A) \[[C{{l}_{2}}]>[PC{{l}_{3}}]\]
B) \[[C{{l}_{2}}]>[{{P}_{4}}]\]
C) \[[{{P}_{4}}]>[C{{l}_{2}}]\]
D) \[[PC{{l}_{3}}]>[{{P}_{4}}]\]
Correct Answer: C
Solution :
For the reaction: \[{{P}_{4}}(s)+6C{{l}_{2}}(g)4PC{{l}_{3}}(g)\] \[\begin{align} & \text{At}\,\,\text{t}=0\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \\ & \text{At}\,\text{eq}\text{.}\,\,\,\,\,\,\,\,\,\,(1-x)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1-6x)\,\,\,\,\,\,4x \\ & \text{As}\,\,\,\,\,\,\,\,\,(1-x)>(1-6x) \\ \end{align}\] Hence, at equilibrium?\[[{{P}_{4}}]>[C{{l}_{2}}]\]You need to login to perform this action.
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