1
$\begingroup$

This reaction is given in my book and the direction of shift of reaction on addition of $\ce{CO}$ and removal of $\ce{CO2}$ at constant volume is asked. $$\ce{2CO(g) + O2(g) ⇌ 2CO2(g) + \Delta H}$$ If "at constant volume" was not specified, I understand that the forward reaction would occur since the reactant is added and the product is removed, but I do not know if this will vary with constant volume. What will happen in the case of constant volume? Does it matter that there is a change in the number of particles comparing reactants and products? Which direction will the equilibrium shift?

$\endgroup$
  • 1
    $\begingroup$ Since all your species are gases, for Le Chatêlier purposes, it's relevant whether you are working at constant pressure or at constant volume. At constant volume, total pressure will be proportional to the total number of gas molecules (of all species). You can figure out how will pressure change in your case; equilibrium will shift to counteract that change. $\endgroup$ – user41033 May 10 '18 at 8:34
2
$\begingroup$

I understand that the forward reaction would occur since the reactant is added and the product is removed

You have to be careful about the physical state of reactants and products when making such a broad statement. In the reaction $$\ce{NaCl(s) <=> Na+(aq) + Cl-(aq)},$$

nothing will change if I remove some of the solid sodium chloride or add some - the equilibrium will not be disturbed. If I add some silver nitrate, though, silver ions and chloride ions will form a precipitate, the chloride concentration will drop, and the equilibrium will be disturbed.

In general, the concentration (or partial pressure) of at least one of the species has to be changed in order to disturb an equilibrium (unless of course you change K by changing the temperature).

What will happen in the case of constant volume?

This, I think, is supposed to make the question easier, not harder. If you add a substance at constant volume, you know that the partial pressure (the concentration) of that substance will increase while the other partial pressures remain the same.

At constant pressure, however, the other partial pressures change too. You could have a reaction where the small changes in partial pressure of the other species has a larger effect on the equilibrium, and a shallow application of Le Chatelier's principle will yield an incorrect answer:

$$\ce{A(g) + 2 B(g) <=> P(s)}$$

If this system is at equilibrium with equal partial pressure of A and B, and I add some A at constant pressure, the partial pressure of B would drop (I am "diluting" B by adding more A). Because B has the higher coefficient (higher exponent of the concentration of B in the equilibrium expression), there would be a reverse reaction toward equilibrium. This is not what you might expect.

So, the question avoids that complication by saying "at constant volume".

$\endgroup$
4
$\begingroup$

Let's see your formula (with the difference of enthalpy being negative):

$$\ce{2CO (g) + O2 (g) <=> 2CO2 (g)}\qquad(\Delta H^\circ<0)$$

At constant temperature, for Le Chatêlier purposes,

  • Adding $\ce{CO}$ produces that it react with $\ce{O2}$ to produce more $\ce{CO2}$ to counteract the effect of the applied change and a new equilibrium is established.
  • Removing $\ce{CO2}$ produce that $\ce{CO}$ react with $\ce{O2}$ to produce more $\ce{CO2}$.

When you are adding or removing a reagent or product, and all components are in a gaseous state, you are changing the pressure and you can study the reaction by $K_p=\dfrac{P^2_\ce{CO2}}{P^2_\ce{CO}P_\ce{O2}}$.

We also have $K_p=K_c$ $RT^{Δn}$:

  • $K_p$ is the equilibrium constant written with partial pressure of gases.

  • $K_c$ is the constant but being written by concentration of substances.

  • $R$ is universal gas constant.

  • $T$ is the absolute temperature (Kelvin degrees).

  • $\Delta n$ are moles $(\Delta n=n_{\text{products}} - n_{\text{reagents}})$.

If you change concentration of $\ce{CO}$ you are changing the pressure it, too.

But you can change also the pressure, by changing the volume. (For Boyle's Law: the absolute pressure exerted by a given mass of an ideal gas is inversely proportional to the volume it occupies if the temperature and amount of gas remain unchanged within a closed system). Boyle's Law.

For example, if you increase the pressure, volume will decrease and the system will shift towards it will be less molecules to counteract the change. In this case, to products. If you in decrease pressure, volume will increase and the system will shift towards reagents.

Other option is to change the pressure by adding or removing an inert gas, but since it don't intervenes in the reaction, the equilibrium is not affected.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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