AP Chemistry Quiz: Catalysts

20 questions · exam conditions

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CatalystsQuestion 1 of 20

The reaction $\text{SO}_2(g)+\text{NO}_2(g)\rightarrow \text{SO}_3(g)+\text{NO}(g)$ is studied in a sealed container at constant temperature. In a second trial, a small amount of $\text{NO}(g)$ is added as a catalyst and is regenerated during the reaction. The overall reactants and products remain the same, and the catalyst is not consumed.

Which statement best explains the increased rate in the presence of the catalyst?

The catalyst increases the final yield of

\text{SO}_3

by shifting the reaction toward products, which makes the reaction faster.

The catalyst provides an alternate sequence of collisions and intermediate encounters that makes product-forming collisions more likely.

The catalyst is consumed in a side reaction that releases heat, raising the temperature and increasing the rate.

The catalyst increases the concentration of

\text{SO}_2

by reacting with it to form more

\text{SO}_2

molecules, raising collision frequency.

The catalyst changes the overall reaction to produce different products that form more quickly than

\text{SO}_3

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AP Chemistry Quiz

AP Chemistry Quiz: Catalysts

Practice Catalysts in AP Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Catalysts, 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.

All questions

Question 1

Which statement best explains the increased rate in the presence of the catalyst?

The catalyst increases the final yield of $\text{SO}_3$ by shifting the reaction toward products, which makes the reaction faster.
The catalyst provides an alternate sequence of collisions and intermediate encounters that makes product-forming collisions more likely. (correct answer)
The catalyst is consumed in a side reaction that releases heat, raising the temperature and increasing the rate.
The catalyst increases the concentration of $\text{SO}_2$ by reacting with it to form more $\text{SO}_2$ molecules, raising collision frequency.
The catalyst changes the overall reaction to produce different products that form more quickly than $\text{SO}_3$ .

Explanation: This question assesses understanding of catalysts. Catalysts increase the reaction rate by enabling an alternative sequence of collisions that makes it more likely for reactants like SO2 and NO2 to form products effectively. With NO(g) as a catalyst, intermediate steps allow for better alignment or interaction during molecular encounters. This results in a larger fraction of collisions successfully producing SO3 and NO. One tempting distractor is choice A, which is wrong due to the misconception that catalysts change equilibrium by increasing final product yield, whereas catalysts speed up the attainment of equilibrium without changing its position. A transferable strategy is to remember that catalysts affect rate, not the final position of a reaction.

Question 2

Aqueous iodide reacts with hydrogen peroxide in acidic solution: $\text{H}_2\text{O}_2(aq)+2\text{I}^-(aq)+2\text{H}^+(aq)\rightarrow \text{I}_2(aq)+2\text{H}_2\text{O}(l)$ . Two experiments use the same initial concentrations and temperature. In Experiment 2, a small amount of $\text{Fe}^{3+}(aq)$ is added and is regenerated during the reaction (not consumed overall). The reactants and products are the same in both experiments.

Why does adding $\text{Fe}^{3+}(aq)$ increase the reaction rate?

Adding the catalyst increases reactant concentrations by supplying additional $\text{I}^-$ ions, leading to more collisions.
Adding the catalyst increases the final amount of $\text{I}_2$ produced, so the reaction must proceed faster to make more product.
Adding the catalyst changes the identity of the overall reactants and products, creating a faster reaction with different products.
Adding the catalyst introduces an alternate pathway in which reacting particles collide in a way that more frequently leads to product formation. (correct answer)
Adding the catalyst increases the average kinetic energy of the solution particles, causing more frequent collisions due to higher temperature.

Explanation: This question assesses understanding of catalysts. Catalysts increase the reaction rate by introducing an alternate pathway that enhances the effectiveness of collisions between reactants like H2O2, I-, and H+. With Fe3+ as a catalyst, the reaction involves intermediate complexes that allow reactant particles to collide in orientations more conducive to product formation. This leads to a higher fraction of successful collisions that produce I2 and H2O. One tempting distractor is choice A, which is wrong because of the misconception that catalysts change equilibrium by increasing final product amounts, whereas catalysts do not alter the equilibrium yield but only speed up reaching it. A transferable strategy is to remember that catalysts affect rate, not the final position of a reaction.

Question 3

Nitrogen monoxide reacts with ozone in the gas phase: $\text{NO}(g)+\text{O}_3(g)\rightarrow \text{NO}_2(g)+\text{O}_2(g)$ . A second run is performed with a small amount of $\text{Cl}(g)$ present as a catalyst; the overall reactants and products are the same in both runs, and $\text{Cl}(g)$ is not consumed.

Which statement best explains why the catalyzed run has a higher reaction rate?

The catalyst is converted into products, so the number of product-forming collisions increases due to added reactant mass.
The catalyst creates an alternative set of steps that makes a greater proportion of collisions between reacting species effective in forming products. (correct answer)
The catalyst increases the partial pressures of NO and $\text{O}_3$ by adding more gas molecules, which increases the rate.
The catalyst shifts the equilibrium toward $\text{NO}_2$ and $\text{O}_2$ , which increases the rate by favoring products.
The catalyst raises the temperature of the reaction mixture, increasing molecular speeds and therefore increasing the rate.

Explanation: This question assesses understanding of catalysts. Catalysts increase the reaction rate by offering an alternative sequence of steps that facilitates more productive interactions between reactant molecules, such as NO and O3. In the catalyzed pathway with Cl(g), the reactants can form temporary intermediates that make it easier for bonds to break and reform effectively during collisions. Consequently, a greater proportion of molecular encounters result in the formation of products like NO2 and O2. One tempting distractor is choice D, which is incorrect due to the misconception that catalysts change equilibrium by favoring products, but in reality, catalysts accelerate both directions of a reversible reaction equally without shifting the equilibrium. A transferable strategy is to remember that catalysts affect rate, not the final position of a reaction.

Question 4

The reaction $2\text{NO}(g)+\text{O}_2(g)\rightarrow 2\text{NO}_2(g)$ is run at the same temperature and initial pressures in two containers. In Container 2, a catalyst is present and is recovered unchanged; the overall reactants and products are the same.

Which statement best explains why the reaction rate is higher in Container 2?

The catalyst increases the rate by increasing the amount of $\text{NO}_2$ produced at completion, requiring faster production.
The catalyst increases the rate by decreasing the concentration of $\text{NO}_2$ , which forces the reaction to proceed faster to replace it.
The catalyst increases the rate by being consumed to form $\text{NO}_2$ , increasing the number of reacting particles.
The catalyst increases the rate by changing the overall reaction to a different product that forms more quickly than $\text{NO}_2$ .
The catalyst increases the rate by enabling a different pathway in which collisions between reacting species more frequently result in product formation. (correct answer)

Explanation: This question assesses understanding of catalysts. Catalysts increase the reaction rate by enabling a different pathway where collisions between reactants like NO and O2 more frequently result in product formation. In the catalyzed container, intermediate steps enhance the effectiveness of molecular encounters for NO2 production. Thus, a larger proportion of collisions are successful. One tempting distractor is choice A, which is incorrect because it embodies the misconception that catalysts change equilibrium by increasing final product amounts, but catalysts only speed up reaching equilibrium. A transferable strategy is to remember that catalysts affect rate, not the final position of a reaction.

Question 5

In acidic solution, bromate reacts with bromide: $\text{BrO}_3^-(aq)+5\text{Br}^-(aq)+6\text{H}^+(aq)\rightarrow 3\text{Br}_2(aq)+3\text{H}_2\text{O}(l)$ . Two runs use the same initial concentrations and temperature. Run 2 includes a small amount of $\text{Br}^-(aq)$ -regenerating catalyst (present at the end). The overall reactants and products are the same.

Which statement best explains why Run 2 proceeds faster?

The catalyst increases the rate by increasing the concentration of $\text{H}^+$ permanently, so collisions occur more frequently.
The catalyst increases the rate by shifting the reaction toward $\text{Br}_2$ , increasing the driving force for product formation.
The catalyst increases the rate by participating in steps that allow reactant species to interact in a way that more often forms products. (correct answer)
The catalyst increases the rate by being consumed to form $\text{Br}_2$ , adding additional reactant mass to the system.
The catalyst increases the rate by supplying energy to the reactants, increasing their average kinetic energy without changing temperature.

Explanation: This question assesses understanding of catalysts. Catalysts increase the reaction rate by participating in steps that allow reactants like BrO3-, Br-, and H+ to interact more effectively during collisions. With the catalyst, the pathway enhances the success rate of encounters leading to Br2 and H2O. This leads to a larger fraction of productive collisions. One tempting distractor is choice B, which is wrong due to the misconception that catalysts change equilibrium by shifting toward products, whereas catalysts do not affect the equilibrium position. A transferable strategy is to remember that catalysts affect rate, not the final position of a reaction.

Question 6

In a lab, students study the reaction $\text{N}_2(g)+3\text{H}_2(g)\rightarrow 2\text{NH}_3(g)$ . Trial 1 is run with an iron surface present; Trial 2 is run without iron. The temperature, pressures, and initial amounts of gases are the same in both trials. The overall reactants and products are the same, and the iron is not used up.

Which statement best explains why the iron surface increases the reaction rate?

The iron surface increases the pressure of the gases by releasing trapped gas, increasing collision frequency.
The iron surface is converted into ammonia, increasing the amount of reactant available and speeding the reaction.
The iron surface increases the equilibrium constant for ammonia formation, which increases the forward reaction rate permanently.
The iron surface supplies energy to break bonds in $\text{N}_2$ and $\text{H}_2$ , increasing the rate by heating the mixture.
The iron surface provides sites that help orient and interact the reactant molecules so a larger fraction of their collisions lead to product formation. (correct answer)

Explanation: This question assesses understanding of catalysts. Catalysts increase the reaction rate by providing a surface or pathway where reactants like N2 and H2 can adsorb and interact more effectively, leading to a higher success rate in collisions. On the iron surface, molecules are oriented in ways that facilitate bond breaking and forming during encounters. Thus, more collisions result in the formation of NH3 compared to the uncatalyzed trial. One tempting distractor is choice B, which is incorrect because it embodies the misconception that catalysts change equilibrium by increasing the equilibrium constant, but catalysts actually do not affect the equilibrium constant or position. A transferable strategy is to remember that catalysts affect rate, not the final position of a reaction.

Question 7

Two trials of the same reaction are performed in water at the same temperature: $\text{CO}_2(aq)+\text{H}_2\text{O}(l)\rightarrow \text{H}_2\text{CO}_3(aq)$ . Trial 2 contains the enzyme carbonic anhydrase, which is not consumed. The overall reactants and products are the same in both trials.

Which statement best explains why the enzyme increases the reaction rate?

The enzyme increases the rate by shifting the reaction toward $\text{H}_2\text{CO}_3$ , increasing product formation and speeding the process.
The enzyme increases the rate by providing a specific environment that helps reactants collide in an orientation that more often leads to product formation. (correct answer)
The enzyme increases the rate by being consumed to form $\text{H}_2\text{CO}_3$ , increasing the amount of reactant available.
The enzyme increases the rate by increasing the temperature of the solution through exothermic binding, increasing collision frequency.
The enzyme increases the rate by increasing the concentration of dissolved $\text{CO}_2$ by converting water into $\text{CO}_2$ .

Explanation: This question assesses understanding of catalysts. Catalysts increase the reaction rate by providing a specific environment that helps reactants like CO2 and H2O collide in orientations more conducive to product formation. With the enzyme carbonic anhydrase, binding sites facilitate effective interactions leading to H2CO3. This results in a higher fraction of successful encounters. One tempting distractor is choice A, which is wrong due to the misconception that catalysts change equilibrium by shifting toward products, whereas catalysts accelerate both directions equally. A transferable strategy is to remember that catalysts affect rate, not the final position of a reaction.

Question 8

A student investigates the decomposition of hydrogen peroxide in water: $2\text{H}_2\text{O}_2(aq) \rightarrow 2\text{H}_2\text{O}(l) + \text{O}_2(g)$ . Two trials are run under identical conditions (same temperature, same initial $[\text{H}_2\text{O}_2]$ , same volume). In Trial 2, a small amount of $\text{MnO}_2(s)$ is added. The same reactants and products are present in both trials, and the $\text{MnO}_2$ is recovered unchanged at the end.

Which statement best explains why Trial 2 proceeds faster than Trial 1?

The catalyst shifts the reaction to favor products by changing the equilibrium position, increasing the forward rate only.
The catalyst provides energy to the reactant molecules, increasing their average kinetic energy and making more collisions occur.
The catalyst is consumed to form additional reactive particles, increasing the concentration of reactants during the reaction.
The catalyst offers an alternative pathway that increases the fraction of collisions that successfully form products during each encounter. (correct answer)
The catalyst increases the amount of product formed at completion, so the reaction must occur faster to reach the higher yield.

Explanation: This question assesses understanding of catalysts. Catalysts increase the reaction rate by providing an alternative pathway for the reaction, which allows reactant molecules to interact in a way that requires less precise orientation or energy for successful product formation. In the presence of a catalyst like MnO2, the decomposition of hydrogen peroxide proceeds through intermediate steps where collisions between reactants and the catalyst lead to more effective encounters. As a result, a larger fraction of the collisions between reacting species successfully form products compared to the uncatalyzed reaction. One tempting distractor is choice A, which is incorrect because it reflects the misconception that catalysts change equilibrium by shifting it toward products, whereas catalysts actually speed up both forward and reverse reactions equally without altering the equilibrium position. A transferable strategy is to remember that catalysts affect the rate of a reaction but not the final position of equilibrium.

Question 9

The same reaction is run twice: $\text{C}(s)+\text{O}_2(g)\rightarrow \text{CO}_2(g)$ . In Trial 2, a catalyst is added that is recovered unchanged, and the overall reactants and products remain the same.

Which statement best explains why Trial 2 has a higher reaction rate?

The catalyst increases the reaction rate by being consumed to form $\text{CO}_2$ , increasing the amount of reactant available.
The catalyst increases the reaction rate by shifting the reaction toward products, which increases the forward reaction rate only.
The catalyst increases the reaction rate by increasing the amount of $\text{CO}_2$ produced at completion, requiring faster formation.
The catalyst increases the reaction rate by increasing the surface area of carbon permanently, creating more carbon atoms to react.
The catalyst increases the reaction rate by providing a pathway that makes collisions between reacting species more effective at forming products. (correct answer)

Explanation: This question assesses understanding of catalysts. Catalysts increase the reaction rate by offering a pathway that enhances the effectiveness of collisions between reactants like C and O2. In the catalyzed reaction, intermediate steps or surfaces make product formation more likely during encounters. This results in a higher fraction of successful collisions producing CO2. One tempting distractor is choice E, which is wrong because it reflects the misconception that catalysts change equilibrium by shifting toward products, whereas catalysts do not alter equilibrium positions. A transferable strategy is to remember that catalysts affect rate, not the final position of a reaction.

Question 10

The reaction $\text{CH}_4(g)+2\text{O}_2(g)\rightarrow \text{CO}_2(g)+2\text{H}_2\text{O}(g)$ is studied at the same temperature and initial pressures in two combustion chambers. Chamber 2 contains a catalyst that is not consumed, and the overall reactants and products are the same.

Which statement best explains why the catalyzed reaction can proceed faster?

The catalyst increases the rate by shifting the reaction toward products, increasing the forward rate and decreasing the reverse rate.
The catalyst increases the rate by being converted into $\text{CO}_2$ , increasing the total number of product molecules formed.
The catalyst increases the rate by increasing the average kinetic energy of reactant molecules by supplying heat without changing temperature.
The catalyst increases the rate by providing an alternate pathway that makes collisions between reactant molecules more effective at forming products. (correct answer)
The catalyst increases the rate by increasing the amount of $\text{CO}_2$ and $\text{H}_2\text{O}$ formed at completion, requiring faster production.

Explanation: This question assesses understanding of catalysts. Catalysts increase the reaction rate by offering an alternate pathway that makes collisions between reactants like CH4 and O2 more effective at forming products. In the catalyzed chamber, interactions enhance the success rate of encounters for CO2 and H2O production. As a result, a greater fraction of collisions are productive. One tempting distractor is choice D, which is incorrect due to the misconception that catalysts change equilibrium by shifting toward products, but catalysts do not affect the equilibrium position. A transferable strategy is to remember that catalysts affect rate, not the final position of a reaction.

Question 11

A student studies the reaction $\text{Zn}(s)+2\text{HCl}(aq)\rightarrow \text{ZnCl}_2(aq)+\text{H}_2(g)$ . Two beakers contain the same mass of zinc and the same concentration and volume of HCl at the same temperature. Beaker 2 also contains a small amount of $\text{Cu}^{2+}(aq)$ that is regenerated during the process (not consumed overall). The overall reactants and products are the same.

Which statement best explains why Beaker 2 produces $\text{H}_2$ gas faster?

The catalyst increases the rate by being consumed to form $\text{H}_2$ , increasing the number of product molecules formed.
The catalyst increases the rate by heating the solution, increasing the average kinetic energy of particles.
The catalyst increases the rate by creating an alternate set of interactions at the metal surface so that electron-transfer encounters more often lead to products. (correct answer)
The catalyst increases the rate by shifting the reaction toward $\text{H}_2$ and $\text{ZnCl}_2$ , increasing the driving force for products.
The catalyst increases the rate by increasing the concentration of HCl, which increases collision frequency in solution.

Explanation: This question assesses understanding of catalysts. Catalysts increase the reaction rate by creating alternate interactions at the surface that make electron-transfer encounters between Zn and HCl more effective. With Cu2+, the pathway facilitates better collision outcomes for H2 and ZnCl2 production. This leads to a higher fraction of successful reactions. One tempting distractor is choice D, which is wrong because of the misconception that catalysts change equilibrium by shifting toward products, whereas catalysts do not affect equilibrium positions. A transferable strategy is to remember that catalysts affect rate, not the final position of a reaction.

Question 12

The hydrolysis of an ester in acidic solution is studied: $\text{RCOOR}'(aq)+\text{H}_2\text{O}(l)\rightarrow \text{RCOOH}(aq)+\text{R'OH}(aq)$ . Two mixtures contain the same amounts of ester and water at the same temperature. Mixture 2 also contains a small amount of $\text{H}^+(aq)$ that is regenerated during the process (not consumed overall). The overall reactants and products are the same.

Which statement best explains why Mixture 2 reacts faster?

The acid catalyst participates in steps that help reactant particles interact in a way that increases the fraction of collisions that lead to products. (correct answer)
The acid catalyst increases the concentration of ester by converting water into ester, increasing collision frequency.
The acid catalyst is consumed to form $\text{RCOOH}$ , increasing the amount of product and thus increasing the rate.
The acid catalyst increases the final percent yield of products, which requires the reaction to proceed faster.
The acid catalyst raises the temperature of the solution by releasing heat, increasing molecular speeds and the rate.

Explanation: This question assesses understanding of catalysts. Catalysts increase the reaction rate by participating in intermediate steps that make collisions between reactants like ester and water more effective in forming products. With H+ as a catalyst, the pathway involves protonation that facilitates better interaction during encounters. This leads to a higher fraction of successful collisions producing RCOOH and R'OH. One tempting distractor is choice D, which is wrong because it reflects the misconception that catalysts change equilibrium by increasing percent yield, whereas catalysts do not affect the equilibrium composition. A transferable strategy is to remember that catalysts affect rate, not the final position of a reaction.

Question 13

A student studies the reaction $\text{H}_2(g)+\text{I}_2(g)\rightarrow 2\text{HI}(g)$ at a fixed temperature. In a second trial, a small amount of platinum is added as a catalyst and is recovered unchanged. The same reactants and products are involved in both trials.

Which statement best explains why the catalyzed trial has a faster rate?

The catalyst provides an alternate way for reactant particles to interact so that a greater fraction of their encounters leads to product formation. (correct answer)
The catalyst increases the pressure inside the container by adding solid mass, increasing collision frequency among gas molecules.
The catalyst increases the amount of HI produced at completion by favoring products, which increases the rate.
The catalyst is converted into HI, increasing the concentration of products and accelerating the reaction.
The catalyst supplies energy directly to break the H–H and I–I bonds, increasing the rate by heating the reactants.

Explanation: This question assesses understanding of catalysts. Catalysts increase the reaction rate by introducing an alternate pathway that enhances the productivity of collisions between reactants like H2 and I2. With platinum as a catalyst, intermediate interactions allow for more effective encounters leading to HI formation. This results in a greater proportion of successful collisions. One tempting distractor is choice C, which is wrong because of the misconception that catalysts change equilibrium by increasing product amounts, whereas catalysts do not alter the equilibrium yield. A transferable strategy is to remember that catalysts affect rate, not the final position of a reaction.

Question 14

The reaction $\text{C}_2\text{H}_4(g)+\text{H}_2(g)\rightarrow \text{C}_2\text{H}_6(g)$ is run at the same temperature and pressure in two sealed flasks. Flask 2 contains a nickel catalyst; the overall reactants and products are the same in both flasks, and the nickel is not consumed.

Which statement best explains why the reaction rate is higher in Flask 2?

The nickel increases the rate by increasing the concentration of $\text{H}_2$ through decomposition of nickel hydride into hydrogen gas.
The nickel provides a surface that allows reactants to interact through a different pathway so that more collisions result in product formation. (correct answer)
The nickel increases the final amount of ethane produced, so the reaction must occur faster to reach the larger amount.
The nickel is consumed as a reactant, increasing the total number of reacting particles and thus increasing the rate.
The nickel shifts the reaction toward products by changing the equilibrium position, which increases the forward rate only.

Explanation: This question assesses understanding of catalysts. Catalysts increase the reaction rate by providing a surface or pathway that allows reactants like C2H4 and H2 to interact more productively during collisions. On the nickel surface, adsorption leads to orientations that make bond formation more likely in each encounter. Consequently, a larger fraction of collisions result in C2H6 formation. One tempting distractor is choice E, which is incorrect due to the misconception that catalysts change equilibrium by shifting toward products, but catalysts speed up both directions equally without changing equilibrium position. A transferable strategy is to remember that catalysts affect rate, not the final position of a reaction.

Question 15

Students compare the same reaction run two ways: $\text{CO}(g)+\text{NO}_2(g)\rightarrow \text{CO}_2(g)+\text{NO}(g)$ . In Trial 2, a platinum catalyst is present; the overall reactants and products are unchanged, and the platinum is not consumed.

Which statement best explains why the platinum increases the reaction rate?

The platinum increases the number of collisions by increasing the volume available to the gases, so collisions occur more often.
The platinum is incorporated into the products, increasing the number of product molecules formed per collision.
The platinum changes the overall stoichiometry so fewer bonds must be broken, increasing the rate.
The platinum provides a surface where reactants can adsorb and collide in orientations that more frequently result in product formation. (correct answer)
The platinum shifts the reaction toward products, increasing the driving force and therefore increasing the rate.

Explanation: This question assesses understanding of catalysts. Catalysts increase the reaction rate by offering a surface where reactants like CO and NO2 can adsorb, allowing collisions to occur in more favorable orientations for product formation. On the platinum surface, these adsorbed molecules interact through pathways that enhance the effectiveness of their encounters. As a result, a greater proportion of collisions lead to CO2 and NO. One tempting distractor is choice E, which is incorrect because of the misconception that catalysts change equilibrium by shifting toward products, but catalysts actually accelerate both forward and reverse rates equally without altering equilibrium. A transferable strategy is to remember that catalysts affect rate, not the final position of a reaction.

Question 16

A student compares two trials of the same aqueous reaction: $\text{S}_2\text{O}_8^{2-}(aq)+2\text{I}^-(aq)\rightarrow 2\text{SO}_4^{2-}(aq)+\text{I}_2(aq)$ . Trial 2 includes a small amount of $\text{Cu}^{2+}(aq)$ that is regenerated (not consumed). The same reactants and products are involved in both trials.

Which statement best explains the cause of the faster rate in Trial 2?

The catalyst increases the rate by increasing the final concentration of $\text{I}_2$ , so the reaction must proceed faster to reach that value.
The catalyst increases the rate by being consumed to form $\text{SO}_4^{2-}$ , increasing the number of reacting species in solution.
The catalyst increases the rate by raising the temperature of the solution, increasing the frequency of collisions.
The catalyst increases the rate by providing a different pathway that increases the fraction of reactant encounters that result in product formation. (correct answer)
The catalyst increases the rate by shifting the reaction toward products, increasing the forward rate and decreasing the reverse rate.

Explanation: This question assesses understanding of catalysts. Catalysts increase the reaction rate by providing a different pathway that makes encounters between reactants like S2O82- and I- more effective at forming products. With Cu2+ as a catalyst, intermediate interactions improve collision outcomes for SO42- and I2 production. As a result, more collisions successfully lead to products. One tempting distractor is choice E, which is incorrect because of the misconception that catalysts change equilibrium by shifting toward products, but catalysts accelerate both rates equally without altering equilibrium. A transferable strategy is to remember that catalysts affect rate, not the final position of a reaction.

Question 17

In the gas phase, $2\text{SO}_2(g)+\text{O}_2(g)\rightarrow 2\text{SO}_3(g)$ . Two trials are conducted at the same temperature and with the same initial pressures. Trial 2 contains a $\text{V}_2\text{O}_5$ catalyst that is not consumed, and the overall reactants and products are the same.

Which statement best explains why Trial 2 proceeds faster?

The catalyst increases the rate by shifting the reaction toward $\text{SO}_3$ , increasing product formation and thus increasing speed.
The catalyst increases the rate by providing a surface and alternate steps that make product-forming encounters between reactants more likely. (correct answer)
The catalyst increases the rate by increasing the pressure inside the container, increasing collision frequency among gas molecules.
The catalyst increases the rate by being consumed to form $\text{SO}_3$ , increasing the amount of product formed per unit time.
The catalyst increases the rate by supplying oxygen atoms directly to $\text{SO}_2$ , eliminating the need for $\text{O}_2$ collisions.

Explanation: This question assesses understanding of catalysts. Catalysts increase the reaction rate by providing a surface and alternate steps that make collisions between reactants like SO2 and O2 more productive. With V2O5, adsorption and interactions lead to better orientations for SO3 formation. Consequently, a greater proportion of encounters result in products. One tempting distractor is choice A, which is incorrect due to the misconception that catalysts change equilibrium by shifting toward products, but catalysts speed up attainment without changing equilibrium. A transferable strategy is to remember that catalysts affect rate, not the final position of a reaction.

Question 18

The reaction $\text{CH}_3\text{CHO}(g)\rightarrow \text{CH}_4(g)+\text{CO}(g)$ is conducted in a closed vessel. In Trial 2, a catalyst is added; the overall reactants and products are identical in both trials, and the catalyst is not consumed.

Which statement best explains why the reaction rate increases when the catalyst is present?

The catalyst increases the rate because it increases the final amount of $\text{CH}_4$ and CO produced, requiring faster formation.
The catalyst increases the rate because it is used up to form CO, increasing the number of product molecules formed.
The catalyst increases the rate because it changes the overall reaction into a different reaction with different products that form faster.
The catalyst increases the rate because it provides an alternate pathway that makes product-forming molecular encounters more effective. (correct answer)
The catalyst increases the rate because it raises the temperature of the vessel, increasing average kinetic energy and collision frequency.

Explanation: This question assesses understanding of catalysts. Catalysts increase the reaction rate by enabling a different pathway where collisions of reactants like CH3CHO are more likely to form products effectively. In the presence of the catalyst, intermediate steps facilitate better molecular interactions during encounters. Thus, a higher fraction of collisions lead to CH4 and CO. One tempting distractor is choice A, which is incorrect because it embodies the misconception that catalysts change equilibrium by increasing final product amounts, but catalysts only accelerate reaching equilibrium without shifting it. A transferable strategy is to remember that catalysts affect rate, not the final position of a reaction.

Question 19

A student studies the reaction $\text{ClO}^-(aq)+2\text{I}^-(aq)+2\text{H}^+(aq)\rightarrow \text{Cl}^-(aq)+\text{I}_2(aq)+\text{H}_2\text{O}(l)$ . Two trials are run with the same initial concentrations and temperature. Trial 2 includes a catalyst that is regenerated and not consumed overall; the overall reactants and products are unchanged.

Which statement best explains why Trial 2 has a higher rate?

The catalyst increases the rate by shifting the reaction toward products, which increases the forward rate and decreases the reverse rate.
The catalyst increases the rate by increasing the final amount of $\text{I}_2$ produced, so the reaction must proceed faster.
The catalyst increases the rate by enabling reactant species to undergo a different set of interactions so that more encounters lead to products. (correct answer)
The catalyst increases the rate by being consumed in the reaction, increasing the number of reacting particles in solution.
The catalyst increases the rate by supplying energy to the reactants, increasing their kinetic energy without changing temperature.

Explanation: This question assesses understanding of catalysts. Catalysts increase the reaction rate by enabling a different set of interactions that make encounters between reactants like ClO-, I-, and H+ more effective. With the catalyst, the pathway improves collision success for Cl-, I2, and H2O formation. Consequently, more collisions lead to products. One tempting distractor is choice E, which is incorrect because of the misconception that catalysts change equilibrium by shifting toward products, but catalysts do not alter equilibrium positions. A transferable strategy is to remember that catalysts affect rate, not the final position of a reaction.

Question 20

In the reaction $\text{A}(aq)+\text{B}(aq)\rightarrow \text{C}(aq)$ , two solutions are prepared with the same initial concentrations of A and B at the same temperature. In Solution 2, a catalyst is added; the same reactants and product are involved in both solutions, and the catalyst is recovered unchanged.

Which statement best explains why Solution 2 forms C faster?

The catalyst increases the rate by providing an alternative pathway that increases the fraction of A–B encounters that successfully form C. (correct answer)
The catalyst increases the rate by shifting the reaction toward C, increasing the amount of C formed per unit time.
The catalyst increases the rate by increasing the temperature of the solution, increasing particle speeds and collision frequency.
The catalyst increases the rate by increasing the concentration of A and B, which increases collision frequency in solution.
The catalyst increases the rate by being consumed to form C, increasing the number of reacting particles present.

Explanation: This question assesses understanding of catalysts. Catalysts increase the reaction rate by providing an alternative pathway that increases the fraction of encounters between A and B that successfully form C. In the catalyzed solution, intermediate steps facilitate more productive collisions. This leads to a higher rate of product formation. One tempting distractor is choice B, which is wrong because it reflects the misconception that catalysts change equilibrium by shifting toward products, whereas catalysts speed up both rates without changing equilibrium. A transferable strategy is to remember that catalysts affect rate, not the final position of a reaction.