A) 25
B) 88
C) 50
D) 71
Correct Answer: D
Solution :
\[{{N}_{2}}{{O}_{4}}(g)2N{{O}_{2}}(g)\] Initial moles 1 0 Moles at equil. \[(1-\alpha )\] \[2\alpha \] (\[\alpha =\] degree of dissociation) Total number of moles at equil. \[=1(1-\alpha )+2\alpha \] \[=(1+\alpha )\] \[p{{N}_{2}}{{O}_{4}}=\frac{(1-\alpha )}{a(1+\alpha )}\times P\] \[{{p}_{N{{O}_{2}}}}=\frac{2\alpha )}{a(1+\alpha )}\times P\] \[{{K}_{P}}=\frac{{{({{p}_{N{{O}_{2}}}})}^{2}}}{{{P}_{{{N}_{2}}{{O}_{4}}}}}=\frac{{{\left( \frac{2\alpha }{(1+\alpha )}\times P \right)}^{2}}}{\left( \frac{1-\alpha }{1+\alpha } \right)\times P}=\frac{4{{\alpha }^{2}}P}{1-{{\alpha }^{2}}}\] Given, \[{{K}_{P}}=2,P=0.5\,atm\] \[\therefore \]\[{{K}_{P}}=\frac{4{{\alpha }^{2}}P}{1-{{\alpha }^{2}}}\] \[=\frac{4{{\alpha }^{2}}\times 0.5}{1-{{\alpha }^{2}}}\] \[\alpha =0.707\approx 0.71\] \[\therefore \]Percentage dissociation\[=0.71\times 100=71\]
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