Physical Chemistry 2 Quiz: Collision Theory
17 questions · exam conditions
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Collision TheoryQuestion 1 of 17

Two reactions have identical activation energies and pre-exponential factors, but reaction X occurs between spherical molecules while reaction Y occurs between linear molecules of similar size. At the same temperature and concentration, which statement best describes the relative reaction rates?

Rate of X equals rate of Y because activation energies are identical and molecular size effects cancel out
Rate of X is greater than rate of Y because spherical molecules have higher collision frequencies per unit volume
Rate of Y is greater than rate of X because linear molecules have more favorable orientational requirements for reaction
Rate of X is greater than rate of Y because spherical molecules require less precise orientational alignment for productive collisions
Rate of Y is greater than rate of X because linear molecules have larger effective collision cross-sections
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Physical Chemistry 2 Quiz

Physical Chemistry 2 Quiz: Collision Theory

Practice Collision Theory in Physical Chemistry 2 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 Collision Theory, giving you a quick way to practice the rules, question types, and explanations that matter most for Physical Chemistry 2.

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

Two reactions have identical activation energies and pre-exponential factors, but reaction X occurs between spherical molecules while reaction Y occurs between linear molecules of similar size. At the same temperature and concentration, which statement best describes the relative reaction rates?

  1. Rate of X equals rate of Y because activation energies are identical and molecular size effects cancel out
  2. Rate of X is greater than rate of Y because spherical molecules have higher collision frequencies per unit volume
  3. Rate of Y is greater than rate of X because linear molecules have more favorable orientational requirements for reaction
  4. Rate of X is greater than rate of Y because spherical molecules require less precise orientational alignment for productive collisions (correct answer)
  5. Rate of Y is greater than rate of X because linear molecules have larger effective collision cross-sections
Explanation: When you encounter questions about molecular shape effects on reaction rates, focus on how geometry influences the probability of successful collisions, especially orientational requirements. Both reactions have identical activation energies and pre-exponential factors in the Arrhenius equation k=Ae−Ea/RTk = Ae^{-E_a/RT}, so you might initially think the rates should be equal. However, the pre-exponential factor A includes a steric factor that accounts for the fraction of collisions with proper orientation for reaction. Spherical molecules have a significant advantage: they present a more uniform surface for collision regardless of their orientation. This means a much larger fraction of their collisions will have the correct geometry for reaction to occur. Linear molecules, in contrast, must align more precisely - often requiring specific end-to-end or side-on approaches depending on where the reactive sites are located. Let's examine why the other options miss the mark: Option A incorrectly assumes molecular shape effects cancel out, ignoring the crucial role of orientational factors in determining reaction probability. Option B focuses on collision frequency, but molecules of similar size should have comparable collision rates - the key difference lies in collision effectiveness, not frequency. Option C suggests linear molecules have more favorable orientational requirements, which contradicts the geometric reality that linear molecules typically need more precise alignment. Remember this pattern: when comparing reaction rates for molecules with different geometries, always consider steric factors. More symmetrical, compact shapes generally lead to higher effective collision rates because they're less demanding about orientational alignment.

Question 2

For a bimolecular reaction A + B → products, collision theory predicts that doubling both [A] and [B] while simultaneously halving the absolute temperature will result in which change to the reaction rate?

  1. Rate increases by factor of 2 because concentration effects dominate over the modest temperature decrease at typical reaction conditions
  2. Rate decreases significantly because the exponential temperature dependence of the rate constant overwhelms concentration effects for most activation energies (correct answer)
  3. Rate remains approximately constant because concentration increases exactly compensate for the temperature-dependent rate constant decrease
  4. Rate increases by factor of 8 because both collision frequency and concentration effects contribute multiplicatively while temperature effects are minimal
  5. Rate decreases by factor of 2 because temperature effects on collision frequency cancel concentration increases but activation energy effects remain
Explanation: When you encounter collision theory problems involving both concentration and temperature changes, remember that temperature effects are exponential while concentration effects are linear, making temperature the dominant factor in most cases. According to collision theory, the rate equation is: Rate=k[A][B]\text{Rate} = k[A][B] where k=Ae−Ea/RTk = Ae^{-E_a/RT}. Doubling both [A] and [B] increases the concentration term by a factor of 4. However, halving the absolute temperature (T → T/2) dramatically affects the rate constant: knew=Ae−Ea/R(T/2)=Ae−2Ea/RTk_{new} = Ae^{-E_a/R(T/2)} = Ae^{-2E_a/RT}. This means knew/koriginal=e−Ea/RTk_{new}/k_{original} = e^{-E_a/RT}, which for typical activation energies (20-100 kJ/mol) at room temperature gives values much less than 1. For example, with Ea=50E_a = 50 kJ/mol at 300 K, this ratio is approximately 0.0001. The overall rate change is 4×0.0001=0.00044 \times 0.0001 = 0.0004, representing a massive decrease despite the 4-fold concentration increase. Choice A incorrectly assumes temperature effects are "modest" - they're actually severe due to the exponential dependence. Choice C wrongly suggests these effects balance out, but the exponential temperature term dominates over linear concentration changes. Choice D completely misses the temperature effect's magnitude and incorrectly claims an 8-fold increase. Study tip: For collision theory problems, always evaluate temperature effects first since they're exponential. A halving of absolute temperature will almost always overwhelm any reasonable concentration changes, leading to significant rate decreases.

Question 3

Two similar reactions have identical collision frequencies and activation energies, but different steric factors: Reaction 1 has P₁ = 0.25 and Reaction 2 has P₂ = 0.0063. If both reactions are run under identical conditions, what is the primary molecular-level reason for this 40-fold difference in steric factors?

  1. Reaction 1 involves smaller molecules with fewer rotational degrees of freedom, requiring less precise orientational alignment for successful collisions
  2. Reaction 2 involves more complex molecules with specific geometric requirements, such as particular bond angles or conformations needed for reaction (correct answer)
  3. The difference reflects varying collision cross-sections, with Reaction 1 having larger effective molecular sizes for collision processes
  4. Reaction 1 occurs through a concerted mechanism while Reaction 2 requires sequential bond-breaking and forming with precise timing
  5. The steric factor difference arises from different solvent effects on molecular orientation during collision, despite identical activation energies
Explanation: When you encounter questions about steric factors in collision theory, focus on the molecular-level geometry and orientation requirements for successful reactions. The steric factor (P) represents the fraction of collisions with proper orientation to lead to reaction, even when molecules have sufficient energy. The 40-fold difference in steric factors (0.25 vs 0.0063) indicates dramatically different geometric requirements. Answer B correctly identifies that Reaction 2 involves complex molecules requiring very specific orientations - particular bond angles, conformations, or precise alignment of reactive sites. Only a tiny fraction (0.63%) of energetically sufficient collisions have the exact geometry needed. In contrast, Reaction 1's higher steric factor (25%) suggests more flexible geometric requirements where successful reaction can occur from multiple orientational approaches. Answer A incorrectly suggests smaller molecules always have higher steric factors. While molecular size can influence orientation requirements, the key factor is geometric complexity, not just size. Answer C confuses steric factors with collision cross-sections. Collision frequency depends on cross-sectional area, but since both reactions have identical collision frequencies, this isn't the distinguishing factor. Answer D incorrectly attributes the difference to reaction mechanisms (concerted vs sequential). While mechanism can influence steric requirements, the steric factor specifically measures orientational constraints, not the timing of bond formation/breaking. Remember: steric factors reflect geometric pickiness. Complex molecules with rigid structures and specific reactive site requirements will have much lower steric factors than flexible molecules that can react from multiple orientations.

Question 4

A gas-phase reaction follows collision theory with a measured pre-exponential factor of 2.1×10122.1 \times 10^{12} M⁻¹s⁻¹. The calculated collision frequency from kinetic theory is 8.4×10138.4 \times 10^{13} M⁻¹s⁻¹. What does this comparison suggest about the reaction mechanism?

  1. The reaction requires highly specific molecular orientations with a steric factor of approximately 0.025, indicating complex geometric requirements (correct answer)
  2. The mechanism involves multiple elementary steps since the pre-exponential factor exceeds typical values for simple bimolecular collisions
  3. Quantum mechanical tunneling effects are significant, reducing the effective collision frequency below classical predictions for this system
  4. The reaction occurs through a pre-equilibrium mechanism where reactant association precedes the rate-determining chemical transformation step
  5. Solvent cage effects are reducing the effective collision frequency, suggesting the reaction occurs in solution rather than gas phase
Explanation: When you encounter collision theory problems comparing experimental and theoretical values, you're analyzing how molecular-level factors affect reaction rates beyond simple collision frequency. The key insight comes from comparing the measured pre-exponential factor (2.1×10122.1 \times 10^{12} M⁻¹s⁻¹) to the calculated collision frequency (8.4×10138.4 \times 10^{13} M⁻¹s⁻¹). The ratio gives us the steric factor: 2.1×10128.4×1013=0.025\frac{2.1 \times 10^{12}}{8.4 \times 10^{13}} = 0.025. This means only about 2.5% of collisions have the proper molecular orientation to lead to reaction, indicating highly specific geometric requirements for successful collisions. Answer A correctly identifies this steric factor interpretation and recognizes that such a low value (0.025) suggests complex geometric requirements where molecules must collide in very specific orientations. Answer B misinterprets the data - the pre-exponential factor is actually much smaller than the collision frequency, not larger, ruling out this multi-step mechanism explanation. Answer C incorrectly invokes quantum tunneling. Tunneling would typically increase reaction rates at lower temperatures, but here we're seeing reduced effective collisions due to orientation requirements, not quantum effects. Answer D suggests a pre-equilibrium mechanism, but this doesn't explain why the pre-exponential factor is lower than the collision frequency. Pre-equilibrium mechanisms affect the rate law form, not necessarily the steric requirements. Remember: when experimental pre-exponential factors are significantly lower than calculated collision frequencies, think steric effects first. The ratio directly gives you the steric factor, revealing how geometrically demanding the reaction is.

Question 5

For the reaction H₂ + I₂ → 2HI, collision theory can be used to estimate reaction rates. Given that H₂ and I₂ have molecular diameters of 2.9 Å and 5.1 Å respectively, and the reaction has an activation energy of 165 kJ/mol at 700 K, which factor most significantly limits the reaction rate?

  1. The collision frequency between H₂ and I₂ molecules because of their different sizes and masses affecting collision cross-sections
  2. The steric factor because H₂ and I₂ must achieve proper orientational alignment for simultaneous bond breaking and forming
  3. The high activation energy creating a very small fraction of molecules with sufficient energy for reaction at this temperature (correct answer)
  4. The reduced mass of the H₂-I₂ collision pair which affects both collision frequency and activation energy requirements
  5. The temperature being insufficient to overcome the endothermic nature of the bond reorganization process during collision
Explanation: When analyzing reaction rates using collision theory, you need to consider three key factors: collision frequency, steric requirements, and the energy barrier. The relative importance of these factors depends on the specific conditions and molecular properties involved. For the H₂ + I₂ reaction at 700 K with an activation energy of 165 kJ/mol, the energy barrier dominates. Using the Arrhenius equation, the fraction of molecules with sufficient energy is proportional to e−Ea/RTe^{-E_a/RT}. Plugging in the values: e−165,000/(8.314×700)≈e−28.3≈5×10−13e^{-165,000/(8.314 × 700)} ≈ e^{-28.3} ≈ 5 × 10^{-13}. This incredibly small fraction means that virtually no collisions have enough energy to react, making this the rate-limiting step. Option A is incorrect because while H₂ and I₂ do have different sizes affecting collision cross-sections, the collision frequency itself isn't severely limiting—molecules still collide frequently at 700 K. Option B misidentifies steric factors as the primary limitation. While proper orientation matters for this reaction, the steric factor (typically 0.01-1.0) is much less restrictive than the energy requirement. Option D incorrectly suggests that reduced mass significantly affects activation energy requirements, when activation energy is primarily determined by the electronic rearrangements needed for bond breaking and forming. Remember that when activation energies exceed about 100 kJ/mol, the exponential energy term in collision theory typically dominates over geometric and steric considerations. Always calculate or estimate the Boltzmann factor e−Ea/RTe^{-E_a/RT} to assess whether energy is the limiting factor in reaction kinetics.

Question 6

Collision theory predicts that pressure effects on gas-phase reaction rates arise from changes in molecular concentrations. For the reaction 2NO + O₂ → 2NO₂, if the pressure is increased by a factor of 3 while temperature remains constant, what is the predicted change in reaction rate?

  1. Rate increases by factor of 3 because the reaction is first-order in total pressure according to collision theory principles
  2. Rate increases by factor of 9 because this termolecular reaction depends on the cube of concentration changes from pressure
  3. Rate increases by factor of 27 because each reactant concentration increases by factor of 3 and collision frequency scales with concentration products (correct answer)
  4. Rate increases by factor of 6 because collision theory accounts for both binary NO-NO and NO-O₂ collision frequency enhancements
  5. Rate remains constant because collision theory shows that pressure effects are exactly compensated by changes in molecular mean free paths
Explanation: When analyzing pressure effects on gas-phase reactions, you need to understand how collision theory connects molecular concentrations to reaction rates. For gas reactions, increasing pressure at constant temperature proportionally increases the concentration of all gaseous species. The reaction 2NO + O₂ → 2NO₂ is termolecular, meaning three molecules must collide simultaneously for the reaction to occur. The rate law is: rate=k[NO]2[O2]\text{rate} = k[\text{NO}]^2[\text{O}_2] When pressure increases by a factor of 3, each gas concentration increases by the same factor of 3. Therefore: new rate=k(3[NO])2(3[O2])=k⋅9[NO]2⋅3[O2]=27k[NO]2[O2]\text{new rate} = k(3[\text{NO}])^2(3[\text{O}_2]) = k \cdot 9[\text{NO}]^2 \cdot 3[\text{O}_2] = 27k[\text{NO}]^2[\text{O}_2] The rate increases by a factor of 27, making answer C correct. Answer A incorrectly treats this as first-order in pressure, ignoring the molecular nature of the elementary reaction. Answer B gets the factor of 9 from squaring the concentration change but fails to account for the O₂ concentration term, missing the additional factor of 3. Answer D arbitrarily combines collision types without proper mathematical basis—collision theory for elementary reactions depends on the stoichiometric coefficients, not separate collision pathways. Remember: For elementary gas-phase reactions, the rate law exponents equal the stoichiometric coefficients. When pressure changes affect all reactant concentrations equally, raise the pressure change factor to the power equal to the sum of all stoichiometric coefficients (here: 2+1=3, so 3³=27).

Question 7

According to collision theory, why do reactions between ions in solution often have much higher rate constants than predicted by simple collision frequency calculations for neutral molecules?

  1. Ionic reactions have lower activation energies because electrostatic attractions reduce the energy barrier for bond formation processes
  2. The collision frequency for ionic species is higher due to long-range electrostatic forces that increase effective collision cross-sections significantly (correct answer)
  3. Solvation effects create favorable orientations that dramatically increase steric factors compared to gas-phase molecular collisions
  4. Ion pairing creates pre-equilibrium concentrations of reactive complexes that bypass normal collision requirements for reaction
  5. The ionic strength of solution modifies collision theory parameters through Debye-Hückel effects that enhance reaction probabilities
Explanation: When analyzing reaction rates using collision theory, you need to consider how electrostatic forces affect molecular interactions differently than simple kinetic collisions between neutral species. For ionic reactions in solution, the key insight is that ions don't behave like neutral molecules that must randomly collide within a small contact distance. Instead, charged species experience long-range electrostatic attractions (or repulsions) that extend far beyond their molecular radii. This dramatically increases the effective collision cross-section - essentially, the "target area" for productive encounters becomes much larger than the physical size of the molecules. Option B correctly identifies this phenomenon. The Coulombic forces between ions can guide them toward each other from distances much greater than typical molecular contact distances, making successful encounters far more frequent than simple collision theory predicts for neutral molecules. Option A confuses kinetics with thermodynamics - while electrostatic effects can influence activation energies, the primary rate enhancement comes from increased collision frequency, not lower energy barriers. Option C overemphasizes solvation effects on molecular orientation, which is secondary to the fundamental increase in collision probability. Option D describes a different mechanism entirely - pre-equilibrium complex formation - which isn't the main collision theory explanation for enhanced ionic reaction rates. Remember that collision theory has three components: collision frequency, activation energy, and steric factors. For ionic reactions, the collision frequency term dominates the rate enhancement due to long-range electrostatic interactions expanding the effective collision cross-section.

Question 8

A reaction mechanism involves three elementary steps with different activation energies. Step 1: A + B → C (Ea1 = 60 kJ/mol), Step 2: C + D → E (Ea2 = 40 kJ/mol), Step 3: E → F + G (Ea3 = 75 kJ/mol). If collision theory applies to each step and all pre-exponential factors are similar, which statement about temperature effects is most accurate?

  1. Increasing temperature will accelerate step 3 most because it has the highest activation energy and strongest exponential temperature dependence
  2. The overall reaction rate will be most sensitive to temperature changes affecting step 2 since it has the lowest activation barrier
  3. Temperature effects will be approximately equal for all steps because the pre-exponential factors are similar and collision frequencies scale identically
  4. Step 1 will show intermediate temperature sensitivity, but the rate-determining step controls overall temperature dependence regardless of individual barriers (correct answer)
  5. The sequence of activation energies creates temperature-dependent shifts in which step controls the overall reaction rate under different conditions
Explanation: When analyzing multi-step reaction mechanisms, you need to consider both individual step kinetics and how they combine to determine overall reaction behavior. The key insight is that the rate-determining step (RDS) - the slowest step - controls the overall reaction rate and its temperature dependence. According to collision theory, each elementary step follows the Arrhenius equation: k=Ae−Ea/RTk = Ae^{-E_a/RT}. While it's true that steps with higher activation energies show stronger exponential temperature dependence individually, this doesn't determine which step controls the overall temperature sensitivity. The correct answer is D because step 1 does show intermediate temperature sensitivity (Ea1 = 60 kJ/mol falls between the other values), but more importantly, whichever step is rate-determining will control how the overall reaction responds to temperature changes. The RDS acts as a "bottleneck" - even if other steps speed up dramatically with temperature, the overall rate can only increase as fast as the slowest step allows. Option A incorrectly assumes that the highest activation energy automatically means the greatest impact on overall rate. Option B makes the opposite error, suggesting the lowest barrier controls temperature sensitivity. Option C ignores the fundamental principle that activation energies, not pre-exponential factors, determine temperature dependence strength. Remember this key principle: in multi-step mechanisms, identify the rate-determining step first. The temperature dependence of that step - regardless of whether it has the highest, lowest, or intermediate activation energy - will dominate the overall reaction's temperature sensitivity.

Question 9

A reaction between gases A and B follows collision theory. When the temperature increases from 300 K to 400 K, the collision frequency increases by a factor of 1.15, but the overall reaction rate increases by a factor of 47. What is the approximate activation energy for this reaction?

  1. 52 kJ/mol because the rate enhancement factor minus collision frequency factor gives the activation barrier contribution
  2. 73 kJ/mol because the logarithmic ratio of rate constants relates directly to the temperature-dependent exponential term (correct answer)
  3. 31 kJ/mol because the collision frequency change must be subtracted from the total rate enhancement before calculating activation energy
  4. 94 kJ/mol because both collision frequency and exponential terms contribute multiplicatively to the overall rate enhancement observed
  5. 18 kJ/mol because the square root relationship between temperature and collision frequency modifies the activation energy calculation significantly
Explanation: When dealing with collision theory and temperature effects on reaction rates, you need to separate two distinct contributions: collision frequency changes and activation energy effects. The Arrhenius equation shows that rate constants follow k=Aexp⁡(−Ea/RT)k = A \exp(-E_a/RT), where the pre-exponential factor A includes collision frequency. The key insight is that the overall rate increase (factor of 47) results from both the collision frequency increase (factor of 1.15) and the exponential term increase due to activation energy. To find the activation energy contribution alone, you calculate the "corrected" rate enhancement: 47/1.15=40.947/1.15 = 40.9. This factor of ~41 represents purely the exponential term's contribution. Using the Arrhenius equation ratio between two temperatures: k2k1=exp⁡[EaR(1T1−1T2)]\frac{k_2}{k_1} = \exp\left[\frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)\right]. Substituting the values: 41=exp⁡[Ea8.314(1300−1400)]41 = \exp\left[\frac{E_a}{8.314}\left(\frac{1}{300} - \frac{1}{400}\right)\right]. Taking the natural logarithm: ln⁡(41)=Ea8.314×8.33×10−4\ln(41) = \frac{E_a}{8.314} \times 8.33 \times 10^{-4}. Solving gives Ea=73 kJ/molE_a = 73 \text{ kJ/mol}. Choice A incorrectly suggests subtracting factors rather than dividing them. Choice C makes the same mathematical error with subtraction. Choice D incorrectly includes collision frequency in the activation energy calculation, double-counting its effect. Remember: when temperature affects reaction rates, always separate collision frequency changes from activation energy effects by dividing the total rate enhancement by the collision frequency factor before applying the Arrhenius equation.

Question 10

Collision theory predicts that for a reaction A₂ + B₂ → 2AB, the rate should depend on molecular orientations during collision. Consider the case where A₂ and B₂ are both linear molecules. Which collision orientation would be LEAST favorable for bond formation according to collision theory principles?

  1. Parallel approach with molecular axes aligned, allowing simultaneous interaction of multiple atoms during the collision event
  2. Perpendicular T-shaped approach with one molecule's axis crossing the other's midpoint for optimal orbital overlap geometry
  3. End-to-end collinear approach maximizing the distance between reactive centers while maintaining proper orbital alignment (correct answer)
  4. Skewed approach with molecular axes at 45° angles, providing intermediate geometrical requirements for bond formation
  5. Side-by-side parallel approach with molecules moving in opposite directions, minimizing contact time for reaction
Explanation: When analyzing collision theory for bimolecular reactions, you need to consider how molecular orientation affects the probability of successful bond formation. The key principle is that atoms must come close enough together during collision for their electron orbitals to overlap effectively and form new bonds. For the reaction A₂ + B₂ → 2AB, successful collisions require that atoms from different molecules can approach within bonding distance. Let's examine each orientation: Option C represents the least favorable arrangement because in an end-to-end collinear approach, the reactive centers (the individual A and B atoms) are positioned at maximum distance from each other during collision. Even though the molecules are properly aligned, the atoms that need to form new A-B bonds are separated by the full length of both molecules, making orbital overlap extremely difficult. Option A is actually quite favorable because parallel alignment allows multiple atoms to interact simultaneously, increasing the probability of successful bond formation. Option B describes an effective T-shaped approach where one molecule's atoms can easily access the other molecule's reactive centers at close range. Option D represents a moderate approach that still allows reasonable access between reactive atoms. The critical insight is that "proper orbital alignment" mentioned in option C is meaningless if the atoms are too far apart to interact. Distance trumps alignment in collision theory. Study tip: In collision theory problems, always visualize the actual atomic positions during collision. The arrangement that keeps reactive atoms farthest apart will be least favorable, regardless of how "aligned" the molecules appear to be.

Question 11

According to collision theory, which modification to a bimolecular gas-phase reaction would produce the largest increase in reaction rate at constant temperature and pressure?

  1. Replacing one spherical reactant with a linear molecule of the same mass to increase collision cross-section and orientational possibilities
  2. Adding a catalyst that provides an alternative pathway with activation energy reduced by 20 kJ/mol from the original 85 kJ/mol (correct answer)
  3. Substituting isotopically heavier atoms to increase the reduced mass and collision frequency while maintaining identical bond strengths
  4. Using conformationally flexible molecules instead of rigid ones to increase the number of reactive geometries during collisions
  5. Increasing molecular sizes by 50% through substituent addition while keeping activation energy and reaction mechanism unchanged
Explanation: Collision theory questions test your understanding of how molecular factors affect reaction rates. The key insight is that reaction rate depends exponentially on activation energy but only linearly or as square roots on other factors. The correct answer is B because catalysts provide the most dramatic rate increases. When activation energy decreases from 85 kJ/mol to 65 kJ/mol, the rate constant changes by a factor of e−(Ea,new−Ea,old)/RT=e20,000/(8.314×298)≈e8≈3000e^{-(E_{a,new} - E_{a,old})/RT} = e^{20,000/(8.314 \times 298)} \approx e^8 \approx 3000 at room temperature. This exponential dependence makes catalysis extraordinarily effective. Option A is incorrect because while linear molecules may have larger collision cross-sections than spherical ones, this only increases the pre-exponential factor linearly. The rate enhancement would be modest compared to the exponential effect of lowering activation energy. Option C contains a fundamental error. Heavier isotopes actually decrease collision frequency because velocity is inversely proportional to mass\sqrt{mass} from kinetic theory. While reduced mass affects collision dynamics, the overall effect would likely decrease the reaction rate. Option D is wrong because conformational flexibility doesn't necessarily increase reactive collisions. Flexible molecules might adopt more geometries, but most would be non-reactive orientations. The fraction of productive collisions (steric factor) might even decrease due to increased non-productive conformations. Remember this hierarchy: activation energy changes (exponential effects) always trump collision frequency or steric factor changes (linear effects) when estimating rate modifications. Focus on energetics first in collision theory problems.

Question 12

In collision theory, the effective collision diameter for unlike molecules is often taken as the sum of their radii. For a reaction between molecules with radii 2.8 Å and 3.6 Å, what is the collision cross-section, and how does this compare to the geometric cross-section of the larger molecule alone?

  1. 1.29×10−191.29 \times 10^{-19} m² for collision cross-section, which is 3.2 times larger than the geometric cross-section of the larger molecule (correct answer)
  2. 4.21×10−194.21 \times 10^{-19} m² for collision cross-section, which is 10.3 times larger than the geometric cross-section of the larger molecule
  3. 8.15×10−208.15 \times 10^{-20} m² for collision cross-section, which is 2.0 times larger than the geometric cross-section of the larger molecule
  4. 2.56×10−192.56 \times 10^{-19} m² for collision cross-section, which is 6.3 times larger than the geometric cross-section of the larger molecule
  5. 1.85×10−191.85 \times 10^{-19} m² for collision cross-section, which is 4.5 times larger than the geometric cross-section of the larger molecule
Explanation: When you encounter collision theory problems, you're dealing with how molecules physically interact during reactions. The key insight is that molecules collide when their surfaces touch, so the effective collision diameter equals the sum of their individual radii. For this problem, you need to calculate the collision cross-section using the combined radius: reff=2.8+3.6=6.4r_{eff} = 2.8 + 3.6 = 6.4 Å = 6.4×10−106.4 \times 10^{-10} m. The collision cross-section is σ=πreff2=π(6.4×10−10)2=1.29×10−19\sigma = \pi r_{eff}^2 = \pi (6.4 \times 10^{-10})^2 = 1.29 \times 10^{-19} m². For comparison, the larger molecule's geometric cross-section uses only its own radius: σlarge=π(3.6×10−10)2=4.07×10−20\sigma_{large} = \pi (3.6 \times 10^{-10})^2 = 4.07 \times 10^{-20} m². The ratio is 1.29×10−194.07×10−20=3.2\frac{1.29 \times 10^{-19}}{4.07 \times 10^{-20}} = 3.2. Answer A correctly gives both values. Answer B uses an incorrect calculation that roughly squares the effective diameter instead of radius, leading to an inflated cross-section and ratio. Answer C appears to use an average of the radii rather than their sum, underestimating the collision cross-section. Answer D contains computational errors in both the cross-section calculation and the comparison ratio. Remember that collision cross-sections are always larger than individual molecular cross-sections because two molecules can collide when their surfaces are still separated by the sum of their radii. This geometric principle is fundamental to understanding reaction rates in gas-phase kinetics.

Question 13

Two reactions have identical activation energies and pre-exponential factors, but reaction 1 involves collision between molecules with reduced masses of 14 amu, while reaction 2 involves molecules with reduced masses of 35 amu. At the same temperature and concentration, what is the approximate ratio of rate constants k₁/k₂?

  1. The ratio is approximately 1.58, since collision frequency increases with decreasing reduced mass (correct answer)
  2. The ratio is approximately 0.63, since heavier molecules have higher collision cross-sections
  3. The ratio is exactly 1.00, since activation energies and pre-exponential factors are identical
  4. The ratio is approximately 2.50, reflecting the inverse relationship between reduced mass and reaction probability
Explanation: The collision frequency in collision theory depends on the average relative velocity, which is inversely proportional to √μ (reduced mass). Since the pre-exponential factors and activation energies are identical, the only difference comes from the collision frequency term Z ∝ 1/√μ. Therefore, k₁/k₂ = √(35/14) = √2.5 ≈ 1.58. The lighter molecules move faster and collide more frequently.

Question 14

In collision theory, the collision cross-section σ for two spherical molecules with radii r₁ and r₂ is π(r₁ + r₂)². If molecule A has twice the radius of molecule B, and the reaction A + A → products is compared to B + B → products under identical conditions, what is the ratio of collision frequencies Z_{AA}/Z_{BB}?

  1. The ratio is 4.0, reflecting only the cross-sectional area difference between the molecules
  2. The ratio is approximately 2.83, accounting for both cross-section and velocity differences (correct answer)
  3. The ratio is 2.0, since collision frequency scales linearly with molecular radius
  4. The ratio is approximately 5.66, considering mass scaling effects on collision dynamics
Explanation: Z ∝ σ√(T/μ) where μ is reduced mass. For A+A: σ_AA = π(2r + 2r)² = π(4r)² = 16πr². For B+B: σ_BB = π(2r)² = 4πr². So σ_AA/σ_BB = 4. Assuming mass scales as volume (m ∝ r³), then m_A = 8m_B, so μ_AA = m_A/2 = 4m_B and μ_BB = m_B/2. Therefore Z_AA/Z_BB = (σ_AA/σ_BB)√(μ_BB/μ_AA) = 4 × √(m_B/2)/(4m_B) = 4 × √(1/8) = 4 × 1/(2√2) = 4/(2√2) = √2 × √2 = 2√2 ≈ 2.83.

Question 15

For the elementary reaction O + O₂ + M → O₃ + M, where M is a third body, collision theory predicts that the rate should depend on the nature of M. Experimental data shows rate constants (in order): M = He < M = N₂ < M = CO₂. Which property of the third body M most likely explains this trend?

  1. The polarizability of M, since larger molecules can better stabilize the transition state complex
  2. The collision cross-section of M, since larger molecules have higher collision frequencies
  3. The heat capacity of M, since energy transfer efficiency depends on internal degrees of freedom (correct answer)
  4. The molecular weight of M, since heavier molecules transfer momentum more effectively
Explanation: In termolecular reactions, the third body M must remove excess energy from the nascent O₃* to stabilize it. The efficiency depends on the number of internal degrees of freedom available for energy transfer. He (monatomic) has no internal degrees of freedom, N₂ (diatomic) has rotational and vibrational modes, and CO₂ (triatomic) has even more vibrational modes. Higher heat capacity correlates with more internal degrees of freedom and better energy transfer efficiency.

Question 16

For a bimolecular gas-phase reaction A + B → products, the steric factor is found to be 0.15. If the collision diameter for both molecules is 3.5 Å and the reaction occurs at 400 K with partial pressures of 0.8 atm for A and 1.2 atm for B, which factor contributes MOST significantly to the low observed reaction rate compared to the collision frequency?

  1. The exponential activation energy term, since most collisions lack sufficient energy for reaction (correct answer)
  2. The steric factor, since only 15% of energetically favorable collisions have proper orientation
  3. The pressure dependence, since the collision frequency scales quadratically with pressure
  4. The molecular diameter, since larger molecules have reduced collision cross-sections per unit volume
Explanation: The rate constant k = Z·P·exp(-Ea/RT) where Z is collision frequency, P is steric factor, and the exponential term is the fraction of collisions with sufficient energy. For most reactions, Ea >> RT, making exp(-Ea/RT) << 0.15. Even with a modest Ea of 50 kJ/mol at 400K, exp(-Ea/RT) ≈ 0.00001, which is much smaller than the steric factor of 0.15. The exponential term typically dominates the rate limitation.

Question 17

A reaction has a collision frequency of 5 × 10¹⁰ M⁻¹s⁻¹ at 298 K. When the temperature is increased to 398 K, the observed rate constant increases by a factor of 850, while the collision frequency increases by only 15%. If the steric factor remains constant, what fraction of the rate enhancement is due to the exponential energy term versus the collision frequency change?

  1. Approximately 98% due to the exponential term and 2% due to collision frequency changes
  2. Approximately 93% due to the exponential term and 7% due to collision frequency changes
  3. Approximately 74% due to the exponential term and 26% due to collision frequency changes
  4. Approximately 87% due to the exponential term and 13% due to collision frequency changes (correct answer)
Explanation: When you encounter collision theory problems involving temperature changes, remember that the rate constant depends on both collision frequency and the exponential Boltzmann factor: k=Z⋅p⋅e−Ea/RTk = Z \cdot p \cdot e^{-E_a/RT}, where Z is collision frequency, p is the steric factor, and the exponential term accounts for molecules having sufficient energy to react. The total rate enhancement is 850-fold, while collision frequency increases by only 15% (factor of 1.15). Since the steric factor remains constant, we can separate these contributions. The collision frequency contributes a factor of 1.15 to the rate increase. The remaining enhancement must come from the exponential energy term: 8501.15=739\frac{850}{1.15} = 739 To find the relative contributions, calculate what fraction each factor represents of the total enhancement. The collision frequency contribution is 1.15−1850−1=0.15849=0.177\frac{1.15 - 1}{850 - 1} = \frac{0.15}{849} = 0.177 or about 13%. The exponential term contribution is 739−1850−1=738849=0.87\frac{739 - 1}{850 - 1} = \frac{738}{849} = 0.87 or about 87%. Answer D correctly identifies this 87%/13% split. Answer A (98%/2%) drastically underestimates the collision frequency effect. Answer B (93%/7%) underestimates the collision frequency contribution by about half. Answer C (74%/26%) significantly overestimates the collision frequency effect, perhaps from incorrectly calculating the relative contributions. Study tip: In collision theory problems, always separate the temperature effects on collision frequency (usually modest, following T\sqrt{T}) from the dramatic exponential energy effects. The exponential term typically dominates rate increases with temperature.