4 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 4
An aqueous solution of acetic acid is at equilibrium at constant temperature: CH₃COOH(aq) + H₂O(l) ⇌ CH₃COO⁻(aq) + H₃O⁺(aq). Which statement best describes the proton transfers occurring in the solution?
Answer and reasoning
AProtons have stopped passing between the species now that equilibrium is reached. A student who thinks reactions stop at equilibrium picks this. The concentrations are constant because protons are transferred in both directions at equal rates, not because transfer has stopped.
BProtons pass from CH₃COOH to H₂O, and pass back only if the solution is cooled. A student who thinks a reversible process goes in reverse only when the conditions change picks this. The reverse transfer, from H₃O⁺ to CH₃COO⁻, takes place continuously at constant temperature.
CProtons pass from CH₃COOH to H₂O, and H₃O⁺ does not pass protons to CH₃COO⁻. A student who thinks proton transfer goes only from the acid to the base picks this. H₃O⁺ is itself an acid and CH₃COO⁻ can accept a proton, so the reverse transfer occurs too.
DProtons pass from CH₃COOH to H₂O and from H₃O⁺ to CH₃COO⁻ at equal rates.Correct The forward reaction transfers protons from CH₃COOH to H₂O, and the reverse reaction transfers protons from H₃O⁺ back to CH₃COO⁻. At equilibrium both transfers continue at equal rates, so the concentrations stay constant.
A sealed flask initially contains only NO₂(g). At constant temperature the reversible reaction N₂O₄(g) ⇌ 2 NO₂(g) occurs in the flask. Which of the numbered diagrams best represents the contents of a small volume of the flask after the system has reached equilibrium?
Answer and reasoning
ADiagram 1 A student who thinks a reaction goes to completion picks this diagram, in which all of the NO₂ has become N₂O₄. As N₂O₄ forms, it also dissociates back to NO₂, so some NO₂ is present at equilibrium.
BDiagram 2 A student who thinks a container that starts with only the substance on the right of the equation cannot react picks this diagram, which still shows only NO₂. NO₂ molecules combine to form N₂O₄, so N₂O₄ is present at equilibrium.
CDiagram 3Correct At equilibrium both N₂O₄ and NO₂ are present, and gas particles are spread evenly through the container. This diagram shows two N₂O₄ and four NO₂ molecules mixed together, accounting for the 8 N atoms and 16 O atoms of 8 NO₂ molecules.
DDiagram 4 A student who thinks reactants and products occupy separate regions at equilibrium picks this diagram. Both gases are present, but gas molecules move freely, so N₂O₄ and NO₂ are mixed evenly throughout the flask.
A saturated solution of NaCl is in contact with excess solid NaCl at constant temperature. A student adds a small amount of solid NaCl in which the sodium ions are radioactive ²⁴Na⁺, stirs the mixture and keeps it at the same temperature. Which outcome after several hours is predicted by the model of dynamic equilibrium?
Answer and reasoning
A²⁴Na⁺ ions have stayed in the solid, and Na⁺ ions have stayed in the solution. A student who thinks dissolving and precipitation have stopped in a saturated solution picks this. Both processes continue at equal rates, so radioactive ions do enter the solution.
B²⁴Na⁺ ions have entered the solution, and equally many Na⁺ ions joined the solid.Correct Dissolving and precipitation continue at equal rates in the saturated solution, so ions from the radioactive solid enter the solution while an equal number of ions from the solution join the solid. The tracer spreads into the solution although the amount of solid stays constant.
C²⁴Na⁺ ions entered the solution, and then those same ions went back into the solid. A student who thinks dissolving and precipitation take turns picks this, expecting the ions that dissolved to be returned afterward. The two processes occur at the same time in different ions, so ²⁴Na⁺ remains spread between the solution and the solid.
D²⁴Na⁺ ions have entered the solution, without Na⁺ ions joining the solid. A student who thinks only one of the two processes is going on picks this. If ions dissolved with none precipitating, the concentration would rise above that of the saturated solution; precipitation occurs at the same rate as dissolving.
A rigid, sealed container initially contains N₂O₄(g) and no NO₂(g). At constant temperature the reaction N₂O₄(g) ⇌ 2 NO₂(g) occurs. The graph shows the partial pressure of N₂O₄ as a function of time. What is the partial pressure of NO₂ in the container once equilibrium has been established?
Answer and reasoning
A0.3 atm A student who thinks the amount of product formed equals the amount of reactant used, whatever the coefficients, picks this. The coefficient 2 means that 0.3 atm of N₂O₄ produces 0.6 atm of NO₂.
B0.7 atm A student who thinks reactants and products are present in equal amounts at equilibrium picks this, giving NO₂ the same partial pressure as N₂O₄. The pressure of NO₂ follows from the change in N₂O₄: 2 × 0.3 atm = 0.6 atm.
C1.4 atm A student who thinks equilibrium amounts are in the ratio of the coefficients picks this, doubling the equilibrium pressure of N₂O₄. The coefficients apply to the change: 0.3 atm of N₂O₄ reacted, giving 0.6 atm of NO₂.
D0.6 atmCorrect The partial pressure of N₂O₄ falls from 1.0 atm to 0.7 atm, so 0.3 atm of N₂O₄ reacts. Each mole of N₂O₄ forms 2 mol of NO₂, and at constant volume and temperature partial pressure is proportional to moles, so P(NO₂) = 0.6 atm.
Working From the graph, P(N₂O₄) falls from 1.0 atm to 0.7 atm, a decrease of 0.3 atm. At constant V and T, partial pressure is proportional to moles, and 2 mol NO₂ form per mole of N₂O₄ that reacts, so P(NO₂) = 2 × 0.3 atm = 0.6 atm.
In preparation: 0 of 4 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
7.1.A.1 Reversible process Fix
Reversible process
A process that can occur in both directions under the same conditions, so that the products can re-form the reactants. Examples include evaporation and condensation of water, absorption and desorption of a gas, dissolution and precipitation of a salt, proton transfer in acid-base reactions and electron transfer in redox reactions.
Forward and reverse reactions
For a reversible reaction written as reactants ⇌ products, the forward reaction converts reactants to products and the reverse reaction converts products to reactants. In a closed system both occur at the same time once both reactants and products are present.
Students often think A reversible process goes forward under one set of conditions and reverses only when the conditions change, for example when the system is heated or cooled. In fact No. A reversible process can occur in both directions under the same conditions; at equilibrium both directions occur at the same time without any change in temperature or pressure.
Students often think In an acid-base reaction protons move from the acid to the base and are not transferred back. In fact No. Proton transfer is reversible: an acid can give a proton to water, and the products H₃O⁺ and the conjugate base can transfer a proton back. In a solution of acetic acid both transfers continue at equal rates at equilibrium.
7.1.A.2 Equilibrium state Fix
Equilibrium state
The state of a closed system in which no observable changes occur: reactants and products are present together, and the concentrations or partial pressures of all species remain constant. The constant values are generally not equal to one another.
Closed system
A system that does not exchange matter with its surroundings, such as a sealed flask. A liquid and its vapor, or a mixture of reacting gases, can come to equilibrium in a sealed container because none of the substances can escape.
Students often think At equilibrium the amounts, concentrations or partial pressures of reactants and products are equal. In fact No. At equilibrium the concentrations are constant, not necessarily equal. Depending on the reaction and the starting amounts, the equilibrium mixture can contain mostly reactants, mostly products or comparable amounts.
Students often think Given enough time, a reaction goes to completion, so all of the starting material is eventually converted. In fact No. In a closed container a reversible reaction reaches an equilibrium state in which reactants and products are both present; the reactants are not completely used up.
7.1.A.3 Dynamic equilibrium Fix
Dynamic equilibrium
At equilibrium the forward and reverse processes continue to occur, at equal rates, so there is no net observable change. Individual particles keep changing from reactant to product and back.
Evidence that equilibrium is dynamic
Tracer experiments show that particles keep moving between reactants and products at equilibrium: for example, radioactive ions added in a solid to a saturated solution of the same salt appear in the solution, although the amount of solid does not change.
Students often think When a system reaches equilibrium the reaction stops, so no particles are being converted in either direction. In fact No. Equilibrium is dynamic: the forward and reverse processes continue at equal rates, so the amounts stay constant although particles keep reacting.
Students often think The forward reaction runs first and the reverse reaction starts only when the forward reaction has finished. In fact No. Once some products have formed, the reverse reaction begins while the forward reaction is still going on; the two occur simultaneously throughout the approach to equilibrium and at equilibrium.
7.1.A.4 Concentration-time and pressure-time graphs Fix
Concentration-time and pressure-time graphs
As a reaction approaches equilibrium, the concentrations (or partial pressures) of reactants and products change, and equilibrium is established when every curve becomes horizontal. The changes in the curves before that point are in the ratio of the coefficients of the balanced equation.
Rate-time graphs for a reversible reaction
When a reaction starts with only reactants, the forward rate is greatest at the start and decreases as reactants are used up, while the reverse rate starts at zero and increases as products form. Equilibrium is established when the two rates become equal; both then remain constant and nonzero.
Students often think The forward reaction speeds up as the reaction proceeds, until it reaches equilibrium. In fact No. When only reactants are present at the start, the forward rate is greatest at the start and decreases as the reactants are used up; it is the reverse rate that increases as products accumulate.
Students often think Each substance in a reaction changes by the same amount, so the amount of product formed equals the amount of reactant used, whatever the coefficients. In fact Not necessarily. The changes in amount are in the ratio of the coefficients: when 1 mol of N₂O₄ reacts, 2 mol of NO₂ form, so the changes are equal only for substances with equal coefficients.
6 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 6
A sealed flask at constant temperature contains a mixture of brown NO₂(g) and colorless N₂O₄(g), which interconvert by the reaction N₂O₄(g) ⇌ 2 NO₂(g). The intensity of the brown color has not changed for an hour. Which of the following best explains why the color stays the same?
Answer and reasoning
ANO₂ is formed and used up at equal rates, so the amount of NO₂ stays constant.Correct The color is due to NO₂. At equilibrium the forward reaction forms NO₂ at the same rate as the reverse reaction uses it up, so the amount of NO₂, and therefore the color, does not change.
BNO₂ has stopped forming and stopped reacting, so the amount of NO₂ stays constant. A student who thinks reactions stop at equilibrium picks this. The equilibrium is dynamic: NO₂ molecules keep forming and reacting, at equal rates.
CNO₂ and N₂O₄ are now present in equal amounts, so the color has stopped changing. A student who thinks equilibrium means equal amounts picks this. The amounts at equilibrium are constant but need not be equal, and equal amounts would not by themselves keep the color from changing.
DThe two rate constants have now become equal, so the color has stopped changing. A student who thinks the rate constants become equal at equilibrium picks this. The rates of the forward and reverse reactions become equal because the concentrations adjust; the two rate constants are fixed at a given temperature and generally differ.
Pure A(g) is placed in a rigid, sealed container at constant temperature, where the reaction A(g) ⇌ 2 B(g) occurs. The graph shows [A] and [B] as functions of time, with three times marked. At which marked time is the system at equilibrium, and why?
Answer and reasoning
AAt t₁, because [A] and [B] have the same value at that time A student who thinks equilibrium means equal concentrations picks this. The curves cross at t₁, but both concentrations are still changing there, so the system is not yet at equilibrium.
BAt t₃, because [A] and [B] have stopped changing by that timeCorrect At equilibrium the concentrations of all species remain constant. Both curves are horizontal at t₃, while at t₁ and t₂ both concentrations are still changing, so t₃ is the only marked time at which the system is at equilibrium.
CAt t₂, because [B] is twice [A], as in the balanced equation A student who thinks the equilibrium concentrations are in the ratio of the coefficients picks this. At t₂ [B] is twice [A], but both are still changing; the coefficients relate the changes in concentration, not the equilibrium values.
DAt t₃, because A and B have stopped reacting by that time A student who thinks reactions stop at equilibrium picks this. The system is at equilibrium at t₃, but the reason is wrong: A and B continue to interconvert at equal rates, which is why their concentrations no longer change.
A sealed flask at constant temperature initially contains 0.70 mol of N₂O₄(g) and no NO₂(g). The reaction N₂O₄(g) ⇌ 2 NO₂(g) occurs. When equilibrium has been established, the flask contains 0.20 mol of NO₂(g). What is the total amount of gas in the flask at equilibrium?
Answer and reasoning
A0.70 mol A student who thinks each substance changes by the same amount picks this, taking 0.20 mol of N₂O₄ to have reacted and 0.50 mol to remain. Only 0.10 mol of N₂O₄ is needed to form 0.20 mol of NO₂, so 0.60 mol remains.
B0.40 mol A student who thinks equal amounts of reactant and product are present at equilibrium picks this, taking 0.20 mol of each gas. The amount of N₂O₄ left is 0.70 − 0.10 = 0.60 mol.
C0.80 molCorrect Forming 0.20 mol of NO₂ uses 0.10 mol of N₂O₄, so 0.60 mol of N₂O₄ remains. Both gases are present at equilibrium: 0.60 mol + 0.20 mol = 0.80 mol of gas in total.
D0.30 mol A student who thinks the equilibrium amounts are in the ratio of the coefficients picks this, taking N₂O₄ as half of 0.20 mol. The coefficient ratio gives the amount of N₂O₄ that reacted (0.10 mol), not the amount remaining.
Working 0.20 mol NO₂ forms from 0.20/2 = 0.10 mol N₂O₄. N₂O₄ remaining = 0.70 − 0.10 = 0.60 mol. Total = 0.60 + 0.20 = 0.80 mol.
A small amount of liquid water is placed in an evacuated flask, which is then sealed and kept at constant temperature. The pressure of water vapor in the flask rises and then becomes constant while some liquid remains. Which of the following best explains why the pressure stops rising?
Answer and reasoning
AWater molecules now return to the liquid at the same rate that they leave it.Correct Evaporation continues, while the rate of condensation increases as vapor builds up. Once molecules return to the liquid as often as they leave it, the number of molecules in the vapor, and so the pressure, stays constant.
BWater molecules have stopped moving between the liquid and the vapor. A student who thinks nothing happens at equilibrium picks this. Molecules keep evaporating and condensing; the pressure is constant because the two processes occur at equal rates.
CWater molecules have filled the space, leaving no room for more vapor. A student who thinks evaporation stops when the space above the liquid is full picks this. The space is mostly empty; the pressure stops rising because condensation has become as fast as evaporation.
DWater molecules leave the liquid and return to it by turns, not at the same time. A student who thinks the forward and reverse processes take turns picks this. Evaporation and condensation occur at the same time, in different molecules, at equal rates.
A reversible reaction is started with only reactants in a closed container at constant temperature. Which of the numbered graphs best represents the rates of the forward and reverse reactions as the system approaches and reaches equilibrium?
Answer and reasoning
AGraph 1 A student who thinks the reverse reaction begins only after the forward reaction has finished picks this graph. The reverse reaction starts as soon as products exist, so both rates change together from the start.
BGraph 2 A student who thinks the forward reaction speeds up as the reaction proceeds picks this graph. The forward rate depends on the reactant concentrations, which are greatest at the start, so the forward rate decreases.
CGraph 3 A student who thinks the reactions stop at equilibrium picks this graph, in which both rates end at zero. At equilibrium the two rates are equal and greater than zero.
DGraph 4Correct The forward rate starts high, when reactant concentrations are greatest, and decreases as reactants are used; the reverse rate starts at zero and rises as products accumulate. The rates become equal at equilibrium and then stay constant, above zero.
A rigid, sealed container at constant temperature initially contains only X(g). The hypothetical reaction 2 X(g) ⇌ 2 Y(g) + Z(g) occurs. The table shows the total pressure in the container at various times. What is the partial pressure of X once equilibrium has been established?
Answer and reasoning
A0.40 atmCorrect The pressure stops changing at 1.30 atm, so that is the equilibrium total. If x is the pressure of Z formed, the total is (1.00 − 2x) + 2x + x = 1.00 + x, so x = 0.30 atm and P(X) = 1.00 − 0.60 = 0.40 atm.
B0.70 atm A student who thinks each substance changes by the same amount picks this, writing the pressures as 1.00 − x, x and x, so that x = 0.30 atm and P(X) = 0.70 atm. Two moles of X react for each mole of Z formed, so P(X) falls by 0.60 atm.
C0.43 atm A student who thinks the partial pressures of all species are equal at equilibrium picks this, dividing 1.30 atm by 3. The equilibrium pressures follow from the changes: 0.40 atm of X, 0.60 atm of Y and 0.30 atm of Z.
D0.52 atm A student who thinks the equilibrium amounts are in the ratio of the coefficients, 2:2:1, picks this, taking two fifths of 1.30 atm. The coefficients relate the changes in pressure, which give P(X) = 0.40 atm.
Working The total pressure is constant at 1.30 atm from 15 min, so the system is at equilibrium. Let the pressure of Z formed be x: P(X) = 1.00 − 2x, P(Y) = 2x, P(Z) = x, total = 1.00 + x = 1.30 atm, so x = 0.30 atm. P(X) = 1.00 − 2(0.30) = 0.40 atm (P(Y) = 0.60 atm, P(Z) = 0.30 atm).
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
