Chemistry - Chemical Equilibrium Question with Solution | TestHub

For${\text{Br}}_2\left(\text{g}\right)\rightleftharpoons 2\text{Br}\left(\text{g}\right)$
$\Delta {\text{S}}_{\text{reac}}^{\circ }=2\times 175-244·4=105.6\text{ J}{\text{ K}}^{-1}$
$\Delta {\text{H}}_{\text{reac}}^{\circ }=82\text{ KJ}{\text{ m}}^{-1}$
$\Delta {\text{G}}_{\text{reac}}^{\circ }=\Delta {\text{H}}_{\text{reac}}^{\circ }-\text{T}\Delta {\text{S}}^{\circ }=82000-500\times 105.6$
$=29·2\times 10^3\text{J}=-\text{RT }\text{ln }{\text{K}}_{\text{P}}$
$\text{ln }{\text{K}}_{\text{P}}=\frac{-29700}{8·314\times 500}=-7.02$
$\therefore {\text{K}}_{\text{P}}=9\times {10}^{-4}$
${\text{K}}_{\text{P}}=\frac{4{\text{α}}^{\text{2}}\text{P}}{1-{\text{α}}^{\text{2}}}$
as$\text{α}<<1\therefore 1-{\text{α}}^{\text{2}}=1$
$\therefore {\text{K}}_{\text{P}}=4\times {\text{α}}^{\text{2}}\text{P}$
$\because \text{P}\text{=}\text{40}$
$\therefore 9\times {10}^{-4}=4\cdot {\alpha }^2\cdot 40$
$\therefore {\text{α}}^{\text{2}}=\frac{9}{160}\times {10}^{-4}$
$=\frac{9}{16}\times {10}^{-5}$
$=2.37\times {10}^{-3}$