A) \[2.5\]
B) \[4\]
C) \[2\]
D) \[3\]
Correct Answer: A
Solution :
Initially, 2 moles of CO are present. At equilibrium, 1 mole of CO is present. Hence, 2-1=1 mole of CO has reacted. 1 mole of CO will react with 1 mole of \[C{{l}_{2}}\] to form 1 mole of\[COC{{l}_{2}}\]. \[3-1=2\] moles of \[C{{l}_{2}}\] remains at equilibrium. The equilibrium constant \[{{K}_{C}}=\frac{[COC{{l}_{2}}]}{[CO][C{{l}_{2}}]}\] \[{{K}_{C}}=\frac{\frac{1mol}{5L}}{\frac{1mol}{5L}\times \frac{2mol}{5L}}\] \[{{K}_{C}}=2.5\]You need to login to perform this action.
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