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.


  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.
  • 3
    $\begingroup$ 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. $\endgroup$ Jan 1, 2019 at 4:16
  • $\begingroup$ 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? $\endgroup$ Jan 1, 2019 at 14:49

1 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 $K_\mathrm{c}$:

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.