4 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 4
The decomposition of hydrogen peroxide in the presence of iodide ions is proposed to occur by the following mechanism. Step 1: H₂O₂(aq) + I⁻(aq) → H₂O(l) + IO⁻(aq). Step 2: H₂O₂(aq) + IO⁻(aq) → H₂O(l) + O₂(g) + I⁻(aq). Which statement correctly identifies the roles of I⁻ and IO⁻ in this mechanism?
Answer and reasoning
AI⁻ acts as a catalyst, and IO⁻ is an intermediate.Correct I⁻ is used up in step 1 and formed again in step 2, so its amount is unchanged: it is a catalyst. IO⁻ is formed in step 1 and used up in step 2, so it is present only while the reaction is occurring: it is an intermediate.
BIO⁻ is a catalyst, and I⁻ is an intermediate. A student who swaps the two definitions picks this. IO⁻ is formed first and then used up, which makes it an intermediate; I⁻ is used up first and then re-formed, which makes it a catalyst.
CI⁻ and IO⁻ are both intermediates in the steps. A student who thinks that a catalyst takes no part in any step, so that I⁻, which is used up in step 1, cannot be one, picks this. A catalyst does react in a step and is regenerated later, as I⁻ is.
DI⁻ and IO⁻ both act as catalysts in the steps. A student who thinks that any species appearing in both steps is a catalyst picks this. IO⁻ is formed in step 1 and used up in step 2, so it is not present before or after the reaction: it is an intermediate.
The table shows a proposed three-step mechanism for the reaction 2 NO(g) + 2 H₂(g) → N₂(g) + 2 H₂O(g). Is the mechanism consistent with the overall equation, and why?
Answer and reasoning
AYes: each of the steps is balanced, and that is all that a mechanism requires. A student who thinks that balanced steps are enough picks this. The verdict is right but the reason is not: the steps must also add up to the overall equation, which these do once N₂O₂ and N₂O cancel.
BNo: N₂O₂ and N₂O form in the steps, so they should appear in the overall equation. A student who thinks that intermediates belong in the overall equation picks this. N₂O₂ and N₂O are each formed in one step and used up in the next, so they cancel when the steps are added.
CYes: adding the steps and cancelling N₂O₂ and N₂O gives the overall equation.Correct Adding the three steps gives 2 NO + N₂O₂ + 2 H₂ + N₂O → N₂O₂ + N₂O + N₂ + 2 H₂O. N₂O₂ and N₂O appear on both sides and cancel, leaving 2 NO + 2 H₂ → N₂ + 2 H₂O, the overall equation.
DNo: N₂O₂ and N₂O are not in the overall equation, so the steps should not include them. A student who thinks that a mechanism may contain only species in the overall equation picks this. Mechanisms normally include intermediates; they cancel when the steps are added.
A hypothetical reaction A → B occurs by a two-step mechanism: A → X, followed by X → B. Which of the numbered graphs shown best represents how [X] changes from the moment A is added until the reaction is complete?
Answer and reasoning
AGraph 1 A student who thinks that an intermediate builds up like a product picks this graph. X is consumed by the second step, so it does not remain at the end; [X] falls back to zero when the reaction is complete.
BGraph 2 A student who swaps the two definitions labels X, which is formed and then used up, a catalyst, and so expects its amount to stay the same throughout; this student picks this graph. X is not present before the reaction starts; it is formed in step 1 and used up in step 2.
CGraph 3 A student who thinks that an intermediate is a transition state that never forms as a species picks this graph. An intermediate is a real species formed in one step and consumed in a later one, so [X] rises above zero while the reaction is occurring.
DGraph 4Correct An intermediate is produced by the first step and consumed by the second, so it is present only while the reaction is occurring: [X] starts at zero, rises as A reacts, and falls back to zero as X is converted to B.
Two mechanisms are proposed for the reaction NO₂(g) + CO(g) → NO(g) + CO₂(g). In mechanism 1, NO₂ and CO collide in a single elementary step. In mechanism 2, the first step is NO₂ + NO₂ → NO₃ + NO and the second step is NO₃ + CO → NO₂ + CO₂. Which observation, if it were made, would support mechanism 2 over mechanism 1?
Answer and reasoning
ANO₃ is detected in the reaction mixture after the reaction has gone to completion. A student who thinks that intermediates build up and remain at the end picks this. In mechanism 2, NO₃ is used up in the second step, so it is present only while the reaction is occurring; finding it at the end would not fit either mechanism.
BNO and CO₂ are found to form in a 1 : 1 mole ratio as the reaction takes place. A student who thinks that product ratios can tell mechanisms apart picks this. Both mechanisms add up to the same overall equation, so both predict a 1 : 1 ratio; the observation does not distinguish them.
CThe amount of NO₂ in the mixture at the end equals the amount present at the start. A student who thinks that NO₂, which takes part in both steps of mechanism 2, is a catalyst picks this. In mechanism 2 two NO₂ are used in step 1 and only one is re-formed in step 2, so NO₂ is consumed overall in both mechanisms.
DNO₃ is detected in the reaction mixture while the reaction is taking place.Correct NO₃ is an intermediate in mechanism 2 and is not formed at all in mechanism 1. Finding it in the mixture while the reaction is occurring is evidence for mechanism 2 over mechanism 1.
In preparation: 0 of 4 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
5.7.A.1 Reaction mechanism Fix
Reaction mechanism
A proposed series of elementary reactions (steps) that occur in sequence and together account for an overall reaction. Its components may include reactants, intermediates, products and catalysts.
Elementary step
One elementary reaction within a mechanism: a single event in which the particles shown on its left side collide or rearrange to give the particles on its right side.
Catalyst (in a mechanism)
A species that is consumed in an earlier step and re-formed in a later step, so its amount is the same before and after the reaction. It takes part in the mechanism but does not appear in the overall equation.
Students often think A catalyst is a species that is formed in one step and used up in a later one, whereas an intermediate is one that is used up first and then formed again. In fact No. That describes an intermediate. A catalyst is used up in an earlier step and formed again in a later step, so it is present before the reaction starts and in the same amount after it ends; an intermediate is formed first and then used up, so it is present only while the reaction is occurring.
Students often think Any species that takes part in more than one step of a mechanism is a catalyst. In fact No. A catalyst is used up in one step and re-formed in a later step, with no net change in its amount. Reactants, products and intermediates can also appear in more than one step; what matters is how much of the species is used and formed overall.
5.7.A.2 Combining elementary steps Fix
Combining elementary steps
The elementary steps of a valid mechanism, each multiplied by the number of times it occurs, add up to the overall balanced equation once species that appear on both sides are cancelled.
Students often think A proposed mechanism is acceptable as long as each elementary step is balanced; the steps do not have to add up to the overall equation. In fact No. Each step must be balanced, but the steps, each counted the number of times it occurs, must also add up to the overall balanced equation. A set of balanced steps whose sum is a different equation is not a mechanism for the reaction.
Students often think Intermediates are formed during the reaction, so they should appear in the overall equation. In fact No. An intermediate is produced in one step and consumed in a later step, so it appears on both sides when the steps are added and cancels out. The overall equation contains only the net reactants and products.
5.7.A.3 Reaction intermediate Fix
Reaction intermediate
A species that is produced by some elementary steps and consumed by others, so that it is present only while the reaction is occurring. It cancels when the steps are added and does not appear in the overall equation.
Students often think An intermediate builds up during the reaction like a product and is still present when the reaction is complete. In fact No. An intermediate is produced by some steps and consumed by others, so it is present only while the reaction is occurring. When the reaction is complete, essentially none of it remains.
Students often think An intermediate is the transition state of a step: it exists only at the top of an energy barrier and never forms as a species that could be detected. In fact No. A transition state is the highest-energy arrangement of atoms within one elementary step and lasts only an instant. An intermediate is a species formed by one step and consumed by a later one; it exists between steps while the reaction is occurring, and it can in principle be detected.
5.7.A.4 Evidence for a mechanism Fix
Evidence for a mechanism
Experimental detection of a species that is an intermediate in one proposed mechanism but not in an alternative mechanism supports the first mechanism over the alternative. Such evidence supports a mechanism; it does not prove it.
Students often think Detecting an intermediate proves that the mechanism containing it is the correct one. In fact No. Detecting a species that is an intermediate in one mechanism but not in an alternative supports the first mechanism over that alternative. Other mechanisms with the same intermediate could also fit, so the evidence supports the mechanism without proving it.
Students often think Measuring the mole ratio of the products formed shows which of two proposed mechanisms is the correct one. In fact No. Any acceptable mechanism must add up to the same overall balanced equation, so every acceptable mechanism predicts the same mole ratios of products. Evidence that distinguishes mechanisms must involve something that differs between them, such as an intermediate present in only one.
5 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 5
The diagram represents the two elementary steps of a proposed mechanism for the reaction 2 NO(g) + O₂(g) → 2 NO₂(g). Which description of the role of N₂O₂ in this mechanism is correct?
Answer and reasoning
ACatalyst: it takes part in both of the steps of the mechanism. A student who thinks that any species taking part in two steps is a catalyst picks this. A catalyst is used up first and then re-formed; N₂O₂ is formed first and then used up, so it is an intermediate.
BIntermediate: it is formed in step 1 and used up in step 2.Correct In step 1 two NO molecules combine to form N₂O₂ (the O–N–N–O molecule), and in step 2 N₂O₂ reacts with O₂ to give two NO₂ molecules. It is produced by one step and consumed by the next, so it is an intermediate and does not appear in the overall equation.
CProduct: it is formed when two NO molecules collide with each other. A student who thinks that everything formed in a step is an overall product picks this. N₂O₂ is consumed in step 2, so none remains; the overall product is NO₂.
DReactant: it collides with an O₂ molecule during step 2. A student who thinks that everything on the left of a step is an overall reactant picks this. N₂O₂ was not present at the start; it is formed in step 1 before it reacts in step 2.
A hypothetical reaction has the overall equation A₂(g) + 2 B(g) → 2 AB(g). A proposed mechanism has two elementary steps. Step 1: A₂(g) → 2 A(g). Step 2: A(g) + B(g) → AB(g). How do these steps combine to give the overall equation?
Answer and reasoning
AEach step occurs once, and the overall equation is the sum of the two steps. A student who thinks that each step occurs exactly once picks this. Adding the steps once each gives A₂ + B → AB + A, which is not the overall equation; step 2 must occur twice.
BThe steps need not add up to the overall equation; each step simply has to be balanced. A student who thinks that balanced steps are enough picks this. The steps of a valid mechanism must add up to the overall equation; here they do when step 2 is counted twice.
CThe second step occurs twice for each first step, so both A atoms formed are used up.Correct Step 1 forms two A atoms, and each A atom is used in one step-2 event, so step 2 occurs twice: A₂ → 2 A plus 2 A + 2 B → 2 AB gives A₂ + 2 B → 2 AB after the two A atoms cancel.
DA is a catalyst, so it does not need to cancel when the steps are added together. A student who swaps the definitions of catalyst and intermediate picks this. A is formed in step 1 and used up in step 2, so it is an intermediate, and it does cancel, provided step 2 is counted twice.
Why does a reaction intermediate not appear in the overall balanced equation for a reaction that occurs by a multistep mechanism?
Answer and reasoning
AIt is the highest-energy point of a step, so it does not form as a separate species. A student who confuses an intermediate with a transition state picks this. An intermediate is a species formed by one step and consumed by the next; a transition state is the energy maximum within a single step.
BIt forms in one step and is used up in a later one, so it cancels when steps are added.Correct An intermediate is produced by one elementary step and consumed by another. When the steps are added it appears on both sides and cancels, so the overall equation contains only the net reactants and products.
CIt is used up in one step and formed again in a later step, so there is no net change in it. A student who swaps the definitions of intermediate and catalyst picks this. Being used up first and then re-formed describes a catalyst; an intermediate is formed first and then used up.
DIt forms in an early step, but the overall equation shows just the last step's products. A student who builds the overall equation from the first step's reactants and the last step's products picks this. The overall equation comes from adding all the steps; an intermediate is absent because it cancels, not because of where it forms.
A proposed mechanism for the reaction NO₂(g) + CO(g) → NO(g) + CO₂(g) has two steps. Step 1: NO₂(g) + NO₂(g) → NO₃(g) + NO(g). Step 2: NO₃(g) + CO(g) → NO₂(g) + CO₂(g). Which statement correctly describes the role of NO₂ in this mechanism?
Answer and reasoning
AA reactant: two are used in step 1, and only one is formed again in step 2.Correct Two NO₂ molecules are used in step 1 and one is re-formed in step 2, so one NO₂ is consumed overall for each overall reaction: NO₂ is a reactant, as the overall equation shows.
BA catalyst: it is used in step 1 and is formed again in step 2 of the mechanism. A student who thinks that any species taking part in both steps is a catalyst picks this. A catalyst is re-formed in the same amount as it is used; here two NO₂ are used and only one is re-formed.
CAn intermediate: it appears on both sides of the arrows in the steps. A student who thinks that any species on both sides of the steps is an intermediate picks this. An intermediate is not present at the start and none remains; NO₂ is present at the start and is consumed overall.
DA product: it is one of the species formed in step 2 of the mechanism. A student who thinks that everything formed in a step is an overall product picks this. More NO₂ is used in step 1 than is formed in step 2, so NO₂ is a reactant overall.
The table shows two mechanisms proposed for the hypothetical reaction 2 A(g) + B₂(g) → 2 AB(g). In experiments, B atoms are detected in the reaction mixture while the reaction is occurring but not after it is complete. Which claim about the mechanisms is best supported by this observation, and why?
Answer and reasoning
AMechanism 1 is proven correct, because detecting an intermediate shows that a mechanism is right. A student who thinks that detecting an intermediate proves a mechanism picks this. The observation supports mechanism 1 over mechanism 2, but another mechanism that also forms B could fit the same observation.
BMechanism 1 is supported, because B is formed and then used up in it but never forms in mechanism 2.Correct In mechanism 1, B is produced in step 1 and consumed in step 2, so it is an intermediate, present only while the reaction occurs. Mechanism 2 never forms B. The observation therefore supports mechanism 1 over mechanism 2.
CMechanism 1 is ruled out, because its intermediate B should still be present after the reaction. A student who thinks that intermediates remain at the end picks this. An intermediate is used up by a later step, so B being absent at the end is exactly what mechanism 1 predicts.
DNeither is supported, because a species seen during a reaction is not an intermediate. A student who thinks that an intermediate is a transition state that never exists as a detectable species picks this. An intermediate is a real species formed in one step and consumed in another, so it can be present, and detected, while the reaction occurs.
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
