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: $\ce{N2O4(g) <=> 2NO2(g)}$; $K_p = 0.1642~\mathrm{atm}$.

Find the percentage dissociation.

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 $6.44~\%$.


1 Answer 1


$\ce{N2O4 (g)}$ 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)

  • 2
    $\begingroup$ 1) Please use units. 2) I don’t get your answer. 3) Please use units. $\endgroup$
    – Jan
    Dec 29, 2016 at 23:30
  • $\begingroup$ I have edited my answer to make it more understandable for the viewers. $\endgroup$
    – Yb609
    Dec 30, 2016 at 5:50

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