At temperature $\mathrm{T}$, a compound $\mathrm{AB}_{2}(\mathrm{~g})$ dissociates according to the reaction…

At temperature $\mathrm{T}$, a compound $\mathrm{AB}_{2}(\mathrm{~g})$ dissociates according to the reaction $2 \mathrm{AB}_{2}(\mathrm{~g})$ $ightleftharpoons$ $2 \mathrm{AB}(\mathrm{g})+\mathrm{B}_{2}(\mathrm{~g})$ with degree of dissociation $\alpha$, which is small compared with unity. The expression for $\mathrm{K}_{\mathrm{p}}$, in terms of $\alpha$ and the total pressure, $\mathrm{P}_{\mathrm{T}}$ is
  1. $\frac{\mathrm{P}_{\mathrm{T}} \alpha^{3}}{2}$
  2. $\frac{\mathrm{P}_{\mathrm{T}} \alpha^{2}}{3}$
  3. $\frac{\mathrm{P}_{\mathrm{T}} \alpha^{3}}{3}$
  4. $\frac{\mathrm{P}_{\mathrm{T}} \alpha^{2}}{2}$

Solution

For the given equilibrium, the equilibrium concentration are
$2 \mathrm{AB}_{2}(\mathrm{~g}) ightleftharpoons 2 \mathrm{AB}(\mathrm{g})+\mathrm{B}_{2}(\mathrm{~g})$
Equilib. conc. $c(1-\alpha) \quad c \alpha \quad \frac{C d}{2}$
$\therefore$ $\mathrm{K}_{\mathrm{P}}=\frac{\left(\mathrm{P}_{\mathrm{B}_{2}}ight)\left(\mathrm{P}_{\mathrm{AB}}ight)^{2}}{\left(\mathrm{P}_{\mathrm{AB}_{2}}ight)^{2}}=\frac{\frac{\mathrm{c} \alpha}{2} \times(\mathrm{c} \alpha)^{2} \times \mathrm{P}_{\mathrm{T}}}{[\mathrm{c}(1-\alpha)]^{2}\left[\mathrm{c}\left(1+\frac{\alpha}{2}ight)ight]} ; \mathrm{K}_{\mathrm{P}}=\frac{\alpha^{3} \times \mathrm{P}_{\mathrm{T}}}{2(1-\alpha)^{2}\left(1+\frac{\alpha}{2}ight)}$
Since, $\alpha$ is small compared to unity,
so $1-\alpha \simeq 1$ and $1+\frac{\alpha}{2} \simeq 1$.
$\mathrm{K}_{\mathrm{P}}=\frac{\alpha^{3} \times \mathrm{P}_{\mathrm{T}}}{2}$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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