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

I had following question on my quiz and but I answer it wrong because I can't seem to solve it. It is related to finding equilibrium constant. Question:

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 $K_p$

$$K_p (300~\mathrm{K}) = \frac{P^2(\ce{NO2})}{P(\ce{N2O4})} = (0.5^2)/0.5 = 0.5$$

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 $K_p$ at $400~\mathrm{K} = 92695.44$

Then (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 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 proportionality

Can some one guide me to 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. $P(\ce{N2O4(g)}) = 1~\mathrm{bar}$; $P(\ce{NO2(g)}) = 1~\mathrm{bar}$
• I just calculated the answer by doing some answer to question maping. essentially, I solved for ln(Kp at 400K) on following equation ln(Kp at 400k / Kp at 300k) = - (delta H / R) ( 1/ 400K - 1/300K) Than substitute all the values from multiple choices. current answer should be 1. But I still would like to know how to actually solve it. – printfmyname Mar 24 '15 at 5:57
• Please verify the given numbers! – Yomen Atassi Mar 24 '15 at 10:02

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?