A) \[(2{{K}_{p}}/p)\]
B) \[{{(2{{K}_{p}}/p)}^{1/3}}\]
C) \[{{(2{{K}_{p}}/p)}^{1/2}}\]
D) \[({{K}_{p}}/p)\]
Correct Answer: B
Solution :
\[\begin{matrix} {} & 2A{{B}_{2}}(g) & 2AB(g)+ & {{B}_{2}}(g) \\ \text{Initial}\,\text{moles} & 1 & 0 & 0 \\ \text{At}\,\text{equ}. & 2(1-x) & 2x & x \\ \end{matrix}\]where, x = degree of dissociation Total moles at equilibrium \[=2-2x+2x+x\] \[=(2+x)\] So, \[{{p}_{A{{B}_{2}}}}=\frac{2(1-x)p}{(2+x)}\] \[{{p}_{AB}}=\frac{2xp}{(2+x)}\] \[{{p}_{{{B}_{2}}}}=\frac{xp}{(2+x)}\] \[{{K}_{p}}=\frac{{{({{p}_{AB}})}^{2}}({{p}_{{{B}_{2}}}})}{{{({{p}_{A{{B}_{2}}}})}^{2}}}\] \[=\frac{{{\left( \frac{2xp}{2+x} \right)}^{2}}\left( \frac{x}{2+x}p \right)}{\left( \frac{2(1-x)}{(2+x)}p \right)}\] \[=\frac{{{x}^{3}}p}{(2+x){{(1-x)}^{2}}}\] \[[\because x<<<1and\,2,\,so\,(1-x)\approx 1,(2+x)\approx 2]\] \[=\frac{{{x}^{3}}p}{2}\] \[x={{\left( \frac{2{{K}_{p}}}{p} \right)}^{1/3}}\]
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