| \[{{N}_{2}}(g)+3{{H}_{2}}(g)\rightleftharpoons 2N{{H}_{3}}(g)\]is given by \[Q=\frac{{{[N{{H}_{3}}]}^{2}}}{[{{N}_{2}}]{{[{{H}_{2}}]}^{3}}}.\] |
A) \[Q>{{K}_{c}}\]
B) \[Q=0\]
C) \[Q={{K}_{c}}\]
D) \[Q<{{K}_{c}}\] where \[{{K}_{c}}\]is the equilibrium constant.
Correct Answer: A
Solution :
\[{{N}_{2}}(g)+3{{H}_{2}}(g)\rightleftharpoons 2N{{H}_{3}}(g)\] \[Q(Quotient)=\frac{{{[N{{H}_{3}}]}^{2}}}{[{{N}_{2}}]{{[{{H}_{2}}]}^{3}}},\Delta n=2-4=-2\] At equilibrium Q is equal to\[{{K}_{c}}\]but for the progress of reaction towards right side \[Q>{{K}_{c}}.\]
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