1 question, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 1
Pure A(g) is placed in a sealed, rigid container at constant temperature, where the reversible reaction A(g) ⇌ B(g) takes place. The diagrams represent the contents of the container at four successive times, t₀ to t₃. Which statement about the system is supported by the diagrams?
Answer and reasoning
AIt is at equilibrium at t₁ alone, as the numbers of A and B are equal at that time A student who thinks equilibrium means equal numbers of reactant and product particles picks this. The composition is still changing after t₁ (5 A and 5 B become 3 A and 7 B), so t₁ is not equilibrium; equilibrium is the unchanging mixture at t₂ and t₃.
BIt has been at equilibrium since t₁, as A and B are both present from then on A student who thinks any mixture containing both reactant and product is at equilibrium picks this. Between t₁ and t₂ the numbers still change, so the system reaches equilibrium only when the composition becomes constant, at 3 A and 7 B.
CIt has not reached equilibrium by t₃, as particles of A are still present at t₃ A student who thinks a reaction reaches equilibrium only when the reactant is used up picks this. In a reversible reaction reactant remains at equilibrium; the unchanged counts at t₂ and t₃ show that equilibrium has been reached.
DIt reached equilibrium by t₂, as its composition is the same in the t₂ and t₃ diagramsCorrect The diagrams show 10 A at t₀, 5 A and 5 B at t₁, and 3 A and 7 B at both t₂ and t₃. A composition that no longer changes, with reactant and product both present, is what a particulate diagram of an equilibrium mixture shows.
In preparation: 0 of 1 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
7.8.A.1 Particulate representation of a reversible reaction Fix
Particulate representation of a reversible reaction
A diagram in which a small, countable number of particles stands for the relative amounts of each reactant and product in a mixture. Diagrams of the same container drawn before reaction and at equilibrium show how the relative numbers of particles change.
Equilibrium mixture in a particulate diagram
At equilibrium the numbers of reactant and product particles no longer change from one diagram to the next, although the forward and reverse reactions continue. Reactant and product particles are both present, in numbers set by K rather than by the coefficients.
Conservation of atoms between diagrams
Diagrams of the same sealed container must contain the same number of atoms of each element at every stage; a reversible reaction rearranges atoms into different molecules but does not create or destroy them.
Equilibrium constant from a particulate diagram
The particle counts in a diagram of an equilibrium mixture are converted to concentrations (or partial pressures) and substituted into the K expression, each raised to the power of its coefficient. When the sum of the coefficients is the same on both sides of the equation, the volume and the amount that each particle stands for cancel, so the counts can be used directly.
Comparing Q with K for a diagram
Substituting the particle counts of any mixture into the same expression gives the reaction quotient Q. If Q = K the diagram shows an equilibrium mixture; if Q < K the net reaction forms more products, and if Q > K it forms more reactants.
Same equilibrium mixture from either direction
At a given temperature, containers of the same volume that hold the same number of atoms of each element reach the same equilibrium composition whether they start from reactants only or from products only.
Students often think At equilibrium the reactant and product particles are present in equal numbers, so a mixture is at equilibrium when the counts match. In fact Not in general. Equilibrium means that the forward and reverse rates are equal, so the numbers of particles stop changing; the numbers themselves are set by K and are usually unequal.
Students often think A reaction is finished, and so at equilibrium, only when the reactant has been used up; if reactant particles remain, equilibrium has not been reached. In fact No. In a reversible reaction the equilibrium mixture contains reactant particles as well as product particles; the numbers stop changing while reactant is still present.
4 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 4
The particulate diagram represents the equilibrium mixture in a rigid container for the hypothetical reaction X₂(g) + Y₂(g) ⇌ 2 XY(g) at a certain temperature. What is the value of Kc for the reaction at this temperature?
Answer and reasoning
A1.5 A student who leaves the coefficient out of the K expression picks this: 3/(1 × 2) = 1.5. The coefficient 2 makes the numerator [XY]², so Kc = 9/2 = 4.5.
B3.0 A student who uses the coefficient as a multiplier picks this: (2 × 3)/(1 × 2) = 3.0. A coefficient is an exponent in the K expression, so the numerator is 3² = 9 and Kc = 4.5.
C4.5Correct Kc = [XY]²/([X₂][Y₂]). With 1 X₂, 2 Y₂ and 3 XY in the same volume, and equal numbers of moles of gas on the two sides so that the volume cancels, Kc = 3²/(1 × 2) = 4.5.
D1.0 A student who takes K as product particles divided by reactant particles picks this: 3/(1 + 2) = 1.0. The reactant terms are multiplied, not added, and [XY] is squared, so Kc = 9/2 = 4.5.
Working The diagram shows 1 X₂, 2 Y₂ and 3 XY. Kc = [XY]²/([X₂][Y₂]). Both sides of the equation have two moles of gas, so the volume and the amount each particle stands for cancel and the counts can be used directly: Kc = 3²/(1 × 2) = 4.5.
For the hypothetical reaction D(g) ⇌ E(g), Kc = 3.0 at a certain temperature. A particulate diagram of a mixture of the two gases in a rigid container at this temperature shows 4 particles of D and 8 particles of E. Which claim about the mixture is correct?
Answer and reasoning
AIt is at equilibrium: E outnumbers D by 8 to 4, as a value of K above 1 requires A student who thinks K > 1 only requires more product than reactant picks this. Kc = 3.0 requires [E]/[D] = 3.0 exactly; here the ratio is 2.0, so the mixture is still changing.
BIt is not at equilibrium: the ratio of E to D is 2.0, not 3.0, so more E formsCorrect Qc = 8/4 = 2.0, which is less than Kc = 3.0, so the net reaction converts D to E until the ratio is 3.0 (3 D and 9 E for these 12 particles).
CIt is not at equilibrium: D and E must reach 6 particles each, so more D will form A student who thinks equilibrium means equal numbers of reactant and product particles picks this. Equal numbers would give Qc = 1.0; Kc = 3.0 requires three E for every D, so the net reaction forms more E.
DIt is at equilibrium: with 4 D and 8 E present, both reactions occur at equal rates A student who thinks any mixture containing both reactant and product is at equilibrium picks this. Both reactions do occur, but their rates are equal only when Qc = Kc; here Qc = 2.0 is less than 3.0, so the forward reaction is faster.
Working Qc = [E]/[D] = 8/4 = 2.0 (the volume cancels). Qc < Kc = 3.0, so the mixture is not at equilibrium and the net reaction converts D to E until [E]/[D] = 3.0, which for 12 particles is 3 D and 9 E.
For the hypothetical reaction X(g) ⇌ Y(g), Kc = 0.50 at a certain temperature. Two identical rigid containers are held at this temperature. Container 1 initially holds only X and container 2 initially holds only Y; a particulate diagram of each initial mixture shows 12 particles. A student then draws a particulate diagram of each container at equilibrium. Which pair of diagrams is correct?
Answer and reasoning
AContainer 1: 8 X, 4 Y; container 2: 8 X, 4 YCorrect Kc = [Y]/[X] = 0.50 applies in both containers, whichever substance was put in. With 12 particles, [Y]/[X] = 0.50 gives 8 X and 4 Y in each container: the same equilibrium mixture is reached from either direction.
BContainer 1: 6 X, 6 Y; container 2: 6 X, 6 Y A student who thinks equilibrium means equal numbers of reactant and product particles picks this. Equal numbers would give [Y]/[X] = 1.0, not 0.50; the equilibrium mixtures contain 8 X and 4 Y.
CContainer 1: 8 X, 4 Y; container 2: 4 X, 8 Y A student who treats the starting substance as the reactant in each container picks this, taking Kc as [X]/[Y] in container 2. Kc belongs to the written equation, so [Y]/[X] = 0.50 in both containers and each holds 8 X and 4 Y.
DContainer 1: 4 X, 8 Y; container 2: 4 X, 8 Y A student who writes K with the reactant in the numerator picks this, taking [X]/[Y] = 0.50. Kc = [Y]/[X] = 0.50, so each equilibrium mixture contains twice as much X as Y: 8 X and 4 Y.
A student represents the reversible reaction 2 NO₂(g) ⇌ N₂O₄(g) with two particulate diagrams of the same sealed container. The initial diagram shows 10 NO₂ molecules. The equilibrium diagram shows 4 NO₂ molecules and 4 N₂O₄ molecules. Which evaluation of the equilibrium diagram is correct?
Answer and reasoning
AIt is not valid: it should show NO₂ and N₂O₄ in the 2 : 1 ratio of the equation A student who thinks the species are present at equilibrium in the ratio of their coefficients picks this. The coefficients give the ratio in which the amounts change; the equilibrium amounts are set by K. The diagram is wrong because it does not conserve atoms.
BIt is valid: it shows equal numbers of NO₂ and N₂O₄, as an equilibrium mixture has A student who thinks equilibrium means equal numbers of reactant and product particles picks this. Equal numbers are not required at equilibrium, and this diagram contains 12 N atoms where the initial diagram has 10.
CIt is not valid: it shows 12 N atoms, but the initial diagram has 10 N atomsCorrect The container is sealed, so atoms are conserved. The initial diagram has 10 N atoms; 4 NO₂ and 4 N₂O₄ contain 4 + 8 = 12 N atoms. If 6 NO₂ molecules react, they form 3 N₂O₄ molecules, not 4.
DIt is valid: it shows both NO₂ and N₂O₄, which is what equilibrium requires A student who thinks any mixture containing both reactant and product represents equilibrium picks this. The diagram must also be consistent with the initial one: 4 NO₂ and 4 N₂O₄ contain 12 N atoms, but the container holds 10.
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
