2 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 2
Four hypothetical gas-phase reactions, each of which occurs in a single elementary step, are listed. Reaction 1: A₂B₂(g) → A₂(g) + B₂(g). Reaction 2: AB(g) + B(g) → A(g) + B₂(g). Reaction 3: A₂(g) + B₂(g) → 2 AB(g). Reaction 4: 2 AB(g) → A₂(g) + B₂(g). Which reaction has the rate law rate = k[AB]²?
Answer and reasoning
AReaction 1 A student who reads the exponent in [AB]² as subscripts picks the step whose reactant is A₂B₂. A₂B₂ is one particle, so that step has rate = k[A₂B₂], which is first order.
BReaction 2 A student who thinks 'second order' means that two different reactants collide picks the step in which AB collides with B. That step is second order overall, but its rate law is rate = k[AB][B], not k[AB]².
CReaction 3 A student who writes the rate law from the products of the step picks the step that forms 2 AB. The rate law of an elementary reaction contains the particles that collide, the reactants: for this step rate = k[A₂][B₂].
DReaction 4Correct In an elementary reaction each reactant's concentration is raised to its coefficient in the step. Two AB particles collide, so rate = k[AB]².
Working For an elementary reaction the exponent of each reactant in the rate law is its coefficient in the step. Rate = k[AB]² needs two AB particles, and nothing else, as reactants: 2 AB(g) → A₂(g) + B₂(g). The other steps give k[A₂B₂], k[AB][B] and k[A₂][B₂].
Each numbered diagram shows, to the left of the arrow, the particles that take part in a proposed elementary reaction and, to the right of the arrow, the particles formed. Which diagram shows the kind of elementary reaction that is least likely to occur?
Answer and reasoning
ADiagram 1Correct Three particles must collide at the same instant in this step. Elementary reactions that need the simultaneous collision of three or more particles are rare, because such a collision is far less probable than one between two particles.
BDiagram 2 A student who thinks a reaction needs two particles to collide picks this step, which has one reactant particle. An elementary reaction can involve a single particle breaking apart; it is steps needing three or more particles to collide at once that are rare.
CDiagram 3 A student who thinks two particles of the same substance do not react with each other picks this step. Two identical particles can collide and react; this is a two-particle collision, which is common.
DDiagram 4 A student who applies the 'three particles' rule to the products picks this step, which forms three particles. Only two particles collide here; the rarity of a step depends on how many reactant particles must meet at once.
In preparation: 0 of 2 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
5.4.A.1 Elementary reaction Fix
Elementary reaction
A reaction that takes place in a single step, in which the reactant particles shown in its equation come together in one collision (or one particle changes on its own) to give the products of that step.
Rate law of an elementary reaction
For an elementary reaction the rate law can be written from the equation of that step: the rate equals a rate constant multiplied by the concentration of each reactant particle raised to its coefficient. For A + B → products, rate = k[A][B]; for 2 A → products, rate = k[A]².
Molecularity
The number of reactant particles that take part in an elementary reaction: one (unimolecular), two (bimolecular) or three (termolecular). It equals the overall order of the rate law of that elementary reaction.
Students often think A subscript and an exponent are interchangeable: a molecule X₂ in an elementary reaction gives [X]² in the rate law, and [X]² in a rate law means that the reactant is X₂. In fact No. A subscript is part of the formula of one particle: X₂ is a single molecule, and its concentration term is [X₂]. An exponent in the rate law of an elementary reaction counts how many particles of that species take part in the collision: [X]² means two separate X particles collide.
Students often think The rate law of an elementary reaction is written from the products of the step (or from every species in its equation), because the rate is the rate at which products form. In fact No. The rate law of an elementary reaction is written from the particles that collide, which are the reactants of that step. The concentrations of the products of the step do not appear.
5.4.A.2 Rarity of three-particle elementary reactions Fix
Rarity of three-particle elementary reactions
Elementary reactions that need three or more particles to collide at the same instant are rare, because such a simultaneous collision is much less probable than a collision between two particles.
Students often think An elementary reaction with one reactant particle is the least likely kind, because a reaction needs two particles to collide. In fact Yes. An elementary reaction can involve one reactant particle that breaks apart or rearranges. Such steps are not rare in the way that three-particle collisions are.
Students often think Two particles of the same substance do not react with each other in a single step; an elementary reaction needs two different reactants. In fact Yes. Two identical particles can collide and react in one step; the rate law is then second order in that substance. What makes an elementary reaction rare is needing three or more particles to collide at the same instant.
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 hypothetical reaction X₂(g) + Y₂(g) → 2 XY(g) occurs in a single elementary step. At a certain temperature the rate constant for the reaction is 0.50 M⁻¹ s⁻¹. At one moment during the reaction, [X₂] = 0.30 M, [Y₂] = 0.20 M and [XY] = 0.10 M. What is the rate of the reaction at that moment?
Answer and reasoning
A1.8 × 10⁻³ M s⁻¹ A student who uses the subscripts in X₂ and Y₂ as exponents picks this: (0.50)(0.30)²(0.20)² = 1.8 × 10⁻³. One X₂ molecule and one Y₂ molecule collide, so each concentration is raised to the first power.
B5.0 × 10⁻³ M s⁻¹ A student who writes the rate law from the product of the step picks this: (0.50)(0.10)² = 5.0 × 10⁻³. The rate law of an elementary reaction contains the reactants of the step, the particles that collide.
C3.0 × 10⁻² M s⁻¹Correct The step is elementary, so its rate law comes from the particles that collide: rate = k[X₂][Y₂] = (0.50 M⁻¹ s⁻¹)(0.30 M)(0.20 M) = 3.0 × 10⁻² M s⁻¹.
D2.5 × 10⁻¹ M s⁻¹ A student who adds the reactant concentrations picks this: (0.50)(0.30 + 0.20) = 2.5 × 10⁻¹. The concentration terms in a rate law are multiplied.
Working Elementary step: rate = k[X₂][Y₂] = (0.50 M⁻¹ s⁻¹)(0.30 M)(0.20 M) = 0.030 M s⁻¹ = 3.0 × 10⁻² M s⁻¹. Distractors: subscripts used as exponents, (0.50)(0.30)²(0.20)² = 1.8 × 10⁻³; products used, (0.50)(0.10)² = 5.0 × 10⁻³; concentrations added, (0.50)(0.30 + 0.20) = 2.5 × 10⁻¹.
A student hypothesizes that the hypothetical reaction A₂(g) + B₂(g) → 2 AB(g) occurs in a single elementary step. The table shows the initial concentrations chosen for three trials at the same temperature and the initial rate measured in trial 1. If the hypothesis is correct, which initial rates will be measured in trials 2 and 3?
Answer and reasoning
ATrial 2: 8.0 × 10⁻⁴ M s⁻¹; trial 3: 7.2 × 10⁻³ M s⁻¹ A student who uses the subscripts in A₂ and B₂ as exponents picks this, with rate = k[A₂]²[B₂]²: 4 times the trial 1 rate in trial 2 and 4 × 9 = 36 times in trial 3. One A₂ molecule and one B₂ molecule collide, so each concentration is raised to the first power.
BTrial 2: 4.0 × 10⁻⁴ M s⁻¹; trial 3: 1.2 × 10⁻³ M s⁻¹Correct For a single elementary step the rate law follows from the particles that collide: rate = k[A₂][B₂]. Doubling [A₂] alone doubles the rate (trial 2); doubling [A₂] and tripling [B₂] makes it 2 × 3 = 6 times as great (trial 3).
CTrial 2: 3.0 × 10⁻⁴ M s⁻¹; trial 3: 5.0 × 10⁻⁴ M s⁻¹ A student who adds the reactant concentrations picks this: the total goes from 0.20 M to 0.30 M (1.5 times) and to 0.50 M (2.5 times). The concentration terms are multiplied in the rate law, not added.
DTrial 2: 2.0 × 10⁻⁴ M s⁻¹; trial 3: 4.0 × 10⁻⁴ M s⁻¹ A student who thinks the reactant in the smaller concentration alone sets the rate picks this: that concentration is still 0.10 M in trial 2 and is 0.20 M in trial 3. In an elementary step between A₂ and B₂ the rate is proportional to both concentrations.
Working If the reaction is elementary, rate = k[A₂][B₂]. Trial 2: [A₂] is doubled and [B₂] is unchanged, so rate = 2 × 2.0 × 10⁻⁴ = 4.0 × 10⁻⁴ M s⁻¹. Trial 3: [A₂] is doubled and [B₂] is tripled compared with trial 1, so rate = 2 × 3 × 2.0 × 10⁻⁴ = 1.2 × 10⁻³ M s⁻¹. Distractors: rate = k[A₂]²[B₂]² gives 4 × and 36 × the trial 1 rate; rate proportional to [A₂] + [B₂] gives 1.5 × and 2.5 ×; rate set by the smaller concentration gives 1 × and 2 ×.
The reaction 2 NO(g) + O₂(g) → 2 NO₂(g) has the experimentally determined rate law rate = k[NO]²[O₂]. A student claims that this rate law shows that the reaction occurs in a single elementary step. Which evaluation of the claim is best?
Answer and reasoning
ADoubtful: a single step would need a rare collision of three particles at once.Correct A single elementary step would give this rate law, so the rate law is consistent with the claim, but it does not show that the claim is true. One step would require two NO molecules and one O₂ molecule to collide at the same instant, and elementary reactions involving three particles are rare.
BJustified: orders equal to the coefficients show that a reaction is a single step. A student who runs the rule for elementary reactions backwards picks this. An elementary step gives orders equal to its coefficients, but orders equal to the coefficients do not show that a reaction is elementary.
CJustified: a balanced equation shows the particles that collide in the reaction. A student who reads a balanced overall equation as a picture of one collision picks this. An overall equation shows only the amounts that react; only the equation of an elementary step shows the particles that collide.
DDoubtful: two molecules of one substance, NO, do not react in the same step. A student who thinks identical particles do not react with each other picks this. Two identical molecules can collide in an elementary step; the difficulty with a single step here is that three particles would have to collide at once.
The hypothetical gas-phase reaction 2 XY₂(g) + Z₂(g) → 2 XY₂Z(g) takes place in more than one elementary step. The diagram represents one of these elementary steps: to the left of the arrow are the particles that collide, and to the right is the particle formed. Which expression is the rate law for the elementary step shown in the diagram (rate = …)?
Answer and reasoning
Ak[XY₂]²[Z₂] A student who reads the balanced overall equation as a picture of the collision picks this, using its coefficients 2 and 1. The overall reaction takes place in more than one step; the step in the diagram involves only one XY₂ particle and one Z₂ particle.
Bk[X][Y]²[Z]² A student who uses the subscripts in XY₂ and Z₂ as exponents picks this. XY₂ and Z₂ are each one particle in the collision, so the rate law contains [XY₂] and [Z₂], each to the first power.
Ck[XY₂Z₂] A student who writes the rate law from the product of the step picks this. The rate law of an elementary step contains the particles that collide, the reactants XY₂ and Z₂, not the particle that is formed.
Dk[XY₂][Z₂]Correct The step shown is a collision between one XY₂ particle and one Z₂ particle. The rate law of an elementary step follows from the particles that collide, each concentration raised to the number of those particles in the collision: rate = k[XY₂][Z₂].
Working The diagram shows one XY₂ particle colliding with one Z₂ particle: XY₂ + Z₂ → XY₂Z₂. For an elementary step the rate law is inferred from the particles that collide, each concentration raised to the number of those particles in the collision: rate = k[XY₂][Z₂]. Distractors: coefficients of the overall equation, k[XY₂]²[Z₂]; subscripts used as exponents, k[X][Y]²[Z]²; product of the step, k[XY₂Z₂].
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
