A) \[\frac{2}{3}\]
B) \[\frac{3}{2}\]
C) \[\frac{4}{5}\]
D) \[\frac{5}{2}\]
Correct Answer: A
Solution :
Given \[\frac{{{r}_{{{O}_{2}}}}}{{{r}_{mix}}}=2.084=\sqrt{\frac{{{M}_{mix}}}{32}}\] \[{{M}_{mix}}=139\,u\] The equilibrium reaction is: \[PC{{l}_{5}}(g)\xrightarrow{{}}PC{{l}_{3}}(g)+C{{l}_{2}}(g)\] \[1-\alpha \] \[\alpha \] \[\alpha \] \[{{M}_{mix}}=\frac{{{M}_{PC{{l}_{5}}}}}{1+\alpha }=\frac{208.5}{1+\alpha }=139,\,\,\alpha =0.5\] \[{{K}_{P}}=\frac{0.5\times 0.5\times 2}{1-0.25}=\frac{2}{3}\]
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