2+(1+\frac{\alpha }{2})}$
Since, $\alpha$ is very small as compared to unity, so $1-\alpha \cong 1$
and $1+\frac{\alpha }{2}\cong 1$
$\therefore K_P=\frac{a^3\times P_T}{2}$