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Current time:0:00Total duration:3:58

Worked example: Using Le Chȃtelier’s principle to predict shifts in equilibrium

AP.Chem:
TRA‑8 (EU)
,
TRA‑8.A (LO)
,
TRA‑8.A.1 (EK)
,
TRA‑8.B (LO)
,
TRA‑8.B.1 (EK)
,
TRA‑8.B.2 (EK)

Video transcript

- [Instructor] Carbon monoxide will react with hydrogen gas to produce methanol. Let's say that the reaction is at equilibrium and our job is to figure out which direction the equilibrium will shift, to the left, to the right, or not at all, as we try to make changes to the reaction at equilibrium. For example, if we add some hydrogen gas to our reaction at equilibrium, we're increasing the concentration of one of our reactants. According to the Le Chatelier's principle, the net reaction will move in the direction that decreases the stress placed on the system. So if the stress is increased amount of one of the reactants, the equilibrium will shift to the right to get rid of some of that reactant. In part B, some methanol is removed. So if we're decreasing the concentration of our product, the equilibrium's gonna shift to make more of our product, therefore, the equilibrium will shift to the right. Next, the volume is increased on the reaction at equilibrium. And if we increase the volume, we decrease the pressure, therefore, we could consider the stress to be decreased pressure. Le Chatelier's principle says the net reaction is gonna go in the direction that relieves the stress. So if the stress is decreased pressure, the net reaction is going to shift to increase the pressure. And we can figure out which direction that is by looking at the balanced equation. On the reactant side, there's one mole of gas and two moles of gas for a total of three moles of gas. On the product side, there's only one mole of gas. So there's three moles of gas on the left and only one mole of gas on the right. Since the net reaction is going to try to increase the pressure, the equilibrium shifts to the left, toward the side that's gonna form more moles of gas, therefore increasing the pressure. Next, we try adding some neon gas to our reaction mixture at equilibrium. Well, neon gas is an inert gas, which means it doesn't react with any of our reactants or products. And if we look at the expression for the reaction quotient Qp, neon gas is not included. Therefore, adding neon gas is not going to change the value for Qp, so the reaction remains at equilibrium. So the answer is there's no shift when an inert gas is added. And that might sound a little strange at first because adding neon gas means that the total pressure would increase, the total pressure since we're adding a gas. However, the partial pressures stay the same. So the partial pressures for methanol and carbon monoxide and hydrogen gas actually stay the same and therefore Q doesn't change. Next, we add a catalyst to our reaction at equilibrium. Catalysts speed up reactions by lowering deactivation energy. However, the catalyst is gonna speed up that the forward and the reverse reactions by the same amount and therefore the reaction remains at equilibrium. So there's no shift when a catalyst is added to a reaction at equilibrium. And then in part F, let's try decreasing the temperature on the reaction at equilibrium. Well, this reaction is exothermic because Delta H is less than 0, so we can treat heat as a product. So we go ahead and write heat on the product side. If we treat heat like a product, decreasing the temperature is like decreasing the amount of our product, therefore, the net reaction will move to the right to make more of the product. Whether that reaction moves to the right, you can think about that being an increase in the amount of products and therefore a decrease in the amount of reactants. And when you increase the products and decrease the reactants, you increase the value for the equilibrium constant. Therefore, lowering the temperature causes an increase in the equilibrium constant for an exothermic reaction. Note that changing the temperature in part F is the only change that actually changed the equilibrium constant. So in all the other ones, in A through E, the equilibrium constant stayed the same value.
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