How can I calculate the percentage of dissociated dinitrogen tetroxide?


> 
> When $\pu{36.8 g}$ $\ce{N2O4(g)}$ is introduced into a 1.0-litre flask at $\pu{27 °C}$ the following equilibrium reaction occurs: $\ce{N2O4(g) <=> 2NO2(g)}$; $K\_p = 0.1642~\mathrm{atm}$.
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> 
> Find the percentage dissociation.
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> 


I tried the following:


Let $x$ be the degree of dissociation and $P$ be the initial pressure of $\ce{N2O4}$ that is $\approx 9\ \mathrm{atm}$.
Then, $$\frac{9Px^2}{1-x} = 0.1642$$


Neglecting $x$ in $(1-x)$, I got $x$ as $\displaystyle \frac{0.4}{3}$ [approx]


But the answer given as $6.44~\%$.




$\ce{N2O4 (g)}$ say initially is $P\ \mathrm{atm}$.


$$PV = nRT$$


$$P(1)= (36.8/92)( 0.082 )(300 K)$$


$$P= 9.84\ \mathrm{atm}$$


Now we have:


$$\ce{N2O4(g) <=>2NO2(g)}$$


Considering initial pressure of N2O4 as p, at equilibrium, assuming 'x' as the degree of dissociation, we will have pressure of N2O4 to be p(1-x) and pressure of NO2 as 2px.


So, $$K\_p = \frac{[\ce{NO2}]^2}{[\ce{N2O4}]}$$


$$K\_p= \frac{(2px)^2}{p(1-x)}$$


$$K\_p= \frac{4px^2}{(1-x)}$$


assuming $1-x$ approximately as $1$,


$$K\_p= 4px^2$$


substituting $p$ as $9.84\ \mathrm{atm}$ and $K\_p$ as $0.1642$:


$$0.1642= 4(9.84)x^2$$


$$x^2= \frac{0.1642}{4(9.84)}$$


$$x^2= 0.1642/39.36$$


$$x^2= 0.00417$$


$$x= 0.0645$$


degree of dissociation is 0.0645


Hence % of dissociation is $0.0645 \times (100) = 6.45\ \%$ (near to your answer)


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