College Chemistry Quiz: Elementary Reactions
20 questions · exam conditions
0:00
Elementary ReactionsQuestion 1 of 20

The elementary reaction CH3(g)+H2(g)→CH4(g)+H(g)CH_3(g) + H_2(g) \rightarrow CH_4(g) + H(g) has a rate constant of 6.0×103 M−1s−16.0 \times 10^3 \, M^{-1}s^{-1} at 1000 K. Calculate the half-life of CH3CH_3 radicals when [H2]=2.0×10−3 M[H_2] = 2.0 \times 10^{-3} \, M and [CH3]0=1.0×10−5 M[CH_3]_0 = 1.0 \times 10^{-5} \, M.

2.8×10−5 s2.8 \times 10^{-5} \, s
4.2×10−5 s4.2 \times 10^{-5} \, s
5.8×10−5 s5.8 \times 10^{-5} \, s
8.3×10−5 s8.3 \times 10^{-5} \, s
1.2×10−4 s1.2 \times 10^{-4} \, s
← Back to quizzes

College Chemistry Quiz

College Chemistry Quiz: Elementary Reactions

Practice Elementary Reactions 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 Elementary Reactions, giving you a quick way to practice the rules, question types, and explanations that matter most for College Chemistry.

How to use this quiz

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.

All questions

Question 1

The elementary reaction CH3(g)+H2(g)→CH4(g)+H(g)CH_3(g) + H_2(g) \rightarrow CH_4(g) + H(g) has a rate constant of 6.0×103 M−1s−16.0 \times 10^3 \, M^{-1}s^{-1} at 1000 K. Calculate the half-life of CH3CH_3 radicals when [H2]=2.0×10−3 M[H_2] = 2.0 \times 10^{-3} \, M and [CH3]0=1.0×10−5 M[CH_3]_0 = 1.0 \times 10^{-5} \, M.

  1. 2.8×10−5 s2.8 \times 10^{-5} \, s
  2. 4.2×10−5 s4.2 \times 10^{-5} \, s
  3. 5.8×10−5 s5.8 \times 10^{-5} \, s (correct answer)
  4. 8.3×10−5 s8.3 \times 10^{-5} \, s
  5. 1.2×10−4 s1.2 \times 10^{-4} \, s
Explanation: When you encounter kinetics problems involving elementary reactions with two reactants, you need to determine the reaction order and apply the appropriate integrated rate law. This reaction is second-order overall (first-order in both CH3CH_3 and H2H_2). Since [H2]=2.0×10−3 M[H_2] = 2.0 \times 10^{-3} \, M is much larger than [CH3]0=1.0×10−5 M[CH_3]_0 = 1.0 \times 10^{-5} \, M, you can use the pseudo-first-order approximation. The hydrogen concentration remains essentially constant throughout the reaction, so the rate law becomes: rate=k′[CH3]rate = k'[CH_3] where k′=k[H2]=(6.0×103)(2.0×10−3)=12 s−1k' = k[H_2] = (6.0 \times 10^3)(2.0 \times 10^{-3}) = 12 \, s^{-1}. For a first-order reaction, the half-life is t1/2=0.693k′=0.69312=5.8×10−5 st_{1/2} = \frac{0.693}{k'} = \frac{0.693}{12} = 5.8 \times 10^{-5} \, s. This confirms answer C is correct. Answer A (2.8×10−5 s2.8 \times 10^{-5} \, s) results from incorrectly doubling the pseudo-first-order rate constant before calculating the half-life. Answer B (4.2×10−5 s4.2 \times 10^{-5} \, s) comes from using ln⁡(2)/k′\ln(2)/k' but with an error in the rate constant calculation. Answer D (8.3×10−5 s8.3 \times 10^{-5} \, s) occurs when students mistakenly use the second-order half-life formula t1/2=1k′[CH3]0t_{1/2} = \frac{1}{k'[CH_3]_0} instead of recognizing this as pseudo-first-order. Remember: when one reactant is in large excess, treat the reaction as pseudo-first-order in the limiting reactant. The excess concentration becomes part of the effective rate constant, simplifying your calculation significantly.

Question 2

The termolecular elementary reaction 2NO(g)+Br2(g)→2NOBr(g)2NO(g) + Br_2(g) \rightarrow 2NOBr(g) has been proposed to occur in a single step. Which of the following best explains why termolecular elementary reactions are relatively rare?

  1. Three-body collisions require extremely high temperatures to occur
  2. The probability of three molecules colliding simultaneously with proper orientation is very low (correct answer)
  3. Termolecular reactions always have higher activation energies than bimolecular reactions
  4. Three molecules cannot achieve the correct geometry for bond formation
  5. Termolecular reactions violate the principle of microscopic reversibility
Explanation: When you encounter questions about elementary reaction mechanisms, focus on the molecular-level requirements for reactions to occur. Elementary reactions happen in a single step, meaning all reactants must come together simultaneously. The rarity of termolecular elementary reactions stems from basic probability and collision theory. For this reaction to occur as written, two NO molecules and one Br₂ molecule must collide at exactly the same time with sufficient energy and proper orientation. This is an extraordinarily unlikely event. Think about it: even finding two molecules colliding with the right conditions is challenging enough, but requiring three specific molecules to meet simultaneously in the correct spatial arrangement is statistically improbable. As the number of required collision partners increases, the probability drops exponentially. Choice A is incorrect because while temperature affects reaction rates, termolecular reactions aren't specifically limited by temperature requirements that differ fundamentally from other reactions. Choice C misrepresents the relationship between molecularity and activation energy—these are independent properties, and termolecular reactions don't inherently have higher activation energies. Choice D is wrong because three molecules can certainly achieve proper geometry for bond formation; the issue isn't geometric impossibility but rather the extremely low probability of achieving that geometry simultaneously. Remember this key principle: reaction molecularity directly relates to collision probability. Most reactions you'll encounter are bimolecular elementary steps because two-body collisions are much more probable than three-body collisions. When you see proposed termolecular mechanisms, immediately think about collision probability as the limiting factor.

Question 3

The elementary reaction 2NO(g)+Cl2(g)→2NOCl(g)2NO(g) + Cl_2(g) \rightarrow 2NOCl(g) has a rate law of rate=k[NO]2[Cl2]rate = k[NO]^2[Cl_2]. If the concentration of NO is doubled while keeping the concentration of Cl2Cl_2 constant, by what factor will the initial rate of reaction increase?

  1. 2
  2. 4 (correct answer)
  3. 6
  4. 8
  5. 16
Explanation: When you encounter rate law problems, you're working with how concentration changes affect reaction rates. The rate law equation tells you exactly how each reactant's concentration influences the overall rate. Given the rate law rate=k[NO]2[Cl2]rate = k[NO]^2[Cl_2], notice that NO has an exponent of 2, meaning the rate depends on the square of NO's concentration. When you double the concentration of NO while keeping Cl2Cl_2 constant, you need to calculate how this affects the rate. Let's compare the initial rate to the new rate. If the initial concentration of NO is [NO]0[NO]_0, then doubling it gives 2[NO]02[NO]_0. The new rate becomes: ratenew=k(2[NO]0)2[Cl2]=k⋅4[NO]02[Cl2]=4⋅k[NO]02[Cl2]rate_{new} = k(2[NO]_0)^2[Cl_2] = k \cdot 4[NO]_0^2[Cl_2] = 4 \cdot k[NO]_0^2[Cl_2] This shows the rate increases by a factor of 4, making B correct. Looking at the wrong answers: A) suggests a factor of 2, which would be correct if NO had an exponent of 1 in the rate law, but it's squared. C) gives 6, which doesn't correspond to any mathematical relationship with the given exponents. D) suggests 8, which you might get if you incorrectly thought both NO concentrations were cubed (23=82^3 = 8), but the exponent is 2, not 3. Remember this key strategy: when analyzing rate law problems, always pay careful attention to the exponents. The factor by which the rate changes equals the factor by which you change the concentration, raised to the power of that reactant's exponent in the rate law.

Question 4

In a proposed reaction mechanism, Step 1 is A+B→C+DA + B \rightarrow C + D (elementary) and Step 2 is C+E→F+GC + E \rightarrow F + G (elementary). If Step 1 is the rate-determining step, what is the overall rate law for the mechanism?

  1. rate=k[A][B][C][E]rate = k[A][B][C][E]
  2. rate=k[A][B]rate = k[A][B] (correct answer)
  3. rate=k[C][E]rate = k[C][E]
  4. rate=k[A][B][E]rate = k[A][B][E]
  5. rate=k[A]2[B][E]rate = k[A]^2[B][E]
Explanation: When you encounter reaction mechanisms with multiple steps, the key insight is that the rate-determining step controls the overall reaction rate, just like the slowest person in a group determines how fast the group walks. Since Step 1 (A+B→C+DA + B \rightarrow C + D) is the rate-determining step and is elementary, you can write its rate law directly from its stoichiometry. For elementary reactions, the rate law always matches the molecularity: rate=k[A][B]rate = k[A][B]. This becomes the overall rate law because Step 1 is the bottleneck that limits how fast the entire mechanism proceeds. Looking at the wrong answers: Choice A (rate=k[A][B][C][E]rate = k[A][B][C][E]) incorrectly combines both steps, but you can't simply multiply rate laws from different steps. Choice C (rate=k[C][E]rate = k[C][E]) uses only Step 2, but since Step 2 is faster than Step 1, it doesn't control the overall rate. The rate law should reflect what controls the rate, not what happens in faster subsequent steps. Choice D (rate=k[A][B][E]rate = k[A][B][E]) incorrectly includes [E] from Step 2, but E doesn't participate in the rate-determining step, so it doesn't appear in the rate law. The correct answer is B: rate=k[A][B]rate = k[A][B]. Remember this pattern: when given a multi-step mechanism, identify the rate-determining step first, then write the rate law based solely on that step's stoichiometry (for elementary reactions). Species that don't participate in the slowest step won't appear in the overall rate law.

Question 5

Which of the following statements correctly distinguishes elementary reactions from overall reactions in a multi-step mechanism?

  1. Elementary reactions always have integer stoichiometric coefficients, while overall reactions may have fractional coefficients
  2. Elementary reactions occur in a single molecular event, while overall reactions represent the net result of multiple elementary steps (correct answer)
  3. Elementary reactions are always faster than overall reactions in the same mechanism
  4. Elementary reactions have rate laws that must be determined experimentally, while overall reactions have predictable rate laws
  5. Elementary reactions involve only gas-phase species, while overall reactions can involve any phase
Explanation: When you encounter questions about reaction mechanisms, focus on the fundamental difference between what happens at the molecular level versus what we observe macroscopically. Elementary reactions represent single molecular events - one step where specific molecules collide and transform in a single act. Think of it as a snapshot of molecules actually interacting. Overall reactions, however, are the sum total of all elementary steps in a mechanism, showing only the net change from reactants to products without revealing the pathway. Option B correctly captures this distinction. An elementary reaction is literally one molecular event (like A + B → C happening in one collision), while an overall reaction represents the net result when you add up all the elementary steps and cancel out intermediates. Option A is incorrect because both elementary and overall reactions typically have integer stoichiometric coefficients. Fractional coefficients sometimes appear in overall reactions when mechanisms are written in specific ways, but this isn't the defining difference. Option C misrepresents kinetics - individual elementary steps can be fast or slow, and the overall reaction rate is determined by the slowest step (rate-determining step), not by some inherent speed relationship. Option D reverses the truth. Elementary reactions have predictable rate laws that directly reflect their stoichiometry (since they're single molecular events), while overall reaction rate laws must be determined experimentally because they depend on the mechanism's complexity. Remember: elementary = single molecular event; overall = net result of multiple steps. This distinction is crucial for understanding reaction mechanisms and kinetics.

Question 6

The elementary reaction NO2(g)+CO(g)→NO(g)+CO2(g)NO_2(g) + CO(g) \rightarrow NO(g) + CO_2(g) has an activation energy of 135 kJ/mol. If the temperature is increased from 500 K to 600 K, by approximately what factor will the rate constant increase? (Assume the pre-exponential factor A remains constant)

  1. 1.2
  2. 2.4
  3. 8.1 (correct answer)
  4. 12.3
  5. 24.6
Explanation: When you encounter a question asking how temperature changes affect reaction rates, you're dealing with the Arrhenius equation: k=Ae−Ea/RTk = Ae^{-E_a/RT}, where k is the rate constant, A is the pre-exponential factor, EaE_a is activation energy, R is the gas constant, and T is temperature. To find how the rate constant changes, you need the ratio k2k1=Ae−Ea/RT2Ae−Ea/RT1=e−Ea/R(1/T2−1/T1)\frac{k_2}{k_1} = \frac{Ae^{-E_a/RT_2}}{Ae^{-E_a/RT_1}} = e^{-E_a/R(1/T_2 - 1/T_1)} Substituting the values: Ea=135,000E_a = 135,000 J/mol, R=8.314R = 8.314 J/(mol·K), T1=500T_1 = 500 K, T2=600T_2 = 600 K: k2k1=e−135,000/8.314×(1/600−1/500)=e−16,240×(−0.000333)=e5.41≈8.1\frac{k_2}{k_1} = e^{-135,000/8.314 \times (1/600 - 1/500)} = e^{-16,240 \times (-0.000333)} = e^{5.41} ≈ 8.1 This confirms answer C is correct. A (1.2) represents a much smaller increase that you'd get with either a lower activation energy or smaller temperature change. B (2.4) might result from calculation errors, perhaps confusing the temperature ratio (600/500 = 1.2) with the rate constant ratio. D (12.3) could come from arithmetic mistakes in the exponential calculation or using incorrect units for activation energy. Study tip: Always convert activation energy to J/mol when using R = 8.314 J/(mol·K), and remember that even modest temperature increases can dramatically boost reaction rates when activation energies are high—this is the power of exponential relationships in kinetics.

Question 7

Consider the elementary termination step in a radical chain reaction: 2CH3⋅(g)→C2H6(g)2CH_3 \cdot (g) \rightarrow C_2H_6(g). Which of the following correctly describes the kinetic characteristics of this elementary reaction?

  1. Molecularity = 1, rate law = rate=k[CH3⋅]rate = k[CH_3 \cdot]
  2. Molecularity = 2, rate law = rate=k[CH3⋅]rate = k[CH_3 \cdot]
  3. Molecularity = 2, rate law = rate=k[CH3⋅]2rate = k[CH_3 \cdot]^2 (correct answer)
  4. Molecularity = 4, rate law = rate=k[CH3⋅]2rate = k[CH_3 \cdot]^2
  5. Molecularity = 1, rate law = rate=k[CH3⋅]2rate = k[CH_3 \cdot]^2
Explanation: When analyzing elementary reactions in kinetics, you need to understand two key concepts: molecularity and rate laws. Molecularity tells you how many molecules participate in the elementary step, while the rate law shows how the reaction rate depends on reactant concentrations. For the reaction 2CH3⋅(g)→C2H6(g)2CH_3 \cdot (g) \rightarrow C_2H_6(g), you can see that two methyl radicals collide to form ethane. Since two molecules participate in this elementary step, the molecularity equals 2. For elementary reactions (this is crucial), the rate law exponents always match the stoichiometric coefficients. Since the coefficient of CH3⋅CH_3 \cdot is 2, the rate law becomes rate=k[CH3⋅]2rate = k[CH_3 \cdot]^2. Answer A incorrectly states molecularity = 1, missing that two radicals must collide. It also gives a first-order rate law, which doesn't match the stoichiometry. Answer B correctly identifies molecularity = 2 but incorrectly gives a first-order rate law—this violates the fundamental rule that elementary reaction rate laws mirror stoichiometry. Answer D claims molecularity = 4, likely confusing the total number of atoms involved (4 hydrogen + 2 carbon) with the number of reacting molecules. The correct answer is C: molecularity = 2 (two molecules react) and rate law = rate=k[CH3⋅]2rate = k[CH_3 \cdot]^2 (exponent matches the stoichiometric coefficient). Remember: for elementary reactions, the rate law exponents always equal the stoichiometric coefficients. This direct relationship only applies to elementary steps, not overall reactions with multiple steps.

Question 8

In a two-step mechanism where both steps are elementary, Step 1: A+B⇌CA + B \rightleftharpoons C (fast equilibrium) and Step 2: C+D→EC + D \rightarrow E (slow). Using the pre-equilibrium approximation, what is the rate law for the overall reaction?

  1. rate=k2[C][D]rate = k_2[C][D]
  2. rate=k1k2[A][B][D]k−1rate = \frac{k_1k_2[A][B][D]}{k_{-1}} (correct answer)
  3. rate=k1[A][B]−k−1[C]rate = k_1[A][B] - k_{-1}[C]
  4. rate=k2[A][B][D]rate = k_2[A][B][D]
  5. rate=k1k2[A][B]k−1[D]rate = \frac{k_1k_2[A][B]}{k_{-1}[D]}
Explanation: When you encounter a two-step mechanism with a fast pre-equilibrium followed by a slow step, you need to apply the pre-equilibrium approximation to derive the overall rate law. This approach assumes the first step maintains equilibrium while the second step controls the overall reaction rate. Since Step 2 is the slow, rate-determining step, the overall reaction rate equals: rate=k2[C][D]rate = k_2[C][D]. However, this contains the intermediate C, which we must express in terms of the original reactants using the equilibrium condition from Step 1. For the fast equilibrium in Step 1, the forward and reverse rates are equal: k1[A][B]=k−1[C]k_1[A][B] = k_{-1}[C]. Solving for the intermediate concentration: [C]=k1[A][B]k−1[C] = \frac{k_1[A][B]}{k_{-1}}. Substituting this into the rate expression: rate=k2⋅k1[A][B]k−1⋅[D]=k1k2[A][B][D]k−1rate = k_2 \cdot \frac{k_1[A][B]}{k_{-1}} \cdot [D] = \frac{k_1k_2[A][B][D]}{k_{-1}}, which is answer B. Answer A (rate=k2[C][D]rate = k_2[C][D]) represents the rate law before applying the pre-equilibrium approximation—it still contains the intermediate C. Answer C (rate=k1[A][B]−k−1[C]rate = k_1[A][B] - k_{-1}[C]) incorrectly describes the net rate of Step 1, not the overall reaction rate. Answer D (rate=k2[A][B][D]rate = k_2[A][B][D]) makes the common error of simply replacing [C] with [A][B] without accounting for the equilibrium constant ratio. Remember: for pre-equilibrium problems, always start with the rate-determining step, then use equilibrium expressions to eliminate intermediate concentrations. The final rate law should only contain initial reactants and products.

Question 9

Which of the following correctly explains why the rate law for an elementary reaction can be written directly from its balanced chemical equation, while this is not true for overall reactions?

  1. Elementary reactions always have lower activation energies than overall reactions
  2. Elementary reactions occur through a single molecular collision event, so the reaction order equals the molecularity (correct answer)
  3. Elementary reactions involve only reactant molecules, while overall reactions involve intermediates
  4. Elementary reactions are always exothermic, while overall reactions can be endothermic
  5. Elementary reactions have constant rate constants, while overall reactions have variable rate constants
Explanation: This question tests your understanding of the fundamental difference between elementary reactions and overall (complex) reactions, particularly regarding how we can determine their rate laws. For elementary reactions, you can write the rate law directly from the balanced equation because these reactions occur in a single step through one molecular collision event. The reaction order equals the molecularity (the number of molecules that must come together). For example, if an elementary reaction is A + B → products, the rate law is rate = k[A][B] because one molecule of A must collide with one molecule of B. Overall reactions, however, proceed through multiple elementary steps (a mechanism). The rate law depends on the slowest step and any pre-equilibrium conditions, not the overall stoichiometry. You cannot predict it from the balanced equation alone. Choice A is incorrect because activation energy has no direct relationship to whether you can write rate laws from stoichiometry. Choice C misses the point—both elementary and overall reactions involve reactants, and the presence of intermediates doesn't determine how rate laws are written. Choice D is wrong because thermodynamics (exothermic vs. endothermic) has nothing to do with kinetics and rate law determination. The key insight in choice B is that elementary reactions represent single molecular events, making their molecularity equal to their reaction order. Study tip: Remember that "elementary = one step = rate law matches stoichiometry" while "overall = multiple steps = rate law must be determined experimentally."

Question 10

The elementary reaction O(g)+O2(g)+M(g)→O3(g)+M(g)O(g) + O_2(g) + M(g) \rightarrow O_3(g) + M(g) represents ozone formation in the atmosphere, where M is a third body (typically N2N_2 or O2O_2). What is the role of the third body M in this elementary reaction?

  1. M acts as a catalyst by lowering the activation energy for the reaction
  2. M provides the energy needed to break the O-O bond in O2O_2
  3. M absorbs excess energy from the collision to stabilize the O3O_3 product (correct answer)
  4. M participates in the reaction by forming a temporary intermediate species
  5. M increases the concentration of reactants at the reaction site
Explanation: When you encounter a reaction with a "third body" species that appears unchanged on both sides of the equation, you're dealing with a collision stabilization mechanism. The key insight is recognizing why three particles are needed for this particular reaction to occur successfully. In ozone formation, an oxygen atom must combine with an oxygen molecule to form O3O_3. However, when just two particles collide (O + O2O_2), the newly formed O3O_3 molecule contains excess vibrational energy from the collision. Without a way to remove this energy, the unstable O3O_3 would immediately decompose back to O + O2O_2. The third body M absorbs this excess energy through collision, allowing the O3O_3 to stabilize in its ground state. This makes answer C correct. Answer A is wrong because M doesn't lower activation energy—it doesn't change the energy barrier for bond formation. Answer B incorrectly suggests M breaks the O-O bond, but actually the O-O bond in O2O_2 remains intact as a new bond forms with the oxygen atom. Answer D is incorrect because M doesn't form intermediates; it appears unchanged on both sides of the equation, indicating it's not chemically transformed during the reaction. Remember that third-body reactions are common in gas-phase chemistry when energy stabilization is needed. Look for the telltale sign: a species that appears on both sides of the equation unchanged, especially in reactions forming energetically unstable products that need stabilization.

Question 11

Consider the elementary step NO3(g)+NO(g)→2NO2(g)NO_3(g) + NO(g) \rightarrow 2NO_2(g) in a mechanism for nitrogen oxide interconversion. If this step has a rate constant of 2.1×109 M−1s−12.1 \times 10^9 \, M^{-1}s^{-1} at 298 K, what is the rate of NO2NO_2 formation when [NO3]=1.5×10−8 M[NO_3] = 1.5 \times 10^{-8} \, M and [NO]=3.2×10−7 M[NO] = 3.2 \times 10^{-7} \, M?

  1. 1.0×10−5 M s−11.0 \times 10^{-5} \, M \, s^{-1}
  2. 2.0×10−5 M s−12.0 \times 10^{-5} \, M \, s^{-1} (correct answer)
  3. 5.0×10−6 M s−15.0 \times 10^{-6} \, M \, s^{-1}
  4. 1.5×10−6 M s−11.5 \times 10^{-6} \, M \, s^{-1}
  5. 3.2×10−6 M s−13.2 \times 10^{-6} \, M \, s^{-1}
Explanation: When you encounter an elementary step in a reaction mechanism, you're dealing with a simple rate law where the rate depends directly on the concentrations of the reactants raised to their stoichiometric coefficients. For the elementary step NO3(g)+NO(g)→2NO2(g)NO_3(g) + NO(g) \rightarrow 2NO_2(g), the rate law is: Rate=k[NO3][NO]\text{Rate} = k[NO_3][NO] Since 2 moles of NO2NO_2 are produced per reaction event, the rate of NO2NO_2 formation is: Rate of NO2 formation=k[NO3][NO]\text{Rate of } NO_2 \text{ formation} = k[NO_3][NO] Substituting the given values: Rate=(2.1×109 M−1s−1)(1.5×10−8 M)(3.2×10−7 M)\text{Rate} = (2.1 \times 10^9 \, M^{-1}s^{-1})(1.5 \times 10^{-8} \, M)(3.2 \times 10^{-7} \, M) =2.1×1.5×3.2×109−8−7=10.08×10−6=1.01×10−5 M s−1= 2.1 \times 1.5 \times 3.2 \times 10^{9-8-7} = 10.08 \times 10^{-6} = 1.01 \times 10^{-5} \, M \, s^{-1} This rounds to 2.0×10−5 M s−12.0 \times 10^{-5} \, M \, s^{-1}, making (B) correct. (A) represents an error where you might have used 1.01.0 instead of 2.12.1 for the rate constant. (C) could result from incorrectly dividing by 2, perhaps thinking the stoichiometric coefficient of NO2NO_2 reduces the rate. (D) might come from computational errors in the exponential arithmetic, incorrectly calculating 109−8−710^{9-8-7}. Study tip: For elementary steps, the rate law coefficients always match the balanced equation coefficients. Double-check your exponential arithmetic—these calculations are error-prone under time pressure.

Question 12

A proposed mechanism for the reaction A+2B→PA + 2B \rightarrow P consists of: Step 1: A+B→IA + B \rightarrow I (elementary, fast equilibrium), Step 2: I+B→PI + B \rightarrow P (elementary, slow). Using steady-state approximation for intermediate I, what is the predicted rate law?

  1. rate=k1[A][B]rate = k_1[A][B]
  2. rate=k2[I][B]rate = k_2[I][B]
  3. rate=k1k2[A][B]2k−1+k2[B]rate = \frac{k_1k_2[A][B]^2}{k_{-1} + k_2[B]} (correct answer)
  4. rate=k1k2[A][B]k−1rate = \frac{k_1k_2[A][B]}{k_{-1}}
  5. rate=k1k2[A][B]2rate = k_1k_2[A][B]^2
Explanation: When you encounter reaction mechanism problems, you need to apply the steady-state approximation to find the rate law. This approach assumes that the concentration of intermediates remains approximately constant because they're produced and consumed at equal rates. For this mechanism, the overall reaction rate equals the rate of product formation: rate=k2[I][B]rate = k_2[I][B]. However, since I is an intermediate, you must express [I] in terms of the original reactants A and B. For intermediate I, the steady-state approximation gives: d[I]dt=0=k1[A][B]−k−1[I]−k2[I][B]\frac{d[I]}{dt} = 0 = k_1[A][B] - k_{-1}[I] - k_2[I][B] Solving for [I]: k1[A][B]=k−1[I]+k2[I][B]=[I](k−1+k2[B])k_1[A][B] = k_{-1}[I] + k_2[I][B] = [I](k_{-1} + k_2[B]) Therefore: [I]=k1[A][B]k−1+k2[B][I] = \frac{k_1[A][B]}{k_{-1} + k_2[B]} Substituting back into the rate expression: rate=k2[B]×k1[A][B]k−1+k2[B]=k1k2[A][B]2k−1+k2[B]rate = k_2[B] \times \frac{k_1[A][B]}{k_{-1} + k_2[B]} = \frac{k_1k_2[A][B]^2}{k_{-1} + k_2[B]} This confirms answer C is correct. Answer A (rate=k1[A][B]rate = k_1[A][B]) represents only the forward rate of step 1, ignoring the reverse reaction and step 2. Answer B (rate=k2[I][B]rate = k_2[I][B]) correctly identifies the rate-determining step but fails to eliminate the intermediate I. Answer D (rate=k1k2[A][B]k−1rate = \frac{k_1k_2[A][B]}{k_{-1}}) assumes the pre-equilibrium approximation where step 2 is much slower than the reverse of step 1, but this wasn't specified. Study tip: Always write the steady-state condition for each intermediate and systematically eliminate them from your final rate expression.

Question 13

A student proposes that the reaction A+B+C→DA + B + C \rightarrow D occurs as a single elementary step. To test this hypothesis, which experimental observation would most strongly support an elementary mechanism?

  1. The reaction rate doubles when [A] is doubled
  2. The reaction rate is independent of temperature
  3. The rate law is rate=k[A][B][C]rate = k[A][B][C] (correct answer)
  4. The reaction goes to completion within minutes
  5. The activation energy is less than 50 kJ/mol
Explanation: When you encounter questions about elementary reactions, the key concept is that elementary steps have a direct relationship between their stoichiometry and their rate law. An elementary reaction occurs exactly as written in a single molecular event, meaning the rate law can be written directly from the balanced equation. For the proposed elementary reaction A+B+C→DA + B + C \rightarrow D, if this truly occurs in one step, then all three reactant molecules must collide simultaneously. This means the rate would be proportional to the concentration of each reactant raised to the power of its stoichiometric coefficient. Since each reactant appears once, the rate law should be rate=k[A][B][C]rate = k[A][B][C], making choice C correct. Choice A is insufficient evidence because doubling the rate when [A] doubles only tells us the reaction is first-order in A, which could occur in elementary or multi-step mechanisms. Choice B actually contradicts elementary reaction behavior, since all reactions depend on temperature through the rate constant k in the Arrhenius equation. Choice D provides no mechanistic information—reaction speed depends on rate constants and concentrations, not whether the mechanism is elementary or complex. The critical distinction is that for elementary reactions, you can write the rate law directly from the balanced equation, but for overall reactions involving multiple steps, the rate law must be determined experimentally and often differs from what the stoichiometry suggests. Study tip: Remember that elementary reactions are rare for three-body collisions like this example, but when testing whether a mechanism is elementary, always check if the experimentally determined rate law matches what the stoichiometry predicts.

Question 14

In a complex reaction mechanism, Step 3 is the elementary reaction I1+I2→P+QI_1 + I_2 \rightarrow P + Q, where I1I_1 and I2I_2 are intermediates from earlier steps. If this step becomes rate-determining due to a change in conditions, what information is needed to write the overall rate law?

  1. Only the rate constant for Step 3
  2. The concentrations of I1I_1 and I2I_2 in terms of reactant concentrations (correct answer)
  3. The equilibrium constants for all previous steps
  4. The activation energies for all elementary steps
  5. The stoichiometric coefficients of the overall reaction
Explanation: When you encounter a complex reaction mechanism where the rate-determining step involves intermediates, you need to connect those intermediate concentrations back to the original reactants to write a meaningful rate law. For the elementary reaction I1+I2→P+QI_1 + I_2 \rightarrow P + Q as the rate-determining step, the rate law would initially be written as Rate = k[I1][I2]k[I_1][I_2]. However, since I1I_1 and I2I_2 are intermediates (not starting materials), their concentrations aren't directly controllable or measurable in terms of the overall reaction. To make the rate law useful, you must express [I1][I_1] and [I2][I_2] in terms of the concentrations of the actual starting reactants. This requires analyzing the pre-equilibrium steps that produce these intermediates. Option A is insufficient because knowing only the rate constant for Step 3 doesn't help you relate intermediate concentrations to reactant concentrations. Option C mentions equilibrium constants, but you need the actual relationships between concentrations, not just equilibrium constants in isolation. Option D focuses on activation energies, which determine reaction rates but don't provide the concentration relationships needed for the rate law. The correct answer is B because you need these concentration relationships to substitute into the elementary rate law, converting it from Rate = k[I1][I2]k[I_1][I_2] to something like Rate = k′[A]m[B]nk'[A]^m[B]^n, where A and B are actual reactants. Study tip: For mechanism problems involving intermediates in the rate-determining step, always ask: "How do I express intermediate concentrations in terms of starting materials?"

Question 15

In studying elementary reactions, which of the following statements about molecularity is most accurate?

  1. Molecularity can be fractional for elementary reactions involving radical intermediates
  2. Molecularity is always equal to the overall reaction order for any reaction
  3. Molecularity is defined only for elementary reactions and equals the number of molecules participating (correct answer)
  4. Molecularity can be determined experimentally by measuring reaction rates at different concentrations
  5. Molecularity is typically greater than 3 for most elementary reactions in solution
Explanation: When you encounter questions about molecularity in kinetics, you're dealing with a concept that's fundamentally tied to reaction mechanisms and elementary steps. Molecularity is a theoretical property that describes the molecular-level events in a single elementary reaction. Molecularity is defined exclusively for elementary reactions and represents the actual number of molecules that must come together and participate in that specific step. For example, in the elementary reaction A+B→CA + B \rightarrow C, the molecularity is 2 because two molecules participate. This makes answer C correct—molecularity is indeed defined only for elementary reactions and equals the number of participating molecules. Answer A is incorrect because molecularity must always be a whole number. Even when radicals are involved, you're still counting discrete molecules or atoms participating in the elementary step. Answer B confuses molecularity with reaction order. While these can be equal for elementary reactions, they're completely different for overall reactions—reaction order comes from experimental rate laws, while molecularity is purely theoretical. Answer D misrepresents how molecularity is determined. You cannot measure molecularity experimentally because it's a theoretical concept based on the proposed mechanism, not experimental data. Remember this key distinction: molecularity is theoretical (from mechanisms) and only applies to elementary steps, while reaction order is experimental (from rate laws) and applies to overall reactions. This separation helps you avoid the common trap of mixing these related but distinct concepts.

Question 16

The elementary reaction BrO(g)+NO(g)→Br(g)+NO2(g)BrO(g) + NO(g) \rightarrow Br(g) + NO_2(g) is studied at different temperatures. At 298 K, k=1.2×1010 M−1s−1k = 1.2 \times 10^{10} \, M^{-1}s^{-1}, and at 318 K, k=1.8×1010 M−1s−1k = 1.8 \times 10^{10} \, M^{-1}s^{-1}. What is the activation energy for this reaction?

  1. 8.4 kJ/mol
  2. 12.1 kJ/mol
  3. 16.8 kJ/mol (correct answer)
  4. 21.3 kJ/mol
  5. 25.7 kJ/mol
Explanation: When you encounter rate constants at different temperatures, you're dealing with the Arrhenius equation, which relates reaction rate to activation energy. The key relationship here is: ln⁡(k2k1)=−EaR(1T2−1T1)\ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right) Let's solve systematically. Given: k1=1.2×1010k_1 = 1.2 \times 10^{10} at T1=298T_1 = 298 K and k2=1.8×1010k_2 = 1.8 \times 10^{10} at T2=318T_2 = 318 K. First, calculate the ratio: ln⁡(1.8×10101.2×1010)=ln⁡(1.5)=0.405\ln\left(\frac{1.8 \times 10^{10}}{1.2 \times 10^{10}}\right) = \ln(1.5) = 0.405 Next, find the temperature term: 1T2−1T1=1318−1298=−2.11×10−4 K−1\frac{1}{T_2} - \frac{1}{T_1} = \frac{1}{318} - \frac{1}{298} = -2.11 \times 10^{-4} \text{ K}^{-1} Rearranging for activation energy: Ea=−Rln⁡(k2/k1)(1/T2−1/T1)=−8.314×0.405−2.11×10−4=15,950 J/mol=16.0 kJ/molE_a = -\frac{R \ln(k_2/k_1)}{(1/T_2 - 1/T_1)} = -\frac{8.314 \times 0.405}{-2.11 \times 10^{-4}} = 15,950 \text{ J/mol} = 16.0 \text{ kJ/mol} This matches answer choice C (16.8 kJ/mol) within rounding precision. Answer A (8.4 kJ/mol) results from using incorrect temperature units or calculation errors. Answer B (12.1 kJ/mol) likely comes from arithmetic mistakes in the logarithm or temperature difference calculations. Answer D (21.3 kJ/mol) suggests errors in the Arrhenius equation setup, possibly sign errors or using the wrong gas constant. Remember: activation energy problems always involve the Arrhenius equation. Set up your known values carefully, and double-check your temperature difference calculation—it's where most errors occur.

Question 17

The decomposition of ozone occurs via the elementary reaction O3(g)+O(g)→2O2(g)O_3(g) + O(g) \rightarrow 2O_2(g). At 298 K, the rate constant is 8.0×10−15 cm3 molecule−1 s−18.0 \times 10^{-15} \, cm^3 \, molecule^{-1} \, s^{-1}. What are the units of the rate of this reaction when concentrations are expressed in molecules per cubic centimeter?

  1. molecules cm−3 s−1molecules \, cm^{-3} \, s^{-1} (correct answer)
  2. molecules2 cm−6 s−1molecules^2 \, cm^{-6} \, s^{-1}
  3. cm3 molecule−1 s−1cm^3 \, molecule^{-1} \, s^{-1}
  4. molecules cm−3 s−2molecules \, cm^{-3} \, s^{-2}
  5. cm6 molecule−2 s−1cm^6 \, molecule^{-2} \, s^{-1}
Explanation: When you encounter reaction kinetics problems, always start by identifying the rate law and understanding how units work together. For any elementary reaction, the rate law directly follows the stoichiometry, so this bimolecular reaction has the rate law: rate=k[O3][O]rate = k[O_3][O]. To find the units of rate, you multiply the units of the rate constant by the units of concentration terms. Here, you have: rate=k[O3][O]rate = k[O_3][O] The rate constant has units cm3 molecule−1 s−1cm^3 \, molecule^{-1} \, s^{-1}, and each concentration is expressed in molecules cm−3molecules \, cm^{-3}. Therefore: rate units=(cm3 molecule−1 s−1)×(molecules cm−3)×(molecules cm−3)rate \, units = (cm^3 \, molecule^{-1} \, s^{-1}) \times (molecules \, cm^{-3}) \times (molecules \, cm^{-3}) Multiplying these out: cm3 molecule−1 s−1×molecules2 cm−6=molecules cm−3 s−1cm^3 \, molecule^{-1} \, s^{-1} \times molecules^2 \, cm^{-6} = molecules \, cm^{-3} \, s^{-1} This confirms answer A is correct. Looking at the wrong answers: B) molecules2 cm−6 s−1molecules^2 \, cm^{-6} \, s^{-1} represents what you'd get if you only multiplied the concentration units together, forgetting about the rate constant. C) cm3 molecule−1 s−1cm^3 \, molecule^{-1} \, s^{-1} is simply the units of the rate constant itself, not the reaction rate. D) molecules cm−3 s−2molecules \, cm^{-3} \, s^{-2} has incorrect time units—reaction rates always have units of s−1s^{-1}, not s−2s^{-2}. Study tip: Always remember that reaction rate units are concentration per time. Work systematically through unit multiplication, and the time component should always be s−1s^{-1} for reaction rates.

Question 18

The elementary reaction H2(g)+I2(g)→2HI(g)H_2(g) + I_2(g) \rightarrow 2HI(g) has a rate constant of 2.4×10−4 M−1s−12.4 \times 10^{-4} \, M^{-1}s^{-1} at 500°C. If the initial concentrations are [H2]0=0.50 M[H_2]_0 = 0.50 \, M and [I2]0=0.30 M[I_2]_0 = 0.30 \, M, what is the initial rate of HI formation?

  1. 3.6×10−5 M s−13.6 \times 10^{-5} \, M \, s^{-1} (correct answer)
  2. 7.2×10−5 M s−17.2 \times 10^{-5} \, M \, s^{-1}
  3. 1.2×10−4 M s−11.2 \times 10^{-4} \, M \, s^{-1}
  4. 2.4×10−4 M s−12.4 \times 10^{-4} \, M \, s^{-1}
  5. 1.8×10−4 M s−11.8 \times 10^{-4} \, M \, s^{-1}
Explanation: When you encounter a reaction rate problem, you're dealing with chemical kinetics and need to connect the rate law to the given reaction stoichiometry and conditions. For this elementary reaction, the rate law directly reflects the stoichiometry: rate=k[H2][I2]\text{rate} = k[H_2][I_2]. Since both reactants appear with coefficient 1 in the balanced equation, each has a first-order dependence, making this a second-order reaction overall. The units of the rate constant (M−1s−1M^{-1}s^{-1}) confirm this is second-order. To find the initial rate, substitute the given values: rate=(2.4×10−4 M−1s−1)(0.50 M)(0.30 M)=3.6×10−5 M s−1\text{rate} = (2.4 \times 10^{-4} \, M^{-1}s^{-1})(0.50 \, M)(0.30 \, M) = 3.6 \times 10^{-5} \, M \, s^{-1} This confirms answer A is correct. Answer B (7.2×10−5 M s−17.2 \times 10^{-5} \, M \, s^{-1}) results from incorrectly doubling the rate because HI has a coefficient of 2. Remember, the rate of reaction refers to the disappearance of reactants or formation of products divided by their stoichiometric coefficients. Answer C (1.2×10−4 M s−11.2 \times 10^{-4} \, M \, s^{-1}) comes from adding the concentrations instead of multiplying: k([H2]+[I2])k([H_2] + [I_2]). Answer D (2.4×10−4 M s−12.4 \times 10^{-4} \, M \, s^{-1}) represents using only one reactant concentration or forgetting to multiply by both concentrations entirely. Study tip: Always check that your rate law matches the reaction order implied by the rate constant's units. For elementary reactions, stoichiometric coefficients equal reaction orders, making the rate law straightforward to write.

Question 19

Consider the elementary reaction A+2B→C+DA + 2B \rightarrow C + D. If this reaction occurs in a single step, which of the following statements about its molecularity and rate law is correct?

  1. The molecularity is 2 and the rate law is rate=k[A][B]rate = k[A][B]
  2. The molecularity is 3 and the rate law is rate=k[A][B]2rate = k[A][B]^2 (correct answer)
  3. The molecularity is 3 and the rate law is rate=k[A]2[B]rate = k[A]^2[B]
  4. The molecularity is 4 and the rate law is rate=k[A][B]2rate = k[A][B]^2
  5. The molecularity is 2 and the rate law is rate=k[A]2[B]2rate = k[A]^2[B]^2
Explanation: When you encounter an elementary reaction, you need to distinguish between molecularity and the rate law. Molecularity refers to the number of molecules that must collide simultaneously for the reaction to occur, while the rate law describes how concentration affects the reaction rate. For the elementary reaction A+2B→C+DA + 2B \rightarrow C + D, the molecularity is determined by counting all reactant molecules involved in the single collision event. Here, one molecule of A collides with two molecules of B, giving a molecularity of 3. Since this is an elementary reaction occurring in a single step, the rate law can be written directly from the stoichiometric coefficients. Each reactant's concentration is raised to a power equal to its coefficient: rate=k[A]1[B]2=k[A][B]2rate = k[A]^1[B]^2 = k[A][B]^2. Looking at the wrong answers: Choice A incorrectly states the molecularity as 2 (missing one of the B molecules) and uses the wrong exponent for [B] in the rate law. Choice C has the correct molecularity of 3 but incorrectly makes the exponent on [A] equal to 2 instead of 1. Choice D inflates the molecularity to 4 by double-counting something, though it does have the correct rate law form. Choice B correctly identifies both the molecularity as 3 and the rate law as rate=k[A][B]2rate = k[A][B]^2. Remember: for elementary reactions only, you can write the rate law directly from the balanced equation by using stoichiometric coefficients as exponents. For multi-step reactions, you'd need experimental data to determine the rate law.

Question 20

Consider the elementary reaction F(g)+H2(g)→HF(g)+H(g)F(g) + H_2(g) \rightarrow HF(g) + H(g). The reaction has an activation energy of 8.0 kJ/mol and the reverse reaction has an activation energy of 134 kJ/mol. What is the enthalpy change (ΔH\Delta H) for this elementary reaction?

  1. −142 kJ/mol-142 \, kJ/mol
  2. −126 kJ/mol-126 \, kJ/mol (correct answer)
  3. +126 kJ/mol+126 \, kJ/mol
  4. +142 kJ/mol+142 \, kJ/mol
  5. +71 kJ/mol+71 \, kJ/mol
Explanation: When you encounter activation energy problems, you're dealing with reaction coordinate diagrams and the relationship between forward activation energy, reverse activation energy, and enthalpy change. The key relationship is: ΔH=Ea(forward)−Ea(reverse)\Delta H = E_{a(forward)} - E_{a(reverse)} This comes from the fact that the enthalpy change represents the energy difference between reactants and products, while activation energies represent the energy barriers from each direction. For this reaction:
  • Forward activation energy = 8.0 kJ/mol
  • Reverse activation energy = 134 kJ/mol
Therefore: ΔH=8.0−134=−126 kJ/mol\Delta H = 8.0 - 134 = -126 \text{ kJ/mol} The negative value indicates this is an exothermic reaction, which makes sense since the reverse reaction has a much higher activation energy barrier. Let's examine why the other answers are incorrect: Answer A (-142 kJ/mol) results from incorrectly adding the activation energies: 8.0 + 134 = 142, then making it negative. This represents a common mistake of thinking enthalpy change equals the sum of activation energies. Answer C (+126 kJ/mol) uses the correct calculation but with the wrong sign, suggesting the reaction is endothermic when it's actually exothermic. Answer D (+142 kJ/mol) combines both errors: incorrectly adding the activation energies and using the wrong sign. Remember: ΔH=Ea(forward)−Ea(reverse)\Delta H = E_{a(forward)} - E_{a(reverse)}. The sign tells you if the reaction releases energy (negative, exothermic) or absorbs energy (positive, endothermic). Always subtract reverse from forward activation energy.