# How to find partial pressures from a given equilibrium constant Kp?

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?

1. $$P(\ce{N2O4(g)}) = 0.0252~\mathrm{bar}$$; $$P(\ce{NO2(g)}) = 1.9748~\mathrm{bar}$$
2. $$P(\ce{N2O4(g)}) = 0.039~\mathrm{bar}$$; $$P(\ce{NO2(g)}) = 1.961~\mathrm{bar}$$
3. $$P(\ce{N2O4(g)}) = 1.45~\mathrm{bar}$$; $$P(\ce{NO2(g)}) = 0.098~\mathrm{bar}$$
4. $$P(\ce{N2O4(g)}) = 1~\mathrm{bar}$$; $$P(\ce{NO2(g)}) = 1~\mathrm{bar}$$
• At the initial conditions, the mole fractions of $\ce{N2O4}$ and $\ce{NO2}$ are both 0.5. Take as a basis 1 mole of gas to start with at the initial conditions. Let $x$ represent the number of moles of $\ce{N2O4}$ that react to form $\ce{NO2}$. In terms of $x$, what are the new numbers of moles of these species? What are the new mole fractions? What are the new partial pressures? – Chet Miller Mar 24 '15 at 15:36
• Please note that in reality $K_p=0.14 bar$ for the reaction in your problem, not 0.5. This of course should not matter assuming this is considered only an exercise. It does matter however whether you provided the right answers to check the validity. – Buck Thorn 4 hours ago