A) \[Q>{{K}_{c}}\]
B) Q = 0
C) \[Q={{K}_{c}}\]
D) \[Q<{{K}_{c}}\]
Correct Answer: A
Solution :
For reaction, \[{{N}_{2}}(g)+3{{H}_{2}}(g)\rightleftharpoons 2N{{H}_{3}}(g)\] 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 \[Q>{{K}_{C}}\] where \[{{K}_{C}}\] is the equilibrium constant.You need to login to perform this action.
You will be redirected in
3 sec