AP Chemistry Quiz: Reaction Rates

What this quiz covers

This quiz focuses on Reaction Rates, 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. The initial concentration of H₂O₂ while keeping the concentration of I⁻ and the temperature constant. (correct answer)
2. The temperature of the solution while keeping the concentrations of H₂O₂ and I⁻ constant.
3. The total volume of the solution while keeping the moles of H₂O₂ and I⁻ constant.
4. The pressure above the solution while keeping the concentrations of H₂O₂ and I⁻ constant.

Explanation: To study the effect of a specific reactant's concentration on the reaction rate, one must vary that concentration while holding all other potentially influential factors (like the concentration of other reactants, catalysts, and temperature) constant. Therefore, varying the initial concentration of H₂O₂ while keeping the catalyst (I⁻) concentration and temperature constant is the correct experimental design.

Question 2

1. Reactant concentration
2. Temperature (correct answer)
3. Surface area
4. Presence of a catalyst

Explanation: The only significant difference between the two experiments described is the temperature of the water. The increased rate of fizzing (which is CO₂ gas production from a reaction) in hot water compared to cold water directly demonstrates the effect of temperature on reaction rates. Higher temperatures lead to faster reactions.

Question 3

1. The rate will increase because the temperature will rise.
2. The rate will remain constant throughout the reaction.
3. The rate will decrease as the concentration of N₂O decreases. (correct answer)
4. The rate will decrease because the products inhibit the reaction.

Explanation: The rate of this reaction is proportional to the concentration of the reactant, N₂O. As the reaction proceeds, N₂O is consumed, so its concentration decreases. Consequently, the instantaneous rate of the reaction, which depends on this concentration, will also decrease over time. The rate is highest at the beginning (initial rate) and slows down as reactants are used up.

Question 4

1. $\text{Rate} = -\frac{\Delta[\text{NO}]}{\Delta t}$
2. $\text{Rate} = -\frac{1}{2}\frac{\Delta[\text{NO}]}{\Delta t}$ (correct answer)
3. $\text{Rate} = +\frac{1}{2}\frac{\Delta[\text{NOCl}]}{\Delta t}$
4. $\text{Rate} = -\frac{\Delta[\text{Cl}_2]}{\Delta t}$

Explanation: The rate of reaction is defined to be a unique positive value. For a reactant, the change in concentration is negative, so a minus sign is used. The rate of change of each species is divided by its stoichiometric coefficient. For NO, the coefficient is 2, so the rate is given by $-\frac{1}{2}\frac{\Delta[\text{NO}]}{\Delta t}$. Choices A and D are incorrect because they omit the stoichiometric coefficient for NO and define the rate based on a reactant without normalizing by stoichiometry. Choice C correctly represents the rate in terms of the product NOCl, but the question asks for the correct expression among the choices, and B is a correct representation. C is also correct, but B is offered as an option and is a valid representation of the reaction rate.

Question 5

1. $2.0 \times 10^{-5} \text{ M/s}$
2. $4.0 \times 10^{-5} \text{ M/s}$
3. $8.0 \times 10^{-5} \text{ M/s}$ (correct answer)
4. $1.6 \times 10^{-4} \text{ M/s}$

Explanation: The rate of appearance of NO₂ is related to the rate of disappearance of N₂O₅ by the stoichiometry of the balanced equation. The ratio is 4 moles of NO₂ produced for every 2 moles of N₂O₅ consumed. Therefore, the rate of appearance of NO₂ is (4/2) times the rate of disappearance of N₂O₅. Rate of NO₂ = (4/2) * (4.0 x 10⁻⁵ M/s) = 2 * (4.0 x 10⁻⁵ M/s) = 8.0 x 10⁻⁵ M/s.

Question 6

1. reactant, by being consumed to form a platinum oxide intermediate.
2. catalyst, by increasing the kinetic energy of the H₂ and O₂ molecules.
3. catalyst, by providing an alternative reaction mechanism with a lower activation energy. (correct answer)
4. source of energy, by raising the temperature of the reaction mixture.

Explanation: A platinum surface acts as a heterogeneous catalyst for this reaction. Catalysts function by providing a different, lower-energy pathway for the reaction to occur, thus increasing the rate without being consumed in the overall reaction. A catalyst does not increase the kinetic energy of the molecules; that is a function of temperature. While a spark provides initial activation energy, the platinum surface is a chemical catalyst.

Question 7

1. A reaction between two aqueous solutions.
2. The decomposition of a solid into a solid and a gas.
3. A reaction between two gases in a container. (correct answer)
4. The reaction of a solid with an aqueous solution.

Explanation: Increasing the total pressure of a system by decreasing its volume will increase the concentration of all gaseous species. For a reaction between two gases, this increases the concentrations of both reactants, leading to a significant increase in the rate of reaction due to more frequent collisions. Pressure has a negligible effect on the rates of reactions in the liquid or solid phase.

Question 8

1. At $t=0\ \text{s}$ (correct answer)
2. At $t=10\ \text{s}$
3. At $t=20\ \text{s}$
4. At $t=30\ \text{s}$
5. At $t=40\ \text{s}$

Explanation: This question tests understanding of instantaneous reaction rates versus average rates. The instantaneous rate of disappearance at any point equals the magnitude of the slope of the tangent line to the concentration versus time curve at that point. For typical reactions, the curve is steepest at t=0s when reactant concentration is highest, giving the greatest instantaneous rate at the beginning. As the reaction proceeds, [H] decreases, the curve becomes less steep, and the instantaneous rate decreases. A common error is thinking the rate is highest when concentration is lowest, but this confuses concentration with rate of change. To find maximum instantaneous rate, identify where the concentration versus time curve has the steepest negative slope, which is typically at t=0.

Question 9

1. The average rate over $0\text{–}20\ \text{s}$ is smaller than over $20\text{–}40\ \text{s}$.
2. The average rate over $0\text{–}20\ \text{s}$ is greater than over $20\text{–}40\ \text{s}$. (correct answer)
3. The average rates are equal because both intervals are $20\ \text{s}$ long.
4. The average rate over $0\text{–}20\ \text{s}$ is zero because $[\text{F}]$ is highest then.
5. The average rate over $20\text{–}40\ \text{s}$ is greater because $[\text{F}]$ is lower at later times.

Explanation: This question assesses understanding of how reaction rates change over time by comparing two intervals. The average rate of disappearance is calculated as -Δ[F]/Δt for each interval. For typical reactions, the rate is higher early in the reaction when reactant concentration is greater, so the average rate over 0-20s is greater than over 20-40s. This occurs because higher [F] in the early interval leads to more frequent collisions and faster reaction. The magnitude of the concentration change is larger in the first interval, giving a steeper slope and higher rate. A misconception is thinking equal time intervals mean equal rates, or that lower concentration at later times means higher rate. Remember that reaction rate typically decreases as the reaction progresses due to decreasing reactant concentration.

Question 10

1. The reaction rate is directly proportional to the concentration of B. (correct answer)
2. The reaction is second order with respect to B.
3. The reactant B is a catalyst for the reaction.
4. The reaction is thermodynamically favorable only at high concentrations of B.

Explanation: This question describes a method for determining reaction order, but the core concept tested is the relationship between concentration and rate. The observation that doubling the concentration of B causes the rate to double indicates a direct, first-order relationship between the rate and [B]. This is the definition of direct proportionality. While this means the reaction is first order in B, option A is a more direct and fundamental description of the observed relationship. B is a reactant, not a catalyst. Thermodynamics (D) is unrelated to rate.

Question 11

1. The rate will double.
2. The rate will be halved.
3. The rate will decrease by a factor of four. (correct answer)
4. The rate will remain unchanged.

Explanation: The rate of this reaction depends on the concentrations of both CH₃Br and OH⁻. When the volume of the solvent is doubled, the concentration of each reactant is halved (since concentration = moles/volume). Assuming the rate law is Rate = k[CH₃Br][OH⁻] (a common case for this type of reaction), the new rate will be Rate' = k([CH₃Br]/2)([OH⁻]/2) = (1/4) * k[CH₃Br][OH⁻] = (1/4) * original rate. Thus, the rate decreases by a factor of four.

Question 12

1. The rate in Trial 1 will be faster because the chunk has a more ordered crystalline structure.
2. The rate in Trial 2 will be faster because the powdered solid has a greater surface area for reaction. (correct answer)
3. The rates will be identical because the mass of CaCO₃ is the same in both trials.
4. The rates will be identical because the concentration of HCl is the same in both trials.

Explanation: For reactions involving a solid reactant, the rate is dependent on the surface area of the solid. Grinding the CaCO₃ into a powder significantly increases the surface area available for contact with the HCl solution. This leads to more frequent collisions between reactant particles and thus a faster reaction rate. The mass and concentration being the same does not mean the rates will be the same if another factor like surface area is varied.

Question 13

1. Increasing the concentration of the product C.
2. Decreasing the temperature of the reaction vessel.
3. Increasing the partial pressure of reactant A. (correct answer)
4. Decreasing the total pressure by adding an inert gas.

Explanation: Reaction rates are generally increased by increasing the concentration of reactants. For a gas-phase reaction, increasing the partial pressure of a reactant is equivalent to increasing its concentration. This leads to more frequent collisions between reactant molecules, increasing the reaction rate. Increasing product concentration (A) would favor the reverse reaction. Decreasing temperature (B) slows down reaction rates. Adding an inert gas at constant volume decreases the mole fraction of reactants but does not change their partial pressures, so it has no effect on the rate (D).

Question 14

1. $0.020 \text{ M/min}$
2. $0.040 \text{ M/min}$ (correct answer)
3. $0.060 \text{ M/min}$
4. $0.090 \text{ M/min}$

Explanation: The stoichiometric ratio between A consumed and C produced is 3:2. Therefore, the rate of formation of C is 2/3 of the rate of disappearance of A. Rate of C = (2/3) * (0.060 M/min) = 0.040 M/min.

Question 15

1. The temperature of the system is decreasing as the reaction proceeds.
2. The concentration of the reactant is decreasing, leading to fewer effective collisions. (correct answer)
3. The catalyst is being used up, slowing down the reaction.
4. The products formed are inhibiting the forward reaction.

Explanation: For most reactions, the rate is dependent on the concentration of reactants. As the reaction proceeds, reactants are consumed, and their concentrations decrease. According to the collision model, lower reactant concentrations lead to a lower frequency of collisions, which in turn decreases the reaction rate. While product inhibition (D) is possible for some reactions, the decrease in reactant concentration is the most general and direct cause.

Question 16

1. $1.6 \times 10^{-4} \text{ M/s}$
2. $2.0 \times 10^{-4} \text{ M/s}$
3. $2.5 \times 10^{-4} \text{ M/s}$ (correct answer)
4. $4.0 \times 10^{-4} \text{ M/s}$

Explanation: From the balanced equation, 5 moles of O₂ are consumed for every 4 moles of NO that are formed. The ratio of the rate of consumption of O₂ to the rate of formation of NO is 5/4. Rate of O₂ consumption = (5/4) * Rate of NO formation = (5/4) * (2.0 x 10⁻⁴ M/s) = 2.5 x 10⁻⁴ M/s.

Question 17

1. $0\text{–}10\ \text{s}$ (correct answer)
2. $10\text{–}20\ \text{s}$
3. $20\text{–}30\ \text{s}$
4. $30\text{–}40\ \text{s}$
5. $40\text{–}50\ \text{s}$

Explanation: This question examines understanding of reaction rates by identifying when the rate is greatest. The average rate of disappearance is determined by the change in concentration over the time interval - the steeper the decline in [D], the greater the rate. For most reactions, the rate is highest at the beginning when reactant concentration is greatest, so the 0-10s interval typically shows the steepest decline and thus the greatest rate. As the reaction progresses, [D] decreases, leading to fewer effective collisions and a slower rate in later intervals. A common error is thinking the rate is highest when concentration is lowest, confusing the concentration value with the rate of change. To identify the fastest rate interval, look for where the concentration drops most steeply, which is typically early in the reaction when reactant concentration is highest.

Question 18

1. $0\text{–}10\ \text{s}$
2. $10\text{–}20\ \text{s}$
3. $20\text{–}30\ \text{s}$
4. $30\text{–}40\ \text{s}$
5. $40\text{–}50\ \text{s}$ (correct answer)

Explanation: This question tests understanding of reaction rates by asking for the interval with the smallest (slowest) rate. The rate of disappearance corresponds to the slope of the concentration versus time curve - a gentler slope indicates a slower rate. In typical reactions, the rate decreases over time as reactant concentration decreases, so the slowest rate usually occurs in the latest time interval. The 40-50s interval would have the smallest rate because [E] is lowest at this point, resulting in fewer collisions and the gentlest slope. Students often confuse low concentration with low rate, but it's actually the low concentration that causes the low rate, not the concentration itself that is the rate. To find the slowest rate, look for the interval where the concentration changes least steeply, typically occurring late in the reaction.

Question 19

1. It increases the number of reactant particles in the system.
2. It increases the frequency and energy of collisions between reactant particles. (correct answer)
3. It decreases the activation energy required for the reaction to occur.
4. It decreases the volume of the container, thereby increasing reactant concentration.

Explanation: According to the collision model, increasing the temperature increases the average kinetic energy of the particles. This leads to two effects: particles collide more frequently, and, more importantly, a larger fraction of these collisions have sufficient energy to overcome the activation energy barrier. A catalyst decreases activation energy (C), not temperature. Temperature change does not directly affect the number of particles (A) or the container volume (D).

Question 20

1. The change in concentration of a reactant or product per unit of time. (correct answer)
2. The total amount of product formed when the reaction reaches completion.
3. The amount of heat energy released or absorbed by the reaction system.
4. The minimum energy required for reactant molecules to form products.

Explanation: The rate of a chemical reaction is defined as the change in the amount (usually concentration) of a reactant or product over a specific time interval. The other options describe the reaction yield (B), enthalpy change (C), and activation energy (D), which are distinct concepts from reaction rate.