# Dissociation percentages of N2O4 at different pressures using Gibbs' free energy

\begin{align} \ce{N2O4(g)&<=>2NO2(g)} & \Delta_\mathrm{r}G^\circ &= \pu{4.7 kJ mol^-1} \end{align}

Knowing that the standard values for pressure and temperature are $$\pu{1 bar}$$ and $$\pu{298 K}$$ respectively, find:

a) the dissociation grade (noted as $$\alpha$$) of $$\ce{N2O4}$$ at equilibrium, in standard conditions

b) the dissociation grade of $$\ce{N2O4}$$ at equilibrium and at $$\pu{298 K}$$ / $$\pu{10 bar}$$

First, we find $$K$$ from $$\Delta G$$

Then we write the partial pressures as the molar fraction of the compound times the total pressure of the system (at equilibrium)

After that, we write the relation for $$K_p$$ using partial pressures

and I'm stuck...

\begin{align} \Delta_\mathrm{r}G^\circ &= -RT \ln K & \Rightarrow K&=0.15 \mathrm{~(unitless)} \\ P_{\ce{N2O4}} &= \frac{1-\alpha}{1+\alpha}\cdot P & P_{\ce{NO2}} &= \frac{2\alpha}{1+\alpha}\cdot P & K_p = \frac{(P_{\ce{NO2}})^2}{P_{\ce{N2O4}}} \\ \Rightarrow K_p &= \frac{4\alpha^2}{1-\alpha^2}\cdot P \\ \end{align}

My question is: how can I convert between $$K$$ (which is unitless) and $$K_p$$ (which is not) so that I can find the dissociation grade at $$\pu{1 bar}$$ and $$\pu{10 bar}$$?

• See my answer to this recent question: chemistry.stackexchange.com/a/40586/16683 The short answer is that the dimensionless $K$ is calculated using $p_i/p^\circ$ instead of $p_i$, where $p^\circ = 1~\mathrm{bar}$. – orthocresol Nov 15 '15 at 16:08
• Thanks for the clarification! I looked it up first, but I somehow managed to miss that post. – L3ul Nov 15 '15 at 16:59
• Don't worry about it, there are a ton of questions here about equilibrium constants and Gibbs free energies which are difficult to search. – orthocresol Nov 15 '15 at 17:00