question_answer

It the equilibrium constant for the reaction, \[{{N}_{2}}+3{{H}_{2}}2N{{H}_{3}}\]then   the equilibrium constant for the reaction \[2{{N}_{2}}+6{{H}_{2}}4N{{H}_{3}}\] would be equal to:

A) \[{{K}^{2}}\]              

B) \[\sqrt{k}\]

C) \[\frac{1}{\sqrt{k}}\]                             

D) \[\frac{1}{{{k}^{2}}}\]