# Finding the value of pressure equilibrium constant

Question:

A rigid vessel containing only $\ce{NO2(g)}$ is heated to $\pu{337^{ \circ}C}$ and allowed to come to equilibrium according to the following reaction: $$\ce{2NO2 <=> 2NO + O2}$$

The density of the resulting mixture is measured to be $\pu{0.52 g/L}$ at a total pressure of $\pu{ 0.75 atm}$. What is the value of $K_\mathrm p$?

The answer key says it should be $\pu{0.65 atm}$, but I can't get that number. I first tried to find mols using the given density and molar masses, and then finding the partial pressure of each component using the mol fractions, but it doesn't work out.

The density is given.From here you will get a relation between initial concentration and volume.From $K_c$ you will get relation between volume,extent of dissociation and initial concentration.From $K_p$ you will get relation between pressure,extent of dissociation and initial concentration.After that use the formulas that Klaus Warzecha gave.I think that will be enough to solve this.
$$K_p = \frac{p_{\ce{NO}}^2 \cdot p_{\ce{O2}}} {p_{\ce{NO2}}^2}$$
1. The relation between $K_p$ and $K_c$ is given as $K_p = K_c\cdot(\mathrm{R}T)^{\Delta n}$, where $\Delta n$ is the difference between the number of moles of product and reactant gases.