# Find the percentage of dissociation of nitrogen tetroxide given pressure, temperature, enthalpy and entropy

Given

\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?

• Ideal gas law, $pV=nRT$, to find $n_0$? Commented Oct 5, 2022 at 22:29
• But we don't know the volume? Commented Oct 6, 2022 at 6:31
• 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 $K_p=P_{NO2}^2/P_{N2O4}$. Commented Oct 6, 2022 at 7:28

Let:

$$A$$ 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 $$A$$:

$$X_{A^o}=1$$

$$X_{C^o}=0$$

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

$$X_A=X_{A^o}-y=1-y$$

$$X_C=X_{C^o}+2y=2y$$

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:

$$K_x=\frac{X_C^2}{X_A}=\frac{(2y)^2}{1-y}$$

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

$$\alpha=\frac{y}{X_{A^o}}$$

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 $$n_0$$.