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At temperature $T$ a compound $AB_2(g)$ dissociates according to the reaction with a degree of dissociation $'x'$ which is very small compared to unity. The value of $x$ is
$2A{B_2}(g) \rightleftharpoons 2AB(g) + {B_2}(g)$
$\sqrt {\frac{{2{K_P}}}{P}} $
$\sqrt[3]{{\frac{{2{K_P}}}{P}}}$
$\sqrt[3]{{\frac{{{K_P}}}{P}}}$
$\sqrt[3]{{{K_P}}}$
Solution
$2A{B_2}(g) \leftrightarrow 2AB(g) + {B_2}(g)$
$1$ $0$ $0$
Initially
$(1-x)$ $x$ $\frac {x}{2}$
At equilibrium
Total moles at equilibrium $=$
$1-x+x+\frac{1}{2}=1+\frac{x}{2}=1$
$[\because \,{\text{x}}$ is small incomparision to unity $]$
${P_{{\text{A}}{{\text{B}}_2}}} = (1 – {\text{x}}){\text{P}}\quad {P_{{\text{AB}}}} = {\text{xP}}\quad {{\text{p}}_{{{\text{B}}_2}}} = \frac{{{\text{xP}}}}{2}$
$\mathrm{K}_{\mathrm{p}}=\frac{\mathrm{x}^{3} \mathrm{P}^{3}}{2(1-\mathrm{x})^{2} \mathrm{P}^{2}}=\frac{\mathrm{x}^{3} \mathrm{P}}{2} \quad[\because(1-\mathrm{x}) \approx 1]$
$K_{p}=\frac{x^{3} P}{2}$
$x^{3}=\frac{2 K_{p}}{P}$
${x}=\sqrt{\frac{2 K_{p}}{P}}$
$(2)$ is correct answer.
