I answered wrong on my quiz the following question on finding an equilibrium constant:
For the following reaction, at $1~\mathrm{bar}$ and $300~\mathrm{K}$, partial pressures of $\ce{NO2}$ and $\ce{N2O4}$ are both $0.5~\mathrm{bar}$. If pressure is increased to $2~\mathrm{bar}$ (this is the final pressure), and temperature is increased to $400~\mathrm{K}$, calculate the final composition of the system. $\Delta H$ for this reaction is $57.2~\mathrm{kJ/mol}$; assume it is independent of temperature.
$$\ce{N2O4 (g) <=> 2NO2 (g)}$$
My attempt:
I first found $K_\textrm{p}$:
$$K_p (300~\mathrm{K}) = \frac{P^2(\ce{NO2})}{P(\ce{N2O4})} = (0.5^2)/0.5 = 0.5$$
Then I used
$$\ln\left(\frac{K_p(400~\mathrm{K})}{K_p(300~\mathrm{K})}\right) = - \frac{\Delta H}{\mathcal{R}} \left( \frac{1}{400~\mathrm{K}} - \frac{1}{300~\mathrm{K}}\right) $$
I solved for $K_\textrm{p}$ at $\pu{400 K} = 92695.44$
This is where everything went wrong. I can not figure out how to get from $K_p$ to partial pressures. All I got to is the following line:
$P_\text{total} = 2P(\ce{NO2}) + P(\ce{N2O4})$, since $n(\ce{NO2})$ to $n(\ce{N2O4})$ ratio is $2:1$. I assumed they exert pressure proportionally.
Can some one guide me in the right direction?
Multiple choice answers:
- $P(\ce{N2O4(g)}) = 0.0252~\mathrm{bar}$; $P(\ce{NO2(g)}) = 1.9748~\mathrm{bar}$
- $P(\ce{N2O4(g)}) = 0.039~\mathrm{bar}$; $P(\ce{NO2(g)}) = 1.961~\mathrm{bar}$
- $P(\ce{N2O4(g)}) = 1.45~\mathrm{bar}$; $P(\ce{NO2(g)}) = 0.098~\mathrm{bar}$
- $P(\ce{N2O4(g)}) = 1~\mathrm{bar}$; $P(\ce{NO2(g)}) = 1~\mathrm{bar}$