| The value of \[{{\operatorname{K}}_{\rho }}\] for the equilibrium reaction |
| \[{{N}_{2}}{{O}_{4}}(g)2N{{O}_{2}}(g)is2.\] |
| The percentage dissociation of \[{{N}_{2}}{{O}_{4}}(g)\] at a pressure of 0.5 atm is |
A) 25
B) 88
C) 50
D) 71
Correct Answer: D
Solution :
| \[{{\operatorname{N}}_{2}}{{O}_{4}}\left( g \right)2N{{O}_{2}}\left( g \right)\] |
| \[\begin{align} & \operatorname{initial}\,moles\,10 \\ & moles\,at\,eqm.(1-\alpha )2\alpha \\ \end{align}\] |
| Where, \[\alpha \]=degree of dissociation |
| Total number of moles at equilibrium. |
| \[=(1-\alpha )+2\alpha =(1+\alpha )\] |
| \[{{P}_{{{N}_{2}}{{O}_{4}}}}=\frac{(1-\alpha )}{(1+\alpha )}\times P\,\,;\,\,{{P}_{N{{O}_{2}}}}=\frac{2a}{(1+\alpha )}\times P\] |
| \[{{K}_{P}}=\frac{{{({{p}_{N{{O}_{2}}}})}^{2}}}{{{p}_{{{N}_{2}}{{O}_{4}}}}}=\frac{{{\left( \frac{2a}{1+a}\times p \right)}^{2}}}{\left( \frac{1-\alpha }{1+\alpha } \right)\times P}=\frac{4{{\alpha }^{2}}P}{{{(1-\alpha )}^{2}}}\] |
| Given, \[{{K}_{p}}=2\], \[P=0.5atm\] |
| \[\therefore \] \[{{K}_{p}}=\frac{4{{\alpha }^{2}}P}{1-{{\alpha }^{2}}};2=\frac{4{{\alpha }^{2}}\times 0.5}{1-{{\alpha }^{2}}}\] |
| \[\alpha =0.707\approx 0.71\] |
| \[\therefore \]Percentage dissociation |
| \[=0.71\times 100=71\] |
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