A) 25
B) 88
C) 50
D) 71
Correct Answer: D
Solution :
[d] \[\begin{align} & \begin{matrix} {} & {} & {} & \,\,\,\,\,\,{{N}_{2}}{{O}_{4}}(g)\rightleftharpoons 2N{{O}_{2}}(g) \\ \end{matrix} \\ & \begin{matrix} \text{Initial}\,\,\text{moles} & {} & 1 & {} \\ \end{matrix}\begin{matrix} {} & 0 \\ \end{matrix} \\ & \begin{matrix} \text{Moles}\,\,\text{at}\,\,\text{equil}\text{.} & (1-\alpha ) & 2\alpha & {} \\ \end{matrix} \\ & \begin{matrix} {} & {} & {} & (\alpha =degree\,\,of\,dissociation) \\ \end{matrix} \\ \end{align}\] Total number of moles at equil. \[=\left( 1-\alpha \right)+2a\] \[=\left( 1+\alpha \right)\] \[{{P}_{{{N}_{2}}{{O}_{4}}}}=\frac{\left( 1-\alpha \right)}{\left( 1+\alpha \right)}\times P\] \[{{P}_{N{{O}_{2}}}}=\frac{2\alpha }{\left( 1+\alpha \right)}\times P\] \[{{K}_{p}}=\frac{{{({{P}_{N{{O}_{2}}}})}^{2}}}{{{P}_{{{N}_{2}}{{O}_{4}}}}}=\frac{{{\left( \frac{2\alpha }{(1+\alpha )}\times P \right)}^{2}}}{\left( \frac{1-\alpha }{1+\alpha } \right)\times P}=\frac{4{{\alpha }^{2}}P}{1-{{\alpha }^{2}}}\] 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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