A) \[0.06\,mo{{l}^{-2}}{{L}^{-2}}\]
B) \[0.59\,mo{{l}^{-2}}{{L}^{-2}}\]
C) \[1.69\,\,mo{{l}^{2}}{{L}^{-2}}\]
D) \[0.03\,mo{{l}^{2}}{{L}^{-2}}\]
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 \[N{{H}_{3}}\]neutralized 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 mole of \[N{{H}_{3}}\]by the reaction between\[{{N}_{2}}\]and\[{{H}_{2}}\]. |
| \[{{N}_{2}}+3H2\,N{{H}_{3}}\] |
| \[\begin{align} & \operatorname{initial}\,cine.130 \\ & at\,quilibrium\,1-0.5\,3-0.5\times 3\,\,\,1 \\ & cine. \\ \end{align}\] |
| \[{{K}_{e}}=\frac{1\times 1}{0.5\times {{(0.5)}^{3}}}=0.592\,\,mo{{1}^{-2}}-{{L}^{2}}\] |
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