Changes in pressure for an equilibrium reaction containing a solid/liquid on one side only

Question

For a reaction where there is only one solid/liquid present, but gases present as both reactants and products (are there any examples of this type of reaction?):

$$\ce{A(s) + B(g) <=> C(g)},$$

if pressure is increased, it will favour the formation of the side not containing the solid/liquid (e.g. the forward reaction favoured), so as to decrease pressure of system – since one mole of a gas molecules has a considerably lower volume than a mole of any solid or liquid (far more spread out).

And the opposite if pressure decreased.

E.g. all that is considered is which side the solid/liquid is on, not the moles of gas present on either side. Is this true?

Answer 1

There are many examples of such reactions, for instance:

$$ \begin{align} \ce{S (s) + O2 (g) &<=> SO2 (g)}\\ \ce{C (s) + 2 H2 (g) &<=> CH4 (g)}\\ \ce{Ni (s) + 4 CO (g) &<=> Ni(CO)4 (g)} \end{align} $$

However, your assumption

all that is considered is which side the solid/liquid is on, not the moles of gas present on either side

is not correct. Since shift of equilibrium according to Le Chatelier’s principle depends on partial pressures, which in turn, rely on molar ratio of gaseous reactants and products, decrease/increase of pressure doesn't depend on which side of the reaction solid/liquid participant is located; instead, it is defined for each system individually based on ratio between gaseous products and reactants.

Answer 2

Well the reaction will favor the side not containing the solid/liquid, this is not true. It will favor the side where number of gaseous moles is less. This will be found giving a balanced chemical reaction.

Answer 3

If you want to discuss using Ideal conditions, then pressure should have no effect on that equilibrium - the reaction produces and consumes an exact equal amount of gas - and with ideal conditions individual molecules have zero volume so the solid part doesn't take any. The assumption is fairly good, the volume of a mole of solid is completely negligible compared to a mole of gas of the same compound.

If you use Real conditions you have to evaluate the compressibility of the gas vs gas+liquid/solid along with some thermodynamic data about the compounds to get the shift of internal energy per atmosphere and compound. The magnitudes of these should show which way the equilibrium goes. And if you are really fond of doing long and complicated integrals and Legendre transforms and whatnot, you could also try to estimate where the new equilibrium would be by finding the minimum of the Helmholtz' free energy with varying the ratio of pressures of C and B but it does not sound like an interesting evening to me.

Answer 4

The equilibrium constant $K_p$ is defined in terms of the free energy of the standard states of the gaseous species at 1 atmosphere pressure so is independent of pressure. This means that if the pressure is changed the degree of dissociation $\alpha$ may change and this is determined by working out the equilibrium in terms of $\alpha$ and the partial pressures.

If the partial pressures are expressed as mole fractions $x$ then $K_x =K_pP^{-\Delta n}$ where $\Delta n$ is the change in the number of moles, products minus reactants. Thus, unless $\Delta n =0 $, which is your case, the mole fractions of the components will vary if the total pressure changes even though $K_p$ does not. If the partial pressures are expressed as concentrations then $K_c=K_P(RT)^{-\Delta n}$ and $K_c$ like $K_p$ is independent of pressure for perfect gasses.