3 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 3
The hypothetical gases A and B react in the elementary reaction A(g) + B(g) → AB(g). A rigid container at constant temperature initially holds equal numbers of A and B molecules. More of each gas is then added so that the number of A molecules is doubled and the number of B molecules is tripled. By what factor is the frequency of collisions between A and B molecules multiplied?
Answer and reasoning
A1.0 A student who thinks that the rate of a reaction has one fixed value at a given temperature picks this, because the temperature is constant. The molecules move no faster, but there are more of them in the same volume, so A–B collisions become more frequent.
B2.0 A student who thinks that the rate is controlled only by the limiting reactant picks this: after the additions A is limiting, and its amount has doubled. Every A molecule now also meets three times as many B molecules, so the factor for B counts as well.
C6.0Correct The frequency of A–B collisions is proportional to [A] × [B]. In the same volume [A] doubles and [B] triples, so the frequency is multiplied by 2 × 3 = 6.0.
D2.5 A student who thinks that collisions are proportional to the total number of molecules picks this: the total goes from 2n to 5n. Only A–B collisions can produce AB, and their frequency is proportional to [A] × [B], which is multiplied by 6.
Working Frequency of A–B collisions ∝ [A] × [B]. In the same volume [A] is doubled and [B] is tripled, so the frequency is multiplied by 2 × 3 = 6.0. Distractors: 1.0 (rate taken as fixed at a given temperature); 2.0 (A, now the limiting reactant, is doubled, so only its factor is counted); 2.5 (total number of molecules goes from 2n to 5n).
The hypothetical elementary reaction X–Y(g) + Z(g) → X(g) + Y–Z(g) occurs in the gas phase. The diagram shows two collisions between an X–Y molecule and a Z atom, with the energy of each collision compared with the activation energy, Eₐ. Which prediction about the two collisions is correct?
Answer and reasoning
ACollision 1 is more likely to give products, because Z meets the Y atom it must bond to.Correct A successful collision needs enough energy and a suitable orientation. Both collisions have energy greater than Eₐ, but only in collision 1 does Z meet the Y atom, so only there can the Y–Z bond form as the X–Y bond breaks.
BCollision 2 is more likely to give products, because it has the greater collision energy. A student who thinks that extra energy makes up for orientation picks this. In collision 2, Z strikes the X end, so the Y–Z bond cannot form in that collision, however much energy it has.
CThe two are equally likely to give products, because both have energy greater than Eₐ. A student who thinks that every collision with enough energy reacts picks this. Energy of at least Eₐ is necessary but not enough; the particles must also meet in an orientation that lets the bonds rearrange.
DNeither gives products, because X–Y must first break apart into separate atoms of X and Y. A student who thinks that reactants must break completely into atoms before new bonds form picks this. In an elementary reaction the Y–Z bond forms during the same collision in which the X–Y bond breaks.
In each of the numbered graphs shown, the solid curve is the Maxwell–Boltzmann distribution of kinetic energies for a sample of a gas at 300 K. Which graph's dashed curve best represents the distribution for the same sample at 400 K?
Answer and reasoning
AGraph 1Correct At 400 K the particles' energies are spread over a wider range: the curve still starts at zero, its peak moves to a higher energy and is lower, and the curve is broader with a longer high-energy tail. The area, which stands for all the particles in the sample, is unchanged.
BGraph 2 A student who thinks that the peak becomes taller at a higher temperature picks this graph. Heating spreads the particles over more energies, so the peak is lower, not taller, although it does move to a higher energy.
CGraph 3 A student who thinks that heating adds the same energy to every particle picks this graph, in which the whole curve slides to higher energy. At 400 K some particles still have very low energies, so the curve starts at zero, and its shape changes: lower and broader.
DGraph 4 A student who thinks that heating puts more particles at every energy picks this graph, which is higher everywhere and so has a larger area. The number of particles does not change, so the area stays the same; the curve flattens and spreads toward higher energies.
In preparation: 0 of 3 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
5.5.A.1 Collision model Fix
Collision model
The model that explains the rate of an elementary reaction in terms of particle collisions: reactant particles must collide for bonds to begin breaking and new bonds to begin forming. The rate depends on how often reactant particles collide and on what fraction of those collisions are successful.
Collision frequency
The number of collisions between reactant particles per second in a given volume. At a fixed temperature, the frequency of collisions between A and B particles is proportional to [A] × [B], so it rises when either concentration rises, for example when the same particles are confined in a smaller volume.
Students often think Crowding reactant particles together, by raising their concentration, makes their collisions more energetic, so a larger fraction of the collisions succeed. In fact No. At a fixed temperature the distribution of particle energies does not depend on concentration. A higher concentration makes collisions between reactant particles more frequent, so more successful collisions occur per second, but the fraction of collisions with enough energy is unchanged.
Students often think The rate of a reaction is controlled only by the limiting reactant, so adding more of a reactant that is in excess does not change the rate. In fact Yes. The rate of the elementary reaction A + B → AB is proportional to [A] × [B], so raising [B] raises the rate even when B is already in excess. The limiting reactant decides how much product can form, not how fast it forms.
5.5.A.2 Successful collision Fix
Successful collision
A collision that leads to reaction. It has energy at least equal to the activation energy and an orientation that allows the bonds to rearrange in the required manner. In most reactions only a small fraction of collisions are successful.
Activation energy, Eₐ
The minimum energy that a collision between reactant particles must have for the particles to rearrange into products. A collision with less energy than Eₐ does not produce products, whatever its orientation.
Orientation requirement
The requirement that colliding particles meet in an arrangement that lets the required bonds break and form; for example, in X–Y + Z → X + Y–Z the Z atom must meet the Y end of the molecule for the Y–Z bond to form. Extra collision energy does not make up for an unsuitable orientation.
Students often think Every collision with energy of at least Eₐ leads to reaction; the orientation of the colliding particles does not matter. In fact No. A successful collision needs both enough energy and an orientation that allows the bonds to rearrange in the required manner. Collisions with enough energy in which the particles meet in an unsuitable orientation do not lead to reaction.
Students often think The more energy a collision has, the more likely it is to react, whatever the orientation of the colliding particles. In fact No. The orientation decides which atoms meet. In X–Y + Z → X + Y–Z, if Z strikes the X end of the molecule the Y–Z bond cannot form in that collision, however much energy the collision has.
5.5.A.3 Maxwell–Boltzmann distribution Fix
Maxwell–Boltzmann distribution
A curve showing how the kinetic energies of the particles in a sample at one temperature are spread out: fraction of particles plotted against energy. The area under the curve beyond Eₐ represents the fraction of particles with at least Eₐ, which gives a qualitative estimate of the fraction of collisions with enough energy to react.
Effect of temperature on the energy distribution
At a higher temperature the Maxwell–Boltzmann curve is lower and broader, with its peak at a higher energy; the total area, which stands for all the particles, is unchanged. The area beyond Eₐ is larger, so a larger fraction of collisions have enough energy to react.
Students often think Raising the temperature increases the rate of a reaction only because the faster-moving particles collide more often. In fact No. Faster particles do collide a little more often, but the larger effect is that a greater fraction of collisions have energy of at least Eₐ. For a reaction with a large Eₐ, a rise of 10 K increases the collision frequency by only a few percent but can roughly double the rate.
Students often think Heating a reaction mixture lowers the activation energy, and that is why the reaction goes faster. In fact No. Eₐ is the minimum collision energy that the reaction requires, and at AP level it is treated as the same at every temperature. Raising the temperature increases the fraction of collisions that have at least that energy.
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 gases A and B react in the elementary reaction A(g) + B(g) → AB(g). The diagram represents the contents of two rigid containers, at the same temperature, just after the gases are mixed. Which statement about the initial rates of reaction in the two containers is correct, with its reasoning?
Answer and reasoning
AThe rate is greater in container 1, because collisions between A and B molecules are more energetic there. A student who thinks that crowding makes collisions more energetic picks this. The containers are at the same temperature, so the energy distribution is the same; the molecules in container 1 collide more often, not harder.
BThe rate is greater in container 1, because A and B molecules collide with each other more frequently there.Correct Container 1 holds the same numbers of A and B molecules in half the volume, so [A] and [B] are both twice as great. The molecules are closer together and A–B collisions happen more often, so more successful collisions occur per second and the rate is greater.
CThe rates are equal, because the two containers hold the same numbers of A and B molecules. A student who thinks that the rate depends on the number of molecules present, whatever volume they occupy, picks this. The rate depends on concentrations: the same molecules in twice the volume collide with each other less often.
DThe rates are equal, because the two containers are at the same temperature as each other. A student who thinks that the rate of a reaction is fixed at a given temperature picks this. The rate constant, k, is the same in both containers, but rate = k[A][B], and the concentrations are greater in container 1.
A student plans to measure the initial rate of a hypothetical elementary reaction that releases heat and has a large activation energy, Eₐ. Trials will be run at 300 K and at 310 K with the same initial concentrations. The student calculates that the 10 K rise increases the frequency of collisions between reactant molecules by only about 2%. Which prediction about the measured rates is consistent with the collision model?
Answer and reasoning
AThe rate at 310 K will be about 2% greater, because the rate increases in step with the collision frequency. A student who thinks that temperature raises the rate only through more frequent collisions picks this. The main effect of heating is the larger fraction of collisions with energy of at least Eₐ, which makes the rate rise by much more than 2%.
BThe rate at 310 K will be much more than 2% greater, because the activation energy will be lower at the higher temperature. A student who thinks that heating lowers Eₐ picks this. The prediction of a large rise is right, but the reasoning is not: Eₐ is the same at both temperatures; more of the collisions reach it at 310 K.
CThe rate at 310 K will be lower than at 300 K, because heating works against a reaction that releases heat. A student who thinks that heating slows a reaction that releases heat picks this. For an elementary reaction a higher temperature gives a larger fraction of collisions with energy of at least Eₐ, so the rate increases whether the reaction releases or absorbs heat.
DThe rate at 310 K will be much more than 2% greater, because more of the collisions will have energy of at least Eₐ.Correct At 310 K the Maxwell–Boltzmann distribution has a larger area beyond Eₐ, so a larger fraction of collisions have enough energy. For a large Eₐ this fraction rises far more steeply than the 2% rise in collision frequency, so the rate rises by much more than 2%.
Working Rate ∝ (collision frequency) × (fraction of collisions with energy ≥ Eₐ and suitable orientation). The collision frequency rises by about 2%, but for a large Eₐ the fraction of collisions with energy ≥ Eₐ rises steeply with temperature, so the rate at 310 K is much more than 2% greater. Whether the reaction releases or absorbs heat does not change this for an elementary reaction.
A student has a graph of the Maxwell–Boltzmann distribution of kinetic energies for the reactant molecules in a gas mixture at 350 K. The student wants to use the graph to estimate what fraction of the collisions in the mixture at 350 K have enough energy to lead to reaction. Which additional quantity does the student need?
Answer and reasoning
AThe energy released by the overall reaction A student who thinks that the energy a reaction releases determines how fast it goes picks this. The fraction of collisions with enough energy depends on Eₐ, which is a different quantity from the energy released by the overall reaction.
BThe minimum collision energy for reactionCorrect The minimum collision energy for reaction is the activation energy, Eₐ. The fraction of collisions with enough energy is estimated from the area under the curve beyond Eₐ: the curve gives the distribution of energies at 350 K, and Eₐ marks where the counted area starts.
CThe concentrations of the reactants A student who thinks that a higher concentration makes collisions more energetic picks this. Concentration changes how often molecules collide, not the distribution of their energies, so it does not change the fraction with enough energy.
DThe molar masses of the reactant molecules A student who thinks that heavier molecules have more kinetic energy at the same temperature picks this. At 350 K all the gases in the mixture have the same distribution of kinetic energies, which the graph already shows.
Working The fraction of collisions with enough energy is estimated from the area under the curve at energies of at least Eₐ, compared with the total area. The curve already describes the molecules at 350 K, so the only extra quantity needed is Eₐ, to locate where that area begins.
The hypothetical gases A and B react in the elementary reaction A(g) + B(g) → AB(g). In a rigid container at constant temperature, [A] and [B] are initially equal. Extra B is then injected so that [B] is tripled while [A] is unchanged. Compared with just before the injection, how does the number of successful collisions per second change just after it, and what happens to the fraction of collisions that are successful?
Answer and reasoning
AIt rises by more than the collision frequency, as more of them succeed. A student who thinks that crowding makes collisions more energetic picks this. At constant temperature the energy distribution is unchanged, so the fraction of successful collisions stays the same; only the collision frequency changes.
BIt increases ninefold, while the fraction that succeed stays the same. A student who applies the overall order (2) to the one reactant that changed picks this, squaring the factor 3. The collision frequency is first order in [B], so tripling [B] triples it.
CIt triples, while the fraction that succeed stays the same.Correct A–B collisions become three times as frequent because [B] triples, and the fraction that succeed depends on the energy distribution and orientation, which are unchanged at constant temperature. So three times as many successful collisions occur per second.
DIt does not change, and neither does the fraction that succeed. A student who thinks that the rate is controlled only by the limiting reactant picks this, because A is now limiting and unchanged. The extra B molecules collide with A molecules, so A–B collisions, and successful collisions, become more frequent.
Working Successful collisions per second = (A–B collision frequency) × (fraction successful). Collision frequency ∝ [A][B] → × 3. The fraction successful depends on the energy distribution (temperature) and orientation, not on concentration → unchanged. So the number of successful collisions per second triples. Distractors: rises by more than the collision frequency (concentration thought to raise collision energy); ninefold (overall order 2 applied to [B]: 3² = 9); unchanged (A, now limiting, is unchanged).
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
