Collision Model
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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 improves mixing and reduces concentration gradients, increasing how often reactant particles encounter each other and collide effectively.
Trial 1 is faster because stirring increases the rate constant by changing the reaction mechanism in solution.
Trial 1 is faster because stirring increases the equilibrium constant, so products are favored sooner.
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 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.
A student studies the same reaction in two sealed flasks: $\text{NO}(g)+\text{O}_3(g)\rightarrow \text{NO}_2(g)+\text{O}_2(g)$. The mechanism is unchanged between trials, and the only difference is the temperature.
Condition 1: The gases are mixed at a lower temperature.
Condition 2: The gases are mixed at a higher temperature.
Which statement best explains why the reaction occurs faster in Condition 2 than in Condition 1, based on collision effectiveness?
At higher temperature, the reaction produces more stable products, so the system speeds up to form them more quickly.
At higher temperature, the equilibrium constant becomes larger, so the forward reaction rate must increase.
At higher temperature, the mechanism changes to a single-step pathway, so fewer collisions are required to form products.
At higher temperature, a larger fraction of collisions have sufficient kinetic energy to be effective, so more collisions lead to reaction per unit time.
At higher temperature, the reactant molecules collide less often because they spread farther apart, so only the strongest collisions react.
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.
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 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 increases the concentration of gases, so collisions occur more frequently.
The reaction is faster because higher temperature shifts equilibrium toward products, which increases the forward reaction rate.
The reaction is faster because heating acts as a catalyst that provides an alternative mechanism while keeping the same reactants.
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 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.
Cutting the zinc decreases the number of collisions needed by changing the products formed, so the reaction finishes sooner.
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.
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 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 increases the energy released by the reaction, so each collision produces products more quickly.
The more concentrated acid causes the reaction to shift toward products, so the reaction must proceed faster to reach equilibrium.
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 lowers the activation energy by acting as a catalyst, allowing more collisions to form products.
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 reaction between sodium thiosulfate and hydrochloric acid in aqueous solution. The mechanism is unchanged between trials, and no catalyst is used.
Condition 1: The reactant solutions are mixed at a lower temperature.
Condition 2: The reactant solutions are mixed at a higher temperature.
Initial concentrations and volumes are the same. Which statement best explains why the reaction occurs faster in Condition 2 than in Condition 1, emphasizing collision effectiveness rather than simply stating that “temperature increases rate”?
At higher temperature, the reaction produces more products at equilibrium, so the forward reaction rate increases.
At higher temperature, the reactant particles collide less often because they spend more time moving in straight lines between collisions.
At higher temperature, the mechanism changes to require fewer steps, so fewer collisions are needed to complete the reaction.
At higher temperature, reactant particles move faster, so a greater fraction of collisions have sufficient kinetic energy to be effective and form products.
At higher temperature, the reactants become catalysts for each other, lowering the activation energy without changing collisions.
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.
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, ions collide less often because electrostatic attractions keep them separated, slowing precipitation.
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, the equilibrium constant is larger, so the forward reaction rate must increase.
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.
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 solution contains more $\text{H}_2\text{O}_2$ molecules per unit volume, so collision frequency increases.
At higher temperature, the reaction changes mechanism to a faster pathway, so fewer collisions are needed to form products.
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.
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.
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 studies the same reaction in two sealed flasks: $\text{NO}(g)+\text{O}_3(g)\rightarrow \text{NO}_2(g)+\text{O}_2(g)$. The mechanism is unchanged between trials, and the only difference is the temperature.
Condition 1: The gases are mixed at a lower temperature.
Condition 2: The gases are mixed at a higher temperature.
Which statement best explains why the reaction occurs faster in Condition 2 than in Condition 1, based on collision effectiveness?
At higher temperature, the reaction produces more stable products, so the system speeds up to form them more quickly.
At higher temperature, the mechanism changes to a single-step pathway, so fewer collisions are required to form products.
At higher temperature, the equilibrium constant becomes larger, so the forward reaction rate must increase.
At higher temperature, the reactant molecules collide less often because they spread farther apart, so only the strongest collisions react.
At higher temperature, a larger fraction of collisions have sufficient kinetic energy to be effective, so more collisions lead to reaction per unit time.
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.
A student compares the reaction of calcium carbonate with hydrochloric acid, producing $\text{CO}_2(g)$. The mechanism is unchanged and no catalyst is used.
Condition 1: A single large chip of $\text{CaCO}_3(s)$ is placed in the acid.
Condition 2: The same mass of $\text{CaCO}_3(s)$ is added as a fine powder to the acid.
The acid concentration and temperature are the same in both conditions. Which statement best explains why the reaction occurs faster in Condition 2 than in Condition 1 in terms of collision frequency?
Powdering lowers the activation energy by creating a catalyst on the surface of the solid, increasing the reaction rate.
Powdered $\text{CaCO}_3$ is a different allotrope with weaker ionic bonds, so it reacts faster even with the same collisions.
Powdering shifts the reaction toward $\text{CO}_2$, so the forward reaction speeds up to reach equilibrium sooner.
Powdered $\text{CaCO}_3$ has greater surface area, so more acid particles can collide with the solid per unit time, increasing the reaction rate.
Powdering increases the temperature of the solid, so collisions have higher energy and always become products.
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.