If you remove some CO2(g) from an equilibrium mixture represented below, which direction would you expect the equilibrium to shift?
C3H8(l) + 5O2(g) → 4H2O(g) + 3CO2(g)
a. right
b. left
c. no change
d. indeterminable based on information given
I think the definitions in texts and lecture notes sounds so very esoteric when discussing Le Chatelier's Principle; however, it seems to confuse most students. Therefore, I've developed my own definition for purposes of Le Chatelier's Principle. In a nutshell, in a system at equilibrium, the reaction will shift so as to UNDO what we do to it. Therefore, if we remove some of the CO2 the reaction will try to make more CO2 to undo what we just did. How can it do that? By shifting to the right to produce more CO2. Adding H2O would make it go to the left with the same kind of reasoning. Adding O2 would make it go to the right with the same reasoning.
To determine the direction in which the equilibrium will shift when some CO2(g) is removed, we need to consider the reaction's stoichiometry.
In this case, the stoichiometry of the reaction shows that four moles of H2O(g) and three moles of CO2(g) are produced for every one mole of C3H8(l) that reacts. According to Le Chatelier's principle, when a reactant or product is removed from an equilibrium system, the system will shift in the direction that minimizes the change caused by the removal.
Therefore, if we remove some CO2(g) from the system, the reaction will shift to replace the lost CO2(g) by producing more CO2(g). Since the forward reaction in the given equation produces CO2(g), the equilibrium will shift to the right to produce more CO2(g).
Hence, the answer is (a) right.