College Chemistry Quiz: Introduction To Reaction Mechanisms
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Introduction To Reaction MechanismsQuestion 1 of 20

In a reaction mechanism, the rate-determining step is:

always the first step in the mechanism
the step with the largest rate constant
the step with the highest activation energy
the slowest step that controls the overall reaction rate
the step that produces the most product molecules
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College Chemistry Quiz

College Chemistry Quiz: Introduction To Reaction Mechanisms

Practice Introduction To Reaction Mechanisms in College Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Introduction To Reaction Mechanisms, giving you a quick way to practice the rules, question types, and explanations that matter most for College Chemistry.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

In a reaction mechanism, the rate-determining step is:

  1. always the first step in the mechanism
  2. the step with the largest rate constant
  3. the step with the highest activation energy
  4. the slowest step that controls the overall reaction rate (correct answer)
  5. the step that produces the most product molecules
Explanation: When you encounter questions about reaction mechanisms, focus on understanding how multiple elementary steps combine to determine the overall reaction rate. The rate-determining step is the bottleneck in a multi-step reaction mechanism—it's the slowest step that limits how fast the overall reaction can proceed. Think of it like traffic flowing through a series of toll booths: no matter how fast cars move through the other booths, the overall flow is limited by the slowest one. This step controls the overall reaction rate because the reaction cannot proceed faster than its slowest component. Choice A is incorrect because the rate-determining step can occur anywhere in the mechanism—first, middle, or last. The position doesn't determine which step is rate-limiting. Choice B confuses rate constants with actual reaction rates. A large rate constant (k) indicates a fast step, but the rate-determining step is actually the slow one with a small rate constant. Choice C creates confusion between thermodynamics and kinetics. While the rate-determining step often has high activation energy (making it slow), this isn't always true—other factors like reactant concentrations and the pre-exponential factor also influence reaction rates. The correct answer is D because it directly defines what "rate-determining" means: the step that controls (limits) the overall reaction rate by being the slowest. Study tip: Remember that "rate-determining" = "rate-limiting" = slowest step. When analyzing mechanisms, look for the step that would create the biggest bottleneck—that's your rate-determining step, regardless of its position in the sequence.

Question 2

For the reaction 2A+B→C2A + B \rightarrow C, the following mechanism is proposed:

Step 1: A+B→DA + B \rightarrow D (fast equilibrium) Step 2: A+D→CA + D \rightarrow C (slow)

What is the predicted rate law for this reaction?

  1. Rate = k[A][B]k[A][B]
  2. Rate = k[A]2[B]k[A]^2[B] (correct answer)
  3. Rate = k[A][D]k[A][D]
  4. Rate = k[D]k[D]
  5. Rate = k[A]2[B]2k[A]^2[B]^2
Explanation: When you encounter a multi-step reaction mechanism, you need to derive the rate law by identifying the rate-determining step and expressing any intermediates in terms of the original reactants. The rate law is determined by the slowest step (Step 2): Rate = k2[A][D]k_2[A][D]. However, since D is an intermediate, you must eliminate it using the equilibrium condition from Step 1. For the fast equilibrium in Step 1, the equilibrium constant is: Keq=[D][A][B]K_{eq} = \frac{[D]}{[A][B]} Solving for [D]: [D]=Keq[A][B][D] = K_{eq}[A][B] Substituting this into the rate expression: Rate = k2[A]⋅Keq[A][B]=k2Keq[A]2[B]k_2[A] \cdot K_{eq}[A][B] = k_2K_{eq}[A]^2[B] Since k2Keqk_2K_{eq} is just a combined constant (k), the final rate law is: Rate = k[A]2[B]k[A]^2[B] Choice A (Rate = k[A][B]k[A][B]) incorrectly suggests a simple bimolecular reaction without considering the mechanism. Choice C (Rate = k[A][D]k[A][D]) fails to eliminate the intermediate D, which must always be done in rate laws. Choice D (Rate = k[D]k[D]) also contains an intermediate and ignores the dependence on A from the rate-determining step. Study tip: For mechanism problems, always follow this sequence: identify the slow step, write its rate law, then use pre-equilibrium approximations to eliminate any intermediates. The final rate law should only contain original reactants, never intermediates.

Question 3

A proposed mechanism for the decomposition of ozone is:

Step 1: O3+Cl→O2+ClOO_3 + Cl \rightarrow O_2 + ClO (fast) Step 2: ClO+O→Cl+O2ClO + O \rightarrow Cl + O_2 (slow)

Which species acts as a catalyst in this mechanism?

  1. O3O_3 because it appears in the first step
  2. ClCl because it is consumed in step 1 and regenerated in step 2 (correct answer)
  3. ClOClO because it is produced in step 1 and consumed in step 2
  4. O2O_2 because it appears as a product in both steps
  5. OO because it only appears in the rate-determining step
Explanation: When you encounter a multi-step reaction mechanism, identifying the catalyst requires understanding how catalysts behave: they participate in the reaction but are neither consumed nor produced overall. Let's trace each species through both steps. Chlorine (ClCl) is consumed in step 1 but then regenerated in step 2, returning to its original form. This is the hallmark of a catalyst - it facilitates the reaction by providing an alternative pathway but emerges unchanged at the end. The catalyst allows ozone decomposition to occur more readily by breaking it into two simpler steps. Option A is incorrect because O3O_3 is a reactant that gets consumed in step 1 and doesn't reappear. Simply appearing in the first step doesn't make something a catalyst. Option C misidentifies ClOClO as the catalyst, but ClOClO is actually an intermediate - it's produced in step 1 and consumed in step 2, so it doesn't survive the overall process. While intermediates are temporary species that facilitate reactions, they're not catalysts because they don't regenerate. Option D incorrectly suggests O2O_2 is the catalyst because it appears in both steps, but O2O_2 is actually a product being formed, not a species that cycles back to participate again. Remember this key distinction: catalysts are regenerated and can participate in multiple reaction cycles, while intermediates are temporary species that are consumed. Look for the species that returns to its original state after all steps are complete.

Question 4

In a reaction mechanism, an intermediate is defined as a species that:

  1. appears in the overall balanced equation as either a reactant or product
  2. is formed in one elementary step and consumed in a subsequent elementary step (correct answer)
  3. decreases the activation energy of the reaction without being consumed
  4. determines the overall rate of the reaction by being in the slowest step
  5. has the highest concentration during the course of the reaction
Explanation: When analyzing reaction mechanisms, you need to distinguish between different types of species based on their roles throughout the multi-step process. An intermediate has a very specific definition that relates to its temporary existence during the reaction pathway. An intermediate is formed in one elementary step and then consumed (used up) in a later elementary step within the same mechanism. Think of it as a "middleman" species that exists temporarily to help the reaction proceed from reactants to final products. Because intermediates are both created and destroyed within the mechanism, they don't appear in the overall balanced equation. Let's examine why the other options are incorrect. Choice A is wrong because intermediates specifically do NOT appear in the overall balanced equation—they cancel out when you add up all the elementary steps. Choice C describes a catalyst, not an intermediate. Catalysts lower activation energy and are regenerated unchanged, while intermediates are consumed. Choice D confuses intermediates with rate-determining steps. While an intermediate might be involved in the rate-determining step, simply being an intermediate doesn't determine the reaction rate. A helpful way to remember this: intermediates are like temporary parking spots for atoms and molecules as they rearrange from reactants to products. They're created, they exist briefly, and then they're consumed to form the final products. Study tip: When analyzing mechanisms, always track what gets made and what gets used up in each step. If a species appears as a product in one step and a reactant in another, it's an intermediate.

Question 5

In the mechanism below, which species is an intermediate?

Step 1: A+B→C+DA + B \rightarrow C + D Step 2: C+E→F+GC + E \rightarrow F + G Step 3: D+F→HD + F \rightarrow H

  1. A, because it appears only in the first step
  2. B, because it is consumed in the first step
  3. C, because it is produced in step 1 and consumed in step 2 (correct answer)
  4. E, because it appears only in step 2
  5. H, because it appears only in the final step
Explanation: When analyzing reaction mechanisms, you need to distinguish between reactants, products, and intermediates. An intermediate is a species that is formed in one step and then consumed in a subsequent step of the mechanism—it appears temporarily during the reaction pathway but doesn't show up in the overall reaction equation. Looking at this mechanism, species C is produced in step 1 (A+B→C+DA + B \rightarrow C + D) and then consumed in step 2 (C+E→F+GC + E \rightarrow F + G). This makes C an intermediate because it exists only temporarily in the reaction sequence. Let's examine why the other choices are incorrect. Choice A suggests that A is an intermediate because it appears only in the first step, but A is actually a starting material (reactant) that gets consumed. Choice B claims B is an intermediate because it's consumed in step 1, but like A, B is simply a reactant—being consumed doesn't make something an intermediate. Choice D identifies E as an intermediate because it appears only in step 2, but E is actually a reactant in step 2, not something produced and then consumed. The key distinction is that intermediates have a "birth and death" within the mechanism—they're created in one step and destroyed in another. Reactants only have a "death" (consumption), while final products only have a "birth" (formation). Study tip: When identifying intermediates in mechanisms, look for species that appear on the product side of one equation and the reactant side of another. This "appears twice" pattern is your signal for intermediates.

Question 6

The reaction A+2B→CA + 2B \rightarrow C has the experimental rate law Rate = k[A]2[B]k[A]^2[B]. This rate law suggests that:

  1. the reaction occurs in a single elementary step
  2. the reaction must involve a pre-equilibrium step (correct answer)
  3. the stoichiometric coefficients determine the reaction orders
  4. B is consumed twice as fast as A
  5. the reaction is second order overall
Explanation: When you encounter rate law problems, the key insight is recognizing the difference between stoichiometric coefficients (from the balanced equation) and reaction orders (from experimental data). These often don't match, which tells you something important about the reaction mechanism. The given reaction A+2B→CA + 2B \rightarrow C has an experimental rate law of Rate = k[A]2[B]k[A]^2[B]. Notice that the order with respect to A is 2, but A's stoichiometric coefficient is 1. This mismatch is the crucial clue that the reaction cannot occur in a single step as written. When reaction orders don't match stoichiometric coefficients, the reaction must proceed through multiple steps, often involving a pre-equilibrium. In this case, there's likely a fast equilibrium step involving A that occurs before the rate-determining step, causing A to appear squared in the rate law even though only one A appears in the overall equation. Let's examine why the other options are incorrect. Option A is wrong because if this were an elementary reaction, the rate law would be Rate = k[A][B]2k[A][B]^2, matching the stoichiometric coefficients. Option C is incorrect because, as we've seen, reaction orders and stoichiometric coefficients often differ due to multi-step mechanisms. Option D addresses consumption rates, which relate to stoichiometry (B is indeed consumed twice as fast as A), but this doesn't explain the rate law discrepancy. Remember: when experimental rate laws don't match stoichiometric coefficients, immediately think "multi-step mechanism." This is a common pattern in kinetics problems.

Question 7

Which of the following correctly describes the relationship between elementary steps and the overall reaction?

  1. The sum of all elementary steps gives the overall reaction equation (correct answer)
  2. The overall rate law is the sum of the rate laws of all elementary steps
  3. The overall rate constant is the product of all elementary step rate constants
  4. Each elementary step must have the same activation energy as the overall reaction
  5. The overall reaction order equals the sum of orders of all elementary steps
Explanation: When you encounter questions about reaction mechanisms, focus on how elementary steps combine to create the overall reaction. A reaction mechanism is a step-by-step pathway showing how reactants transform into products at the molecular level. Option A is correct because elementary steps must add up to give the overall balanced equation. When you sum all elementary steps in a mechanism (canceling out intermediates that appear on both sides), you get the net reaction. This is a fundamental requirement - the mechanism must be consistent with the observed stoichiometry. Option B is incorrect because the overall rate law depends only on the rate-determining step (the slowest step), not the sum of all rate laws. Adding rate laws would give meaningless units and doesn't reflect how reactions actually proceed through bottleneck steps. Option C is wrong because rate constants don't simply multiply together to give the overall rate constant. The relationship between elementary step rate constants and the overall rate constant is complex and depends on the specific mechanism, often involving ratios and equilibrium expressions. Option D is false because each elementary step has its own unique activation energy based on the specific bond-breaking and bond-forming processes involved. The overall activation energy corresponds to the highest energy barrier along the reaction pathway, typically associated with the rate-determining step. Remember this key principle: mechanisms must be stoichiometrically consistent with the overall reaction, but kinetic properties (rate laws, rate constants, activation energies) follow more complex relationships determined by the slowest step and energy profile.

Question 8

A mechanism is proposed where the second step is rate-determining. If a catalyst is added that specifically speeds up only the first step, what happens to the overall reaction rate?

  1. The overall rate increases significantly because the first step is now faster
  2. The overall rate decreases because the mechanism is disrupted
  3. The overall rate remains essentially unchanged because the second step still limits the rate (correct answer)
  4. The overall rate doubles because catalysts always double reaction rates
  5. The overall rate becomes infinite because there is no longer a rate-limiting step
Explanation: When you encounter multi-step reaction mechanisms, understanding the rate-determining step is crucial for predicting how changes affect the overall reaction rate. The rate-determining step acts like a bottleneck—it controls how fast the entire process can proceed, regardless of what happens in other steps. In this mechanism, the second step is rate-determining, meaning it's the slowest step that limits the overall reaction rate. When a catalyst speeds up only the first step, you're essentially making a fast step even faster while leaving the bottleneck unchanged. Think of it like widening the entrance to a tunnel while keeping the narrow middle section the same—traffic still moves at the speed of the narrowest point. The correct answer is C because the second step still determines the overall rate. Making the first step faster doesn't relieve the bottleneck created by the slower second step. Choice A incorrectly assumes that speeding up any step will increase the overall rate. This ignores the concept that only the slowest step matters for the overall kinetics. Choice B suggests the mechanism is disrupted, but catalysts don't change reaction pathways—they only lower activation barriers for existing steps. Choice D reflects the misconception that all catalysts have the same quantitative effect, which is false. Remember this key principle: catalysts only increase the overall reaction rate if they speed up the rate-determining step. If they affect other steps, the overall rate remains essentially unchanged because the bottleneck persists.

Question 9

In the mechanism:

Step 1: A+B→I1A + B \rightarrow I_1 (slow) Step 2: I1+C→I2I_1 + C \rightarrow I_2 (fast) Step 3: I2→ProductsI_2 \rightarrow Products (fast)

What is the rate law for the overall reaction?

  1. Rate = k[A][B][C]k[A][B][C]
  2. Rate = k[A][B]k[A][B] (correct answer)
  3. Rate = k[I1][C]k[I_1][C]
  4. Rate = k[I2]k[I_2]
  5. Rate = k[A][B][I1][I2]k[A][B][I_1][I_2]
Explanation: When you encounter a multi-step reaction mechanism, the key principle is that the overall reaction rate is determined by the slowest step - the rate-determining step. This is because the entire process can only proceed as fast as its bottleneck allows. In this mechanism, Step 1 is labeled as "slow" while Steps 2 and 3 are "fast." This means Step 1 controls the overall rate. Since the rate law describes how fast the overall reaction proceeds, you write it based solely on the rate-determining step: A+B→I1A + B \rightarrow I_1. The rate law is therefore Rate = k[A][B]k[A][B], making choice B correct. Let's examine why the other options are wrong. Choice A (Rate = k[A][B][C]k[A][B][C]) incorrectly includes all reactants from the entire mechanism. While C does participate in the overall reaction, it's not involved in the rate-determining step, so it doesn't appear in the rate law. Choice C (Rate = k[I1][C]k[I_1][C]) focuses on Step 2, but even though this step involves the intermediate I1I_1, it's fast and doesn't control the overall rate. Additionally, rate laws are typically written in terms of starting materials, not intermediates. Choice D (Rate = k[I2]k[I_2]) similarly focuses on Step 3 and an intermediate, neither of which determines the overall rate. Remember this key strategy: in mechanism problems, identify the slowest step first, then write the rate law using only the reactants from that step. Ignore fast steps and avoid including intermediates in your final rate law.

Question 10

A student proposes the mechanism:

Step 1: A→B+CA \rightarrow B + C (slow) Step 2: B+D→EB + D \rightarrow E (fast) Step 3: C+D→FC + D \rightarrow F (fast)

If this mechanism is correct, which relationship must hold between reaction rates?

  1. Rate of A consumption = Rate of B formation = Rate of C formation
  2. Rate of E formation = Rate of F formation at all times
  3. Rate of D consumption = 2 × Rate of A consumption (correct answer)
  4. Rate of step 1 = Rate of step 2 + Rate of step 3
  5. All steps must have equal rates at steady state
Explanation: When you encounter a multi-step reaction mechanism, you need to apply conservation principles and consider how species are produced and consumed across all steps. Let's trace each species through this mechanism. In step 1, one molecule of A disappears and produces one molecule each of B and C. In step 2, B is consumed to make E. In step 3, C is consumed to make F. Importantly, both steps 2 and 3 consume one molecule of D each. Since steps 2 and 3 are both fast compared to step 1, they essentially keep up with whatever step 1 produces. This means that for every A molecule that reacts in step 1, one D molecule is consumed in step 2 and another D molecule is consumed in step 3. Therefore, two D molecules are consumed for every one A molecule consumed, making the rate of D consumption exactly twice the rate of A consumption. This confirms answer C. Answer A is incorrect because while A consumption does equal B and C formation individually, the rates aren't necessarily equal at all times due to the different reaction steps. Answer B is wrong because although E and F formation rates might be similar, they depend on the concentrations of B and D versus C and D, which can differ. Answer D incorrectly suggests that step 1's rate equals the sum of steps 2 and 3, but rate relationships depend on stoichiometry, not simple addition. Remember: in mechanism problems, always track stoichiometry across all steps and apply conservation of mass to find rate relationships.

Question 11

A research team investigates the reaction between compounds X and Y to form product Z. They propose the following mechanism:

Step 1: X+Y⇌WX + Y \rightleftharpoons W (fast equilibrium, K1=2.5K_1 = 2.5) Step 2: W+X→Z+QW + X \rightarrow Z + Q (slow, k2=0.040 M−1s−1k_2 = 0.040 \text{ M}^{-1}\text{s}^{-1}) Step 3: Q→RQ \rightarrow R (fast)

The team wants to determine if this mechanism is consistent with their experimental observations.

Based on the proposed mechanism described in the passage, what is the predicted rate law for the formation of product Z?

  1. Rate = k[X][Y]k[X][Y] where k is the effective rate constant
  2. Rate = k[X]2[Y]k[X]^2[Y] where k is the effective rate constant (correct answer)
  3. Rate = k[W][X]k[W][X] where k is the effective rate constant
  4. Rate = k[X][Y][W]k[X][Y][W] where k is the effective rate constant
  5. Rate = k[X]2[Y]2k[X]^2[Y]^2 where k is the effective rate constant
Explanation: When you encounter a multi-step reaction mechanism, you need to identify the rate-determining step and derive the rate law based on the concentrations of reactants that participate in forming the final product. Since Step 2 is labeled as "slow," it's the rate-determining step. The rate of the overall reaction equals the rate of this slowest step. Step 2 shows: W+X→Z+QW + X \rightarrow Z + Q, so the rate appears to be k2[W][X]k_2[W][X]. However, W is an intermediate species, and rate laws must be expressed in terms of the original reactants (X and Y). To eliminate [W], you use the fast equilibrium in Step 1. Since Step 1 reaches equilibrium quickly, K1=[W][X][Y]=2.5K_1 = \frac{[W]}{[X][Y]} = 2.5. Solving for [W]: [W]=K1[X][Y][W] = K_1[X][Y]. Substituting this into the rate expression: Rate = k2[W][X]=k2K1[X][Y][X]=k2K1[X]2[Y]k_2[W][X] = k_2K_1[X][Y][X] = k_2K_1[X]^2[Y]. This gives Rate = k[X]2[Y]k[X]^2[Y] where k=k2K1k = k_2K_1. Choice A incorrectly assumes first-order dependence on both X and Y, missing that X participates in both steps. Choice C keeps the intermediate W in the rate law, which violates the requirement to express rate laws in terms of initial reactants. Choice D incorrectly includes three concentration terms, suggesting a termolecular process that doesn't match the mechanism. Remember: for mechanisms with fast pre-equilibria, always substitute equilibrium expressions to eliminate intermediates from your final rate law. The rate-determining step gives you the framework, but you must express everything in terms of the original reactants.

Question 12

In a proposed mechanism, step 2 is identified as rate-determining. If experimental data shows that doubling the concentration of a reactant from step 3 doubles the overall reaction rate, what can be concluded?

  1. Step 3 must actually be the rate-determining step instead of step 2
  2. The reactant from step 3 must also appear in step 2 (correct answer)
  3. The proposed mechanism is definitely incorrect
  4. Step 3 must be in pre-equilibrium with step 2
  5. The reactant from step 3 must be a catalyst
Explanation: When you encounter questions about reaction mechanisms and rate-determining steps, remember that the overall reaction rate depends on the slowest step, but concentrations of species from other steps can still affect the rate if those species participate in the rate-determining step. If step 2 is rate-determining but doubling a reactant from step 3 doubles the overall rate, this tells you that the step 3 reactant must also be present in step 2's rate expression. This happens when the same species participates in multiple steps of a mechanism. Since step 2 controls the overall rate, any species that appears in step 2 will affect the reaction rate when its concentration changes. Looking at the wrong answers: (A) incorrectly assumes that only the rate-determining step can influence reaction rate - but if a species appears in multiple steps, it can affect the rate even when it's not from the slowest step. (C) is too extreme; the mechanism could still be correct if the same reactant appears in both steps. (D) suggests a pre-equilibrium relationship, but this doesn't explain why changing the step 3 reactant's concentration affects the rate - pre-equilibrium would actually make step 3 fast compared to step 2. The correct answer is (B): the reactant from step 3 must also appear in step 2, making it part of the rate-determining step's kinetics. Study tip: In mechanism problems, always check whether species appear in multiple steps. A reactant can influence the overall rate even if it's not exclusively part of the rate-determining step.

Question 13

Which statement about elementary steps in a reaction mechanism is correct?

  1. Elementary steps can involve any number of reactant molecules colliding simultaneously
  2. The molecularity of an elementary step equals the sum of stoichiometric coefficients of reactants (correct answer)
  3. Elementary steps always have the same rate constant as the overall reaction
  4. The order of reaction in an elementary step can differ from its stoichiometric coefficients
  5. Elementary steps must always be reversible under reaction conditions
Explanation: When analyzing reaction mechanisms, you need to understand the fundamental properties of elementary steps - the individual molecular events that make up an overall reaction. The molecularity of an elementary step is defined as the number of molecules that participate in that specific step. This always equals the sum of the stoichiometric coefficients of the reactants in that elementary step. For example, in the elementary step 2A+B→C2A + B \rightarrow C, the molecularity is 3 (2 + 1). This makes option B correct. Let's examine why the other options are incorrect: Option A is wrong because elementary steps are limited by molecular collision probability. While theoretically any number of molecules could collide, steps involving more than three molecules are extremely rare because the probability of simultaneous collision decreases dramatically with each additional molecule. Option C is incorrect because elementary steps and overall reactions have completely different rate constants. The overall rate constant is a complex function of all elementary step rate constants and doesn't equal any individual step's rate constant. Option D contains a crucial misconception. Unlike overall reactions, elementary steps have a special property: their reaction order with respect to each reactant always equals the stoichiometric coefficient of that reactant. This is because elementary steps represent actual molecular collisions, so the rate directly depends on the concentration of each participating molecule raised to the power of its stoichiometric coefficient. Remember: elementary steps follow stoichiometry exactly for both molecularity and reaction order, while overall reactions often don't. This distinction frequently appears on exams.

Question 14

For the mechanism:

Step 1: A⇌BA \rightleftharpoons B (fast equilibrium, K1=0.50K_1 = 0.50) Step 2: B+C→DB + C \rightarrow D (slow, k2=2.0×10−3 M−1s−1k_2 = 2.0 \times 10^{-3} \text{ M}^{-1}\text{s}^{-1})

If [A]0=0.40 M[A]_0 = 0.40 \text{ M} and [C]=0.20 M[C] = 0.20 \text{ M}, what is the initial rate of formation of D?

  1. 1.6×10−4 M/s1.6 \times 10^{-4} \text{ M/s}
  2. 8.0×10−5 M/s8.0 \times 10^{-5} \text{ M/s} (correct answer)
  3. 4.0×10−5 M/s4.0 \times 10^{-5} \text{ M/s}
  4. 2.0×10−4 M/s2.0 \times 10^{-4} \text{ M/s}
  5. 3.2×10−4 M/s3.2 \times 10^{-4} \text{ M/s}
Explanation: When you encounter a multi-step reaction mechanism, you need to identify the rate-determining step and use equilibrium relationships for fast pre-equilibrium steps. Since Step 1 is a fast equilibrium, you can use the equilibrium constant to find the concentration of intermediate B. The equilibrium expression is K1=[B][A]=0.50K_1 = \frac{[B]}{[A]} = 0.50. With [A]0=0.40[A]_0 = 0.40 M, you get [B]=K1×[A]=0.50×0.40=0.20[B] = K_1 \times [A] = 0.50 \times 0.40 = 0.20 M. Step 2 is the slow step, making it rate-determining. The rate law for this step is: rate=k2[B][C]\text{rate} = k_2[B][C]. Substituting the values: rate=(2.0×10−3)(0.20)(0.20)=8.0×10−5 M/s\text{rate} = (2.0 \times 10^{-3})(0.20)(0.20) = 8.0 \times 10^{-5} \text{ M/s} Answer A (1.6×10−41.6 \times 10^{-4} M/s) likely results from incorrectly using [A]0[A]_0 directly in the rate expression instead of calculating [B][B] from the equilibrium. Answer C (4.0×10−54.0 \times 10^{-5} M/s) probably comes from using K1K_1 incorrectly, perhaps as [A][B]\frac{[A]}{[B]} instead of [B][A]\frac{[B]}{[A]}. Answer D (2.0×10−42.0 \times 10^{-4} M/s) might result from using the wrong equilibrium expression or miscalculating the intermediate concentration. Remember: for mechanisms with fast pre-equilibria followed by a slow step, always use the equilibrium constant to find intermediate concentrations, then apply the rate law for the rate-determining step. The overall rate depends on the slowest step, not the initial reactants directly.

Question 15

For a unimolecular elementary step A→ProductsA \rightarrow Products, which statement is true?

  1. The reaction order with respect to A is always 2
  2. The reaction order with respect to A is always 1 (correct answer)
  3. The reaction order with respect to A depends on the mechanism
  4. The reaction order with respect to A is always 0
  5. The reaction order with respect to A cannot be determined
Explanation: When you encounter questions about elementary reactions, remember that the key distinction is between elementary steps (which occur exactly as written) and overall reactions (which may involve multiple steps). For an elementary step, the reaction order with respect to each reactant directly corresponds to its stoichiometric coefficient in the balanced equation. Since this unimolecular elementary step shows one molecule of A converting to products (A→ProductsA \rightarrow Products), the reaction occurs through the transformation of individual A molecules. The rate law for this elementary step is therefore rate=k[A]1rate = k[A]^1, making the reaction first-order with respect to A. Choice A incorrectly suggests the reaction order is always 2. This would only be true for a bimolecular elementary step like A+A→ProductsA + A \rightarrow Products or if we had 2A→Products2A \rightarrow Products. Choice C wrongly implies that the reaction order depends on the mechanism. While this is true for overall reactions that proceed through multiple elementary steps, for a single elementary step, the reaction order is determined solely by the stoichiometry shown. Choice D suggests zero-order kinetics, which would mean the rate is independent of [A]. This contradicts the fundamental principle that elementary reactions depend directly on the concentrations of the reactants involved. Study tip: For elementary steps, always match the exponent in the rate law to the coefficient in front of each reactant. This 1:1 correspondence only applies to elementary reactions, not complex multi-step mechanisms where you must determine rate laws experimentally.

Question 16

Consider the mechanism:

Step 1: 2A⇌B2A \rightleftharpoons B (fast equilibrium) Step 2: B+C→DB + C \rightarrow D (slow)

If [A] is tripled, by what factor does the initial rate change?

  1. 3
  2. 6
  3. 9 (correct answer)
  4. 27
  5. 1
Explanation: When you encounter multi-step reaction mechanisms, the key is identifying which step controls the overall rate and how concentration changes affect that rate-determining step. Since Step 2 is slow, it determines the overall reaction rate. The rate law for Step 2 is: rate=k[B][C]\text{rate} = k[B][C]. However, B is an intermediate formed in the fast equilibrium of Step 1, so you need to express [B] in terms of the original reactant [A]. For the fast equilibrium in Step 1: Keq=[B][A]2K_{eq} = \frac{[B]}{[A]^2}, which means [B]=Keq[A]2[B] = K_{eq}[A]^2 Substituting this into the rate law: rate=k⋅Keq[A]2[C]=k′[A]2[C]\text{rate} = k \cdot K_{eq}[A]^2[C] = k'[A]^2[C] The overall reaction is second-order in A. When [A] is tripled, the rate changes by a factor of (3)2=9(3)^2 = 9. Looking at the wrong answers: A) 3 assumes the reaction is first-order in A, ignoring that two A molecules are needed to form one B. B) 6 might come from incorrectly thinking the rate is proportional to 3×2=63 \times 2 = 6 because of the coefficient 2 in Step 1. D) 27 would result from incorrectly assuming the reaction is third-order in A, perhaps by misinterpreting the two-step mechanism. Remember: for mechanisms with fast pre-equilibrium steps, always express intermediate concentrations in terms of the original reactants, then determine how the overall rate depends on those reactant concentrations. The stoichiometry in the pre-equilibrium determines the order of dependence.

Question 17

For the elementary step 2A+B→C2A + B \rightarrow C, the rate law is:

  1. Rate = k[A][B]k[A][B]
  2. Rate = k[A]2[B]k[A]^2[B] (correct answer)
  3. Rate = k[A][B]2k[A][B]^2
  4. Rate = k[A]2[B]2k[A]^2[B]^2
  5. Cannot be determined without experimental data
Explanation: When you encounter a question about elementary steps in reaction mechanisms, the key principle is that for an elementary step, the rate law can be written directly from the stoichiometric coefficients in the balanced equation. For the elementary step 2A+B→C2A + B \rightarrow C, you can determine the rate law by using each reactant's stoichiometric coefficient as its exponent in the rate expression. Since 2 molecules of A participate in this elementary step, the concentration of A is raised to the second power. Since 1 molecule of B participates, the concentration of B is raised to the first power. This gives us Rate = k[A]2[B]k[A]^2[B], which is answer choice B. Let's examine why the other options are incorrect. Choice A (Rate = k[A][B]k[A][B]) ignores the stoichiometric coefficient of A, treating it as if only one molecule of A were involved. Choice C (Rate = k[A][B]2k[A][B]^2) incorrectly applies the coefficient of A to species B instead. Choice D (Rate = k[A]2[B]2k[A]^2[B]^2) applies a squared term to both reactants, which would only be correct if the reaction were 2A+2B→C2A + 2B \rightarrow C. Remember this crucial distinction: you can only write the rate law directly from stoichiometric coefficients for elementary steps. For overall reactions (which may involve multiple steps), you must determine the rate law experimentally. Always check whether the question specifies "elementary step" versus "overall reaction" before applying this rule.

Question 18

Which factor does NOT affect the rate of an elementary step?

  1. Temperature of the reaction mixture
  2. Concentrations of the reactant molecules in that step
  3. Presence of a catalyst that lowers the activation energy
  4. The overall stoichiometry of the complete reaction (correct answer)
  5. The activation energy of that particular elementary step
Explanation: When analyzing factors that affect reaction rates, you need to distinguish between what influences an individual elementary step versus the overall multi-step reaction mechanism. The correct answer is D because the overall stoichiometry of the complete reaction doesn't directly affect the rate of any individual elementary step. Each elementary step has its own rate law that depends only on the specific molecules participating in that particular step, not on the broader reaction's balanced equation. The overall stoichiometry is simply the sum of all elementary steps and doesn't influence the kinetics of individual molecular collisions. Let's examine why the other options DO affect elementary step rates: A is incorrect because temperature directly affects reaction rates through the Arrhenius equation - higher temperatures increase molecular kinetic energy and collision frequency. B is incorrect because the rate law for any elementary step depends directly on the concentrations of reactants in that specific step, following the principle that reaction rate is proportional to the probability of molecular collisions. C is incorrect because catalysts lower activation energy barriers, making it easier for reactant molecules to reach the transition state in that elementary step. The key distinction here is between local versus global effects. Temperature, reactant concentrations, and catalysts all operate at the molecular level where the elementary step occurs. However, the overall reaction stoichiometry is a macroscopic description that emerges from combining all elementary steps - it doesn't reach back down to influence individual step kinetics. Remember: elementary steps respond to immediate molecular environment, not to the bigger picture reaction scheme.

Question 19

For the reaction mechanism:

Step 1: NO+O2⇌NO3NO + O_2 \rightleftharpoons NO_3 (fast equilibrium) Step 2: NO3+NO→2NO2NO_3 + NO \rightarrow 2NO_2 (slow)

What is the overall balanced equation?

  1. NO+O2→NO3NO + O_2 \rightarrow NO_3
  2. 2NO+O2→2NO22NO + O_2 \rightarrow 2NO_2 (correct answer)
  3. NO3+NO→2NO2NO_3 + NO \rightarrow 2NO_2
  4. NO+O2+NO3→2NO2+NO3NO + O_2 + NO_3 \rightarrow 2NO_2 + NO_3
  5. 3NO+O2→2NO2+NO3NO + O_2 \rightarrow 2NO_2 + NO
Explanation: When you encounter a reaction mechanism, you need to find the overall reaction by adding all the elementary steps together and canceling out any intermediates that appear on both sides. Let's work through this systematically. First, write out both steps:
  • Step 1: NO+O2⇌NO3NO + O_2 \rightleftharpoons NO_3
  • Step 2: NO3+NO→2NO2NO_3 + NO \rightarrow 2NO_2
To find the overall reaction, add these steps together: NO+O2+NO3+NO→NO3+2NO2NO + O_2 + NO_3 + NO \rightarrow NO_3 + 2NO_2 Now identify and cancel the intermediates. NO3NO_3 appears as a product in Step 1 and a reactant in Step 2, so it cancels out completely. This leaves: 2NO+O2→2NO22NO + O_2 \rightarrow 2NO_2 This matches answer choice B. Looking at the wrong answers: Choice A (NO+O2→NO3NO + O_2 \rightarrow NO_3) is just Step 1 alone, not the complete mechanism. Choice C (NO3+NO→2NO2NO_3 + NO \rightarrow 2NO_2) is only Step 2, ignoring the first step entirely. Choice D (NO+O2+NO3→2NO2+NO3NO + O_2 + NO_3 \rightarrow 2NO_2 + NO_3) represents a common mistake where students add the steps but forget to cancel the intermediate NO3NO_3. Remember this key strategy: when finding overall reactions from mechanisms, always add all steps together, then eliminate any species that appears on both sides of the arrow. These intermediates are produced in one step and consumed in another, so they don't appear in the net reaction. This approach works for any multi-step mechanism you'll encounter.

Question 20

A proposed mechanism has the following steps:

Step 1: X+Y⇌ZX + Y \rightleftharpoons Z (fast equilibrium) Step 2: Z+W→ProductsZ + W \rightarrow Products (slow)

If the concentration of Y is doubled while keeping X and W constant, how does the rate change?

  1. Rate increases by a factor of 2 (correct answer)
  2. Rate increases by a factor of 4
  3. Rate decreases by a factor of 2
  4. Rate remains unchanged
  5. Rate increases by a factor of 8
Explanation: When you encounter reaction mechanism problems, focus on identifying the rate-determining step and expressing concentrations of intermediates in terms of initial reactants. Since Step 2 is slow, it determines the overall reaction rate: Rate=k2[Z][W]\text{Rate} = k_2[Z][W]. However, Z is an intermediate from the fast equilibrium in Step 1, so you need to express [Z] in terms of the original reactants. For the fast equilibrium: Keq=[Z][X][Y]K_{eq} = \frac{[Z]}{[X][Y]}, which gives us [Z]=Keq[X][Y][Z] = K_{eq}[X][Y] Substituting into the rate expression: Rate=k2⋅Keq[X][Y][W]=kobs[X][Y][W]\text{Rate} = k_2 \cdot K_{eq}[X][Y][W] = k_{obs}[X][Y][W] When [Y] doubles while [X] and [W] remain constant, the rate doubles because [Y] appears to the first power in the rate law. Looking at the wrong answers: B) suggests the rate increases by a factor of 4, which would occur if [Y] appeared squared in the rate law, but it doesn't. C) claims the rate decreases, which makes no sense since increasing reactant concentration should increase rate (assuming positive reaction order). D) suggests no change, which would only be true if the reaction were zero-order in Y. The correct answer is A) - the rate increases by a factor of 2. Study tip: For mechanism problems, always identify the slow step first, then use equilibrium expressions to substitute intermediate concentrations with initial reactant concentrations. The overall rate law will show how each original reactant affects the rate.