Collision Model
Help Questions
AP Chemistry › Collision Model
In a kinetics lab, a student reacts aqueous iodide ions with aqueous hydrogen peroxide under acidic conditions to produce iodine. Two trials are run with the same total volume, the same temperature, and the same reaction mechanism (no catalyst; same species present). Trial 1 is vigorously stirred throughout, while Trial 2 is left unstirred. The iodine color develops faster in Trial 1. Which statement best explains why Trial 1 proceeds faster in terms of collision frequency or collision effectiveness?
Trial 1 is faster because stirring adds energy to the system in the form of heat, which guarantees that every collision leads to reaction.
Trial 1 is faster because stirring increases the rate constant by changing the reaction mechanism in solution.
Trial 1 is faster because stirring improves mixing and reduces concentration gradients, increasing how often reactant particles encounter each other and collide effectively.
Trial 1 is faster because unstirred solutions cause reactant particles to lose energy permanently, making collisions ineffective after a short time.
Trial 1 is faster because stirring increases the equilibrium constant, so products are favored sooner.
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.
A student investigates the reaction $\text{R(g)} + \text{S(g)} \rightarrow \text{products}$ in a container where the mechanism is unchanged. In Trial 1, the gases are at a lower temperature. In Trial 2, the gases are at a higher temperature, while the volume and the number of moles of each gas are kept the same. Which statement best explains why Trial 2 is faster, focusing specifically on collision effectiveness (not just collision frequency)?
The reaction is faster because higher temperature increases the concentration of gases, so collisions occur more frequently.
The reaction is faster because heating acts as a catalyst that provides an alternative mechanism while keeping the same reactants.
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.
The reaction is faster because higher temperature increases the fraction of collisions with sufficient energy to result in reaction, making collisions more effective.
The reaction is faster because higher temperature shifts equilibrium toward products, which increases the forward reaction rate.
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.
A student compares two trials of the reaction between zinc metal and aqueous copper(II) sulfate, producing copper metal. The mechanism is unchanged and the temperature and solution concentration are the same.
Condition 1: A single strip of Zn(s) is placed into the solution.
Condition 2: The same mass of Zn(s) is used, but it is cut into many small pieces before being placed into the solution.
Which statement best explains why the reaction proceeds faster in Condition 2 than in Condition 1 using collision frequency at the solid–solution interface?
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.
Cutting the zinc introduces a catalyst on the fresh metal surface that lowers the activation energy and changes the mechanism.
Cutting the zinc decreases the number of collisions needed by changing the products formed, so the reaction finishes sooner.
Cutting the zinc makes each collision more energetic, so nearly every collision becomes products even at the same temperature.
Cutting the zinc increases the equilibrium constant for the reaction, so the forward reaction must occur faster.
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.
A student investigates the reaction between aqueous hydrochloric acid and magnesium metal, which produces hydrogen gas and dissolved magnesium ions. Two trials use the same total volume of solution and the same mass of Mg(s), and no catalyst is present.
Condition 1: Mg(s) is added to a dilute HCl(aq) solution.
Condition 2: Mg(s) is added to a more concentrated HCl(aq) solution.
The student is told that the reaction mechanism is unchanged between the two conditions and that the only difference is how often reactant particles collide at the metal surface. Which statement best explains why the reaction occurs faster in Condition 2 than in Condition 1, using collision-model reasoning?
The more concentrated acid makes the Mg atoms vibrate faster, so the Mg bonds break without needing collisions from acid particles.
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.
The more concentrated acid causes the reaction to shift toward products, so the reaction must proceed faster to reach equilibrium.
The more concentrated acid lowers the activation energy by acting as a catalyst, allowing more collisions to form products.
The more concentrated acid increases the energy released by the reaction, so each collision produces products more quickly.
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.
A student investigates the precipitation reaction that occurs when aqueous solutions of $\text{AgNO}_3$ and $\text{NaCl}$ are mixed, forming $\text{AgCl}(s)$. The mechanism is unchanged, and the only difference between trials is how frequently reactant ions encounter each other after mixing.
Condition 1: Both solutions are relatively dilute before mixing.
Condition 2: Both solutions are more concentrated before mixing.
Both mixtures are stirred in the same way and kept at the same temperature. Which statement best explains why the precipitate forms faster in Condition 2 than in Condition 1 using collision-model reasoning?
In the more concentrated mixture, the products are more stable, so the reaction pathway becomes faster even if collision frequency is unchanged.
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 $\text{AgCl}(s)$.
In the more concentrated mixture, ions collide less often because electrostatic attractions keep them separated, slowing precipitation.
In the more concentrated mixture, the activation energy is lower because concentration acts like a catalyst, making every collision effective.
In the more concentrated mixture, the equilibrium constant is larger, so the forward reaction rate must increase.
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.
A student examines the decomposition of hydrogen peroxide in water, which produces oxygen gas. Two samples contain the same volume of solution and no catalyst is present. The mechanism is unchanged between conditions; only collision-related factors differ.
Condition 1: The $\text{H}_2\text{O}_2(aq)$ solution is kept at a lower temperature.
Condition 2: The $\text{H}_2\text{O}_2(aq)$ solution is kept at a higher temperature.
Which statement best explains why oxygen gas is produced faster in Condition 2 than in Condition 1, focusing on collision effectiveness rather than memorized rules?
At higher temperature, the reaction changes mechanism to a faster pathway, so fewer collisions are needed to form products.
At higher temperature, the system approaches equilibrium faster, so the forward rate increases because the reverse rate decreases.
At higher temperature, the products have lower potential energy, so the reaction becomes faster to release energy sooner.
At higher temperature, the solution contains more $\text{H}_2\text{O}_2$ molecules per unit volume, so collision frequency increases.
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.
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.
A student reacts calcium carbonate with hydrochloric acid: $\text{CaCO}_3(s)+2\text{HCl}(aq)\rightarrow \text{CaCl}_2(aq)+\text{CO}_2(g)+\text{H}_2\text{O}(l)$. Trial 1 uses a single marble chip of $\text{CaCO}_3$; Trial 2 uses the same mass of $\text{CaCO}_3$ but finely crushed into powder. The acid concentration and temperature are the same in both trials, and no catalyst is used, so the mechanism is unchanged. Trial 2 produces $\text{CO}_2$ gas faster. Which statement best explains this difference in terms of collisions?
Powdered $\text{CaCO}_3$ exposes more surface area, increasing the frequency of effective collisions between acid particles and the solid surface.
Powdered $\text{CaCO}_3$ contains weaker bonds than a marble chip, so every collision forms products regardless of orientation.
Powdered $\text{CaCO}_3$ increases the acid concentration near the surface, which changes the mechanism and speeds the reaction.
Powdered $\text{CaCO}_3$ shifts the equilibrium toward $\text{CO}_2$, so the reaction must occur faster to produce more gas.
Powdering the solid increases the kinetic energy of the acid particles, so a greater fraction of collisions has enough energy to react.
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.
A student reacts iron(III) nitrate with potassium thiocyanate: $\text{Fe}^{3+}(aq)+\text{SCN}^-(aq)\rightarrow \text{FeSCN}^{2+}(aq)$. Two trials are run at the same temperature with no catalyst, so the mechanism is unchanged. In Trial 1, the student mixes a dilute $\text{Fe}^{3+}$ solution with a fixed $\text{SCN}^-$ concentration. In Trial 2, the student mixes a more concentrated $\text{Fe}^{3+}$ solution with the same $\text{SCN}^-$ concentration. The red color forms faster in Trial 2. Which statement best explains the faster rate in Trial 2 using collision reasoning?
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.
The higher $\text{Fe}^{3+}$ concentration introduces a catalyst-like effect by nitrate ions, changing the mechanism to a faster pathway.
The higher $\text{Fe}^{3+}$ concentration changes the equilibrium position, and the reaction rate increases because more product is favored.
The higher $\text{Fe}^{3+}$ concentration increases the temperature due to stronger ionic attractions, so collisions become more energetic.
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.
A student compares two trials of the same reaction between a metal oxide solid and an acid solution. The mechanism is unchanged; only collision-related factors differ.
Condition 1: The solid metal oxide is present as large pellets in the acid.
Condition 2: The same mass of the solid metal oxide is ground into a fine powder before being added to the acid.
Acid concentration, temperature, and stirring are the same. Which statement best explains why the reaction is faster in Condition 2 than in Condition 1, focusing on collision frequency at the solid surface?
Grinding the solid increases the energy of the acid molecules, so collisions become effective more often even without changing surface area.
Grinding the solid increases surface area, providing more sites where acid particles can collide with the solid per unit time, increasing the reaction rate.
Grinding the solid shifts the reaction toward products, so the forward reaction rate increases to reestablish equilibrium.
Grinding the solid decreases the number of reactant particles in solution, so fewer collisions occur but each collision forms more product.
Grinding the solid changes the reaction mechanism to a faster one because freshly exposed atoms act as a catalyst.
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.
A student compares the reaction of aqueous potassium iodide with aqueous lead(II) nitrate: $2\text{KI}(aq)+\text{Pb(NO}_3)_2(aq)\rightarrow \text{PbI}_2(s)+2\text{KNO}_3(aq)$. In Trial 1, both solutions are at $15^\circ$C. In Trial 2, both solutions are at $35^\circ$C. The concentrations are the same in both trials and no catalyst is used, so the mechanism is unchanged. The yellow precipitate appears faster in Trial 2. Which statement best explains the faster reaction at the higher temperature in terms of collision effectiveness?
At higher temperature, the solutions become more concentrated because water molecules disappear, increasing collision frequency.
At higher temperature, the mechanism changes to a different pathway that produces $\text{PbI}_2$ without requiring collisions.
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.
At higher temperature, ions move more slowly due to increased viscosity, so fewer collisions occur but each collision is more likely to react.
At higher temperature, the equilibrium constant increases, and the reaction rate increases because more precipitate is favored.
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.