A) \[P=3{{K}_{p}}\]
B) \[P={{K}_{p}}\]
C) \[P=8{{K}_{p}}\]
D) \[P=4{{K}_{p}}\]
Correct Answer: C
Solution :
\[A{{B}_{(g)}}{{A}_{(g)}}+{{B}_{(g)}}\] \[\begin{align} & \text{Initial}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{0}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{0} \\ & \text{Moles}\,\text{at}\,\text{equilibrium}\,\,\,\,\,1-0.33\,\,\,\,\,\,\,\,0.33\,\,\,\,\,0.33 \\ \end{align}\] \[{{P}_{A}}=\left( \frac{0.33}{1.33} \right)P.{{P}_{B}}=\left( \frac{0.33}{1.33} \right)P\] \[{{P}_{AB}}=\left( \frac{0.67}{1.33} \right)P\] \[{{K}_{P}}=\frac{{{P}_{A}}\times {{P}_{B}}}{{{P}_{AB}}}\] \[{{K}_{P}}=\frac{0.33\times 0.33\times P}{0.67\times 1.33}\] \[\frac{P}{{{K}_{P}}}=8\] \[P=8{{K}_{P}}.\] Hence, the correct option is [c].
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