A) \[0.0625\]
B) \[0.625\]
C) \[6.280\]
D) \[6.250\]
Correct Answer: A
Solution :
\[\underset{4}{\mathop{2A}}\,+\underset{2}{\mathop{B}}\,\underset{2}{\mathop{B{{A}_{2}}}}\,_{Equilibrium\text{ }concentration}^{{}}\] \[{{K}_{c}}=\frac{[B{{A}_{2}}]}{{{[A]}^{2}}[B]}\] \[=\frac{2}{{{(4)}^{2}}\times 2}=\frac{1}{16}=0.0625\]You need to login to perform this action.
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