$$ \begin{align} \Delta_\mathrm f H^\circ(\ce{N2O4}) &= \pu{9.16 kJ mol^-1} &\quad \Delta_\mathrm f H^\circ(\ce{NO2}) &= \pu{33.18 kJ mol^-1} \\ S^\circ(\ce{N2O4}) &= \pu{304.3 J K^{-1} mol^{-1}} &\quad S^\circ(\ce{NO2}) &= \pu{204.1 J K^-1 mol^-1} \end{align} $$

for the reaction $\ce{N2O4 <=> 2 NO2},$ find the percentage of dissociation of $\ce{N2O4}$ at $\pu{1 bar}$ and $T = \pu{333 K}.$

I formed the equilibrium:

$$ \begin{array}{ccc} \ce{&N2O4 &<=> &2 NO2,} \tag{1} \\ &n_0\left(1 - \frac{\alpha}{2}\right) & & n_0\alpha \end{array} $$

where $n_0 $ are the initial amount of $\ce{N2O4}$ and $\alpha$ the dissociation value:

$$\alpha = \frac{2x}{n_0}. \tag{2}$$

After doing

$$\Delta G^\circ = -RT\ln K_p \quad\implies\quad K_p = \exp{\frac{-\Delta G^\circ}{RT}} = \pu{7.32E21}, \tag{3}$$

and having to find $\alpha,$ I get stuck with two unknown variables $n_0$ and $\alpha$:

$$K_p = p_\mathrm{tot}\frac{x(\ce{NO2})^2}{x(\ce{N2O4})} = \ldots = p_\mathrm{tot}\frac{n_0\alpha^2}{1 - \alpha/2}. \tag{4}$$

How to proceed?

  • 1
    $\begingroup$ Ideal gas law, $pV=nRT$, to find $n_0$? $\endgroup$
    – Domen
    Commented Oct 5, 2022 at 22:29
  • $\begingroup$ But we don't know the volume? $\endgroup$
    – Acedium 20
    Commented Oct 6, 2022 at 6:31
  • 1
    $\begingroup$ Your $K_p$ is vast but $K_p \approx 0.14$ so it looks like you forget that some values are in kJ and others in J . It is easier if you use $1-\alpha$ and $2\alpha$ and get partial pressure as $P_{NO2}=2\alpha P_{tot}/(1+\alpha) $ and $P_{N2O4}=(1-\alpha)P_{tot}/(1+\alpha)$ and $K_p=P_{NO2}^2/P_{N2O4}$. $\endgroup$
    – porphyrin
    Commented Oct 6, 2022 at 7:28

2 Answers 2



$A$ represent $\ce{N_2O_4}$

$C$ represent $\ce{NO_2}$

Then the reaction becomes:

$$\ce{A <=> 2C}$$

Since no initial amounts are given, I'd suggest using mole fractions for the ICE table rather than moles. Assuming we start with pure $A$:



Writing equations for the equilibrium mole fractions of $A$ and $C$:



Then, we calculate $K_x$ by using its relationship with $K_p$:

$K_x=\frac{K_p}{P^{\Delta n}}$

($\Delta n=1$ for this reaction)

Then we can use the equilibrium expression for $K_x$ and substitute the expressions of the equilibrium molar fractions:


We can then solve for $y$, and finally use its value to find $\alpha$:


But since $X_{A^o}=1$, $\alpha$ is simply:

$\alpha = y$

Note 1: The answer will depend on whatever units your calculated $K_p$ value has, so make sure it matches the units of total pressure $P$ when calculating $K_x$.

Note 2: The assumption that we initially start with pure $A$ can be found to be correct or incorrect by evaluating: $X_A+X_C≈1$


The total number of moles of gas is $n_0(1+\frac{\alpha}{2})$. So the mole fraction of $\ce{NO_2}$ is $\frac{\alpha}{(1+\frac{\alpha}{2})}$ and the mole fraction of $\ce{N_2O_4}$ is $\frac{(1-\frac{\alpha}{2})}{(1+\frac{\alpha}{2})}$. The corresponding partial pressures are equal to these mole fractions times the total pressure, and do not contain $n_0$.


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