A) 3/4 \[mo{{l}^{2}}\,litr{{e}^{-2}}\]
B) 27/64 \[mo{{l}^{2}}\,litr{{e}^{-2}}\]
C) 27/32 \[mo{{l}^{2}}\,litr{{e}^{-2}}\]
D) 27/1 \[mo{{l}^{2}}\,litr{{e}^{-2}}\]
Correct Answer: D
Solution :
\[_{\underset{at.equil.}{\mathop{{{_{{}}}_{_{initial}}}}}\,}\underset{\begin{smallmatrix} 2 \\ 1 \end{smallmatrix}}{\mathop{2N{{H}_{3}}}}\,\] ⇌\[\underset{\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}}{\mathop{{{N}_{2}}}}\,+\underset{\begin{smallmatrix} 0 \\ 3 \end{smallmatrix}}{\mathop{3{{H}_{2}}}}\,\]; \[K\frac{[{{N}_{2}}]{{[{{H}_{2}}]}^{3}}}{{{[N{{H}_{3}}]}^{2}}}=\frac{1\times {{3}^{3}}}{1}=27\]
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