MCQ
If ${K_c}$ is the equilibrium constant for the formation of $N{H_3}$, the dissociation constant of ammonia under the same temperature will be
- A${K_c}$
- B$\sqrt {{K_c}} $
- C$K_c^2$
- ✓$1/{K_c}$
${{K}_{c}}=\frac{{{[{{N}_{2}}]}^{{1}/{2}\;}}{{[{{H}_{2}}]}^{{3}/{2}\;}}}{N{{H}_{3}}}$ and $\frac{1}{2} {{N}_{2}}+\frac{3}{2}{{H}_{2}}$ $ \rightleftharpoons $ $N{{H}_{3}}$
${{K}_{c}} =\frac{[N{{H}_{3}}]}{{{[{{N}_{2}}]}^{{1}/{2}\;}}{{[{{H}_{2}}]}^{{3}/{2}\;}}}$
So for dissociation $=\frac{1}{{{K}_{c}}}$
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