A) \[\frac{p}{2}\]
B) \[\frac{p}{3}\]
C) \[\frac{p}{4}\]
D) \[\frac{p}{6}\]
Correct Answer: B
Solution :
| [b] \[{{N}_{2}}+3{{H}_{2}}2\,N{{H}_{3}}\] |
| Initially 1 3 0 |
| At equilibrium \[\underset{0.5}{\mathop{(1-x)}}\,\] \[\underset{1.5}{\mathop{(3-3x)}}\,\]\[\underset{1.0}{\mathop{2x}}\,\] \[(\because \,\,x=0.5)\] |
| Mole fraction of \[N{{H}_{3}}=\frac{{{n}_{N{{H}_{3}}}}}{{{n}_{Total}}}=\frac{1}{0.5+1.5+1}=\frac{1}{3}\] |
| Thus, partial pressure of \[N{{H}_{3}},\,{{P}_{N{{H}_{3}}}}={{P}_{Total}}\times mole\] |
| fraction of \[N{{H}_{3}}=p\times \frac{1}{3}=\frac{p}{3}\] |
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