1. Tardigrade
  2. Question
  3. Chemistry
  4. If the equilibrium constant for the reaction N2(g)+3H2(g)leftharpoons 2NH3(g) at 750 K is 49, then the equilibrium constant for the reaction NH3(g)leftharpoons (1/2)N2(g)+(3/2)H2(g) at the same temperature is

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Q. If the equilibrium constant for the reaction $ {{N}_{2}}(g)+3{{H}_{2}}(g)\rightleftharpoons 2N{{H}_{3}}(g) $ at 750 K is 49, then the equilibrium constant for the reaction $ N{{H}_{3}}(g)\rightleftharpoons \frac{1}{2}{{N}_{2}}(g)+\frac{3}{2}{{H}_{2}}(g) $ at the same temperature is

CMC MedicalCMC Medical 2008

Solution:

For the reaction $ {{N}_{2}}(g)+3{{H}_{2}}(g)\rightleftharpoons 2N{{H}_{3}}(g) $ $ {{K}_{c}}=\frac{{{[N{{H}_{3}}]}^{2}}}{[{{N}_{2}}]\,{{[{{H}_{2}}]}^{3}}} $ ?(i) For the reaction: $ N{{H}_{3}}(g)\rightleftharpoons \frac{1}{2}{{N}_{2}}(g)+\frac{3}{2}{{H}_{2}}(g) $ $ {{K}_{c}}=\frac{{{[{{N}_{2}}]}^{1/2}}{{[{{H}_{2}}]}^{3/2}}}{[N{{H}_{3}}]} $ ?(ii) From Eqs (i) and (ii) $ {{K}_{c}}=\sqrt{\frac{1}{{{K}_{c}}}} $ $ \therefore \,\,{{K}_{c}}=49 $ $ \because $ $ {{K}_{c}}=\sqrt{\frac{2}{49}} $ $ \because \,{{K}_{c}}=\frac{1}{7} $