Equilibrium constant from mole ratio

Question

Chapter 4, problem 13 from the Chemical Priciples [1, p. 170]:

Equilibrium concentrations

Experiments have shown that at $\pu{60 °C}$ and $\pu{1 atm}$ total pressure, the equilibrium ratio of $\ce{NO2}$ to $\ce{N2O4}$ in moles in a closed vessel is exactly $2:1$.

a) Calculate the equilibrium constant, $K_\mathrm{c}$, for the dissociation of $1$ mole of $\ce{N2O4}$ into $2$ moles of $\ce{NO2}$.

My answer:

The chemical equation is $\ce{N2O4 <=> 2 NO2}$ and at equilibrium the ratio is $2:1$ so shouldn't

$$K_\mathrm{c} = \frac{[\ce{NO2}]^2}{[\ce{N2O4}]} = \frac{(\pu{2 mol L-1})^2}{\pu{1 mol L-1}} = \pu{4 mol L-1}?$$

The answer in the back of the text book is $K_\mathrm{c}= \pu{0.0488 mol L-1}$. I do not understand how this answer was calculated from the information provided.

References

1. Dickerson, R. E.; Gray, H. B.; Haight, G. P. Chemical Principles, 3d ed.; Benjamin/Cummings Pub. Co: Menlo Park, California, 1979. ISBN-13: 978-0-8053-2398-6.

Answer 1

I suggest to pick one gaseous component, say, $\ce{N2O4}$, and, applying ideal gas law, express $K_\mathrm{c}$ as a function of the partial pressure of the chosen gas. Since at the equilibrium the ratio $n(\ce{N2O4}):n(\ce{NO2})=1:2$ is exact (e.g. molar fractions are $1/3$ and $2/3$, respectively), and the volume doesn't change, you can immediately deduce that

$$[\ce{N2O4}] = 2[\ce{NO2}];$$ $$p(\ce{N2H4}) = \frac{p_\mathrm{tot}}{3}$$

where $p_\mathrm{tot}$ is the known total pressure of the system. Using ideal gas law

$$[\ce{N2O4}] = \frac{p(\ce{N2H4})}{RT} = \frac{p_\mathrm{tot}}{3RT}$$

and the law of mass action

$$K_\mathrm{c} = \frac{[\ce{NO2}]^2}{[\ce{N2O4}]} = \frac{(2[\ce{N2O4}])^2}{[\ce{N2O4}]} = 4[\ce{N2O4}] = \frac{4p_\mathrm{tot}}{3RT},$$

one can plug in the parameters and calculate $K_\mathrm{c}$:

$$K_\mathrm{c} = \frac{4\cdot\pu{101325 Pa}}{3\cdot\pu{8.314 J mol-1 K-1}\cdot\pu{333 K}} = \pu{48.8 mol m-3} = \pu{4.88e-2 mol L-1}$$