A) \[\frac{P{{x}^{3}}}{2}\]
B) \[\frac{P{{x}^{2}}}{3}\]
C) \[\frac{P{{x}^{3}}}{3}\]
D) \[\frac{P{{x}^{2}}}{2}\]
Correct Answer: A
Solution :
| [a] \[\underset{1}{\mathop{2A{{B}_{2(g)}}}}\,\underset{0}{\mathop{2A{{B}_{(g)}}}}\,+\underset{0}{\mathop{{{B}_{2(g)}}}}\,\] |
| Initially |
| At equilibrium \[\left( 1-x \right)~\] x x/2 |
| Total no. of moles \[=(1-x)+x+\frac{x}{2}=1+\frac{x}{2}=\frac{2+x}{2}\] |
| Partial pressure = mole fraction x total pressure |
| Applying \[{{K}_{p}}=\frac{p_{AB}^{2}\times {{p}_{{{B}_{2}}}}}{p_{A{{B}_{2}}}^{2}}\] |
| \[=\frac{{{\left( \frac{x}{\frac{2+x}{2}}\times p \right)}^{2}}\times \left( \frac{\frac{x}{2}}{\frac{2+x}{2}}\times p \right)}{\left( \frac{1-x}{\frac{2+x}{2}}\times p \right)}=\frac{P{{x}^{3}}}{(2+x){{(1-x)}^{2}}}\] |
| Since x << 1 so \[{{(1-x)}^{2}}\] can be neglected and (2 + x) can be taken as 2. |
| \[\therefore \,{{K}_{P}}=\frac{P{{x}^{3}}}{2}\] |
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