1 question, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 1
A proposed mechanism for a hypothetical reaction has two steps. Step 1: X + X ⇌ X₂ (fast, reversible). Step 2: X₂ + Y → Z (slow). The rate law for the slow step, rate = k₂[X₂][Y], contains the concentration of the intermediate X₂. Which relationship does the pre-equilibrium approximation use to express [X₂] in terms of [X]?
Answer and reasoning
AThe rate of the forward reaction of step 1 equals the rate of the reverse reaction of that step.Correct Because step 1 is fast in both directions compared with step 2, its forward and reverse rates are taken to be equal: k₁[X]² = k₋₁[X₂], so [X₂] = (k₁/k₋₁)[X]².
BThe rate of the forward reaction of step 1 equals the rate of the slow reaction of step 2. A student who thinks all the steps of a mechanism go at the same rate picks this. Step 1 is much faster than step 2 in both directions; the approximation balances the forward reaction of step 1 against its own reverse reaction.
CThe concentration of X₂ in the reaction mixture equals the concentration of X in the mixture. A student who thinks equal forward and reverse rates mean equal concentrations picks this. Equal rates give k₁[X]² = k₋₁[X₂]; the concentrations are related through the rate constants and are generally not equal.
DThe concentration of X₂ in the reaction mixture equals half the initial concentration of X. A student who thinks a fast step goes to completion picks this, converting all of the X into X₂ by the 2 : 1 mole ratio. Step 1 is reversible, so [X₂] is fixed by the balance of the forward and reverse rates.
Working Pre-equilibrium: rate of forward step 1 = rate of reverse step 1, so k₁[X]² = k₋₁[X₂] and [X₂] = (k₁/k₋₁)[X]². Substituting: rate = k₂(k₁/k₋₁)[X]²[Y].
In preparation: 0 of 1 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
5.9.A.1 Fast, reversible first step Fix
Fast, reversible first step
An elementary step, written with the arrows ⇌, in which the forward and the reverse reactions both occur quickly compared with the slow step that follows. It forms an intermediate that the slow step then uses.
Pre-equilibrium approximation
The approximation used when a fast, reversible step comes before the rate-limiting step: the rate of the forward reaction of the fast step is taken to be equal to the rate of its reverse reaction. Setting the two rate laws equal gives the concentration of the intermediate in terms of reactant concentrations.
Rate law for a mechanism whose first step is not rate limiting
Write the rate law of the slow step, which contains the concentration of an intermediate; then use the pre-equilibrium approximation to replace that concentration. For A + A ⇌ A₂ (fast) followed by A₂ + B → C (slow): rate = k₂[A₂][B] and k₁[A]² = k₋₁[A₂], so rate = (k₂k₁/k₋₁)[A]²[B].
Rate constant of the overall rate law
In a rate law obtained with the pre-equilibrium approximation, the rate constant k is a combination of the rate constants of the steps, for example k = k₂k₁/k₋₁, where k₁ and k₋₁ are for the forward and reverse reactions of the fast step and k₂ is for the slow step.
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. The rate law is found from the mechanism: the rate law of the slow step, with the concentration of any intermediate replaced by using the fast, reversible step before it. Reactants used only after the slow step do not appear, so the orders need not equal the overall coefficients.
Students often think The rate law of a reaction is always the rate law of the first step of its mechanism, so reactants that are not in the first step do not appear. In fact No. The first step sets the rate law only when it is the rate-limiting step. When the first step is fast and reversible and a later step is slow, the rate law starts from the slow step, and the first step is used to replace the intermediate's concentration.
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) + 3 Y(g) → XY₃(g). The first step is reversible. What is the overall order of the rate law that this mechanism predicts for the reaction?
Answer and reasoning
AOrder 4 A student who writes the rate law from the coefficients of the overall equation picks this: rate = k[X][Y]³, with 1 + 3 = 4. The third Y reacts in step 3, after the slow step, so it does not appear in the rate law.
BOrder 2 A student who takes the rate law from the first step whatever its speed picks this: rate = k[X][Y]. Here the first step is fast; the slow step is step 2, in which another Y reacts, so the rate law is k[X][Y]².
COrder 1 A student who drops the intermediate XY from the slow step's rate law picks this, leaving rate = k[Y]. The concentration of XY must be replaced by (k₁/k₋₁)[X][Y], not left out.
DOrder 3Correct The slow step gives rate = k₂[XY][Y]. XY is an intermediate, and equal forward and reverse rates for the fast first step give [XY] = (k₁/k₋₁)[X][Y]. So rate = k[X][Y]², which is third order overall. Step 3 comes after the slow step and does not affect the rate law.
Working Slow step: rate = k₂[XY][Y]. XY is an intermediate. Pre-equilibrium for step 1: k₁[X][Y] = k₋₁[XY], so [XY] = (k₁/k₋₁)[X][Y]. Rate = (k₂k₁/k₋₁)[X][Y]²: overall order 1 + 2 = 3. Distractors: overall coefficients, 1 + 3 = 4; first step only, rate = k[X][Y], 2; slow step with the intermediate dropped, rate = k[Y], 1.
A proposed mechanism for the hypothetical reaction 2 A(g) + B(g) → C(g) has two steps. Step 1: A + A ⇌ A₂ (fast, reversible). Step 2: A₂ + B → C (slow). The rate law predicted by the mechanism is rate = k[A]²[B]. Which statement explains why [A] appears in this rate law although A is not a reactant in the slow step?
Answer and reasoning
AStep 1 comes before step 2, and the first step of a mechanism sets the rate law. A student who thinks the first step always sets the rate law picks this. If step 1 alone set the rate law, [B] would not appear; [A] enters because the concentration of the intermediate A₂ in the slow step depends on [A].
BA is a reactant in the overall equation, and overall reactants appear in the rate law. A student who writes rate laws from the overall equation picks this. A rate law is derived from the mechanism; a reactant that is used only after the slow step would not appear, even though it is in the overall equation.
CStep 1 ties the amount of A₂ to the amount of A, and A₂ is a reactant in the slow step.Correct The slow step gives rate = k₂[A₂][B]. Because the forward and reverse rates of the fast first step are equal, k₁[A]² = k₋₁[A₂], so [A₂] is proportional to [A]². Substituting puts [A]² into the rate law.
DStep 1 and step 2 both count, and the rate law combines the reactants of both steps. A student who thinks every step adds its reactants to the rate law picks this. Combining the reactants of both steps would also put the intermediate A₂ into the rate law; [A]² appears because it replaces [A₂].
A student proposes a three-step mechanism for the hypothetical reaction 2 A(g) + 2 B(g) → E(g). Step 1: A + B ⇌ C (fast, reversible). Step 2: C + A → D (slow). Step 3: D + B → E (fast). The table shows the initial concentrations for two trials at the same temperature and the initial rate measured in trial 1. If the mechanism is correct, what initial rate should be measured in trial 2?
Answer and reasoning
A1.6 × 10⁻³ M s⁻¹ A student who writes the rate law from the coefficients of the overall equation picks this: rate = k[A]²[B]², so 16 times the rate. The second B reacts in step 3, after the slow step, so the rate law is first order in B.
B8.0 × 10⁻⁴ M s⁻¹Correct The slow step gives rate = k₂[C][A], and equal forward and reverse rates for the fast first step give [C] = (k₁/k₋₁)[A][B], so rate = k[A]²[B]. Doubling both concentrations makes the rate 2² × 2 = 8 times as great.
C4.0 × 10⁻⁴ M s⁻¹ A student who takes the rate law from the first step whatever its speed picks this: rate = k[A][B], so 4 times the rate. Step 1 is fast; the slow step is step 2, in which a second A reacts.
D2.0 × 10⁻⁴ M s⁻¹ A student who drops the intermediate C from the slow step's rate law picks this: rate = k[A], so twice the rate. [C] must be replaced by (k₁/k₋₁)[A][B], which brings another [A] and a [B] into the rate law.
Working Slow step: rate = k₂[C][A]. Pre-equilibrium for step 1: k₁[A][B] = k₋₁[C], so [C] = (k₁/k₋₁)[A][B] and rate = k[A]²[B]. In trial 2 both concentrations are doubled: rate = 2² × 2 × 1.0 × 10⁻⁴ = 8.0 × 10⁻⁴ M s⁻¹. Distractors: overall coefficients, k[A]²[B]², 16 × 1.0 × 10⁻⁴ = 1.6 × 10⁻³; first step only, k[A][B], 4.0 × 10⁻⁴; slow step with the intermediate dropped, k[A], 2.0 × 10⁻⁴.
The reaction 2 NO(g) + 2 H₂(g) → N₂(g) + 2 H₂O(g) has the experimentally determined rate law rate = k[NO]²[H₂]. A proposed mechanism has three steps. Step 1: NO + NO ⇌ N₂O₂ (fast, reversible). Step 2: N₂O₂ + H₂ → N₂O + H₂O (slow). Step 3: N₂O + H₂ → N₂ + H₂O (fast). A student claims that the mechanism must be rejected, because two H₂ molecules react in it but the rate law is first order in H₂. Which evaluation of the student's claim is best?
Answer and reasoning
AIncorrect: step 3 follows the slow step, so its H₂ does not enter the predicted rate law.Correct The slow step gives rate = k₂[N₂O₂][H₂], and the fast first step gives [N₂O₂] = (k₁/k₋₁)[NO]², so the mechanism predicts rate = k[NO]²[H₂], in agreement with experiment. The H₂ that reacts in step 3 comes after the slow step and does not affect the rate law.
BCorrect: both H₂ molecules react, so the mechanism predicts a rate law second order in H₂. A student who takes the orders from the total number of each reactant used, as in the overall equation, picks this. Only reactants in the slow step, or in the fast step before it, enter the rate law; the H₂ in step 3 does not.
CCorrect: step 1 has no H₂ in it, so the mechanism predicts a rate law zero order in H₂. A student who thinks the first step always sets the rate law picks this. Step 1 is fast; the slow step is step 2, in which one H₂ reacts, so the predicted rate law is first order in H₂.
DIncorrect: the steps add up to the overall equation, so the mechanism needs no further checks. A student who thinks a mechanism only has to add up to the overall equation picks this. A mechanism must also predict the experimental rate law; this one does, which is the reason the claim is wrong.
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
