Equilibrium Question 769
Question: The value of $ K_{p} $ for the equilibrium reaction $ N_2O_4(g)\rightarrow 2NO_2(g) $ is 2. The percentage dissociation of $ N_2O_4(g) $ at a pressure of 0.5 atm is
Options:
A) 25
B) 88
C) 50
D) 71
Show Answer
Answer:
Correct Answer: D
Solution:
$ \begin{aligned} & \begin{matrix} {} & {} & {} & N_2O_4(g)\rightarrow 2NO_2(g) \\ \end{matrix} \\ & \begin{matrix} Initialmoles & {} & 1 & {} \\ \end{matrix}\begin{matrix} {} & 0 \\ \end{matrix} \\ & \begin{matrix} Molesatequil\text{.} & (1-\alpha ) & 2\alpha & {} \\ \end{matrix} \\ & \begin{matrix} {} & {} & {} & (\alpha =degreeof,dissociation) \\ \end{matrix} \\ \end{aligned} $ Total number of moles at equil. $ =( 1-\alpha )+2a $
$ =( 1+\alpha ) $
$ {P_{N_2O_4}}=\frac{( 1-\alpha )}{( 1+\alpha )}\times P $
$ {P_{NO_2}}=\frac{2\alpha }{( 1+\alpha )}\times P $
$ K_{p}=\frac{{{({P_{NO_2}})}^{2}}}{{P_{N_2O_4}}}=\frac{{{( \frac{2\alpha }{(1+\alpha )}\times P )}^{2}}}{( \frac{1-\alpha }{1+\alpha } )\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 $
