What will happen to the $K_\text{eq}$ in the following reaction if volume is increased?

$$\ce{2NO(g) + O2(g) <=> 2NO2(g) + energy}$$

From le Chatelier's principle I know that it must be constant unless temperature is changed, however it doesn't make sense because if

$$K_\text{eq}= \frac{[\ce{NO2}]^2}{[\ce{O2}][\ce{NO}]^2},$$

then by increasing the volume we know that equilibrium will shift left and favor reactants. Hence concentration of reactants would increase while concentration of products would decrease. If we plug that back into the equilibrium constant equation we see that if numerator decrease and denominator increase that must yield a smaller number than the initial number. Hence the equilibrium is not constant. Why this occurs?


marked as duplicate by airhuff, paracetamol, Pritt Balagopal, M.A.R. ಠ_ಠ, jerepierre Aug 2 '17 at 19:12

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


First step: Don't use Kc... use Kp!

You are focusing on concentration, but as we're working with gases it becomes easier to work with partial pressures.

$K_\text{p}= \frac{\ce{p(NO2)}^2}{\ce{p(O2)}\ce{p(NO)}^2}$

Now, if we remember that partial pressure is pressure $\times$ mole fraction ($p_{a}=P \times \chi_{a}$), and substitute these in we get (and cancelling out the pressures):

$K_\text{p}= \frac{\chi_{(NO_{2})}^2}{\chi_{(O_{2})}\chi_{(NO)}^{2}P}$

Now, if we consider increasing the volume. By doing so (and assuming that temperature is kept constant), then we have decreased the pressure (ideal gas law: PV=nRT).

Now we can see that although the number of moles changes (and hence the mole fractions), our equilibrium constant will remain unchanged thanks to the effect of pressure.

A good guide is here.


It is called STP or standard temperature and pressure(If I remember correctly there are two STP's one for thermo and one for practical lab conditions). If PV=nRT than ideally pressure is inversely related to volume. The algebra is simple enough. Remember Keq is always reported at ''standard conditions'', or STP. Typically 25 degrees celsius and 1 atm(I think SI units would be 1.01 bars but I could be wrong) at a 1 molar concentration. according to that definition n, T, and P are all fixed and R is a constant. It is only constant under a given set of conditions that you may or may not choose. As an aside Keq is strictly speaking a thermodynamic parameter so don't forget about entropy as well (there are three or four laws after all)


Volume??? Wow. You may get better answers if you learn about a thermodynamic property called "pressure". After you learn about it, I suggest you use it when asking about the concentration of gas phase reactions. For gas phase reactions usually concentration is measured not in "moles per liter" but as pressure, invoking Dalton's Law of partial pressures. So, your Keq would be in units of (pressure units)-1, say Pa-1.
Lets say Keq = a²/(bc²) and a+b+c= P. You seem to be saying that the only way you can change P is to change Keq. Well, no. You have three equations and 3 unknowns, I'm not seeing the problem. (The unknowns are the new a, b, & c...you could call them a', b', c') (Assuming you knew the new P').

  • 1
    $\begingroup$ Firstly by measuring the gas in terms of pressure that doesn't solve the problem. Secondly your answer does not address the actual question. Thirdly I'm not saying P is the only way to change Keq. I am saying TEMPERATURE is the ONLY way not P. Lastly there is no equations or variables involved in this question. This is a conceptual question not a quantitative question. $\endgroup$ – coderhk Aug 1 '17 at 21:41

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