A) \[0.06\text{ }mo{{l}^{-2}}{{L}^{2}}\]
B) \[0.59\text{ }mo{{l}^{-2}}{{L}^{2}}\]
C) \[1.69\text{ }mo{{l}^{2}}{{L}^{-2}}\]
D) \[0.03\text{ }mo{{l}^{2}}{{L}^{-2}}\]
Correct Answer: B
Solution :
[b] Moles of \[{{N}_{2}}=\frac{28}{28}=1,\] Moles of \[{{H}_{2}}=\frac{6}{2}=3\] Moles of \[{{H}_{2}}S{{O}_{4}}\] required \[=\frac{500\times 1}{1000}=0.5\] Moles of \[N{{H}_{3}}\] neutralised by \[{{H}_{2}}S{{O}_{4}}=1.0\] \[\left( 2N{{H}_{3}}+{{H}_{2}}S{{O}_{4}}\xrightarrow{{}}{{\left( N{{H}_{4}} \right)}_{2}}S{{O}_{4}} \right)\] Hence 1 mol of \[N{{H}_{3}}\] by the reaction between \[{{N}_{2}}\]and\[{{H}_{2}}\]. \[\begin{align} & \begin{matrix} {} & {} & {} & {{N}_{2}}+3{{H}_{2}}\rightleftharpoons 2N{{H}_{3}} \\ \end{matrix} \\ & \,\,\,\begin{matrix} \text{initial} & {} & 1 & {} \\ \end{matrix}\begin{matrix} 3 & {} & \,\,\,\,\,\, & 0 \\ \end{matrix} \\ & \begin{matrix} \text{at}\,\text{eqm} & 1-0.5 & 3-0.5\times 3 & 1 \\ \end{matrix} \\ \end{align}\] \[{{K}_{c}}=\frac{1\times 1}{0.5\times {{(1.5)}^{3}}}=0.592\]
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