5 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 5
The graph shows the distribution of the kinetic energies of the reactant particles in a gas mixture at a constant temperature. Lines P and Q mark the activation energies of the reaction with a catalyst and without a catalyst, not necessarily in that order. When the catalyst is present, which particles have enough energy to react by the catalyzed path?
Answer and reasoning
AOnly the particles with energies at or above line Q A student who thinks a catalyst does not change the energy a collision needs, only how often collisions happen, picks this. The catalyzed path has its own, lower activation energy, so particles between the two lines can also react.
BAll of the particles with energies at or above line PCorrect Reaction by the catalyzed path needs a collision energy of at least its activation energy, so every particle at or above line P can react by that path. The catalyzed path has the lower activation energy, so its threshold is line P, the line at lower energy. The catalyst does not change the distribution; it moves the threshold to a lower energy, so this region is larger than the region beyond line Q.
CEvery particle in the sample, whatever its energy A student who thinks a catalyst removes the energy barrier altogether picks this. The catalyzed path still has an activation energy, so particles below line P cannot react by it.
DThe particles with kinetic energies below line P A student who reads 'lowers the energy needed' as 'lets low-energy particles react' picks this. An activation energy is a minimum, so the particles that can react are to the right of line P, not to the left.
The decomposition of hydrogen peroxide in a solution containing iodide ions is thought to occur by the following mechanism.
Step 1: H₂O₂(aq) + I⁻(aq) → H₂O(l) + IO⁻(aq) (slow)
Step 2: H₂O₂(aq) + IO⁻(aq) → H₂O(l) + O₂(g) + I⁻(aq) (fast)
Which of the numbered graphs shown best represents [I⁻] as a function of time while the reaction occurs?
Answer and reasoning
AGraph 1 A student who thinks a catalyst is used up because it reacts picks this. I⁻ reacts in step 1, but step 2 re-forms one I⁻ for each one used, so [I⁻] does not fall to zero.
BGraph 2 A student who thinks each step happens for the whole sample before the next begins picks this dip and recovery. Both steps occur at the same time in different molecules, so I⁻ is re-formed as fast as it is used and its net concentration stays constant.
CGraph 3Correct I⁻ is consumed in step 1 and regenerated in step 2, and both steps occur at the same time in different molecules, so I⁻ is a catalyst whose net concentration stays constant: a horizontal line.
DGraph 4 A student who treats every species on the right of a step as a product of the reaction picks this rising line. I⁻ formed in step 2 replaces the I⁻ used in step 1, so it cancels from the overall equation, 2 H₂O₂(aq) → 2 H₂O(l) + O₂(g), and does not build up.
An enzyme, E, catalyzes the conversion of a substrate, S, into a product, P. In the mechanism, S binds to the active site of E to form an enzyme–substrate complex, ES, which then forms P and releases E. Which statement best explains how the enzyme increases the rate of the reaction?
Answer and reasoning
AThe enzyme passes energy to the substrate, so more substrate molecules reach the original activation energy A student who thinks a catalyst works by supplying energy to the reactants picks this. The enzyme does not raise the substrate's kinetic energy; it provides a path that needs less energy.
BThe enzyme binds S as ES, which holds S in a good orientation and reacts with a lower activation energyCorrect Binding forms a new intermediate, ES, in which the substrate is held in an orientation that favors reaction, and ES reacts by a path whose activation energy is lower than that of the uncatalyzed reaction, so a larger fraction of encounters lead to product.
CThe enzyme makes the substrate molecules move faster, so they collide with each other more often A student who thinks a catalyst works only by increasing how often reactant particles collide picks this. The enzyme does not change how fast the substrate molecules move; it changes the path they react by.
DThe enzyme lowers the energy of the product, so the reaction gives out more energy and goes faster A student who thinks a catalyst changes the overall energy change picks this. The catalyzed and uncatalyzed reactions start and end at the same energies; the enzyme lowers the barrier, not the energy of the product.
In acidic solution, a hypothetical compound, R, is converted into an isomer, P. The proposed mechanism is shown. Without acid, R is converted into P by a single, much slower elementary step.
Step 1: R + H₃O⁺ → RH⁺ + H₂O (fast)
Step 2: RH⁺ → PH⁺ (slow)
Step 3: PH⁺ + H₂O → P + H₃O⁺ (fast)
Which statement correctly describes the acid-catalyzed reaction compared with the uncatalyzed reaction?
Answer and reasoning
AIt gives out more energy overall, as H₃O⁺ lowers the energy of the product P A student who thinks a catalyst changes the overall energy change picks this. Both reactions convert R into the same P, so the overall energy change is the same; the acid changes the path, not its ends.
BIt turns a larger amount of R into P by the time that the reaction is complete A student who thinks a catalyst increases the amount of product picks this. The amount of P that forms is set by the amount of R; the acid makes P form faster, not in greater amount.
CIt stops once all the H₃O⁺ has been used up by reacting with R in step 1 A student who thinks a catalyst is used up picks this. H₃O⁺ reacts in step 1 but is re-formed in step 3, so its net concentration stays constant and it does not run out.
DIt goes through protonated intermediates, RH⁺ and PH⁺, absent without acidCorrect In acid catalysis a reactant gains a proton: R accepts H⁺ from H₃O⁺ in step 1, forming RH⁺, which rearranges to PH⁺ before PH⁺ loses the proton in step 3. These protonated intermediates and the steps that form and use them are new, and H₃O⁺ is regenerated.
The decomposition of a gas on the surface of a solid metal catalyst is carried out twice at the same temperature: once with a single lump of the metal and once with the same mass of the metal as a fine powder. The reaction is faster with the powder. Which statement best explains this observation?
Answer and reasoning
AThe powder lowers the activation energy further, because its particles are smaller than the lump A student who explains every change in a catalyzed rate by a change in activation energy picks this. The same metal surface provides the same path and activation energy; the powder simply provides more of that surface.
BThe powder transfers more heat to the gas, so the gas molecules gain more kinetic energy A student who thinks a catalyst works by giving energy to the reactants picks this. Both trials are at the same temperature, and a catalyst does not raise the kinetic energy of the gas molecules.
CThe powder exposes more metal surface atoms, so more gas molecules can bind to the catalyst at one timeCorrect In surface catalysis the gas molecules react while bound to the metal surface. The same mass as a powder has far more surface atoms exposed, so more molecules are bound and reacting at any moment, and the rate of the whole sample is greater.
DThe powder spreads through the gas, so the gas molecules collide with each other more often A student who thinks a catalyst speeds up a reaction by making reactant particles collide with each other more often picks this. The reaction occurs on the metal surface, and the powder does not change how often gas molecules collide with each other.
In preparation: 0 of 5 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
5.11.A.1 Catalyst Fix
Catalyst
A substance that increases the rate of a reaction and is not used up overall. It does this by providing a reaction path with a lower activation energy and/or by increasing the number of effective collisions; because it is regenerated, it does not appear in the overall equation.
Effective collision
A collision between reactant particles that has at least the activation energy for the path being followed and an orientation that allows the bonds to rearrange, so that it leads to reaction.
Catalyzed reaction path
The different mechanism a reaction follows when a catalyst is present. It starts at the same reactants and ends at the same products as the uncatalyzed reaction, so the overall energy change is unchanged, but its highest barrier is lower or more of its collisions are effective.
Students often think A catalyst speeds up a reaction by giving energy to the reactant particles, raising their kinetic energy so that more of them reach the activation energy. In fact No. A catalyst does not supply energy to the particles: at the same temperature the distribution of their kinetic energies is unchanged. The catalyst provides a different path, usually with a lower activation energy, so a larger fraction of collisions have enough energy; some catalysts also make more collisions effective by holding reactants in favorable orientations.
Students often think A catalyst speeds up a reaction only by making the reactant particles move faster or collide with each other more often; the energy a collision needs in order to be effective stays the same. In fact No. A catalyst does not make the reactant particles move faster or collide with each other more often. It increases the rate by providing a path with a lower activation energy and/or by making a larger share of collisions effective, for example by binding reactants in a favorable orientation.
5.11.A.2 Regeneration of a catalyst Fix
Regeneration of a catalyst
In a mechanism containing a catalyst, the catalyst is frequently consumed in the rate-determining step and re-formed in a later step, so its net concentration stays constant while the reaction occurs.
Catalyst in the rate law
When a catalyst reacts in the rate-determining step, its concentration appears in the rate law obtained from that step (for example rate = k[A][X] for a slow step A + X → AX), even though the catalyst does not appear in the overall equation.
Students often think A catalyst is used up as the reaction proceeds, because it reacts with the reactants, so it runs out or must be replaced. In fact No. A catalyst may be consumed in one elementary step, often the rate-determining step, but it is regenerated in a later step, so its net concentration is the same throughout the reaction as at the start.
Students often think The steps of a mechanism happen one after the other for the whole sample, so a catalyst is first used up in step 1 and then re-formed in step 2. In fact No. All the steps occur at the same time in different molecules. A catalyst that is consumed in one step is being regenerated in a later step at the same time, so its net concentration stays constant.
5.11.A.3 Enzyme and enzyme–substrate complex Fix
Enzyme and enzyme–substrate complex
An enzyme is a biological catalyst. Many enzymes bind the reactant (substrate) at an active site to form an enzyme–substrate complex, a new intermediate in which the substrate is held in a favorable orientation and reacts by a path with a lower activation energy; the enzyme is released when the product forms.
Catalyst-bound intermediate
A species formed when a catalyst binds to one or more reactants. It is a new reaction intermediate that does not form in the uncatalyzed reaction, and its formation and reaction are new elementary steps in the catalyzed mechanism.
5.11.A.4 Acid–base catalysis Fix
Acid–base catalysis
Catalysis in which a reactant or intermediate gains or loses a proton, forming a covalent bond to (or breaking one from) the proton donated by the catalyst. This creates a new protonated or deprotonated intermediate and new elementary steps, and the acid or base catalyst is regenerated in a later step.
5.11.A.5 Surface catalysis Fix
Surface catalysis
Catalysis in which a reactant or intermediate binds to, or forms covalent bonds with, the surface of a solid catalyst (often a metal). The bound species are new intermediates; a larger exposed surface provides more sites at which reactant molecules can bind at one time.
Students often think Breaking a catalyst into smaller pieces makes it lower the activation energy further, so the activation energy depends on how much catalyst surface there is. In fact No. The same kind of catalyst surface provides the same reaction path, with the same activation energy, whatever the size of the pieces. A powder of the same mass exposes more surface, so more reactant molecules can bind at one time and the rate is greater.
Students often think Species bound to a catalyst surface are the transition state of the reaction, the highest-energy arrangement along the path. In fact No. Atoms or molecules bound to a catalyst surface are intermediates: species that last for a time and sit at an energy minimum. A transition state is the highest-energy arrangement in one step and sits at a maximum.
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
The energy profiles shown represent a hypothetical reaction without a catalyst (solid curve) and with a catalyst (dashed curve). Based on the profiles, what is the activation energy of the rate-determining step of the catalyzed path?
Answer and reasoning
A50 kJ/molCorrect The dashed path has two steps. Step 1 rises from the reactants at 40 kJ/mol to a maximum at 90 kJ/mol, a barrier of 50 kJ/mol; step 2 rises from the intermediate at 60 kJ/mol to 70 kJ/mol, a barrier of 10 kJ/mol. Step 1 has the larger barrier and is rate-determining: 50 kJ/mol.
B90 kJ/mol A student who reads the activation energy as the energy of the transition state on the axis picks this. Activation energy is a difference: 90 kJ/mol minus the 40 kJ/mol of the reactants that react in step 1.
C30 kJ/mol A student who measures a barrier from the maximum down to the species the step forms picks this: 90 − 60 = 30 kJ/mol. That drop is the barrier for the reverse of step 1; the forward barrier is the rise from the reactants, 50 kJ/mol.
D10 kJ/mol A student who thinks the last step, which forms the products, sets the rate picks this: 70 − 60 = 10 kJ/mol for step 2. Step 2 has the smaller barrier and is fast; step 1, with a 50 kJ/mol barrier, is rate-determining.
Working Catalyzed path: step 1 rises from the reactants (40 kJ/mol) to 90 kJ/mol, Ea = 90 − 40 = 50 kJ/mol; step 2 rises from the intermediate (60 kJ/mol) to 70 kJ/mol, Ea = 10 kJ/mol. The rate-determining step is step 1 (larger activation energy; its maximum is also the higher one): 50 kJ/mol. Distractors: maximum read from the axis, 90 kJ/mol; barrier measured down to the intermediate, 90 − 60 = 30 kJ/mol; last step taken as rate-determining, 70 − 60 = 10 kJ/mol.
A hypothetical reaction, A + B₂ → AB₂, is catalyzed by X. The proposed mechanism is shown.
Step 1: A + X → AX (slow)
Step 2: AX + B₂ → AB₂ + X (fast)
Which rate law is consistent with the proposed mechanism?
Answer and reasoning
Arate = k[A], first order overall A student who thinks a catalyst cannot appear in the rate law, because it is not used up overall, picks this. X reacts in the rate-determining step, so its concentration is part of the rate law.
Brate = k[A][B₂], second order overall A student who writes the rate law from the overall equation, A + B₂ → AB₂, picks this. The rate law comes from the slow step, in which A and X react; B₂ reacts only in the fast step.
Crate = k[AX][B₂], second order overall A student who thinks the last step, which forms the product, sets the rate picks this. Step 2 is fast; the slow step, A + X → AX, sets the rate law.
Drate = k[A][X], second order overallCorrect The rate law is set by the slow step, A + X → AX, a bimolecular elementary step: rate = k[A][X]. X is consumed in this step and regenerated in step 2, so it appears in the rate law even though it is absent from the overall equation.
Working The rate-determining step is step 1, an elementary bimolecular step, so rate = k[A][X], second order overall. The catalyst X is consumed in this step and regenerated in step 2, so it appears in the rate law but not in the overall equation. Distractors: catalyst left out because it is not consumed overall, rate = k[A]; rate law from the overall equation, rate = k[A][B₂]; rate law from the last step, rate = k[AX][B₂].
The reaction X₂(g) + 2 Y(g) → 2 XY(g) is catalyzed by a metal surface. The particulate diagrams represent one elementary step of the catalyzed mechanism. Which statement correctly describes the step shown?
Answer and reasoning
AThe X–X bond breaks, and each X atom forms a bond to an atom of the metal surfaceCorrect Before the step, X₂ is a single molecule above the surface; after it, the two X atoms are separate and each is joined by a covalent bond to a metal atom. The bound X atoms are new intermediates that later react with Y to form XY.
BMetal atoms combine with X to form a new product, so the metal surface gets used up A student who thinks a catalyst is consumed picks this. All six metal atoms are still present, and the bound X atoms later react with Y and leave the surface as XY, freeing the metal atoms again.
CTwo X atoms form on the surface, and these are products of the overall reaction A student who treats every species formed in a step as a product of the overall reaction picks this. The bound X atoms are intermediates: they are used up when they react with Y, and the product of the overall reaction is XY.
DThe bound X atoms are the transition state, the highest-energy arrangement A student who mixes up intermediates and transition states picks this. The bound X atoms are species that last until they react with Y, so they are an intermediate at an energy minimum, not a transition state at a maximum.
Adding a catalyst lowers the activation energy of the forward reaction of a hypothetical elementary reaction, A → B, by 30 kJ/mol. How does adding the catalyst affect the reverse reaction, B → A?
Answer and reasoning
AIts rate is unchanged, because a catalyst speeds up only the forward reaction A student who thinks a catalyst speeds up only the forward reaction picks this. The lower maximum is lower whichever side it is approached from, so the reverse reaction is also faster.
BIts rate decreases, because the catalyst lowers the energy of the product B A student who thinks a catalyst changes the energies of the products picks this. A and B are at the same energies with and without the catalyst; only the maximum between them is lower, so the reverse barrier also falls.
CIts rate increases, because the catalyst gives the B molecules more kinetic energy A student who thinks a catalyst works by supplying energy to the reacting particles picks this. The reverse rate does increase, but because the barrier is lower, not because the molecules gain energy; their energies are set by the temperature.
DIts rate increases, as its activation energy falls by the same amountCorrect The catalyzed path joins the same A and B, with a maximum lower by 30 kJ/mol. Because A and B are at the same energies as before, the barrier measured from B is also 30 kJ/mol lower, so the reverse reaction is faster as well.
Solid manganese(IV) oxide, MnO₂(s), is added to a solution of hydrogen peroxide, which decomposes: 2 H₂O₂(aq) → 2 H₂O(l) + O₂(g). A student wants to collect evidence that MnO₂ acts as a catalyst for this reaction. Which procedure would provide that evidence?
Answer and reasoning
ATime the collection of O₂ with and without MnO₂, then filter, dry and reweigh the MnO₂ to check its massCorrect A catalyst increases the rate and is not used up. Timing the O₂ collection with and without MnO₂ tests the rate; recovering, drying and reweighing the MnO₂ tests whether any was consumed. A faster rate and an unchanged mass are the evidence.
BMeasure the total volume of O₂ with and without MnO₂ to check that MnO₂ makes more O₂ form A student who thinks a catalyst increases the amount of product picks this. The total O₂ is set by the amount of H₂O₂; a catalyst makes it form faster, so the totals would match and would not show catalysis.
CMeasure the temperature change with and without MnO₂ to check that MnO₂ supplies energy A student who thinks a catalyst works by supplying energy picks this. MnO₂ does not supply the energy for the reaction, and a temperature reading cannot show that MnO₂ is unchanged at the end.
DTime the collection of O₂, adding fresh MnO₂ partway through to replace the MnO₂ that is used up A student who thinks a catalyst is used up picks this. MnO₂ is regenerated, so adding more is unnecessary, and the procedure never tests whether the MnO₂ is unchanged at the end.
A student sketches energy profiles for a reaction with and without a catalyst. Compared with the uncatalyzed path, the student's catalyzed path (i) starts at the same energy, (ii) ends at a lower energy, (iii) has two maxima instead of one, and (iv) has a highest maximum that is lower than the uncatalyzed maximum. Which feature of the sketch is NOT consistent with catalysis?
Answer and reasoning
AFeature (i), a catalyzed path that starts at the same energy A student who thinks a catalyst gives energy to the reactant particles picks this, expecting the catalyzed path to start higher. A catalyst does not supply energy; both paths start at the same reactants, at the same energy.
BFeature (ii), a catalyzed path that ends at lower energyCorrect The catalyzed reaction forms the same products from the same reactants, so both paths must end at the same energy. A new path with more steps (iii) and a lower highest barrier (iv), starting at the same reactants (i), is what catalysis provides.
CFeature (iii), a catalyzed path that has two maxima A student who thinks a catalyst only lowers the barrier of the original single step picks this. A catalyst provides a new path, which can have extra steps and intermediates, so a second maximum is consistent.
DFeature (iv), a highest maximum that is lower than before A student who thinks a catalyst works only by making particles collide more often, not by changing the energy needed, picks this. Providing a path with a lower activation energy is one of the ways a catalyst increases a rate, so a lower highest maximum is consistent.
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
