Introduction to Le Chatelier's Principle - AP Chemistry

A container holds an equilibrium mixture: $\mathrm{Fe^{3+}(aq) + SCN^-(aq) \rightleftharpoons FeSCN^{2+}(aq)}$ . Some $\mathrm{Fe^{3+}(aq)}$ is removed from the mixture. In response to this stress, how will the system shift to re-establish equilibrium?
Answer: Shift toward reactants This question tests Le Châtelier's principle. The equilibrium Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq) is disturbed by removing some Fe³⁺(aq), which decreases the concentration of a reactant. When a reactant is removed from an equilibrium system, the equilibrium shifts to replace what was taken away by favoring the reverse reaction. This shift toward reactants will dissociate some FeSCN²⁺ back into Fe³⁺ and SCN⁻ to partially restore the Fe³⁺ concentration. A common misconception is thinking that removing a species always causes the reaction to proceed away from it, but the system actually shifts to replace what was removed. To apply Le Châtelier correctly, identify the stress (removed Fe³⁺ = decreased reactant) and predict the shift that opposes this change (make more reactant by shifting left).
A piston-cylinder apparatus contains an equilibrium mixture for the reaction $\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g)$ . Once equilibrium is established, the volume is increased by pulling the piston outward. As the system responds to this stress to re-establish equilibrium, in which direction does the equilibrium shift?
Answer: The equilibrium shifts toward products. Le Châtelier's principle states that a system at equilibrium will respond to a stress by shifting to minimize the effect of that stress. Increasing the volume decreases the pressure, disrupting the equilibrium since there are fewer gas moles on the reactant side (1) than the product side (2). To counteract this, the system shifts toward the products, increasing the number of gas molecules and thus raising the pressure. This shift re-establishes equilibrium by favoring the side with more moles. A tempting distractor is choice B, which suggests the equilibrium shifts toward reactants, based on the misconception that volume increases always favor fewer moles without considering the pressure relief. Always identify the stress first, then predict the shift that reduces its effect.
A rigid, sealed vessel contains an equilibrium mixture for the reaction $\text{CO}(g) + \text{Cl}_2(g) \rightleftharpoons \text{COCl}_2(g)$ . After equilibrium is established, some $\text{Cl}_2(g)$ is removed from the vessel. As the system responds to this stress to re-establish equilibrium, in which direction does the equilibrium shift?
Answer: The equilibrium shifts toward reactants. Le Châtelier's principle states that a system at equilibrium will respond to a stress by shifting to minimize the effect of that stress. Removing some Cl₂(g) decreases the concentration of a reactant, disrupting the equilibrium by making Q larger than K for the forward reaction. To counteract this, the system shifts toward the reactants, producing more Cl₂ and CO by decomposing some COCl₂. This shift re-establishes equilibrium by restoring the removed Cl₂. A tempting distractor is choice A, which suggests the equilibrium shifts toward products, based on the misconception that removing a reactant always drives the reaction forward without considering the reverse shift. Always identify the stress first, then predict the shift that reduces its effect.
A reaction mixture is initially at equilibrium in a rigid, sealed container: $\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\,\text{HI}(g)$ . The stress applied is the removal of some $\text{HI}(g)$ from the container. How will the system shift to re-establish equilibrium?
Answer: The equilibrium shifts toward products. This question tests Le Châtelier's principle. The equilibrium H₂(g) + I₂(g) ⇌ 2HI(g) is disrupted when some HI(g) is removed from the container, decreasing the concentration of a product. According to Le Châtelier's principle, the system will shift to counteract this stress by producing more HI to replace what was removed, which means the equilibrium shifts toward the products (right). A common misconception is thinking that removing a substance always causes the reaction to shift away from that side, but the system actually shifts to replace what was removed. To apply Le Châtelier's principle correctly, identify the stress (HI removal) and predict the shift that opposes it (making more HI by shifting right).
A system is initially at equilibrium in a sealed container: $\mathrm{CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g)}$ . Additional $\mathrm{H_2O(g)}$ is added to the container. In response to this stress, how will the system shift to re-establish equilibrium?
Answer: Shift toward products This question tests Le Châtelier's principle. The equilibrium CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g) is disturbed by adding more H₂O(g), which increases the concentration of a reactant. When a reactant concentration is increased, the system shifts to consume some of that excess reactant by favoring the forward reaction. This shift toward products will convert the added H₂O (along with some CO) into CO₂ and H₂ until equilibrium is restored. A common misconception is thinking that adding water has no effect because it's often a solvent, but here H₂O(g) is a reactant that participates in the equilibrium. To solve Le Châtelier problems, identify the role of the added substance (H₂O = reactant) and predict the shift that uses it up (forward to products).
A gaseous system is initially at equilibrium under a movable piston: $2\,\text{NO}(g) + \text{O}_2(g) \rightleftharpoons 2\,\text{NO}_2(g)$ . The piston is pushed inward, decreasing the volume of the container. How will the system shift to re-establish equilibrium?
Answer: The equilibrium shifts toward products. This question tests Le Châtelier's principle. The equilibrium 2NO(g) + O₂(g) ⇌ 2NO₂(g) has 3 moles of gas on the left and 2 moles on the right. When the volume decreases (pressure increases), the system shifts to counteract by reducing the total moles of gas to decrease pressure, which means shifting toward the products (right) where there are fewer moles. A common misconception is thinking that pressure changes affect all equilibria equally, but the shift depends on the difference in gas moles between sides. To solve pressure problems, count moles of gas on each side and remember that increased pressure favors the side with fewer moles.
A closed container holds an equilibrium mixture: $2\,\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\,\text{SO}_3(g)$ . The stress applied is an increase in volume (decreasing the pressure). How will the system shift to re-establish equilibrium?
Answer: The equilibrium shifts toward reactants. This question tests Le Châtelier's principle. The equilibrium 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) has 3 moles of gas on the left and 2 moles on the right. When volume increases (pressure decreases), the system shifts to counteract by producing more moles of gas to increase pressure, which means shifting toward the reactants (left) where there are more moles. A common misconception is confusing volume and pressure effects—remember that increasing volume decreases pressure, so the system shifts toward the side with more moles. To solve pressure/volume problems, count gas moles on each side and remember that the system shifts toward more moles when pressure decreases.
A sealed container is at equilibrium for the reaction $\mathrm{CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g)}$ . Additional $\mathrm{CO_2(g)}$ is added to the container. As the system responds to this stress to re-establish equilibrium, which direction does the equilibrium shift?
Answer: Shifts toward reactants This question tests Le Châtelier's principle. When CO₂(g) is added to the equilibrium CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g), the concentration of a product increases, creating stress. To counteract this stress, the system shifts toward reactants (left) to consume some of the added CO₂ and reduce its concentration. This shift converts CO₂ and H₂ back into CO and H₂O. Students sometimes think adding a product always increases the forward reaction, but the system actually shifts away from the added substance to reduce its concentration. When a product is added, predict a shift toward reactants to consume the excess.
A reaction mixture is initially at equilibrium in a sealed container: $\mathrm{CH_3COOH(aq) \rightleftharpoons H^+(aq) + CH_3COO^-(aq)}$ . A student adds additional $\mathrm{CH_3COO^-(aq)}$ to the solution. As the system responds to this stress to re-establish equilibrium, which direction does the equilibrium shift?
Answer: Shifts toward reactants This question tests Le Châtelier's principle. When CH₃COO⁻(aq) is added to the equilibrium CH₃COOH(aq) ⇌ H⁺(aq) + CH₃COO⁻(aq), the concentration of a product increases, creating stress. To counteract this stress, the system shifts toward reactants (left) to consume some of the added acetate ion. This shift combines CH₃COO⁻ with H⁺ to reform CH₃COOH, reducing both product concentrations. Students might think this is just about pH, but it's a straightforward application of Le Châtelier's principle to any product addition. When a product is added to any equilibrium, the system shifts toward reactants to consume the excess.
A sealed flask contains the equilibrium mixture for $\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}$ . After equilibrium is established, some $\mathrm{I_2(g)}$ is removed from the flask. In response to this stress, which direction will the system shift to re-establish equilibrium?
Answer: Shift toward reactants This question tests Le Châtelier's principle. When I₂(g) is removed from the equilibrium H₂(g) + I₂(g) ⇌ 2HI(g), the concentration of a reactant decreases, disrupting the equilibrium. According to Le Châtelier's principle, the system will shift to counteract this stress by producing more I₂, which means the equilibrium shifts toward the reactants (left). This shift converts some HI back into H₂ and I₂, replacing some of the removed reactant. A common misconception is that removing a reactant cannot affect equilibrium (choice E), but removing any component definitely causes a shift. To solve Le Châtelier problems, identify the stress (removing a reactant), then predict the shift that counteracts it (producing more of that reactant by shifting left).
A system is initially at equilibrium: $\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g)$ . The stress applied is the removal of some $\text{CO}_2(g)$ from the container. How will the system shift to re-establish equilibrium?
Answer: The equilibrium shifts toward products. This question tests Le Châtelier's principle. The equilibrium CaCO₃(s) ⇌ CaO(s) + CO₂(g) is disrupted when some CO₂(g) is removed from the container, decreasing the concentration of a product. According to Le Châtelier's principle, the system will shift to counteract this stress by producing more CO₂ to replace what was removed, which means the equilibrium shifts toward the products (right) by decomposing more CaCO₃. A common misconception is thinking that solids don't participate in shifts, but while their concentrations are constant, they can still be consumed or produced as the equilibrium shifts. To solve heterogeneous equilibria problems, focus on the gases and aqueous species that can change concentration.
A sealed container at constant temperature contains the system at equilibrium: $\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)$ . After equilibrium is established, additional $\text{NO}_2(g)$ is injected into the container. As the system responds to this stress to re-establish equilibrium, in which direction does the equilibrium shift?
Answer: The equilibrium shifts toward reactants. Le Châtelier's principle states that a system at equilibrium will respond to a stress by shifting to minimize the effect of that stress. In this case, injecting additional NO₂(g) increases the concentration of a product, disrupting the equilibrium by making the reaction quotient Q larger than the equilibrium constant K. To counteract this, the system shifts toward the reactants, consuming some of the added NO₂ by converting it back to N₂O₄. This shift re-establishes equilibrium by reducing the excess NO₂ concentration. A tempting distractor is choice B, which suggests the equilibrium shifts toward products, based on the misconception that adding a substance always favors the side it's on without considering the counteraction. Always identify the stress first, then predict the shift that reduces its effect.
A reaction system is initially at equilibrium: $\mathrm{CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)}$ . Some solid $\mathrm{CaCO_3(s)}$ is added to the container while keeping it sealed. As the system responds to this stress to re-establish equilibrium, which direction does the equilibrium shift?
Answer: No shift occurs This question tests Le Châtelier's principle. When solid CaCO₃(s) is added to the equilibrium CaCO₃(s) ⇌ CaO(s) + CO₂(g), no shift occurs because solids do not appear in the equilibrium expression. The concentration of a pure solid is constant and doesn't change when more is added, so there's no stress on the equilibrium. Only changes in gas or aqueous species concentrations cause shifts. A common misconception is that adding any reactant shifts equilibrium toward products, but this only applies to gases and aqueous species, not pure solids or liquids. Remember that pure solids and liquids don't affect equilibrium position—only gases and aqueous species do.
A reaction mixture is at equilibrium for the process $\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq)$ . The student then adds solid $\text{AgCl}(s)$ to the container without changing the volume of solution. As the system responds to this stress to re-establish equilibrium, in which direction does the equilibrium shift?
Answer: The equilibrium does not shift. Le Châtelier's principle states that a system at equilibrium will respond to a stress by shifting to minimize the effect of that stress. Adding solid AgCl(s) introduces more of a pure solid, but solids do not appear in the equilibrium expression, so their concentration is constant and does not disrupt the equilibrium. Therefore, the system does not need to shift to counteract this addition, as the activities of solids remain unity. This maintains the equilibrium without any change in the position. A tempting distractor is choice A, which suggests the equilibrium shifts toward reactants, based on the misconception that adding solids affects equilibrium like solutes. Always identify the stress first, then predict the shift that reduces its effect.
A student has a reaction mixture at equilibrium in a closed flask: $\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)$ . The student then removes some $\text{HI}(g)$ from the flask. As the system responds to this stress to re-establish equilibrium, in which direction does the equilibrium shift?
Answer: The equilibrium shifts toward products. Le Châtelier's principle states that a system at equilibrium will respond to a stress by shifting to minimize the effect of that stress. Removing some HI(g) decreases the concentration of a product, disrupting the equilibrium by making Q smaller than K. To counteract this, the system shifts toward the products, producing more HI by converting H₂ and I₂. This shift re-establishes equilibrium by replenishing the removed HI. A tempting distractor is choice C, which suggests the equilibrium shifts toward reactants, based on the misconception that removing a product reduces the forward reaction rate without increasing it to compensate. Always identify the stress first, then predict the shift that reduces its effect.
A closed flask contains the equilibrium system: $\mathrm{2\,NO(g) + O_2(g) \rightleftharpoons 2\,NO_2(g)}$ . Some $\mathrm{O_2(g)}$ is removed from the flask. In response to this stress, how will the system shift to re-establish equilibrium?
Answer: Shift toward reactants This question tests Le Châtelier's principle. The equilibrium 2NO(g) + O₂(g) ⇌ 2NO₂(g) is disturbed by removing some O₂(g), which decreases the concentration of a reactant. When a reactant is removed from an equilibrium system, the equilibrium shifts to replace what was taken away by favoring the reverse reaction. This shift toward reactants will convert some NO₂ back into NO and O₂ to partially restore the O₂ concentration. A common misconception is thinking that removing a reactant always drives the reaction forward, but the system actually shifts to replace what was removed. To apply Le Châtelier correctly, identify the stress (removed O₂ = decreased reactant) and predict the shift that opposes this change (make more reactant by shifting left).
A system is initially at equilibrium: $\text{CO}(g) + \text{H}_2\text{O}(g) \rightleftharpoons \text{CO}_2(g) + \text{H}_2(g)$ . The volume of the container is decreased (by moving a piston inward) at constant temperature. In response to this stress, how will the system shift to re-establish equilibrium?
Answer: The equilibrium shifts toward neither side. This question tests Le Châtelier's principle. The equilibrium CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g) has 2 moles of gas on each side, so when the volume decreases (pressure increases), there is no favored direction based on mole numbers. Since both sides have equal moles of gas, the system cannot reduce pressure by shifting, so the equilibrium position remains unchanged and shifts toward neither side. A common misconception is thinking that any pressure change must cause a shift, but this only occurs when the mole numbers differ between reactants and products. To solve pressure/volume problems, count the moles of gas on each side first—if equal, no shift occurs.
A closed container holds the equilibrium system $\mathrm{2NO_2(g) \rightleftharpoons N_2O_4(g)}$ . After equilibrium is reached, the volume of the container is increased (by pulling a piston outward). In response to this stress, which direction will the system shift to re-establish equilibrium?
Answer: Shift toward reactants This question tests Le Châtelier's principle. When the volume increases for 2NO₂(g) ⇌ N₂O₄(g), the pressure decreases, which is the stress on the system. The left side has 2 moles of gas while the right side has 1 mole of gas. According to Le Châtelier's principle, the system shifts to counteract the pressure decrease by favoring the side with more gas molecules—shifting toward reactants (left). This shift increases the total number of gas particles and partially counteracts the pressure decrease. A common misconception is that pressure changes do not affect gas equilibria (choice E), but they definitely do when the number of gas moles differs on each side. To solve pressure/volume problems, count gas moles on each side, then predict the shift toward more moles when pressure decreases.
A system is initially at equilibrium in a closed container: $\text{Fe}^{3+}(aq) + \text{SCN}^-(aq) \rightleftharpoons \text{FeSCN}^{2+}(aq)$ . The stress applied is the removal of some $\text{SCN}^-(aq)$ from the solution. How will the system shift to re-establish equilibrium?
Answer: The equilibrium shifts toward reactants. This question tests Le Châtelier's principle. The equilibrium Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq) is disrupted when some SCN⁻(aq) is removed from the solution, decreasing the concentration of a reactant. According to Le Châtelier's principle, the system will shift to counteract this stress by producing more SCN⁻ to replace what was removed, which means the equilibrium shifts toward the reactants (left) to regenerate SCN⁻ and Fe³⁺. A common misconception is thinking that removing a reactant always drives the reaction forward, but the system actually shifts to replace what was removed. To solve removal problems, identify what was removed (SCN⁻) and predict the shift that replaces it (shifting left).
A sealed flask contains the system at equilibrium: $\mathrm{N_2O_4(g) \rightleftharpoons 2\,NO_2(g)}$ . After equilibrium is established, additional $\mathrm{NO_2(g)}$ is injected into the flask. In response to this stress, how will the system shift to re-establish equilibrium?
Answer: Shift toward reactants This question tests Le Châtelier's principle. The equilibrium N₂O₄(g) ⇌ 2NO₂(g) is disturbed by adding more NO₂(g), which increases the concentration of products. According to Le Châtelier's principle, when a product concentration is increased, the system shifts to consume some of that excess product by favoring the reverse reaction. This shift toward reactants will convert some NO₂ back into N₂O₄ until a new equilibrium is established. A common misconception is thinking that adding more of any substance always pushes the reaction forward, but the direction depends on whether you're adding reactants or products. To solve Le Châtelier problems, first identify what was changed (added NO₂ = product), then predict the shift that counteracts this change (consume excess product by shifting left).

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