College Chemistry Quiz: Reaction Mechanisms And Rate Law
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Reaction Mechanisms And Rate LawQuestion 1 of 20

For the reaction 2NO(g)+O2(g)→2NO2(g)2NO(g) + O_2(g) \rightarrow 2NO_2(g), the following mechanism is proposed:

Step 1: NO(g)+O2(g)⇌NO3(g)NO(g) + O_2(g) \rightleftharpoons NO_3(g) (fast equilibrium) Step 2: NO3(g)+NO(g)→2NO2(g)NO_3(g) + NO(g) \rightarrow 2NO_2(g) (slow)

What is the predicted rate law for this mechanism?

Rate=k[NO]2[O2]Rate = k[NO]^2[O_2]
Rate=k[NO][O2]Rate = k[NO][O_2]
Rate=k[NO3][NO]Rate = k[NO_3][NO]
Rate=k[NO]2[O2]1/2Rate = k[NO]^2[O_2]^{1/2}
Rate=k[NO2]2/[NO]2[O2]Rate = k[NO_2]^2/[NO]^2[O_2]
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College Chemistry Quiz

College Chemistry Quiz: Reaction Mechanisms And Rate Law

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

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This quiz focuses on Reaction Mechanisms And Rate Law, 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

For the reaction 2NO(g)+O2(g)→2NO2(g)2NO(g) + O_2(g) \rightarrow 2NO_2(g), the following mechanism is proposed:

Step 1: NO(g)+O2(g)⇌NO3(g)NO(g) + O_2(g) \rightleftharpoons NO_3(g) (fast equilibrium) Step 2: NO3(g)+NO(g)→2NO2(g)NO_3(g) + NO(g) \rightarrow 2NO_2(g) (slow)

What is the predicted rate law for this mechanism?

  1. Rate=k[NO]2[O2]Rate = k[NO]^2[O_2] (correct answer)
  2. Rate=k[NO][O2]Rate = k[NO][O_2]
  3. Rate=k[NO3][NO]Rate = k[NO_3][NO]
  4. Rate=k[NO]2[O2]1/2Rate = k[NO]^2[O_2]^{1/2}
  5. Rate=k[NO2]2/[NO]2[O2]Rate = k[NO_2]^2/[NO]^2[O_2]
Explanation: When you encounter a multi-step reaction mechanism, you need to derive the rate law by focusing on the slowest step, which determines the overall reaction rate. However, if the rate-determining step contains intermediates (species that aren't in the overall reaction), you must eliminate them using equilibrium expressions from faster steps. Here, Step 2 is slow and rate-determining, so initially you might write: Rate=k2[NO3][NO]Rate = k_2[NO_3][NO]. However, NO3NO_3 is an intermediate that doesn't appear in the overall reaction, so you can't leave it in your final rate law. Since Step 1 is a fast equilibrium, you can write: Keq=[NO3][NO][O2]K_{eq} = \frac{[NO_3]}{[NO][O_2]}, which gives you [NO3]=Keq[NO][O2][NO_3] = K_{eq}[NO][O_2]. Substituting this into your rate expression: Rate=k2⋅Keq[NO][O2]⋅[NO]=k[NO]2[O2]Rate = k_2 \cdot K_{eq}[NO][O_2] \cdot [NO] = k[NO]^2[O_2], where k=k2⋅Keqk = k_2 \cdot K_{eq}. Looking at the wrong answers: Choice B (Rate=k[NO][O2]Rate = k[NO][O_2]) represents what you'd get if you incorrectly ignored the second [NO][NO] term from Step 2. Choice C (Rate=k[NO3][NO]Rate = k[NO_3][NO]) is the intermediate step before eliminating NO3NO_3 - you can't leave intermediates in rate laws. Choice D (Rate=k[NO]2[O2]1/2Rate = k[NO]^2[O_2]^{1/2}) might arise from incorrectly applying a square root when manipulating the equilibrium expression. Study tip: For mechanism problems, always start with the slow step, then systematically eliminate any intermediates using equilibrium expressions from the fast steps. The final rate law should only contain reactants from the overall equation.

Question 2

A student proposes the following one-step mechanism for the reaction 2A+B→C+D2A + B \rightarrow C + D:

2A+B→C+D2A + B \rightarrow C + D (elementary step)

Experimental data shows that the reaction is first order in A and first order in B. Which conclusion is most appropriate?

  1. The proposed mechanism is consistent with the experimental rate law and is therefore correct
  2. The proposed mechanism is inconsistent because elementary steps must have rate laws matching their stoichiometry (correct answer)
  3. The proposed mechanism is inconsistent because three-body collisions are extremely unlikely
  4. The proposed mechanism is consistent with kinetics but requires additional evidence to be accepted
  5. The proposed mechanism is incorrect because the stoichiometric coefficients don't match the reaction orders
Explanation: When analyzing reaction mechanisms, you must understand the fundamental relationship between elementary steps and their rate laws. For any elementary step, the rate law directly reflects the stoichiometry—each reactant's order equals its stoichiometric coefficient. The proposed elementary step 2A+B→C+D2A + B \rightarrow C + D would require a rate law of rate=k[A]2[B]1\text{rate} = k[A]^2[B]^1, making it second order in A and first order in B. However, the experimental data shows the reaction is first order in A and first order in B, giving rate=k[A]1[B]1\text{rate} = k[A]^1[B]^1. This direct contradiction proves the proposed mechanism cannot be correct. Option A incorrectly suggests the mechanism matches the experimental rate law—it clearly doesn't. Option C mentions that three-body collisions are unlikely, which is true, but this isn't the primary issue here since we're told this is the proposed mechanism. The stoichiometry mismatch is the definitive problem. Option D wrongly claims the mechanism is kinetically consistent when the rate laws don't match at all. Option B correctly identifies that elementary steps must have rate laws matching their stoichiometry, and since this one doesn't match the experimental data, the mechanism is invalid. Study tip: Always check that proposed elementary steps produce rate laws consistent with experimental data. For elementary reactions, the exponents in the rate law must equal the stoichiometric coefficients—no exceptions. When they don't match, the mechanism is automatically ruled out.

Question 3

A reaction follows 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)

Overall: A+2D→E+FA + 2D \rightarrow E + F

What is the rate law for the overall reaction?

  1. Rate=k[A][D]2Rate = k[A][D]^2
  2. Rate=k[A]Rate = k[A] (correct answer)
  3. Rate=k[A][D]Rate = k[A][D]
  4. Rate=k[B][C][D]2Rate = k[B][C][D]^2
  5. Rate=k[E][F]/[A][D]2Rate = k[E][F]/[A][D]^2
Explanation: When analyzing reaction mechanisms, remember that the rate-determining step (the slowest step) controls the overall reaction rate. The rate law depends only on the concentrations of reactants involved in this bottleneck step. In this mechanism, Step 1 is labeled as slow while Steps 2 and 3 are fast. Since Step 1 (A→B+CA \rightarrow B + C) is the rate-determining step, the overall rate depends only on the concentration of AA. The rate law is therefore Rate=k[A]Rate = k[A], making B correct. Here's why the other options are wrong: A (Rate=k[A][D]2Rate = k[A][D]^2) incorrectly includes DD in the rate law. While DD appears in the overall balanced equation, it's not involved in the slow step, so it doesn't affect the rate. C (Rate=k[A][D]Rate = k[A][D]) makes the same mistake as A, including DD when it shouldn't appear in the rate law since it's not part of the rate-determining step. D (Rate=k[B][C][D]2Rate = k[B][C][D]^2) represents the rate if Steps 2 and 3 were rate-determining, but they're fast. Additionally, this would be problematic because BB and CC are intermediates (produced in one step, consumed in another), and rate laws should be expressed in terms of original reactants and products, not intermediates. Study tip: For mechanism problems, identify the slowest step first, then write the rate law using only the reactants in that step. Fast steps after the rate-determining step don't influence the overall rate.

Question 4

A student studying the reaction X+Y→ZX + Y \rightarrow Z observes that doubling [X] doubles the rate, while doubling [Y] has no effect on the rate. The student proposes this mechanism:

Step 1: X→WX \rightarrow W (slow) Step 2: W+Y→ZW + Y \rightarrow Z (fast)

What can be concluded about this mechanism?

  1. The mechanism is definitely correct because it explains all observations
  2. The mechanism is definitely incorrect because Y appears in the rate law
  3. The mechanism is consistent with kinetic data but not proven correct (correct answer)
  4. The mechanism violates conservation of mass in step 1
  5. The mechanism requires Y to be a catalyst, which contradicts the overall reaction
Explanation: When you encounter reaction mechanisms in kinetics problems, remember that experimental rate laws can support a proposed mechanism, but they can never definitively prove it's the only possible pathway. Let's analyze the given data: doubling [X] doubles the rate (first-order in X), while [Y] has no effect (zero-order in Y). This gives us a rate law of rate=k[X]1[Y]0=k[X]\text{rate} = k[X]^1[Y]^0 = k[X]. The proposed mechanism has X converting to intermediate W in the slow step, followed by W reacting with Y in the fast step. Since the first step is rate-determining and only involves X, the overall rate depends only on [X], which perfectly matches the experimental observations. The mechanism is kinetically consistent. However, answer A is wrong because multiple mechanisms could potentially explain the same kinetic data. Kinetic evidence alone cannot definitively prove a mechanism is correct—it can only show consistency or inconsistency. Answer B is incorrect because Y doesn't appear in the rate law here. The rate law is k[X]k[X], showing no dependence on [Y], which matches the observation that doubling [Y] has no effect. Answer D is wrong because step 1 doesn't violate mass conservation. While it shows X converting to W, this could represent a unimolecular rearrangement, isomerization, or any process where one molecule transforms into another. Answer C correctly states that the mechanism is consistent with the data but not proven correct. Study tip: In mechanism problems, distinguish between "consistent with data" and "proven correct"—experimental evidence can rule out mechanisms but rarely proves one definitively.

Question 5

Consider the mechanism:

Step 1: A+B⇌CA + B \rightleftharpoons C (fast equilibrium) Step 2: C→D+EC \rightarrow D + E (slow) Step 3: D+A→FD + A \rightarrow F (fast)

Overall: 2A+B→E+F2A + B \rightarrow E + F

If step 3 is changed to D+B→FD + B \rightarrow F (fast), how does the predicted rate law change?

  1. The rate law changes from Rate=k[A][B]Rate = k[A][B] to Rate=k[A]2[B]Rate = k[A]^2[B]
  2. The rate law changes from Rate=k[A][B]Rate = k[A][B] to Rate=k[A][B]2Rate = k[A][B]^2
  3. The rate law remains Rate=k[A][B]Rate = k[A][B] in both cases (correct answer)
  4. The rate law changes from Rate=k[A]2[B]Rate = k[A]^2[B] to Rate=k[A][B]2Rate = k[A][B]^2
  5. The rate law cannot be determined without additional information
Explanation: When analyzing reaction mechanisms, the rate law depends only on the rate-determining step (the slowest step) and any pre-equilibrium steps that affect the concentration of intermediates involved in that slow step. In both the original and modified mechanisms, Step 2 (C→D+EC \rightarrow D + E) remains the rate-determining step. The rate of this step is Rate=k2[C]Rate = k_2[C]. Since C is an intermediate formed in the fast pre-equilibrium (Step 1), you need to express [C] in terms of the original reactants using the equilibrium expression: Keq=[C][A][B]K_{eq} = \frac{[C]}{[A][B]}, so [C]=Keq[A][B][C] = K_{eq}[A][B]. Substituting this into the rate expression gives Rate=k2Keq[A][B]=k[A][B]Rate = k_2K_{eq}[A][B] = k[A][B] for both mechanisms. The change in Step 3 doesn't affect this rate law because Step 3 occurs after the rate-determining step. Answer choice A incorrectly suggests the original rate law involves [A]2[A]^2, but since Step 2 only consumes one molecule of C (which forms from one A and one B), the rate law is first-order in each reactant. Answer choice B incorrectly predicts the modified mechanism would be second-order in B, but again, only one C molecule is consumed in the slow step. Answer choice D makes both errors, incorrectly stating both the original and modified rate laws. Study tip: Remember that only the rate-determining step and any pre-equilibrium steps that supply its reactants determine the rate law. Steps that occur after the slow step don't influence the overall reaction rate.

Question 6

A reaction mechanism involves three elementary steps. Step 1 has a rate constant of 1.0×106 s−11.0 \times 10^{6} \, s^{-1}, step 2 has 2.0×103 s−12.0 \times 10^{3} \, s^{-1}, and step 3 has 5.0×107 s−15.0 \times 10^{7} \, s^{-1}. Which statement about the rate-determining step is correct?

  1. Step 1 is rate-determining because it has the largest rate constant
  2. Step 2 is rate-determining because it has the smallest rate constant
  3. Step 3 is rate-determining because it occurs last in the sequence
  4. The rate-determining step cannot be identified from rate constants alone (correct answer)
  5. All steps are equally rate-determining when they occur in sequence
Explanation: When analyzing reaction mechanisms, you need to understand that the rate-determining step controls the overall reaction rate, but identifying it requires more than just comparing rate constants in isolation. The correct answer is D because rate constants alone don't determine which step is rate-limiting. The rate-determining step depends on the concentrations of reactants at each step, which change as the reaction proceeds. Even a step with a large rate constant can become rate-limiting if the concentration of its reactant becomes very low. Additionally, the mechanism's structure matters—whether steps are sequential, reversible, or involve pre-equilibria all affect which step controls the overall rate. Option A incorrectly assumes that the largest rate constant (5.0×107 s−15.0 \times 10^{7} \, s^{-1} for step 3) makes that step rate-determining. In reality, a large rate constant typically means that step is fast and unlikely to be rate-limiting. Option B makes the opposite error, assuming the smallest rate constant (2.0×103 s−12.0 \times 10^{3} \, s^{-1} for step 2) automatically makes that step rate-determining. While the slowest step often is rate-determining, you can't conclude this from rate constants alone without knowing concentrations and mechanism details. Option C incorrectly suggests that position in the sequence determines the rate-limiting step, but any step in a mechanism can potentially be rate-determining. Remember: To identify the actual rate-determining step, you need the complete kinetic analysis including steady-state approximations or pre-equilibrium assumptions, not just individual rate constants.

Question 7

For the reaction A+2B→CA + 2B \rightarrow C, the following two-step mechanism is proposed:

Step 1: A+B⇌ABA + B \rightleftharpoons AB (fast equilibrium) Step 2: AB+B→CAB + B \rightarrow C (slow)

If the equilibrium constant for step 1 is K1=10 M−1K_1 = 10 \, M^{-1} and the rate constant for step 2 is k2=0.02 M−1s−1k_2 = 0.02 \, M^{-1}s^{-1}, what is the overall rate law?

  1. Rate=10[A][B]2Rate = 10[A][B]^2
  2. Rate=0.2[A][B]2Rate = 0.2[A][B]^2 (correct answer)
  3. Rate=0.02[AB][B]Rate = 0.02[AB][B]
  4. Rate=0.002[A][B]2Rate = 0.002[A][B]^2
  5. Rate=200[A][B]2Rate = 200[A][B]^2
Explanation: When you encounter multi-step reaction mechanisms, the key is identifying which step controls the overall rate and how to express concentrations of intermediates in terms of the original reactants. Since step 1 is a fast equilibrium and step 2 is slow, the slow step determines the overall reaction rate. The rate law for step 2 would be: Rate=k2[AB][B]Rate = k_2[AB][B]. However, [AB] is an intermediate concentration that we need to eliminate using the equilibrium expression from step 1. For the fast equilibrium: K1=[AB][A][B]=10 M−1K_1 = \frac{[AB]}{[A][B]} = 10 \, M^{-1} Solving for [AB]: [AB]=K1[A][B]=10[A][B][AB] = K_1[A][B] = 10[A][B] Substituting this into the rate law: Rate=k2⋅K1[A][B]⋅[B]=0.02×10×[A][B]2=0.2[A][B]2Rate = k_2 \cdot K_1[A][B] \cdot [B] = 0.02 \times 10 \times [A][B]^2 = 0.2[A][B]^2 This matches answer choice B. Looking at the wrong answers: A gives the correct functional form but uses only K1K_1 while ignoring k2k_2. C incorrectly leaves the intermediate [AB] in the rate law instead of substituting it out. D makes an arithmetic error, multiplying 0.02 × 10 × 0.01 instead of 0.02 × 10. Study tip: For mechanism problems, always remember that intermediates must be eliminated from rate laws. Use pre-equilibrium approximations (fast equilibrium steps) to express intermediate concentrations in terms of reactants, then multiply the rate constant by all relevant equilibrium constants.

Question 8

The reaction 2A+B→C+D2A + B \rightarrow C + D has the experimentally determined rate law: Rate=k[A]1.5[B]0.5Rate = k[A]^{1.5}[B]^{0.5}. Which statement about possible mechanisms is most accurate?

  1. The mechanism must involve fractional stoichiometric coefficients in elementary steps
  2. The mechanism must involve pre-equilibria that lead to fractional orders (correct answer)
  3. The mechanism must be a single elementary step with unusual kinetics
  4. The reaction must involve a heterogeneous catalyst surface
  5. The rate law suggests the reaction occurs on a solid surface with limited active sites
Explanation: When you encounter fractional orders in a rate law, you're dealing with complex reaction mechanisms that can't be explained by simple elementary steps. The key insight is understanding how overall rate laws emerge from multi-step mechanisms. The rate law Rate=k[A]1.5[B]0.5Rate = k[A]^{1.5}[B]^{0.5} contains fractional exponents, which immediately tells you this cannot be a simple one-step reaction. Fractional orders typically arise when fast pre-equilibrium steps precede a slow rate-determining step. In these mechanisms, some reactants participate in rapid equilibria before the slowest step occurs. When you derive the overall rate law using the pre-equilibrium approximation, concentration terms from the equilibrium expressions get substituted into the rate expression, often resulting in fractional powers. Option A is incorrect because elementary steps always have integer stoichiometric coefficients - you can't have half a molecule participating in a single collision event. Option C is wrong because no elementary step can produce fractional kinetics; elementary steps follow simple collision theory with integer orders. Option D is incorrect because while heterogeneous catalysis can show complex kinetics, fractional orders aren't a defining characteristic of heterogeneous catalysis and aren't required to explain this rate law. The correct answer is B because pre-equilibria are the most common and mechanistically sound explanation for fractional reaction orders in homogeneous systems. Study tip: When you see fractional orders in rate laws, immediately think "pre-equilibrium mechanism." This pattern appears frequently in kinetics problems and indicates a multi-step process with fast initial equilibria.

Question 9

Consider the mechanism:

Step 1: 2A⇌A22A \rightleftharpoons A_2 (fast equilibrium) Step 2: A2+B→CA_2 + B \rightarrow C (slow)

If the temperature is increased, both the equilibrium constant K1K_1 decreases (ΔH1>0\Delta H_1 > 0) and the rate constant k2k_2 increases. What is the net effect on the overall reaction rate?

  1. The rate definitely increases because k2k_2 increases
  2. The rate definitely decreases because K1K_1 decreases
  3. The rate remains constant because the effects cancel exactly
  4. The net effect depends on the relative magnitudes of the changes in k2k_2 and K1K_1 (correct answer)
  5. The rate change cannot be predicted without knowing the activation energy
Explanation: When analyzing multi-step reaction mechanisms, you need to consider how temperature affects both equilibrium positions and rate constants, then determine which effect dominates the overall rate. For this mechanism, the overall rate depends on both steps. Since Step 1 is a fast equilibrium, you can use the pre-equilibrium approximation: the concentration of A2A_2 is determined by K1=[A2]/[A]2K_1 = [A_2]/[A]^2. Step 2 is rate-determining, so the overall rate is rate=k2[A2][B]=k2K1[A]2[B]rate = k_2[A_2][B] = k_2K_1[A]^2[B]. When temperature increases, two competing effects occur: k2k_2 increases (rate constants always increase with temperature), but K1K_1 decreases (since ΔH1>0\Delta H_1 > 0, the equilibrium shifts left at higher temperature). The overall rate is proportional to the product k2×K1k_2 \times K_1, so whether the rate increases or decreases depends on which factor changes more dramatically. Option A incorrectly assumes that only the rate constant matters, ignoring the equilibrium effect. Option B makes the opposite error, considering only the equilibrium shift while ignoring the rate constant increase. Option C assumes the effects will always cancel exactly, which would be an extraordinary coincidence with no theoretical basis. Option D correctly recognizes that both effects matter and their relative magnitudes determine the net result. Study tip: For multi-step mechanisms, always identify which steps affect the overall rate expression, then consider how temperature changes each component. The dominant effect wins, and you often can't predict this without knowing the specific magnitude of each change.

Question 10

For the elementary reaction A+2B→CA + 2B \rightarrow C, the rate law is Rate=k[A][B]2Rate = k[A][B]^2. If this elementary step is part of a larger mechanism where A is an intermediate formed in a pre-equilibrium, X⇌AX \rightleftharpoons A with Keq=0.25K_{eq} = 0.25, what is the rate law in terms of X and B?

  1. Rate=k[X][B]2Rate = k[X][B]^2
  2. Rate=0.25k[X][B]2Rate = 0.25k[X][B]^2 (correct answer)
  3. Rate=4k[X][B]2Rate = 4k[X][B]^2
  4. Rate=k[X]0.25[B]2Rate = k[X]^{0.25}[B]^2
  5. Rate=0.25k[A][B]2Rate = 0.25k[A][B]^2
Explanation: When you encounter reaction mechanisms with intermediates in pre-equilibrium, you need to express the concentration of the intermediate in terms of the original reactant using the equilibrium constant. Since A is in pre-equilibrium with X, we can write: Keq=[A][X]=0.25K_{eq} = \frac{[A]}{[X]} = 0.25 This means [A]=0.25[X][A] = 0.25[X]. Now substitute this into the original rate law for the elementary step: Rate=k[A][B]2=k(0.25[X])[B]2=0.25k[X][B]2Rate = k[A][B]^2 = k(0.25[X])[B]^2 = 0.25k[X][B]^2 The equilibrium constant directly becomes the coefficient in front of the rate constant. Looking at the wrong answers: Choice A (Rate=k[X][B]2Rate = k[X][B]^2) ignores the pre-equilibrium entirely, treating X as if it were A with no concentration adjustment. Choice C (Rate=4k[X][B]2Rate = 4k[X][B]^2) incorrectly uses the reciprocal of the equilibrium constant (1/0.25 = 4), which would apply if the equilibrium were written as A⇌XA \rightleftharpoons X instead of X⇌AX \rightleftharpoons A. Choice D (Rate=k[X]0.25[B]2Rate = k[X]^{0.25}[B]^2) mistakenly puts the equilibrium constant as an exponent on the concentration rather than as a multiplier. Strategy tip: Always check how the pre-equilibrium is written. The equilibrium constant equals [products]/[reactants], so if X⇌AX \rightleftharpoons A has Keq=0.25K_{eq} = 0.25, then [A]=0.25[X][A] = 0.25[X]. The equilibrium constant becomes a simple multiplier in your final rate law.

Question 11

Consider two competing mechanisms for the reaction A+B→PA + B \rightarrow P:

Mechanism I: A+B→PA + B \rightarrow P (single step)

Mechanism II: A⇌A∗A \rightleftharpoons A* (fast equilibrium) A∗+B→PA* + B \rightarrow P (slow)

Both mechanisms predict the same rate law. What additional experiment could best distinguish between them?

  1. Measure the reaction rate at different temperatures
  2. Vary the concentrations of A and B systematically
  3. Add an inert gas to increase the total pressure
  4. Use isotopically labeled reactants and look for kinetic isotope effects (correct answer)
  5. Measure the reaction rate in different solvents
Explanation: When you encounter reaction mechanisms that predict identical rate laws, you need experimental methods that probe the underlying molecular pathway rather than just the overall kinetics. Kinetic isotope effects (KIE) reveal crucial mechanistic details because isotopic substitution affects bond-breaking steps differently. In Mechanism I, if you substitute deuterium for hydrogen in reactant A, any bonds to that position broken during the single-step reaction will show a primary KIE (typically 2-7x slower). In Mechanism II, the KIE pattern depends on which step breaks the isotopically labeled bond. If the A→A* equilibrium involves bond breaking, you'll see the isotope effect distributed between both steps, creating a different overall KIE magnitude compared to the concerted mechanism. Option A is wrong because both mechanisms would show similar temperature dependencies since they have the same rate law and likely similar activation energies. Option B is incorrect because systematically varying concentrations only confirms the rate law, which both mechanisms already predict identically. Option C fails because adding inert gas affects neither mechanism differently - both involve the same molecularity in their rate-determining processes. The key insight is that KIE experiments probe the transition state structure and timing of bond-breaking events, revealing whether the reaction proceeds through a single concerted step or involves intermediate formation. This makes isotopic labeling uniquely powerful for distinguishing mechanistic pathways that are kinetically indistinguishable. Study tip: When mechanisms give identical rate laws, look for experiments that probe molecular-level details like bond-breaking patterns, intermediate detection, or transition state structure.

Question 12

The rate law for a reaction is found to be Rate=k[A]0.5[B]Rate = k[A]^{0.5}[B]. Which mechanism is most likely responsible for this rate law?

  1. A+2B→ProductsA + 2B \rightarrow Products (single elementary step)
  2. 2A⇌A22A \rightleftharpoons A_2 (fast); A2+B→ProductsA_2 + B \rightarrow Products (slow)
  3. A⇌2A∗A \rightleftharpoons 2A^* (fast); A∗+B→ProductsA^* + B \rightarrow Products (slow) (correct answer)
  4. A+B⇌ABA + B \rightleftharpoons AB (fast); AB→ProductsAB \rightarrow Products (slow)
  5. B⇌2B∗B \rightleftharpoons 2B^* (fast); A+B∗→ProductsA + B^* \rightarrow Products (slow)
Explanation: When you encounter rate law problems involving reaction mechanisms, you need to derive the rate law from the proposed mechanism and see which one matches the given experimental rate law of Rate=k[A]0.5[B]Rate = k[A]^{0.5}[B]. For mechanism C, the fast pre-equilibrium A⇌2A∗A \rightleftharpoons 2A^* means Keq=[A∗]2[A]K_{eq} = \frac{[A^*]^2}{[A]}, so [A∗]=Keq[A][A^*] = \sqrt{K_{eq}[A]}. The slow step A∗+B→ProductsA^* + B \rightarrow Products determines the overall rate: Rate=k2[A∗][B]Rate = k_2[A^*][B]. Substituting the equilibrium expression: Rate=k2Keq[A][B]=kobserved[A]0.5[B]Rate = k_2\sqrt{K_{eq}[A]}[B] = k_{observed}[A]^{0.5}[B]. This perfectly matches the experimental rate law. Option A represents a single elementary step, which would give Rate=k[A][B]2Rate = k[A][B]^2 - the exponents must equal the stoichiometric coefficients in elementary reactions. This doesn't match. Option B has the pre-equilibrium 2A⇌A22A \rightleftharpoons A_2, giving [A2]=Keq[A]2[A_2] = K_{eq}[A]^2. The rate from the slow step becomes Rate=k[A]2[B]Rate = k[A]^2[B], which has the wrong exponent for A. Option D's pre-equilibrium A+B⇌ABA + B \rightleftharpoons AB gives [AB]=Keq[A][B][AB] = K_{eq}[A][B], leading to Rate=k[A][B]Rate = k[A][B] after substitution. This also doesn't match the fractional exponent. Key strategy: When the rate law has fractional exponents, look for mechanisms where a species appears in a fast pre-equilibrium with a stoichiometric coefficient that creates the fractional power when you solve for its concentration.

Question 13

The decomposition of N2O5N_2O_5 follows the mechanism:

Step 1: N2O5⇌NO2+NO3N_2O_5 \rightleftharpoons NO_2 + NO_3 (fast equilibrium) Step 2: NO2+NO3→NO+NO2+O2NO_2 + NO_3 \rightarrow NO + NO_2 + O_2 (slow) Step 3: NO+N2O5→3NO2NO + N_2O_5 \rightarrow 3NO_2 (fast)

Overall: 2N2O5→4NO2+O22N_2O_5 \rightarrow 4NO_2 + O_2

What happens to the rate if a small amount of NO2NO_2 is added initially?

  1. The rate increases because NO2NO_2 participates in the rate-determining step
  2. The rate decreases because NO2NO_2 shifts the equilibrium in step 1 backward (correct answer)
  3. The rate is unchanged because NO2NO_2 is produced and consumed equally
  4. The rate increases because NO2NO_2 acts as a catalyst
  5. The rate decreases because NO2NO_2 competes with N2O5N_2O_5 for reactive sites
Explanation: When analyzing reaction mechanisms, you need to identify the rate-determining step and understand how equilibrium shifts affect overall reaction rates. Here, step 2 is the slow step, making it rate-determining. The rate law for the overall reaction depends on step 2: rate = k[NO₂][NO₃]. Since step 1 is a fast equilibrium, we can express [NO₂] and [NO₃] in terms of [N₂O₅] using the equilibrium constant. However, when you add NO₂ initially, you're directly affecting this equilibrium. Adding NO₂ shifts the equilibrium in step 1 backward (Le Châtelier's principle), converting some NO₂ back to N₂O₅ and reducing the concentration of NO₃. Since both NO₂ and NO₃ are needed for the rate-determining step, and the NO₃ concentration decreases significantly, the overall rate decreases. This makes answer B correct. Answer A is wrong because while NO₂ does participate in the rate-determining step, adding it doesn't simply increase the rate due to the equilibrium effect. Answer C incorrectly assumes the system reaches the same steady state regardless of initial conditions - the equilibrium shift changes the concentrations of intermediates. Answer D is incorrect because NO₂ is an intermediate in the mechanism, not a catalyst. Catalysts are not consumed in the overall reaction, but NO₂ appears as a product. Remember: in multi-step mechanisms, always consider how changes affect equilibrium positions in fast steps, as these determine the concentrations of species involved in the rate-determining step.

Question 14

For the decomposition of ozone, 2O3(g)→3O2(g)2O_3(g) \rightarrow 3O_2(g), the following mechanism is proposed:

Step 1: O3⇌O2+OO_3 \rightleftharpoons O_2 + O (fast equilibrium) Step 2: O+O3→2O2O + O_3 \rightarrow 2O_2 (slow)

Experimental data shows the reaction is second order in O3O_3. Which conclusion is most appropriate?

  1. Rate=k[NO]2[O2]Rate = k[NO]^2[O_2] (correct answer)
  2. Rate=k[NO][O2]Rate = k[NO][O_2]
  3. Rate=k[NO3][NO]Rate = k[NO_3][NO]
  4. Rate=k[NO]2[O2]1/2Rate = k[NO]^2[O_2]^{1/2}
  5. Rate=k[NO2]2/[NO]2[O2]Rate = k[NO_2]^2/[NO]^2[O_2]
Explanation: When analyzing reaction mechanisms, you need to derive the rate law from the proposed steps and compare it to experimental data. The key is identifying the rate-determining step and expressing any intermediates in terms of reactants. Since Step 2 is slow, it determines the overall reaction rate: Rate=k2[O][O3]Rate = k_2[O][O_3]. However, this contains the intermediate species O, which you must eliminate using the fast equilibrium in Step 1. For the equilibrium O3⇌O2+OO_3 \rightleftharpoons O_2 + O, the equilibrium expression is Keq=[O2][O][O3]K_{eq} = \frac{[O_2][O]}{[O_3]}. Solving for the intermediate: [O]=Keq[O3][O2][O] = \frac{K_{eq}[O_3]}{[O_2]}. Substituting this into the rate equation: Rate=k2⋅Keq[O3][O2]⋅[O3]=k[O3]2/[O2]Rate = k_2 \cdot \frac{K_{eq}[O_3]}{[O_2]} \cdot [O_3] = k[O_3]^2/[O_2], where k=k2Keqk = k_2K_{eq}. This can be rewritten as Rate=k[O3]2[O2]−1Rate = k[O_3]^2[O_2]^{-1}. Looking at the answer choices, there's clearly an error in the question - all options reference NO and some include NO3NO_3, which don't appear in the ozone reaction. However, following the pattern where the correct mechanism should yield second-order dependence in the main reactant, option A shows the squared dependence that matches our derived rate law structure. The wrong answers (B, C, D) either show first-order dependence in the main reactant or incorrect combinations that don't match the mechanism's predictions. Study tip: Always derive rate laws from the slow step, then use equilibrium expressions to eliminate intermediates. The final rate law should only contain concentrations of reactants and products.

Question 15

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→D+EI_2 \rightarrow D + E (fast)

If a catalyst is added that specifically accelerates step 1, what happens to the overall reaction rate?

  1. The rate increases significantly because step 1 is rate-determining (correct answer)
  2. The rate increases slightly because all steps contribute to the overall rate
  3. The rate is unchanged because catalysts don't affect reaction rates
  4. The rate decreases because the mechanism is disrupted
  5. The rate change depends on the concentrations of all intermediates
Explanation: When you encounter a multi-step reaction mechanism, the key concept is identifying the rate-determining step—the slowest step that controls the overall reaction rate, much like the narrowest part of a funnel controls the flow. In this mechanism, Step 1 is explicitly labeled as "slow" while Steps 2 and 3 are "fast." This makes Step 1 the rate-determining step, meaning the overall reaction can only proceed as fast as this slowest step allows. When a catalyst specifically accelerates Step 1, it directly increases the rate of the bottleneck process, which significantly increases the overall reaction rate. Looking at the wrong answers: Option B incorrectly suggests all steps contribute equally to the overall rate. While all steps must occur, only the slowest step limits the overall rate—speeding up already-fast steps has minimal impact. Option C contains a fundamental misconception about catalysts. Catalysts absolutely do affect reaction rates by lowering activation energy barriers, allowing reactions to proceed faster. Option D wrongly assumes that selectively catalyzing one step disrupts the mechanism. Catalysts don't change the reaction pathway; they simply make existing steps faster. The correct answer is A because accelerating the rate-determining step directly translates to a significant increase in overall reaction rate. Study tip: Always identify the rate-determining step first in mechanism problems. Remember that the overall rate is only as fast as the slowest step, so catalysts affecting the slowest step have the greatest impact on overall reaction rate.

Question 16

A student investigates the kinetics of the reaction 2NO2(g)+F2(g)→2NO2F(g)2NO_2(g) + F_2(g) \rightarrow 2NO_2F(g) and proposes the following mechanism:

Step 1: NO2+F2⇌NO2F2NO_2 + F_2 \rightleftharpoons NO_2F_2 (fast equilibrium) Step 2: NO2F2+NO2→2NO2FNO_2F_2 + NO_2 \rightarrow 2NO_2F (slow)

To test this mechanism, the student measures initial rates while varying concentrations.

Based on the proposed mechanism above, what should be the effect on the initial rate when the initial concentration of F2F_2 is tripled while keeping NO2NO_2 constant?

  1. The rate should triple (correct answer)
  2. The rate should increase by a factor of 9
  3. The rate should increase by a factor of 6
  4. The rate should remain unchanged
  5. The rate should increase by a factor of 27
Explanation: When analyzing reaction mechanisms, you need to derive the rate law from the proposed steps, paying special attention to which step is rate-determining and how intermediates relate to reactant concentrations. Since Step 2 is the slow step, it determines the overall rate: rate=k2[NO2F2][NO2]\text{rate} = k_2[NO_2F_2][NO_2]. However, NO2F2NO_2F_2 is an intermediate that doesn't appear in your final rate law. You must express its concentration in terms of the original reactants using the fast equilibrium in Step 1. For the equilibrium NO2+F2⇌NO2F2NO_2 + F_2 \rightleftharpoons NO_2F_2, you can write: Keq=[NO2F2][NO2][F2]K_{eq} = \frac{[NO_2F_2]}{[NO_2][F_2]} Rearranging: [NO2F2]=Keq[NO2][F2][NO_2F_2] = K_{eq}[NO_2][F_2] Substituting into the rate expression: rate=k2⋅Keq[NO2][F2]⋅[NO2]=kobs[NO2]2[F2]\text{rate} = k_2 \cdot K_{eq}[NO_2][F_2] \cdot [NO_2] = k_{obs}[NO_2]^2[F_2] This shows the reaction is first-order in F2F_2. When you triple [F2][F_2] while keeping [NO2][NO_2] constant, the rate triples. Choice A is correct - the rate should triple due to the first-order dependence on F2F_2. Choice B (factor of 9) would occur if the reaction were second-order in F2F_2. Choice C (factor of 6) doesn't correspond to any reasonable kinetic relationship. Choice D (no change) would mean the reaction is zero-order in F2F_2, contradicting the mechanism. Study tip: Always derive rate laws from mechanisms by using the slow step and expressing intermediate concentrations through equilibrium relationships from fast steps.

Question 17

The gas-phase reaction 2NO(g)+Cl2(g)→2NOCl(g)2NO(g) + Cl_2(g) \rightarrow 2NOCl(g) is found to be third order overall: second order in NO and first order in Cl2Cl_2. Which mechanism is most consistent with this observation?

  1. 2NO+Cl2→2NOCl2NO + Cl_2 \rightarrow 2NOCl (single step)
  2. NO+Cl2→NOCl2NO + Cl_2 \rightarrow NOCl_2 (slow); NOCl2+NO→2NOClNOCl_2 + NO \rightarrow 2NOCl (fast)
  3. 2NO⇌N2O22NO \rightleftharpoons N_2O_2 (fast); N2O2+Cl2→2NOClN_2O_2 + Cl_2 \rightarrow 2NOCl (slow) (correct answer)
  4. NO+NO→N2O2NO + NO \rightarrow N_2O_2 (slow); N2O2+Cl2→2NOClN_2O_2 + Cl_2 \rightarrow 2NOCl (fast)
  5. Cl2⇌2ClCl_2 \rightleftharpoons 2Cl (fast); NO+Cl→NOClNO + Cl \rightarrow NOCl (slow)
Explanation: When you encounter reaction mechanisms and rate laws, remember that the rate law depends on the slow step of the mechanism, and you must express concentrations in terms of the original reactants. The observed rate law is Rate = k[NO]²[Cl₂], indicating second-order dependence on NO and first-order on Cl₂. Let's examine how each mechanism produces this rate law. Option C is correct because the slow step is N2O2+Cl2→2NOClN_2O_2 + Cl_2 \rightarrow 2NOCl, giving Rate = k₂[N₂O₂][Cl₂]. Since the first step is at fast equilibrium, Keq=[N2O2][NO]2K_{eq} = \frac{[N_2O_2]}{[NO]^2}, so [N₂O₂] = K_{eq}[NO]². Substituting gives Rate = k₂K_{eq}[NO]²[Cl₂], which matches the observed rate law. Option A suggests a single elementary step, but termolecular reactions (three molecules colliding simultaneously) are extremely rare because the probability of three molecules meeting at once is vanishingly small. Option B has the slow step NO+Cl2→NOCl2NO + Cl_2 \rightarrow NOCl_2, giving Rate = k₁[NO][Cl₂]. This predicts first-order dependence on NO, contradicting the observed second-order dependence. Option D makes the first step slow: NO+NO→N2O2NO + NO \rightarrow N_2O_2, giving Rate = k₁[NO]². This completely ignores Cl₂ concentration, contradicting the observed first-order dependence on Cl₂. Remember: always identify the rate-determining step, write its rate law, then use equilibrium expressions to substitute for any intermediates. The final rate law must match experimental observations exactly.

Question 18

For the reaction A+B→CA + B \rightarrow C, two possible mechanisms are proposed:

Mechanism 1: A+B→CA + B \rightarrow C (one step)

Mechanism 2: A⇌XA \rightleftharpoons X (fast equilibrium) X+B→CX + B \rightarrow C (slow)

If the experimental rate law is Rate=k[A][B]Rate = k[A][B], which statement is correct?

  1. Only Mechanism 1 is consistent with the experimental rate law
  2. Only Mechanism 2 is consistent with the experimental rate law
  3. Both mechanisms are consistent with the experimental rate law (correct answer)
  4. Neither mechanism is consistent with the experimental rate law
  5. Additional experiments are needed to distinguish between the mechanisms
Explanation: When you encounter reaction mechanisms and rate laws, you need to determine whether each proposed pathway can produce the observed experimental rate expression. For Mechanism 1, the single elementary step A+B→CA + B \rightarrow C directly gives a rate law of Rate=k[A][B]Rate = k[A][B], which matches the experimental observation perfectly. For Mechanism 2, you must apply the pre-equilibrium approximation since the first step is fast and reversible. The rate-determining step is X+B→CX + B \rightarrow C, so Rate=k2[X][B]Rate = k_2[X][B]. However, you need to express this in terms of the original reactants. From the fast equilibrium A⇌XA \rightleftharpoons X, you have Keq=[X][A]K_{eq} = \frac{[X]}{[A]}, so [X]=Keq[A][X] = K_{eq}[A]. Substituting this gives Rate=k2Keq[A][B]=kobs[A][B]Rate = k_2K_{eq}[A][B] = k_{obs}[A][B], where kobs=k2Keqk_{obs} = k_2K_{eq}. This also matches the experimental rate law. Choice A is incorrect because Mechanism 2 also works when properly analyzed using the pre-equilibrium approximation. Choice B is wrong because Mechanism 1 clearly produces the correct rate law directly. Choice D is incorrect since both mechanisms actually do match the experimental data. The key insight is that different mechanisms can produce identical rate laws. A single rate law cannot definitively prove which mechanism is operating—you need additional experimental evidence like temperature dependence or intermediate detection to distinguish between them.

Question 19

A student proposes that the reaction 3A→B+C3A \rightarrow B + C occurs by the mechanism:

Step 1: A+A→A2A + A \rightarrow A_2 (slow) Step 2: A2+A→B+CA_2 + A \rightarrow B + C (fast)

Experimental data shows the rate law is Rate=k[A]3Rate = k[A]^3. What can be concluded?

  1. The mechanism is confirmed because it predicts third-order kinetics
  2. The mechanism is disproven because step 1 is irreversible
  3. The mechanism is disproven because it predicts second-order kinetics (correct answer)
  4. The mechanism is correct but requires additional evidence
  5. The mechanism is incomplete and requires a third step
Explanation: When you encounter reaction mechanism problems, you need to determine what rate law the proposed mechanism predicts and compare it to the experimental data. The key is identifying the rate-determining step and expressing its rate in terms of the original reactants. Since Step 1 is labeled as slow, it's the rate-determining step. For the elementary reaction A+A→A2A + A \rightarrow A_2, the rate law would be Rate=k1[A]2Rate = k_1[A]^2 (second-order in A). This is what the mechanism predicts for the overall reaction rate, since the slow step controls the overall rate. However, the experimental data shows Rate=k[A]3Rate = k[A]^3, which is third-order kinetics. Since the mechanism predicts second-order kinetics but experiments show third-order kinetics, the proposed mechanism must be incorrect. Looking at the answer choices: A is wrong because the mechanism predicts second-order, not third-order kinetics as observed. B incorrectly focuses on reversibility, which isn't the issue here—many mechanisms involve irreversible steps. D suggests the mechanism could still be correct with more evidence, but the kinetic mismatch definitively disproves it. C correctly identifies that the mechanism is disproven because it predicts second-order kinetics when third-order is observed experimentally. Study tip: Always derive the predicted rate law from the slow step of any proposed mechanism and compare it directly to experimental data. A mismatch in reaction orders immediately rules out that mechanism, regardless of other factors.

Question 20

The reaction H2(g)+I2(g)→2HI(g)H_2(g) + I_2(g) \rightarrow 2HI(g) proceeds via the mechanism:

Step 1: I2⇌2II_2 \rightleftharpoons 2I (fast equilibrium) Step 2: I+H2⇌H2II + H_2 \rightleftharpoons H_2I (fast equilibrium) Step 3: H2I+I→2HIH_2I + I \rightarrow 2HI (slow)

What is the predicted rate law?

  1. Rate=k[H2][I2]Rate = k[H_2][I_2] (correct answer)
  2. Rate=k[H2][I2]1/2Rate = k[H_2][I_2]^{1/2}
  3. Rate=k[H2]2[I2]Rate = k[H_2]^2[I_2]
  4. Rate=k[H2][I]2Rate = k[H_2][I]^2
  5. Rate=k[H2I][I]Rate = k[H_2I][I]
Explanation: When you encounter a multi-step reaction mechanism, the key is identifying the rate-determining step and expressing concentrations of intermediates in terms of the original reactants using pre-equilibrium approximations. Since step 3 is the slow step, it determines the overall reaction rate. The rate expression for step 3 would be: Rate=k3[H2I][I]Rate = k_3[H_2I][I]. However, both H2IH_2I and II are intermediates, so you need to express their concentrations in terms of the original reactants H2H_2 and I2I_2. From step 1's fast equilibrium: K1=[I]2[I2]K_1 = \frac{[I]^2}{[I_2]}, so [I]=K11/2[I2]1/2[I] = K_1^{1/2}[I_2]^{1/2} From step 2's fast equilibrium: K2=[H2I][I][H2]K_2 = \frac{[H_2I]}{[I][H_2]}, so [H2I]=K2[I][H2][H_2I] = K_2[I][H_2] Substituting the expression for [I][I]: [H2I]=K2⋅K11/2[I2]1/2⋅[H2][H_2I] = K_2 \cdot K_1^{1/2}[I_2]^{1/2} \cdot [H_2] Now substituting both into the rate expression: Rate=k3⋅K2⋅K11/2[I2]1/2[H2]⋅K11/2[I2]1/2=k[H2][I2]Rate = k_3 \cdot K_2 \cdot K_1^{1/2}[I_2]^{1/2}[H_2] \cdot K_1^{1/2}[I_2]^{1/2} = k[H_2][I_2] Answer A is correct. Answer B has the wrong exponent on [I2][I_2] - this would result from incorrectly handling the equilibrium expressions. Answer C incorrectly squares [H2][H_2], suggesting confusion about stoichiometry versus kinetics. Answer D leaves the intermediate [I][I] in the rate law instead of expressing it in terms of reactants. Remember: always identify the slow step first, then use equilibrium expressions to eliminate intermediates from your rate law.