A) \[1.333\text{ }at{{m}^{-2}}\]
B) \[0.75\text{ }at{{m}^{2}}\]
C) \[0.75\text{ }at{{m}^{-2}}\]
D) \[1.333\text{ }at{{m}^{2}}\]
Correct Answer: B
Solution :
\[{{N}_{2}}(g)+3{{H}_{2}}(g)2N{{H}_{3}}(g)\] Mole fraction of \[N{{H}_{3}}({{X}_{N{{H}_{3}}}})=0.6\] Mole fraction of \[{{N}_{2}}\]and \[{{H}_{2}}=1-0.6=0.4\] Total no. of mole of \[{{N}_{2}}\]and \[{{H}_{2}}=1+3=4\] Then, \[{{X}_{{{N}_{2}}}}=\frac{1}{4}\times 0.4=0.1\] \[{{X}_{{{H}_{2}}}}=\frac{3}{4}\times 0.4=0.3\] Partial pressure of \[{{N}_{2}}({{P}_{{{N}_{2}}}})=0.1\times 10=1\,\text{atm}\] Partial pressure of \[{{H}_{2}}({{P}_{{{H}_{2}}}})\] \[=0.3\times 10=3\,\text{atm}\] Partial pressure of \[N{{H}_{3}}\] \[({{P}_{N{{H}_{3}}}})\] \[=0.6\times 10=6\,\text{atm}\] Dissociation of ammonia \[2N{{H}_{3}}{{N}_{2}}+3{{H}_{2}}\] \[{{K}_{P}}=\frac{{{P}_{{{N}_{2}}}}\times P_{{{H}_{2}}}^{3}}{P_{N{{H}_{3}}}^{2}}\] \[=\frac{1\times {{(3)}^{3}}}{{{(6)}^{2}}}\] \[=\frac{27}{36}=0.75\,\text{at}{{\text{m}}^{\text{2}}}\]
You need to login to perform this action.
You will be redirected in
3 sec
