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 :
[c]:\[{{P}_{4(g)}}+6C{{l}_{2(g)}}4PC{{l}_{3(g)}}\] One mole of \[{{P}_{4}}\]reacts with 6 moles of \[C{{l}_{2}}\] i.e., at equilibrium \[C{{l}_{2}}\]is consumed more than \[{{P}_{4}}\]. If we start the reaction with equal number of moles of \[{{P}_{4}}\]and \[C{{l}_{2}}\]then obviously at equilibrium\[[{{P}_{4}}]>[C{{l}_{2}}]\].You need to login to perform this action.
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