AP Chemistry Quiz: Collision Model

What this quiz covers

This quiz focuses on Collision Model, giving you a quick way to practice the rules, question types, and explanations that matter most for AP 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.

Question 1

1. Trial 1 is faster because stirring increases the rate constant by changing the reaction mechanism in solution.
2. Trial 1 is faster because stirring increases the equilibrium constant, so products are favored sooner.
3. Trial 1 is faster because stirring improves mixing and reduces concentration gradients, increasing how often reactant particles encounter each other and collide effectively. (correct answer)
4. Trial 1 is faster because stirring adds energy to the system in the form of heat, which guarantees that every collision leads to reaction.
5. Trial 1 is faster because unstirred solutions cause reactant particles to lose energy permanently, making collisions ineffective after a short time.

Explanation: This question assesses the collision model, which explains reaction rates based on the frequency and effectiveness of molecular collisions. In Trial 1, stirring promotes better mixing of the reactants, reducing local concentration gradients that could limit encounters between iodide and hydrogen peroxide ions. This improved distribution increases the frequency of collisions throughout the solution, leading to more effective collisions overall. As a result, the iodine color develops faster due to these enhanced molecular interactions. A tempting distractor is choice D, which falsely claims stirring adds heat energy to guarantee effective collisions, misconstruing mechanical action with thermal effects. Faster reactions result from more frequent or more energetic effective collisions.

Question 2

1. The reaction is faster because higher temperature increases the fraction of collisions with sufficient energy to result in reaction, making collisions more effective. (correct answer)
2. The reaction is faster because higher temperature increases the concentration of gases, so collisions occur more frequently.
3. The reaction is faster because higher temperature shifts equilibrium toward products, which increases the forward reaction rate.
4. The reaction is faster because heating acts as a catalyst that provides an alternative mechanism while keeping the same reactants.
5. The reaction is faster because at higher temperature particles have less energy available for bonding, so only the strongest collisions can occur and those always react.

Explanation: This question tests understanding of the collision model. At higher temperature, gas particles have a broader distribution of kinetic energies, with more particles possessing energy above the activation energy threshold. While collision frequency does increase slightly with temperature, the more important effect is that a much larger fraction of R-S collisions have sufficient energy to break bonds and form products - this is collision effectiveness. At lower temperature, many collisions occur but most bounce off without reacting due to insufficient energy. Choice B incorrectly claims temperature changes concentration in a fixed volume with fixed moles. The strategy is that temperature primarily increases reaction rate by increasing the fraction of collisions with energy exceeding the activation barrier.

Question 3

1. Cutting the zinc creates more surface area, allowing more $\text{Cu}^{2+}$ ions to collide with Zn atoms per unit time, increasing the reaction rate. (correct answer)
2. Cutting the zinc increases the equilibrium constant for the reaction, so the forward reaction must occur faster.
3. Cutting the zinc makes each collision more energetic, so nearly every collision becomes products even at the same temperature.
4. Cutting the zinc introduces a catalyst on the fresh metal surface that lowers the activation energy and changes the mechanism.
5. Cutting the zinc decreases the number of collisions needed by changing the products formed, so the reaction finishes sooner.

Explanation: This question tests surface area effects in the collision model for heterogeneous reactions. Cutting the zinc into many pieces in Condition 2 dramatically increases the total surface area compared to the single strip in Condition 1. Since the reaction between Zn atoms and Cu²⁺ ions only occurs at the metal-solution interface, more surface area provides more sites for collisions per unit time. This increased collision frequency at the interface speeds up copper deposition and zinc dissolution. Choice C incorrectly suggests cutting changes collision energy—temperature determines kinetic energy, not the physical subdivision of reactants. For solid-liquid reactions, reaction rate is proportional to the collision frequency at the interface, which depends on surface area.

Question 4

1. The more concentrated acid causes the reaction to shift toward products, so the reaction must proceed faster to reach equilibrium.
2. The more concentrated acid contains more reacting particles per unit volume, increasing the frequency of collisions with the Mg surface and thus increasing the reaction rate. (correct answer)
3. The more concentrated acid increases the energy released by the reaction, so each collision produces products more quickly.
4. The more concentrated acid lowers the activation energy by acting as a catalyst, allowing more collisions to form products.
5. The more concentrated acid makes the Mg atoms vibrate faster, so the Mg bonds break without needing collisions from acid particles.

Explanation: This question tests understanding of the collision model for reaction rates. In Condition 2, the more concentrated HCl solution contains more H⁺ ions per unit volume compared to the dilute solution in Condition 1. Since the reaction occurs at the magnesium surface, having more acid particles in the same volume means more frequent collisions between H⁺ ions and Mg atoms per unit time. This increased collision frequency directly increases the reaction rate, producing hydrogen gas faster. Choice A incorrectly confuses kinetics with equilibrium—reaction rates don't depend on equilibrium position but on collision frequency and effectiveness. When analyzing reaction rates, remember that faster reactions result from more frequent or more energetic effective collisions.

Question 5

1. In the more concentrated mixture, ions collide less often because electrostatic attractions keep them separated, slowing precipitation.
2. In the more concentrated mixture, there are more $\text{Ag}^+$ and $\text{Cl}^-$ ions per unit volume, increasing the frequency of their encounters and speeding formation of $\text{AgCl}(s)$. (correct answer)
3. In the more concentrated mixture, the products are more stable, so the reaction pathway becomes faster even if collision frequency is unchanged.
4. In the more concentrated mixture, the equilibrium constant is larger, so the forward reaction rate must increase.
5. In the more concentrated mixture, the activation energy is lower because concentration acts like a catalyst, making every collision effective.

Explanation: This question applies the collision model to precipitation reactions. In Condition 2, both solutions are more concentrated, meaning there are more Ag⁺ and Cl⁻ ions per unit volume compared to the dilute solutions in Condition 1. When these solutions mix, the higher ion concentrations result in more frequent encounters between Ag⁺ and Cl⁻ ions throughout the solution volume. This increased collision frequency leads to faster AgCl precipitate formation. Choice A incorrectly suggests electrostatic attractions reduce collisions—in reality, attractions between oppositely charged ions enhance their collision rate. For reactions in solution, higher concentration always increases collision frequency between dissolved species.

Question 6

1. At higher temperature, the solution contains more $\text{H}_2\text{O}_2$ molecules per unit volume, so collision frequency increases.
2. At higher temperature, a greater fraction of molecular collisions have enough kinetic energy to overcome the energy barrier, so more collisions lead to decomposition per unit time. (correct answer)
3. At higher temperature, the products have lower potential energy, so the reaction becomes faster to release energy sooner.
4. At higher temperature, the reaction changes mechanism to a faster pathway, so fewer collisions are needed to form products.
5. At higher temperature, the system approaches equilibrium faster, so the forward rate increases because the reverse rate decreases.

Explanation: This question examines temperature effects on collision model for decomposition reactions. At the higher temperature in Condition 2, H₂O₂ molecules have greater kinetic energy on average than in Condition 1. This means a larger fraction of molecular collisions possess enough energy to break the O-O bond and initiate decomposition. While collision frequency also increases slightly with temperature, the primary effect is the exponential increase in the fraction of collisions that are energetically capable of reaction. Choice A incorrectly claims temperature changes concentration—the same solution at different temperatures has the same number of molecules per volume. When temperature increases, focus on how it affects the energy distribution of collisions, not just their frequency.

Question 7

1. Powdered $\text{CaCO}_3$ exposes more surface area, increasing the frequency of effective collisions between acid particles and the solid surface. (correct answer)
2. Powdering the solid increases the kinetic energy of the acid particles, so a greater fraction of collisions has enough energy to react.
3. Powdered $\text{CaCO}_3$ contains weaker bonds than a marble chip, so every collision forms products regardless of orientation.
4. Powdered $\text{CaCO}_3$ increases the acid concentration near the surface, which changes the mechanism and speeds the reaction.
5. Powdered $\text{CaCO}_3$ shifts the equilibrium toward $\text{CO}_2$, so the reaction must occur faster to produce more gas.

Explanation: This question tests understanding of the collision model. Powdering the calcium carbonate dramatically increases its surface area compared to a single marble chip of the same mass. Since the reaction occurs at the solid-liquid interface, more surface area means more sites where HCl particles can collide with CaCO₃. This increases the frequency of collisions between acid particles and the solid surface, leading to more effective collisions per second and faster CO₂ production. Choice D incorrectly suggests that powdering affects the kinetic energy of acid particles, but physical subdivision of the solid doesn't change the temperature or energy of particles in solution. The strategy is that increasing surface area of a solid reactant increases collision frequency at the interface, speeding up heterogeneous reactions.

Question 8

1. The higher $\text{Fe}^{3+}$ concentration increases the frequency of collisions between $\text{Fe}^{3+}$ and $\text{SCN}^-$ ions, increasing the number of effective collisions per unit time. (correct answer)
2. The higher $\text{Fe}^{3+}$ concentration increases the temperature due to stronger ionic attractions, so collisions become more energetic.
3. The higher $\text{Fe}^{3+}$ concentration changes the equilibrium position, and the reaction rate increases because more product is favored.
4. The higher $\text{Fe}^{3+}$ concentration introduces a catalyst-like effect by nitrate ions, changing the mechanism to a faster pathway.
5. The higher $\text{Fe}^{3+}$ concentration makes ions collide less often, but each collision is guaranteed to react because ions run out of energy more slowly.

Explanation: This question tests understanding of the collision model. When Fe³⁺ concentration increases while SCN⁻ concentration stays constant, there are more Fe³⁺ ions available to collide with SCN⁻ ions per unit volume. For the red FeSCN²⁺ complex to form, Fe³⁺ and SCN⁻ ions must collide with proper orientation. With more Fe³⁺ ions present, these collisions occur more frequently, leading to more effective collisions per second and faster formation of the red complex. Choice C incorrectly conflates equilibrium position with reaction rate; while higher Fe³⁺ concentration may shift equilibrium, this doesn't explain the faster initial rate. The strategy is that increasing the concentration of one reactant increases its collision frequency with other reactants, speeding up the reaction.

Question 9

1. Grinding the solid increases the energy of the acid molecules, so collisions become effective more often even without changing surface area.
2. Grinding the solid increases surface area, providing more sites where acid particles can collide with the solid per unit time, increasing the reaction rate. (correct answer)
3. Grinding the solid shifts the reaction toward products, so the forward reaction rate increases to reestablish equilibrium.
4. Grinding the solid changes the reaction mechanism to a faster one because freshly exposed atoms act as a catalyst.
5. Grinding the solid decreases the number of reactant particles in solution, so fewer collisions occur but each collision forms more product.

Explanation: This question applies the collision model to surface area effects in heterogeneous reactions. Grinding the metal oxide into powder in Condition 2 creates vastly more surface area than the large pellets in Condition 1. Since the acid can only react with oxide atoms at the solid surface, this increased surface area provides many more sites where acid molecules can collide with the solid per unit time. The higher collision frequency at the expanded interface directly increases the reaction rate. Choice A incorrectly suggests grinding affects acid molecule energy—grinding only changes the solid's surface area, not the kinetic energy of solution particles. For reactions involving solids, remember that reaction rate is proportional to surface area because collisions can only occur at the interface.

Question 10

1. At higher temperature, ions have higher average kinetic energy, so a larger fraction of collisions is energetic enough to result in reaction when ions collide. (correct answer)
2. At higher temperature, the solutions become more concentrated because water molecules disappear, increasing collision frequency.
3. At higher temperature, the equilibrium constant increases, and the reaction rate increases because more precipitate is favored.
4. At higher temperature, the mechanism changes to a different pathway that produces $\text{PbI}_2$ without requiring collisions.
5. At higher temperature, ions move more slowly due to increased viscosity, so fewer collisions occur but each collision is more likely to react.

Explanation: This question tests understanding of the collision model. At 35°C compared to 15°C, the K⁺, I⁻, Pb²⁺, and NO₃⁻ ions have higher average kinetic energy due to the temperature increase. When these ions collide with more kinetic energy, a larger fraction of Pb²⁺-I⁻ collisions has sufficient energy to overcome electrostatic barriers and form the PbI₂ precipitate. Even though ions must still collide with proper orientation, the increased kinetic energy makes more collisions effective at forming the solid product. Choice B incorrectly claims water molecules disappear at higher temperature, which would violate mass conservation. The strategy is that higher temperature increases the fraction of collisions with enough energy to react, making precipitation reactions occur faster.

Question 11

1. Stirring improves contact between reactant particles by bringing fresh acid to the solid surface more often, increasing the frequency of effective collisions at the surface. (correct answer)
2. Stirring makes bicarbonate particles chemically more reactive by weakening ionic bonds, so collisions no longer require proper orientation.
3. Stirring increases the average kinetic energy of particles, so a greater fraction of collisions is energetic enough to react.
4. Stirring changes the reaction mechanism by mechanically lowering the activation energy of the reaction.
5. Stirring shifts the equilibrium toward $\text{CO}_2$ by removing gas, so the reaction rate increases to restore equilibrium.

Explanation: This question tests understanding of the collision model. Stirring the mixture continuously brings fresh acetic acid solution to the sodium bicarbonate surface and removes products from the surface. Without stirring, a layer of products can accumulate near the solid surface, limiting contact between unreacted acid and bicarbonate. Stirring maintains good contact between reactants by constantly refreshing the liquid layer at the solid surface, increasing the frequency of effective collisions between CH₃COOH and NaHCO₃. This leads to faster CO₂ production. Choice A incorrectly suggests stirring increases kinetic energy, but mechanical stirring at constant temperature doesn't significantly change the thermal energy of particles. The strategy is that stirring improves mixing and maintains fresh reactant contact at interfaces, increasing collision frequency in heterogeneous reactions.

Question 12

1. The more concentrated solution has more $\text{H}_2\text{O}_2$ molecules per unit volume, increasing collision frequency and thus the number of effective collisions per second. (correct answer)
2. The more concentrated solution has a larger equilibrium constant, which forces the decomposition to occur faster.
3. The more concentrated solution increases the activation energy barrier, which paradoxically speeds up the reaction by storing energy in bonds.
4. The more concentrated solution changes the mechanism by creating a new intermediate that is not present in dilute solution.
5. The more concentrated solution makes each molecule move faster at the same temperature, so every collision is more energetic.

Explanation: This question tests understanding of the collision model. In the more concentrated H₂O₂ solution, there are more H₂O₂ molecules per unit volume compared to the dilute solution. For decomposition to occur, H₂O₂ molecules must collide with each other with proper orientation and sufficient energy. With more molecules per unit volume, these collisions happen more frequently, leading to more effective collisions per second and faster O₂ production. Choice E incorrectly claims concentration affects molecular speed, but at constant temperature, average molecular speed depends only on temperature, not concentration. The strategy is that higher concentration increases collision frequency between reactant molecules, which increases the rate of effective collisions.

Question 13

1. The reaction is faster because crushing the solid increases its surface area, allowing more frequent collisions between HCl particles and reactive sites on $\text{CaCO}_3$. (correct answer)
2. The reaction is faster because powdering the solid increases the kinetic energy of the acid particles, making each collision more energetic.
3. The reaction is faster because smaller particles shift the equilibrium toward $\text{CO}_2$ formation, increasing the reaction rate.
4. The reaction is faster because crushing the solid changes the reaction mechanism to one with fewer steps.
5. The reaction is faster because the powdered solid contains more total moles of $\text{CaCO}_3$ than the single large piece.

Explanation: This question tests understanding of the collision model. Crushing a solid into powder dramatically increases its surface area while maintaining the same total mass. With more surface area exposed, there are many more sites where HCl particles can collide with CaCO₃, increasing the collision frequency between reactants. The energy of individual collisions and the fraction that are effective remain unchanged since temperature is constant. Choice E incorrectly assumes powdering changes the amount of substance - the same mass means the same number of moles regardless of particle size. The strategy is that for heterogeneous reactions involving solids, increasing surface area increases collision frequency by providing more contact points for reaction.

Question 14

1. The smaller container increases the number density of gas particles, increasing collision frequency between $\text{NH}_3$ and HCl and thus increasing the number of effective collisions per unit time. (correct answer)
2. The smaller container increases the temperature of the gases, so collisions become more energetic and therefore more effective.
3. The smaller container increases the equilibrium constant for forming $\text{NH}_4\text{Cl}$, so the forward reaction must occur faster.
4. The smaller container changes the reaction mechanism because wall collisions convert kinetic energy into chemical energy more efficiently.
5. The smaller container slows the reaction because particles have less distance to travel and therefore collide less often.

Explanation: This question tests understanding of the collision model. In the smaller container, the same number of NH₃ and HCl molecules occupy less volume, increasing their concentration (number density). Higher concentration means the gas molecules are closer together on average, leading to more frequent collisions between NH₃ and HCl molecules per unit time. Temperature is constant, so the kinetic energy distribution and fraction of effective collisions remain unchanged—only collision frequency increases due to the higher number density. Choice B incorrectly claims volume affects temperature, confusing compression at constant temperature with adiabatic compression. When gases react in smaller volumes at constant temperature, faster reactions result from increased collision frequency due to higher concentration.

Question 15

1. At higher temperature, a larger fraction of collisions have sufficient kinetic energy to be effective, so more collisions lead to reaction per unit time. (correct answer)
2. At higher temperature, the reaction produces more stable products, so the system speeds up to form them more quickly.
3. At higher temperature, the equilibrium constant becomes larger, so the forward reaction rate must increase.
4. At higher temperature, the reactant molecules collide less often because they spread farther apart, so only the strongest collisions react.
5. At higher temperature, the mechanism changes to a single-step pathway, so fewer collisions are required to form products.

Explanation: This question examines the collision model's temperature dependence. At the higher temperature in Condition 2, reactant molecules have greater average kinetic energy compared to Condition 1. This means a larger fraction of NO-O₃ collisions possess sufficient energy to overcome the activation energy barrier and form products. While collision frequency also increases slightly with temperature, the dominant effect is the dramatic increase in the fraction of effective collisions. Choice C incorrectly relates equilibrium constants to reaction rates—a reaction can be fast regardless of its equilibrium position. To predict temperature effects on rates, focus on how temperature affects the fraction of collisions with sufficient energy to react.

Question 16

1. At higher temperature, reactant particles move faster, so a greater fraction of collisions have sufficient kinetic energy to be effective and form products. (correct answer)
2. At higher temperature, the reaction produces more products at equilibrium, so the forward reaction rate increases.
3. At higher temperature, the reactants become catalysts for each other, lowering the activation energy without changing collisions.
4. At higher temperature, the reactant particles collide less often because they spend more time moving in straight lines between collisions.
5. At higher temperature, the mechanism changes to require fewer steps, so fewer collisions are needed to complete the reaction.

Explanation: This question tests temperature effects on collision model for solution reactions. At the higher temperature in Condition 2, reactant particles have greater average kinetic energy than in Condition 1. This increased kinetic energy means a larger fraction of collisions between sodium thiosulfate and HCl particles possess sufficient energy to overcome the activation barrier and form products. The exponential relationship between temperature and the fraction of effective collisions explains why even modest temperature increases can dramatically speed reactions. Choice B incorrectly connects equilibrium position to reaction rate—a reaction can be fast or slow regardless of its equilibrium constant. When analyzing temperature effects, focus on how higher kinetic energy increases the fraction of collisions that are energetically capable of reaction.

Question 17

1. Powdering lowers the activation energy by creating a catalyst on the surface of the solid, increasing the reaction rate.
2. Powdered $\text{CaCO}_3$ has greater surface area, so more acid particles can collide with the solid per unit time, increasing the reaction rate. (correct answer)
3. Powdering shifts the reaction toward $\text{CO}_2$, so the forward reaction speeds up to reach equilibrium sooner.
4. Powdering increases the temperature of the solid, so collisions have higher energy and always become products.
5. Powdered $\text{CaCO}_3$ is a different allotrope with weaker ionic bonds, so it reacts faster even with the same collisions.

Explanation: This question tests how surface area affects collision model predictions. The powdered CaCO₃ in Condition 2 has much greater surface area than the single chip in Condition 1, despite having the same total mass. This increased surface area provides many more sites where HCl molecules can collide with calcium carbonate per unit time. Since the reaction only occurs at the solid-liquid interface, more surface area directly translates to higher collision frequency and faster CO₂ production. Choice B incorrectly suggests powdering changes the chemical identity—powdered and chunk CaCO₃ are the same compound with identical bond strengths. When solid reactants are involved, remember that reaction rate depends on collision frequency at the surface, which increases with surface area.

Question 18

1. At higher temperature, a larger fraction of collisions have sufficient kinetic energy to be effective, so more collisions lead to reaction per unit time. (correct answer)
2. At higher temperature, the reaction produces more stable products, so the system speeds up to form them more quickly.
3. At higher temperature, the equilibrium constant becomes larger, so the forward reaction rate must increase.
4. At higher temperature, the reactant molecules collide less often because they spread farther apart, so only the strongest collisions react.
5. At higher temperature, the mechanism changes to a single-step pathway, so fewer collisions are required to form products.

Explanation: This question examines the collision model's temperature dependence. At the higher temperature in Condition 2, reactant molecules have greater average kinetic energy compared to Condition 1. This means a larger fraction of NO-O₃ collisions possess sufficient energy to overcome the activation energy barrier and form products. While collision frequency also increases slightly with temperature, the dominant effect is the dramatic increase in the fraction of effective collisions. Choice C incorrectly relates equilibrium constants to reaction rates—a reaction can be fast regardless of its equilibrium position. To predict temperature effects on rates, focus on how temperature affects the fraction of collisions with sufficient energy to react.

Question 19

1. The smaller volume increases the frequency of $\text{H}_2$–$\text{I}_2$ collisions by placing more molecules per unit volume, increasing the number of effective collisions per second. (correct answer)
2. The smaller volume increases the average kinetic energy of molecules, so collisions are more effective even though temperature is unchanged.
3. The smaller volume shifts equilibrium toward HI, so the reaction rate increases to produce the favored product more quickly.
4. The smaller volume changes the mechanism because molecules collide with the walls more often, creating a new surface-catalyzed pathway.
5. The smaller volume reduces the need for proper molecular orientation, so nearly all collisions form HI regardless of alignment.

Explanation: This question tests understanding of the collision model. In the smaller volume flask, the same number of H₂ and I₂ molecules occupy less space, resulting in higher gas concentrations. With molecules packed more closely together, H₂ and I₂ molecules collide with each other more frequently per unit time. Since HI formation requires H₂-I₂ collisions with proper orientation and sufficient energy, the increased collision frequency leads to more effective collisions per second. This causes HI to form faster in the smaller volume. Choice B incorrectly claims volume affects kinetic energy, but at constant temperature, average kinetic energy depends only on temperature, not volume or pressure. The strategy is that decreasing volume increases gas concentration, which increases collision frequency and reaction rate.

Question 20

1. At higher temperature, the solution contains more particles, so collisions happen more often because concentration increases.
2. At higher temperature, a larger fraction of particle collisions has sufficient kinetic energy to be effective, increasing the reaction rate. (correct answer)
3. At higher temperature, the equilibrium constant becomes larger, forcing the reaction to proceed faster to make more products.
4. At higher temperature, the mechanism changes so that sulfur forms in a single step rather than multiple steps.
5. At higher temperature, reactant particles run out of energy less quickly, so the reaction continues longer and therefore seems faster.

Explanation: This question tests understanding of the collision model. At 40°C compared to 20°C, the reactant particles have higher average kinetic energy due to the increased temperature. When particles collide with more kinetic energy, a larger fraction of these collisions exceeds the activation energy barrier needed for reaction. This means more collisions are effective at forming products, even though the total collision frequency also increases slightly. The cloudiness appears sooner because sulfur precipitate forms faster when more collisions per second are effective. Choice C incorrectly relates equilibrium constants to reaction rates, but equilibrium position and reaction rate are independent concepts. The strategy is that higher temperature increases the fraction of collisions with sufficient energy to react, making more collisions effective.