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AP Chemistry · Unit 7 Equilibrium

7.9 Introduction to Le Châtelier’s Principle

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2 questions, one for each idea where we can. Answer them, then see which ideas to fix.

Question 1 of 2

The hypothetical reaction D(g) + E(g) ⇌ G(g) is at equilibrium in a rigid container with [D] = 0.20 M. More D(g) is then injected, which raises [D] to 0.50 M at the moment of injection. Which statement describes [D] once equilibrium is re-established at the same temperature?

Answer and reasoning
  1. AIt stays at 0.50 M, its value just after the injection
    A student who thinks an added substance simply stays in the mixture, with no net reaction, picks this. The injection makes Q less than K, so a net forward reaction consumes some of the D.
  2. BIt settles at a value between 0.20 M and 0.50 M Correct
    The added D makes Q less than K, so a net forward reaction uses up some D and [D] falls below 0.50 M. The stress is only partly counteracted: if [D] fell all the way to 0.20 M, with less E and more G than before, Q would exceed K.
  3. CIt rises to some value that is greater than 0.50 M
    A student who thinks an equilibrium shifts toward the side to which a substance is added picks this, expecting still more D to form. Adding D causes the net forward reaction, which consumes D.
  4. DIt returns to 0.20 M, its value before the injection
    A student who thinks the system completely undoes a stress picks this. The net reaction removes only part of the added D; a return to 0.20 M with less E and more G than before would make Q greater than K.

Working Just after the injection Q < K, so a net forward reaction consumes D and E and forms G: [D] falls below 0.50 M. It cannot fall as far as 0.20 M: with [D] back at 0.20 M, [E] lower and [G] higher than originally, Q = [G]/([D][E]) would be greater than K. So 0.20 M < [D] < 0.50 M.

CED 7.9.A.1 · Read this in Fix

Question 2 of 2

A solution of ammonia is at equilibrium at 25°C: NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq). A small amount of NH₄Cl(s) is dissolved in the solution, with no change in volume or temperature. How does the pH of the solution change, and why?

Answer and reasoning
  1. AIt increases, as adding a product shifts the equilibrium toward products
    A student who thinks an equilibrium shifts toward the side to which a substance is added picks this, expecting more OH⁻. Added NH₄⁺ is consumed by the net reverse reaction, which also consumes OH⁻, so the pH decreases.
  2. BIt stays the same, as K is constant when the temperature is constant
    A student who thinks a constant K means unchanged equilibrium concentrations picks this. K is unchanged, but the added NH₄⁺ makes Q greater than K; the net reverse reaction lowers [OH⁻], so the pH decreases.
  3. CIt increases, as the equilibrium shifts toward NH₃, which is a base
    A student who thinks the pH depends on how much base is present picks this. The shift toward NH₃ consumes OH⁻; the pH is set by [OH⁻], which falls, so the pH decreases.
  4. DIt decreases, as the equilibrium shifts toward NH₃ and so [OH⁻] falls Correct
    Dissolving NH₄Cl adds NH₄⁺, a product. The net reverse reaction that follows consumes OH⁻, so [OH⁻] falls, the solution becomes less basic and its pH decreases.

CED 7.9.A.2 · Read this in Fix

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In preparation: 0 of 2 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.

7.9.A.1 Le Châtelier's principle

Le Châtelier's principle
When a system at equilibrium is disturbed by a stress, a net reaction takes place in the direction that partially counteracts the stress, and the system reaches a new equilibrium state.
Stress: addition or removal of a chemical species
Adding a gaseous or dissolved reactant (or removing such a product) at constant volume and temperature causes a net forward reaction; adding a gaseous or dissolved product (or removing such a reactant) causes a net reverse reaction. When every species in the equilibrium is a gas or a solute, the concentration of an added species ends between its original value and its value just after the addition.
Stress: change in volume or pressure of a gas-phase system
Decreasing the volume of a gas-phase equilibrium mixture raises the pressure, and the net reaction goes toward the side of the equation with fewer moles of gas; increasing the volume has the opposite effect. If both sides have the same number of moles of gas, no net reaction follows.
Stress: change in temperature
Raising the temperature of an equilibrium mixture causes a net reaction in the endothermic direction, and lowering it causes a net reaction in the exothermic direction; heat can be treated as a reactant of an endothermic reaction and as a product of an exothermic one.
Stress: dilution of a reaction system
Adding solvent to a solution at equilibrium lowers the concentration of every dissolved species, and the net reaction goes toward the side of the equation with more dissolved particles.
Pure solids and liquids in an equilibrium system
Changing the amount of a pure solid or pure liquid that takes part in an equilibrium does not change its concentration, so it causes no net reaction as long as some of the solid or liquid remains.

Students often think Concentration follows the number of particles alone, so a species whose particles are used up in the shift has a lower concentration at the new equilibrium, whatever happened to the volume. In fact Not necessarily. Concentration is amount divided by volume. When the volume is decreased, a species can decrease in amount and still increase in concentration; in 2 A(g) ⇌ A₂(g), compression raises both [A] and [A₂] although the number of A particles falls.

Students often think K is constant at constant temperature, so the concentrations (and properties such as pH) at equilibrium are the same after a disturbance as before it. In fact No. At constant temperature K keeps its value, but many different sets of concentrations satisfy the same K. After a stress that makes Q differ from K, the system reaches a new equilibrium, in general with different concentrations, and the same K.

7.9.A.2 Measurable effects of a stress

Measurable effects of a stress
The net reaction that follows a stress changes properties that can be measured, such as the color of a solution (when a colored species is formed or consumed), its pH (when H₃O⁺ or OH⁻ is formed or consumed) and its temperature.

Students often think The pH of a solution of a base depends on how much of the base is present, so a shift that forms more NH₃ raises the pH. In fact No. The pH of the solution is set by [OH⁻] (and so [H₃O⁺]). In NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq), a shift toward NH₃ consumes OH⁻, so the solution becomes less basic and its pH falls.

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3 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.

Question 1 of 3

The diagrams represent the same sample of gas at equilibrium for the hypothetical reaction 2 A(g) ⇌ A₂(g), before and after the volume of the container is decreased at constant temperature. How do the concentrations of A and A₂ in the new equilibrium mixture compare with those in the original equilibrium mixture?

Answer and reasoning
  1. A[A] is 0.67 times and [A₂] is 1.33 times as great
    A student who judges concentration by the number of particles alone picks this: A falls from 6 to 4 particles (4/6 = 0.67) and A₂ rises from 3 to 4 (4/3 = 1.33). The volume has fallen from 3.0 L to 1.0 L, so [A] has doubled and [A₂] has become 4 times as great.
  2. B[A] is 3.00 times and [A₂] is 3.00 times as great
    A student who thinks compression only pushes the particles closer together picks this, since one-third of the volume would triple each concentration. The diagrams show a net reaction as well: 6 A and 3 A₂ become 4 A and 4 A₂.
  3. C[A] is 2.00 times and [A₂] is 4.00 times as great Correct
    Concentration is number of particles divided by volume. [A] goes from 6/3.0 = 2.0 to 4/1.0 = 4.0 particles per liter, and [A₂] from 3/3.0 = 1.0 to 4/1.0 = 4.0. Compression caused a net reaction toward A₂, yet both concentrations are greater than before.
  4. D[A] is 1.00 times and [A₂] is 1.00 times as great
    A student who thinks a constant K means constant concentrations picks this, expecting each concentration to be unchanged. K is the same in both diagrams ([A₂]/[A]² = 0.25 in particles per liter), but the concentrations are not: [A] has doubled and [A₂] is 4 times as great.

Working Original: 6 A and 3 A₂ in 3.0 L, so [A] ∝ 6/3.0 = 2.0 and [A₂] ∝ 3/3.0 = 1.0. New: 4 A and 4 A₂ in 1.0 L, so [A] ∝ 4.0 and [A₂] ∝ 4.0. [A] is 2.00 times and [A₂] is 4.00 times as great. (The number of A particles has fallen from 6 to 4 because the net reaction formed A₂, the side with fewer particles; [A₂]/[A]² = 0.25 in both diagrams.)

CED 7.9.A.1 · Read this in Fix

Question 2 of 3

A sealed tube contains a solution in which the equilibrium Co(H₂O)₆²⁺(aq) + 4 Cl⁻(aq) ⇌ CoCl₄²⁻(aq) + 6 H₂O(l) is established. Co(H₂O)₆²⁺(aq) is pink and CoCl₄²⁻(aq) is blue. The tube was held at each temperature shown in the table, in turn, until its color stopped changing. Which claim about the forward reaction is supported by the results?

Answer and reasoning
  1. AIt is endothermic, as each heating shifts the equilibrium toward CoCl₄²⁻ Correct
    The solution is blue (more CoCl₄²⁻) at 90°C and pink (more Co(H₂O)₆²⁺) at 0°C, and the changes reverse. Heating causes a net reaction in the direction that absorbs heat, so the forward reaction, which forms CoCl₄²⁻, is endothermic.
  2. BIt is exothermic, as each heating shifts the equilibrium toward CoCl₄²⁻
    A student who links a product-rich hot mixture with an exothermic reaction picks this. Heating favors the direction that absorbs heat; the shift toward blue CoCl₄²⁻ at 90°C shows that the forward reaction is endothermic.
  3. CIt is faster when heated, so more CoCl₄²⁻ forms with no shift in equilibrium
    A student who thinks temperature affects only the rate picks this. Each color was recorded after it stopped changing, and cooling to 0°C turned the blue solution pink: the equilibrium position itself changes with temperature.
  4. DIt may be endothermic or exothermic, as heating favors products in either case
    A student who thinks heating drives every reaction toward products picks this. Heating favors products only for an endothermic reaction, so the shift toward CoCl₄²⁻ on heating identifies the forward reaction as endothermic.

CED 7.9.A.2 · Read this in Fix

Question 3 of 3

CaCO₃(s), CaO(s) and CO₂(g) are at equilibrium in a cylinder fitted with a movable piston at a constant high temperature: CaCO₃(s) ⇌ CaO(s) + CO₂(g). Which change, made at the same temperature with both solids still present afterward, increases the number of moles of CO₂(g) in the cylinder once equilibrium is re-established?

Answer and reasoning
  1. AAdding more powdered CaCO₃(s) to the cylinder
    A student who thinks adding more of a solid reactant shifts the equilibrium toward products picks this. The concentration of a pure solid does not change when more is added, so no net reaction follows and the amount of CO₂ stays the same.
  2. BLowering the piston so as to decrease the volume
    A student who thinks a decrease in volume favors the side with more moles of gas picks this. Compression causes a net reaction toward the side with fewer moles of gas: CO₂ combines with CaO, and the cylinder holds fewer moles of CO₂.
  3. CRaising the piston so as to increase the volume Correct
    Increasing the volume lowers the pressure of CO₂, and the net reaction goes toward the side with more moles of gas: more CaCO₃ decomposes until the concentration of CO₂ returns to its equilibrium value, now in a larger volume, so the cylinder holds more moles of CO₂.
  4. DInjecting CO₂(g) with the piston held in place
    A student who thinks an added substance simply stays in the mixture picks this. The added CO₂ reacts with CaO until the concentration of CO₂ returns to its equilibrium value; at the same volume and temperature that is the same number of moles of CO₂ as before.

CED 7.9.A.1 · Read this in Fix

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This stop covered multiple choice only, which is 50% of your AP Chemistry exam score. The rest is free response. Practice 7.9 next on the past free-response questions College Board publishes.

← 7.8 Representations of Equilibrium 7.10 Reaction Quotient and Le Châtelier’s Principle →

Compiled from the AP Chemistry Course and Exam Description (effective Fall 2024) and our question bank · Specialist review in progress. How these pages are made · Free, no account