The density of an equilibrium mixture of $\ce{N2O4}$ and $\ce{NO2}$ at $\pu{1 atm}$ and $\pu{373.5 K}$ is $\pu{2.0 g/L}$. Calculate $K_c$ for the dissociation reaction

I did this:

  • Let initial amount of substance be $a$.
  • Initial vapor density $=(\text{mass of }\ce{N2O4})/2 =46$
  • For final vapour density I find effective molecular mass of the mixture $\ce{N2O4}$ + $\ce{NO2}$ at equilibrium using ideal gas equation which comes out to be $61.33$ and thus final vapor density is $61.33/2=30.665$.
  • Now supposing degree of dissociation to be $x$ , total amount of substance at equilibrium is $a\times(1+x)$.
  • As $\text{Vapor density} \times \text{total amount of substance} = \text{constant}$ at same temperature, $46\times a=30.665\times a\times(1+x)$
  • $x=0.5$ Using this in the equation $$K_p=\left[\frac{4x^2}{1-x^2}\right]\times P$$ for this reaction I get $K_p=\frac43\pu{ atm}$ and then $K_c=0.043$.
  • The answer is 2 but how?
  • 1
    $\begingroup$ Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. I have updated your post with chemistry markup. If you want to know more, please also have a look here. I have tried to make your post more readable, but I cannot be sure, since I do not understand in the slightest what you have tried there. $\endgroup$ Commented Jul 21, 2017 at 12:57

2 Answers 2


I confirm your answer.

The molar density of the mixture is

$$\frac{n}{V}=\frac{p}{RT} = \pu{0.0326 mol/L}$$

So, the average molecular weight of the mixture is

$$\frac{2}{0.0326} = \pu{61.35 g/mol}$$

If $x$ is the mole fraction of $\ce{NO2}$ and $(1-x)$ is the mole fraction of $\ce{N2O4}$, then the average molecular weight of the mixture is also

$$46x + 92 (1 - x) = 61.35$$

Solving for $x$ gives $x = 1/3$. So, $(1-x)=2/3$. These are also the partial pressures of $\ce{NO2}$ and $\ce{N2O4}$, respectively (in atm). This leads to the values of $K_p$ and $K_c$ that you calculated.

  • 1
    $\begingroup$ A very elegant solution! I also like how you used the "molar density" term, not the "concentration". Kudos for that! $\endgroup$
    – andselisk
    Commented Jul 23, 2017 at 13:15

I don't entirely understand your way of solving, but I obtained pretty much the same results calculating $K_p$ and $K_C$

$K_p = \pu{1.21e5 Pa}$; $K_C = \pu{3.9e-2 mol L^{-1}}$

in the following way. I assumed the following dissociation process:

$$\ce{N2H4 (g) <=> 2NO2 (g)}$$

Further I denote $\ce{N2H4}$ as compound 1, and $\ce{NO2}$ as compound 2. Ideal gas equilibrium constants are linked via the equation:

$$K_C = \frac{K_p}{(RT)^{\Delta n}},$$

where $\Delta n = \sum{n_j(\text{products})} - \sum{n_i(\text{reactants})}$. In this case $\Delta n = 2 - 1 = 1$. Also at equilibrium $K_p$ is

$$K_p = \frac{p_2^2}{p_1}.$$

Density $d$ for the mixture can be written as:

$$d = \frac{m_1 + m_2}{V} = \frac{m_1}{V} + \frac{m_2}{V}.$$

These terms can be obtained from the ideal gas law:

$$\frac{m_i}{V} = \frac{M_i p_i}{RT},$$

which in combination with partial pressures brings us to the following system of equations:

$$\begin{cases} d = \frac{1}{RT}(M_1 p_1 + M_2 p_2) \\ p = p_1 + p_2 \end{cases}$$

Considering that $p_1 = p - p_2$ and $M_1 = 2M_2$ ($\ce{N2O4}$ has twice the molar mass of $\ce{NO2}$) we have:

$$d = \frac{1}{RT}(2M_2 (p - p_2) + M_2 p_2) = \frac{M_2}{RT}(2p - p_2)$$


$$p_2 = 2p - \frac{dRT}{M_2}$$

from where we can calculate $p_2$ using given values:

$$p_2 = \pu{2e5 Pa} - \frac{\pu{2 kg m^{-3}} \cdot \pu{8.31 J mol^{-1} K^{-1}} \cdot \pu{373.5 K}}{\pu{46e-3 kg mol^{-1}}} = \pu{6.5e4 Pa},$$

$$p_1 = \pu{1e5 Pa} - \pu{6.5e4 Pa} = \pu{3.5e4 Pa}.$$

Now we can calculate $K_p$:

$$K_p = \frac{(\pu{6.5e4 Pa})^2}{\pu{3.5e4 Pa}} = \pu{1.21e5 Pa},$$

which is about $\frac{4}{3}$ atm, and, finally, $K_C$:

$$K_C = \frac{K_p}{RT} = \frac{\pu{1.21e5 Pa}}{\pu{8.31 J mol^{-1} K^{-1}} \cdot \pu{373.5 K}} = \pu{39 mol m^{-3}} = \pu{3.9e-2 mol L^{-1}}.$$

I haven't practiced these problems for a long time, so my solution may not be the optimal one.


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