1 question, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 1
In a two-step mechanism, step 1 forms an intermediate and step 2 converts the intermediate into the final product. Step 1 is much slower than step 2. Why is the rate at which the final product forms set by step 1?
Answer and reasoning
AStep 1 is written first, and the first step of a mechanism sets the rate at any speed. A student who thinks the first step is rate limiting because of its position picks this. Step 1 sets the rate here because it is the slowest step, not because it comes first.
BStep 1 uses the limiting reactant, and that reactant sets how much product forms. A student who merges 'rate-limiting step' with 'limiting reactant' picks this. The limiting reactant decides how much product can form; the rate-limiting step decides how fast it forms.
CStep 1 makes the intermediate, and step 2 is held to the rate of that supply.Correct Step 2 needs the intermediate. However fast step 2 could be, it can use the intermediate only as quickly as the slow step 1 forms it, so the product forms at the rate of step 1.
DStep 1 runs at one fixed rate, and that rate is the same at every concentration. A student who thinks a rate-limiting step runs at a fixed rate picks this. The rate of step 1 depends on the concentrations of its reactants; it limits the overall rate because it is the slowest step.
In preparation: 0 of 1 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
5.8.A.1 Rate-limiting step Fix
Rate-limiting step
The slowest elementary step in a reaction mechanism. The overall reaction cannot form products faster than this step occurs, so it limits the rate of the overall reaction.
Rate law predicted by a mechanism whose first step is rate limiting
When the first elementary step is the slow one, the rate law of the overall reaction is the rate law of that step: a rate constant multiplied by the concentration of each reactant particle of the slow step, raised to its coefficient in that step. Reactants that take part only in later, faster steps do not appear.
Molecularity of an elementary step
The number of reactant particles that take part in the step. For a rate-limiting first step, it sets the overall order of the predicted rate law: one particle gives first order overall, two particles give second order overall.
Comparing a mechanism with an experimental rate law
A proposed mechanism is acceptable only if the rate law it predicts agrees with the rate law found from experiment. Agreement supports the mechanism but does not prove it, because different mechanisms can predict the same rate law.
Students often think The rate law of a reaction is written from the overall balanced equation, with each reactant's coefficient as its order, whatever the mechanism. In fact No. For a mechanism whose first step is rate limiting, the rate law is written from the reactant particles of that slow step. Reactants that are used only in later, faster steps do not appear, so the rate law need not match the coefficients of the overall equation.
Students often think The rate law of a reaction is set by the last step of the mechanism, because that is the step in which the final product is formed. In fact No. The rate at which the final product forms is limited by the slowest step. When the first step is the slow one, later steps can go only as fast as the first step supplies their reactants, so the rate law comes from the first step, not the last.
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 table shows a proposed mechanism for the hypothetical reaction X(g) + 2 Y(g) → Z(g) + Q(g). What is the overall order of the rate law that this mechanism predicts for the reaction?
Answer and reasoning
AOrder 3 A student who writes the rate law from the coefficients of the overall equation picks this: rate = k[X][Y]², with 1 + 2 = 3. The second Y reacts in step 2, after the slow step, so it does not appear in the rate law.
BOrder 1 A student who takes the rate law from the last step, in which the products Z and Q form, picks this, because one particle of V reacts in step 3. The rate is limited by the slow step, step 1, in which two particles collide.
COrder 5 A student who counts the reactant particles of every step picks this: 2 + 2 + 1 = 5. Only the reactant particles of the slow step set the rate law.
DOrder 2Correct The first step is the slow step, so the rate law of the reaction is set by the molecularity of that step: one X and one Y collide, giving rate = k[X][Y], which is second order overall.
Working The first step is the slow (rate-limiting) step, so the rate law is that of step 1: rate = k[X][Y]. Overall order = 1 + 1 = 2. Distractors: overall coefficients, 1 + 2 = 3; last step (one reactant particle), 1; molecularities of all three steps added, 2 + 2 + 1 = 5.
A proposed mechanism for the hypothetical reaction X(aq) + Y(aq) → Z(aq) has two steps. Step 1: X → W (slow). Step 2: W + Y → Z (fast). To test the mechanism, a student measures the initial rate in one trial and then runs a second trial at the same temperature with the same initial [X] and twice the initial [Y]. If the mechanism is correct, which prediction for the second trial, with its reasoning, is correct?
Answer and reasoning
AThe initial rate is the same, as Y reacts only after the slow step has taken place.Correct With a slow first step, the rate law is that of step 1: rate = k[X]. Y takes part only in the fast second step, which can go no faster than step 1 supplies W, so doubling [Y] leaves the initial rate unchanged.
BThe initial rate doubles, as Y is a reactant in the equation for the overall reaction. A student who writes the rate law from the overall equation, rate = k[X][Y], picks this. The mechanism predicts rate = k[X]: Y reacts only in the fast step after the slow step, so [Y] is not in the rate law.
CThe initial rate doubles, as Y reacts in the step in which the product Z forms. A student who thinks the product-forming last step sets the rate law picks this. Step 2 is fast and can go only as fast as step 1 supplies W, so the rate is set by step 1, which does not involve Y.
DThe initial rate is the same, as a slow step holds the reaction to one fixed rate. A student who thinks a rate-limiting step runs at a fixed rate picks this. The rate of the slow step depends on [X]; the rate is unchanged here because Y is not a reactant in the slow step, not because the rate is fixed.
A proposed mechanism for the hypothetical reaction 2 P(g) + Q₂(g) → 2 PQ(g) has two steps. Step 1: P + Q₂ → PQ + Q. Step 2: P + Q → PQ. One of the steps is much slower than the other. The table shows initial-rate data for the reaction at a constant temperature. Which claim about the mechanism is best supported by the data?
Answer and reasoning
AStep 2 is rate limiting: it is the last step, and the last step sets the rate law. A student who thinks the last, product-forming step sets the rate law picks this. The data give rate = k[P][Q₂], which matches the particles that collide in step 1; Q₂ does not react in step 2.
BStep 1 is rate limiting: the orders match the particles that collide in step 1.Correct Doubling [P] alone doubles the rate, and doubling [Q₂] alone doubles the rate, so rate = k[P][Q₂]. That is the rate law of step 1, in which one P and one Q₂ collide, so the data support a slow first step.
CStep 1 is rate limiting: it is the first step, and the first step sets the rate. A student who thinks the first step is always rate limiting picks this. The slow step is identified from evidence, here the rate law found from the data, not from the position of the step.
DStep 2 is rate limiting: it uses the intermediate Q, and little Q is present. A student who thinks a step that uses an intermediate must be slow picks this. A reactive intermediate can be used as fast as it forms; the rate law from the data, rate = k[P][Q₂], matches step 1.
A proposed mechanism for the hypothetical reaction 2 A(g) + B(g) → C(g) has two steps. Step 1: A + A → A₂ (slow). Step 2: A₂ + B → C (fast). According to this mechanism, by what factor does the initial rate of the reaction increase when the initial concentrations of A and of B are both doubled at constant temperature?
Answer and reasoning
A8 A student who writes the rate law from the overall equation, rate = k[A]²[B], picks this: 2² × 2 = 8. B reacts only after the slow step, so [B] is not in the rate law.
B2 A student who takes the rate law from the last step, in which C forms, picks this, because doubling [B] would double the rate of step 2. Step 2 can go only as fast as the slow step 1 supplies A₂, and the rate of step 1 is proportional to [A]².
C1 A student who thinks a rate-limiting step runs at a fixed rate picks this, expecting no change. The rate of the slow step depends on [A]: two A particles collide, so doubling [A] makes the rate 4 times as great.
D4Correct With a slow first step the rate law is that of step 1: rate = k[A]². Doubling [A] makes the rate 2² = 4 times as great, and doubling [B] has no effect because B reacts only in the fast step.
Working The first step is slow, so rate = k[A]². Doubling [A] gives 2² = 4 times the rate; [B] is not in the rate law. Distractors: overall coefficients, rate = k[A]²[B], 2² × 2 = 8; last step, rate proportional to [B], 2; a fixed rate for the slow step, 1.
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
