AP Chemistry Quiz: Energy Diagrams

What this quiz covers

This quiz focuses on Energy Diagrams, 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. $\Delta H<0$ for the forward reaction. (correct answer)
2. $\Delta H>0$ for the forward reaction.
3. $\Delta H=0$ for the forward reaction.
4. $\Delta H<0$ because $E_a$ is positive.
5. $\Delta H>0$ because the peak is above reactants.

Explanation: This question tests the skill of determining the sign of ΔH in single-step reaction energy diagrams. If products are lower in energy than reactants, the reaction is exothermic with ΔH < 0, as ΔH is calculated as the energy of products minus energy of reactants. This reflects the release of energy in exothermic processes, consistent with the diagram's depiction. Thus, option A correctly states ΔH < 0 for the forward reaction. A tempting distractor is option B, which claims ΔH > 0, but this is incorrect due to the misconception of inverting the comparison of reactant and product energies. Always calculate ΔH as E_products - E_reactants to determine if the reaction is exothermic (negative) or endothermic (positive) from the diagram.

Question 2

1. Overall endothermic, and $E_a$(reverse) is from products to the highest peak. (correct answer)
2. Overall exothermic, and $E_a$(reverse) is from reactants to the highest peak.
3. Overall endothermic, and $E_a$(reverse) is from reactants to products.
4. Overall exothermic, and $E_a$(reverse) is from products to reactants.
5. Overall thermoneutral, and $E_a$(reverse) is from intermediate to TS1.

Explanation: This question tests the skill of analyzing overall reaction thermodynamics and reverse activation energy in multi-step energy diagrams. With products higher than reactants, the overall reaction is endothermic (ΔH > 0), and the reverse activation energy Ea(reverse) is the difference from products to the highest peak, as that is the barrier for the reverse pathway. This follows from the principle that Ea for any direction is measured from the starting point to the maximum energy point. Thus, option A correctly describes the reaction as overall endothermic with Ea(reverse) from products to the highest peak. A tempting distractor is option B, which incorrectly states the reaction is exothermic, due to the misconception of misreading the relative energies of reactants and products. For reverse reactions in diagrams, reverse the direction and measure Ea from products back to the highest transition state encountered.

Question 3

1. Overall $\Delta H<0$ because products are lower in energy than reactants. (correct answer)
2. Overall $\Delta H>0$ because there are two peaks.
3. Overall $\Delta H=0$ because the intermediate is present.
4. Overall $\Delta H<0$ because TS1 is higher than TS2.
5. Overall $\Delta H>0$ because $E_a$ is positive.

Explanation: This question tests the skill of determining overall ΔH in multi-step reaction energy diagrams. With products much lower than reactants, the overall ΔH is negative, as ΔH is the net energy change from reactants to products, indicating an exothermic reaction. This holds regardless of the number of steps or transition state heights. Thus, option A correctly states overall ΔH < 0 because products are lower than reactants. A tempting distractor is option B, which claims ΔH > 0 because there are two peaks, but this is incorrect due to the misconception that mechanism complexity affects thermodynamics. For overall ΔH in any diagram, calculate solely as E_products - E_reactants, ignoring internal features like peaks or valleys.

Question 4

1. $\Delta H<0$ when the horizontal axis is longer.
2. $\Delta H<0$ when $E_a$ is larger.
3. $\Delta H<0$ when the peak is lower than the reactants.
4. $\Delta H<0$ when products are lower in potential energy than reactants. (correct answer)
5. $\Delta H<0$ when there is an intermediate.

Explanation: This question tests the skill of determining conditions for negative ΔH in single-step reaction energy diagrams. ΔH is negative when products have lower potential energy than reactants, indicating an exothermic reaction. This condition is directly observable from the relative positions on the vertical axis. Thus, option A correctly identifies the condition for ΔH < 0 as products lower than reactants. A tempting distractor is option B, which claims ΔH < 0 when the peak is lower than reactants, but this is incorrect due to the misconception that a low barrier implies exothermicity, confusing Ea with ΔH. To assess ΔH, compare only the energies of reactants and products, disregarding the transition state for thermodynamic sign determination.

Question 5

1. The peak must be above the reactants, indicating a positive $E_a$. (correct answer)
2. The peak must be below the reactants, indicating a negative $E_a$.
3. The peak must be equal to the products, indicating $\Delta H=0$.
4. The peak must be above the products, indicating $\Delta H>0$.
5. The peak position on the x-axis determines $\Delta H$.

Explanation: This question tests the skill of understanding requirements for reaction feasibility in single-step energy diagrams. For the reaction to proceed at a measurable rate, the peak (transition state) must be above the reactants, indicating a positive Ea that can be surmounted by thermal energy. A peak below reactants would imply a negative Ea, which is unphysical for forward reactions. Thus, option A correctly states that the peak must be above reactants, indicating positive Ea. A tempting distractor is option B, which claims the peak below reactants for negative Ea, but this is incorrect due to the misconception that reactions can have negative barriers, whereas Ea is always positive. Ensure the transition state is higher than both reactants and products for realistic diagrams, but always above the starting point for positive Ea.

Question 6

1. $\Delta H$ is the energy difference from reactants to the peak.
2. $\Delta H$ is the energy difference from products to the peak.
3. $\Delta H$ is the energy difference between products and reactants. (correct answer)
4. $\Delta H$ is the horizontal distance between reactants and products.
5. $\Delta H$ is the difference between the two axes labels.

Explanation: This question tests the skill of identifying ΔH in single-step reaction energy diagrams. ΔH is the vertical energy difference between products and reactants, representing the net enthalpy change of the reaction. This is a direct consequence of enthalpy being a state function, dependent only on initial and final states. Thus, option C correctly identifies ΔH as the energy difference between products and reactants. A tempting distractor is option A, which claims ΔH is from reactants to the peak, but this is incorrect due to the misconception of confusing ΔH with activation energy Ea. To distinguish ΔH from Ea, remember ΔH connects endpoints, while Ea connects the starting point to the transition state peak.

Question 7

1. Reactant energy decreases, making $\Delta H$ more negative
2. Product energy increases, making $\Delta H$ more positive
3. The highest peak decreases while reactant and product energies stay the same (correct answer)
4. The horizontal distance between reactants and products decreases
5. The peak increases because catalysts add energy to the reaction

Explanation: This question tests understanding of how catalysts affect reaction energy diagrams. A catalyst provides an alternative reaction pathway with a lower activation energy, which appears on the diagram as a lower peak (or series of lower peaks) compared to the uncatalyzed reaction. Crucially, catalysts do not change the energy levels of reactants or products, so the starting and ending points remain the same, preserving the same ΔH for the reaction. The catalyst only affects the pathway between reactants and products, lowering the highest energy barrier that must be overcome. A common misconception (choice E) is thinking catalysts add energy to increase the peak, but catalysts actually lower activation energy by providing a more favorable pathway. When analyzing catalyzed reactions, verify that reactant and product energies remain unchanged while the transition state peak(s) decrease.

Question 8

1. $\Delta H>0$ because products are at higher potential energy than reactants (correct answer)
2. $\Delta H<0$ because the peak is above the reactants
3. $\Delta H=0$ because there is only one transition state
4. $\Delta H>0$ because the activation energy is positive
5. $\Delta H<0$ because products are at higher potential energy than reactants

Explanation: This question tests the ability to determine the sign of ΔH from an energy diagram. The enthalpy change (ΔH) is determined by comparing the potential energy levels of reactants and products: if products are higher in energy than reactants, energy must be absorbed, making the reaction endothermic (ΔH > 0). The diagram shows products at a higher potential energy level than reactants, confirming that ΔH > 0. The presence of an activation energy peak is irrelevant to the sign of ΔH; all reactions have activation energy regardless of being endothermic or exothermic. A common misconception (choice E) is misreading the diagram and thinking ΔH < 0 when products are higher, which would violate energy conservation. When determining ΔH from an energy diagram, focus only on the vertical difference between reactant and product energy levels, ignoring transition states.

Question 9

1. Reverse has smaller $E_a$ and $\Delta H>0$
2. Reverse has larger $E_a$ and $\Delta H<0$
3. Reverse has the same $E_a$ and $\Delta H=0$
4. Reverse has larger $E_a$ and $\Delta H>0$ (correct answer)
5. Reverse has smaller $E_a$ and $\Delta H<0$

Explanation: This question tests understanding of the relationship between forward and reverse reactions on an energy diagram. Since the diagram shows products at a higher energy level than reactants, the forward reaction is endothermic (ΔH > 0), and the reverse reaction is exothermic (ΔH < 0 for reverse). The activation energy for the reverse reaction is the vertical distance from products up to the transition state peak, while for the forward reaction it's from reactants to the same peak. Because products are higher than reactants, the reverse reaction has a larger vertical distance to climb to reach the transition state, giving it a larger activation energy. A common misconception (choice A) is thinking the reverse reaction of an endothermic process would have a smaller activation energy, but this ignores the actual energy levels shown. When analyzing reverse reactions, remember that ΔH changes sign and activation energies depend on the vertical distances from starting points to the transition state.

Question 10

1. $\Delta H$ is positive because the intermediate is higher than the reactants.
2. $\Delta H$ is negative because products are lower in potential energy than reactants. (correct answer)
3. $\Delta H$ is zero because the diagram has two steps that cancel.
4. $\Delta H$ is positive because there are two transition states.
5. $\Delta H$ is negative because the second peak is lower than the first peak.

Explanation: This question tests the skill of determining the overall enthalpy change (ΔH) from multi-step energy diagrams, focusing on initial and final states. Despite the intermediate being higher than reactants, the products are lower than reactants, indicating a negative ΔH as the net energy change is exothermic. ΔH is a state function, so it depends only on the energy difference between reactants and products, ignoring the intermediate's position. The two-step profile with peaks and the intermediate above reactants does not affect the overall ΔH, which remains negative. A tempting distractor is choice A, which states ΔH is positive due to the high intermediate, based on the misconception that intermediates contribute to the net enthalpy rather than recognizing ΔH's path-independence. For transferable analysis, always compute ΔH directly from the endpoint energies on the diagram, regardless of steps or intermediates involved.

Question 11

1. $\Delta H$ is positive, and $E_a$ is the vertical distance from products to the peak.
2. $\Delta H$ is negative, and $E_a$ is the vertical distance from reactants to the peak. (correct answer)
3. $\Delta H$ is negative, and $E_a$ is the vertical distance from products to reactants.
4. $\Delta H$ is positive, and $E_a$ is the vertical distance from reactants to products.
5. $\Delta H$ is zero, and $E_a$ is the vertical distance from the peak to the x-axis.

Explanation: This question tests the skill of interpreting reaction energy diagrams to determine the enthalpy change (ΔH) and activation energy (Ea). In the diagram, reactants are at a higher potential energy than products, indicating an exothermic reaction where ΔH is negative because energy is released as the reaction progresses. The activation energy Ea is the energy barrier that must be overcome, represented by the vertical distance from the reactants' energy level to the peak of the transition state. The curve rises to this single peak before falling to the lower energy products, confirming that the reaction requires an initial input of energy (Ea) but overall releases energy (negative ΔH). A tempting distractor is choice A, which incorrectly states ΔH is positive, stemming from the misconception of confusing the higher energy of reactants with an endothermic process instead of properly subtracting product energy from reactant energy. To analyze energy diagrams effectively, always identify ΔH as the vertical difference between reactants and products, and Ea as the rise from reactants to the transition state, regardless of the overall energy change.

Question 12

1. The catalyst lowers $E_a$ but does not change the energy difference between reactants and products. (correct answer)
2. The catalyst lowers $\Delta H$ by lowering the products.
3. The catalyst increases $E_a$ because it introduces an intermediate.
4. The catalyst changes the reaction coordinate axis scale.
5. The catalyst makes the reaction endothermic.

Explanation: This question tests the skill of understanding catalyst effects on exothermic reaction energy diagrams. A catalyst lowers the activation energy Ea by providing a lower-energy pathway, often with intermediates, but does not change the energy difference between reactants and products, thus preserving ΔH. This separates the kinetic role of catalysts from thermodynamic outcomes. Thus, option A correctly states that the catalyst lowers Ea but does not change ΔH. A tempting distractor is option B, which claims it lowers ΔH by lowering products, but this is incorrect due to the misconception that catalysts alter equilibrium positions, whereas they only accelerate rates. When a catalyst is added to a diagram, look for reduced peak heights but unchanged endpoints to confirm its effect on kinetics without impacting thermodynamics.

Question 13

1. The intermediate is the horizontal distance between the peaks.
2. The intermediate is the local minimum between TS1 and TS2. (correct answer)
3. The intermediate is the second transition state.
4. The intermediate is the energy difference between reactants and products.
5. The intermediate is the highest point on the diagram.

Explanation: This question tests the skill of identifying intermediates in multi-step reaction energy diagrams. In the two-step diagram, the intermediate is depicted as a local minimum in potential energy between the two transition states (TS1 and TS2), and it is closer in energy to the products, reflecting its position along the reaction coordinate. This local minimum represents a relatively stable species formed after the first step and consumed in the second. Therefore, option B accurately describes the intermediate as the local minimum between TS1 and TS2. A tempting distractor is option A, which claims the intermediate is the highest point, but this is incorrect due to the misconception of confusing intermediates with transition states, which are energy maxima. When examining multi-step energy diagrams, identify intermediates as valleys between peaks and transition states as the peaks themselves to understand reaction mechanisms.

Question 14

1. Overall $\Delta H$ must be negative because there are two steps.
2. Step 1 cannot be rate-determining unless products are lower than reactants.
3. Overall $\Delta H$ must be positive because the intermediate is high.
4. Step 1 is rate-determining because the activation barrier from reactants to TS1 is largest. (correct answer)
5. Step 2 is rate-determining because the intermediate is high in energy.

Explanation: This question tests the skill of identifying the rate-determining step in a multi-step reaction energy diagram. In the two-step diagram, the first peak (TS1) is higher than the second, meaning the activation energy for Step 1 (from reactants to TS1) is larger than for Step 2, even though the intermediate is high in energy. The rate-determining step is the one with the highest activation barrier relative to its starting point, as it limits the overall reaction rate. Therefore, option B correctly identifies Step 1 as rate-determining due to its larger activation barrier. A tempting distractor is option A, which suggests Step 2 is rate-determining because the intermediate is high in energy, but this is incorrect due to the misconception of equating intermediate energy with activation barrier height. To determine the rate-determining step in energy diagrams, compare the vertical differences from each minimum to the subsequent peak, selecting the largest as the slowest step.

Question 15

1. $\Delta H$ is the energy difference between products and reactants. (correct answer)
2. $\Delta H$ is the energy difference between TS1 and TS2.
3. $\Delta H$ is the energy difference between reactants and TS1.
4. $\Delta H$ is the energy difference between intermediate and products.
5. $\Delta H$ is the horizontal distance from reactants to products.

Explanation: This question tests the skill of identifying overall ΔH in multi-step reaction energy diagrams. The overall ΔH is the vertical energy difference between products and reactants, independent of intermediates or transition states, as it is a state function. This ensures that regardless of the mechanism, ΔH depends only on initial and final states. Thus, option A correctly identifies ΔH as the energy difference between products and reactants. A tempting distractor is option C, which suggests from reactants to TS1, but this is incorrect due to the misconception of equating ΔH with the first step's Ea. In multi-step diagrams, ignore intermediates and peaks for ΔH; focus solely on the net change from start to end.

Question 16

1. Energy difference from intermediate to TS2. (correct answer)
2. Energy difference from reactants to TS2.
3. Energy difference from products to TS2.
4. Energy difference from reactants to products.
5. Energy difference from TS1 to intermediate.

Explanation: This question tests the skill of identifying activation energy for individual steps in multi-step reaction energy diagrams. For Step 2 in a two-step reaction, the activation energy is the vertical difference from the intermediate (the local minimum) to TS2 (the second peak), as this represents the barrier for that specific step. This is based on the principle that each step's Ea is independent and measured from its starting species to its transition state. Thus, option A correctly identifies Ea for Step 2 as the energy difference from intermediate to TS2. A tempting distractor is option B, which suggests from reactants to TS2, but this is incorrect due to the misconception of using the overall Ea instead of the step-specific barrier. When dissecting multi-step diagrams, isolate each step by measuring Ea from the preceding minimum to the immediate peak for accurate mechanistic analysis.

Question 17

1. It decreases $E_a$ but does not change $\Delta H$. (correct answer)
2. It decreases $\Delta H$ but does not change $E_a$.
3. It increases $\Delta H$ because the peak is lower.
4. It increases $E_a$ because the products are unchanged.
5. It makes $\Delta H=0$ because the curve is lower.

Explanation: This question tests the skill of understanding the effects of catalysts on reaction energy diagrams. Lowering the peak while keeping reactant and product energies the same represents the action of a catalyst, which decreases the activation energy Ea by providing an alternative pathway but does not alter the overall enthalpy change ΔH. This is because ΔH is a state function depending only on the difference between initial and final states, independent of the path. Thus, option A correctly states that it decreases Ea but does not change ΔH. A tempting distractor is option B, which claims it decreases ΔH, but this is incorrect due to the misconception that changing the barrier affects the net energy change, confusing kinetics with thermodynamics. When evaluating catalytic effects on energy diagrams, note that catalysts lower peaks (Ea) but leave endpoint differences (ΔH) unchanged to distinguish between rate and equilibrium influences.

Question 18

1. Pathway 1 must be slower because it has more steps.
2. Pathway 2 must be faster because it is single-step.
3. Whichever pathway has the lower highest peak has the lower overall $E_a$. (correct answer)
4. Pathway 1 has a different $\Delta H$ because it has an intermediate.
5. Both pathways have $\Delta H=0$ because they share endpoints.

Explanation: This question tests the skill of comparing activation energies and rates for alternative reaction pathways in energy diagrams. The overall Ea for each pathway is determined by the height of its highest peak relative to reactants, with the pathway having the lower highest peak expected to be faster, regardless of the number of steps. This is because the rate is limited by the largest barrier, and both pathways share the same ΔH since they have identical reactants and products. Thus, option C correctly states that whichever pathway has the lower highest peak has the lower overall Ea. A tempting distractor is option A, which claims Pathway 1 is slower because it has more steps, but this is incorrect due to the misconception that the number of steps directly determines rate, ignoring activation energy heights. To compare pathways, evaluate the maximum Ea for each, as the lowest maximum barrier indicates the kinetically favored route.

Question 19

1. The rate-determining step corresponds to the larger activation energy barrier from the preceding minimum to the next peak. (correct answer)
2. The rate-determining step corresponds to the deeper valley (lowest intermediate).
3. The rate-determining step is always Step 1 because it occurs first.
4. The rate-determining step is always the step with the highest products.
5. The rate-determining step cannot be inferred from an energy diagram.

Explanation: This question tests the skill of identifying the rate-determining step in reaction energy diagrams with intermediates. The rate-determining step is the one with the largest activation energy barrier, measured as the vertical difference from the preceding energy minimum to the next peak. This step limits the overall rate due to its highest energy requirement. Thus, option A correctly states that the rate-determining step corresponds to the larger barrier from the preceding minimum to the peak. A tempting distractor is option C, which claims it's always Step 1 because it occurs first, but this is incorrect due to the misconception that sequence determines rate, ignoring energy barriers. To find the rate-determining step, compare all step-wise Ea values and select the maximum for kinetic analysis.

Question 20

1. Path 2 has a larger $E_a$ than Path 1, and $\Delta H$ is larger for Path 2
2. Path 2 has a smaller $E_a$ than Path 1, and $\Delta H$ is the same for both paths (correct answer)
3. Path 2 has a smaller $E_a$ than Path 1, and $\Delta H$ is smaller for Path 2
4. Path 2 has the same $E_a$ as Path 1, and $\Delta H$ is the same for both paths
5. Path 2 has a larger $E_a$ than Path 1, and $\Delta H$ is the same for both paths

Explanation: This question tests understanding of how catalysts affect energy diagrams. A catalyst provides an alternative pathway with a lower activation energy, so Path 2 (catalyzed) has a smaller Ea than Path 1 (uncatalyzed). Crucially, catalysts do not change the thermodynamics of the reaction - they only affect the kinetics. Therefore, ΔH (the energy difference between reactants and products) remains the same for both pathways. A tempting misconception (choice C) is thinking that because the catalyst lowers the activation energy, it also changes the reaction enthalpy, but catalysts only affect the pathway, not the starting and ending points. When analyzing catalyzed reactions on energy diagrams, remember that the catalyst changes the height of the energy barrier but not the relative positions of reactants and products.