# Equilibrium constant from mole ratio

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 $$\ce{NO2}$$.

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.
• Where did you get your concentrations of $\pu{2 mol/L}$ and $\pu{1 mol/L}$ from? The question only told you the ratio is 2:1, so in general the concentrations should be represented as $2x$ and $x$, where $x$ is not necessarily equal to $\pu{1 mol/L}$. Therefore $K_c = (2x)^2/x = 4x$. The real task for you is to find $x$. To find the concentration $x$, you need to find (1) the amount of substance, i.e. how many moles of each substance there are; and (2) the total volume. The former was, of course, already given to you, but the latter wasn't. – orthocresol Jan 1 '19 at 4:16
• From the ideal gas law, what is the molar volume of an ideal gas mixture at 60 C and 1 atm pressure? From this result, what is the moles/liter of the gas? – Chet Miller Jan 1 '19 at 14:49

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 $$K_\mathrm{c}$$:

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