3
$\begingroup$

\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 $\pu{10 bar}$?

Improve this question
$\endgroup$
  • 2
    $\begingroup$ 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 $p^\circ = 1~\mathrm{bar}$. $\endgroup$ – orthocresol♦ Nov 15 '15 at 16:08
  • $\begingroup$ Thanks for the clarification! I looked it up first, but I somehow managed to miss that post. $\endgroup$ – L3ul Nov 15 '15 at 16:59
  • 1
    $\begingroup$ Don't worry about it, there are a ton of questions here about equilibrium constants and Gibbs free energies which are difficult to search. $\endgroup$ – orthocresol♦ Nov 15 '15 at 17:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.