A) 0.00358 M
B) 0.0358 M
C) 3.58 M
D) 0.358 M
Correct Answer: D
Solution :
\[\begin{matrix} {} & {{N}_{2}}+ & 3{{H}_{2}}\rightleftharpoons & 2N{{H}_{3}} \\ \text{at}\,\,\text{equilibrium} & 2M & 3M & {} \\ \end{matrix}\] From the law of mass-action, \[{{K}_{c}}=\frac{{{[N{{H}_{3}}]}^{2}}}{{{[{{H}_{2}}]}^{3}}\,[{{N}_{2}}]}\] \[2.37\times {{10}^{-3}}=\frac{{{[N{{H}_{3}}]}^{2}}}{{{(3)}^{2}}(2)}\] \[\therefore \] \[{{[N{{H}_{3}}]}^{2}}=2.37\times {{10}^{-3}}\times 27\times 2=0.127\] \[\therefore \] \[[N{{H}_{3}}]=0.3577\approx 0.358\,M\]
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