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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.

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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.

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Just a hint to get you started

  1. From the balanced reaction equation , it seems that

$$K_p = \frac{p_{\ce{NO}}^2 \cdot p_{\ce{O2}}} {p_{\ce{NO2}}^2}$$

  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.
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