College Chemistry Quiz: Multistep Reaction Energy Profile

18 questions · exam conditions

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Multistep Reaction Energy ProfileQuestion 1 of 18

A reaction mechanism consists of two elementary steps:

Step 1: $A + B \rightleftharpoons C$ (fast equilibrium) Step 2: $C + D \rightarrow E + F$ (slow)

On the energy profile for this mechanism, the first transition state is 85 kJ/mol above the reactants, the intermediate C is 25 kJ/mol above the reactants, and the second transition state is 140 kJ/mol above the reactants. What is the activation energy for the reverse of the first step?

25 kJ/mol

60 kJ/mol

85 kJ/mol

115 kJ/mol

140 kJ/mol

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College Chemistry Quiz

College Chemistry Quiz: Multistep Reaction Energy Profile

Practice Multistep Reaction Energy Profile in College 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 Multistep Reaction Energy Profile, giving you a quick way to practice the rules, question types, and explanations that matter most for College 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

A reaction mechanism consists of two elementary steps:

Step 1: $A + B \rightleftharpoons C$ (fast equilibrium) Step 2: $C + D \rightarrow E + F$ (slow)

25 kJ/mol
60 kJ/mol (correct answer)
85 kJ/mol
115 kJ/mol
140 kJ/mol

Explanation: When analyzing reaction mechanisms with energy profiles, you need to understand that activation energy is the energy difference between reactants and the transition state for any given step, whether forward or reverse. For the reverse of step 1 (

C \rightarrow A + B

), you're starting from intermediate C and going back to the original reactants through the same transition state. The activation energy for this reverse process is the energy difference between C (at 25 kJ/mol above reactants) and the first transition state (at 85 kJ/mol above reactants). This gives us

85 - 25 = 60

kJ/mol. Think of it this way: if you're climbing from C back over the energy barrier, you need enough energy to reach the peak from your current position. Choice A (25 kJ/mol) represents the energy of intermediate C above the reactants, not an activation energy. Choice C (85 kJ/mol) is the activation energy for the forward direction of step 1, not the reverse. Choice D (115 kJ/mol) incorrectly adds the intermediate energy to the forward activation energy, which has no physical meaning. The correct answer is B (60 kJ/mol). Remember this key principle: for any reversible elementary step, if you know the forward activation energy and the energy difference between reactants and products, you can find the reverse activation energy. The relationship is:

E_{a,reverse} = E_{a,forward} - \Delta E_{rxn}

. Always identify your starting point and ending point when calculating activation energies.

Question 2

Consider a reaction mechanism where the rate-determining step involves the conversion of intermediate X to intermediate Y. The energy profile shows that X is 40 kJ/mol above the original reactants, the transition state for this conversion is 115 kJ/mol above the original reactants, and Y is 25 kJ/mol above the original reactants. What is the activation energy for the reverse conversion of Y back to X?

15 kJ/mol
40 kJ/mol
75 kJ/mol
90 kJ/mol (correct answer)
115 kJ/mol

Explanation: When analyzing reaction mechanisms and energy profiles, you need to understand that activation energy is always measured from the starting point to the transition state. For a reverse reaction, the starting point becomes the product of the forward reaction. To find the activation energy for Y converting back to X, you start from Y's energy level and measure up to the transition state. Y sits at 25 kJ/mol above the original reactants, while the transition state is at 115 kJ/mol above the original reactants. The activation energy for the reverse reaction is therefore:

115 - 25 = 90 \text{ kJ/mol}

Let's examine why the other answers are incorrect. Choice A (15 kJ/mol) represents the energy difference between X and Y (40 - 25 = 15), which is simply the energy difference between intermediates, not an activation energy. Choice B (40 kJ/mol) is the energy of intermediate X relative to the original reactants, but this isn't relevant to the reverse reaction starting from Y. Choice C (75 kJ/mol) represents the activation energy for the forward reaction (115 - 40 = 75), which measures from X to the transition state, not from Y. Remember this key principle: activation energy always equals the transition state energy minus the starting material energy. When working with reverse reactions, your "starting material" becomes the product of the forward reaction. Drawing or visualizing the energy diagram helps you identify the correct starting and ending points for your calculation.

Question 3

A reaction mechanism shows that the rate-determining step involves breaking a C-H bond with an activation energy of 105 kJ/mol, while a competing side reaction involves breaking a C-C bond with an activation energy of 125 kJ/mol. If both pathways start from the same intermediate that is 25 kJ/mol above the original reactants, what is the difference in activation energy between these two pathways when measured from the original reactants?

20 kJ/mol (correct answer)
25 kJ/mol
80 kJ/mol
105 kJ/mol
125 kJ/mol

Explanation: When analyzing reaction mechanisms with competing pathways, you need to carefully track energy changes from a common reference point. Both pathways here start from the same intermediate, but the question asks for activation energies measured from the original reactants. The intermediate sits 25 kJ/mol above the original reactants. From this intermediate, the C-H bond breaking requires an additional 105 kJ/mol to reach its transition state, while the C-C bond breaking needs 125 kJ/mol. To find the total activation energy from the original reactants, you add the energy to reach the intermediate plus the additional energy needed: C-H pathway = 25 + 105 = 130 kJ/mol, and C-C pathway = 25 + 125 = 150 kJ/mol. The difference is 150 - 130 = 20 kJ/mol. Choice A (20 kJ/mol) correctly accounts for both energy steps. Choice B (25 kJ/mol) represents only the energy difference between reactants and intermediate, ignoring the bond-breaking energies entirely. Choice C (80 kJ/mol) might result from incorrectly subtracting 25 from 105, misunderstanding how to combine the energy values. Choice D (105 kJ/mol) gives just the activation energy for C-H bond breaking from the intermediate, not the difference between pathways. Remember that activation energies in multi-step mechanisms are cumulative from your reference point. Always identify what energy level you're measuring from, then systematically add each energy barrier to reach the transition state.

Question 4

A biochemical reaction pathway involves three consecutive enzymatic steps, each with different activation energies. The energy profile shows that the second enzyme-substrate complex is the most stable intermediate, lying 30 kJ/mol below the original reactants. If this step has an activation energy of 95 kJ/mol for product formation, and this represents the rate-determining step, what is the height of this transition state above the original reactants?

65 kJ/mol (correct answer)
95 kJ/mol
125 kJ/mol
155 kJ/mol
185 kJ/mol

Explanation: When analyzing enzyme kinetics and energy profiles, you need to carefully distinguish between the stability of intermediates and the height of transition states. The key is understanding that activation energy is measured from the starting point of each step, not from the original reactants. Here, the enzyme-substrate complex lies 30 kJ/mol below the original reactants, making it a stable intermediate. The activation energy of 95 kJ/mol represents the energy barrier from this intermediate to the transition state. To find the transition state height above the original reactants, you add these values: the intermediate is 30 kJ/mol below the reactants, so the transition state is at (-30 + 95) = 65 kJ/mol above the original reactants. Choice A (65 kJ/mol) correctly accounts for both the intermediate's stability and the activation barrier. Choice B (95 kJ/mol) incorrectly assumes the activation energy is measured from the original reactants rather than from the intermediate. Choice C (125 kJ/mol) makes the error of adding 30 + 95, wrongly treating the intermediate's stability as if it raises rather than lowers the energy level. Choice D (155 kJ/mol) likely results from confusion about energy reference points and incorrectly combining values. Remember: activation energy is always measured from the immediate starting point of each elementary step, not from the overall reaction's beginning. When intermediates are more stable than reactants, transition states can actually be lower than if calculated from the original energy level.

Question 5

In the energy profile for a two-step reaction mechanism, the first step has an activation energy of 95 kJ/mol and is exothermic by 30 kJ/mol. The second step has an activation energy of 110 kJ/mol relative to the intermediate and is exothermic by 45 kJ/mol. What is the energy of the second transition state relative to the original reactants?

35 kJ/mol
65 kJ/mol
80 kJ/mol (correct answer)
110 kJ/mol
140 kJ/mol

Explanation: When analyzing energy profiles for multi-step reactions, you need to track energy changes relative to your starting point—the original reactants. Each step's activation energy tells you how much energy is needed to reach that transition state from the preceding species. To find the second transition state's energy, work through the mechanism step by step. The first step requires 95 kJ/mol activation energy, so the first transition state sits 95 kJ/mol above the reactants. After this transition state, the reaction releases 30 kJ/mol (exothermic), placing the intermediate at

95 - 30 = 65

kJ/mol above the original reactants. The second step has an activation energy of 110 kJ/mol relative to this intermediate. Therefore, the second transition state is

65 + 110 = 180

kJ/mol above the original reactants. Wait—that's not among the options! Let me recalculate: the intermediate is at

0 - 30 = -30

kJ/mol relative to reactants (30 kJ/mol lower). The second transition state is then

-30 + 110 = 80

kJ/mol above the original reactants. Choice A (35 kJ/mol) incorrectly subtracts the first step's energy release from the second activation energy. Choice B (65 kJ/mol) gives only the intermediate's position, forgetting to add the second activation energy. Choice D (110 kJ/mol) uses only the second step's activation energy without accounting for the intermediate's position. Strategy tip: Always track energies relative to your reference point (usually initial reactants) and draw a rough energy diagram as you calculate—it prevents sign errors and missed steps.

Question 6

A reaction mechanism involves a pre-equilibrium followed by a slow step:

$A + B \rightleftharpoons C$ (fast, $K_{eq} = 0.8$ ) $C + D \rightarrow E + F$ (slow)

The energy profile shows that the equilibrium between A + B and C slightly favors the reactants, with C being 5 kJ/mol above A + B. If the slow step has an activation energy of 90 kJ/mol, what is the overall activation energy for the formation of products?

85 kJ/mol
90 kJ/mol
95 kJ/mol (correct answer)
180 kJ/mol
185 kJ/mol

Explanation: When analyzing multi-step reaction mechanisms, you need to construct an energy profile that accounts for all intermediates and transitions. The overall activation energy is the energy difference between the starting materials and the highest energy point along the entire reaction pathway. Here's how to build this energy profile step by step. Start with A + B as your reference point (0 kJ/mol). The pre-equilibrium tells us that C lies 5 kJ/mol higher in energy than A + B. Now for the slow step: C must overcome a 90 kJ/mol activation barrier to form products. This means the transition state for the slow step sits at 5 + 90 = 95 kJ/mol above the initial reactants A + B. Since this transition state represents the highest energy point in the entire mechanism, the overall activation energy is 95 kJ/mol. Looking at the wrong answers: (A) 85 kJ/mol incorrectly subtracts the intermediate's energy from the activation energy (90 - 5), missing that energies are additive when moving upward from the starting point. (B) 90 kJ/mol only considers the activation energy of the slow step while ignoring that this step starts from the elevated intermediate C, not from the original reactants. (D) 180 kJ/mol incorrectly doubles the activation energy, perhaps treating each step as contributing separately to a sum. Remember: in multi-step mechanisms, always trace the complete energy pathway from initial reactants to identify the highest point. The overall activation energy is measured from the starting materials to this highest transition state, regardless of which step it occurs in.

Question 7

A proposed reaction mechanism involves a rapid pre-equilibrium followed by two slower consecutive steps:

$A \rightleftharpoons B$ (rapid, $K = 2.5$ ) $B \rightarrow C$ (slow, $E_a = 75$ kJ/mol) $C \rightarrow D$ (slow, $E_a = 85$ kJ/mol)

If intermediate B is 8 kJ/mol above reactant A, and intermediate C is 15 kJ/mol above A, what is the overall activation energy for the conversion of A to D?

75 kJ/mol
83 kJ/mol
85 kJ/mol
100 kJ/mol (correct answer)
160 kJ/mol

Explanation: When analyzing multi-step reaction mechanisms, you need to identify the rate-determining step and construct an energy diagram to find the overall activation energy. The overall activation energy is the energy difference between the initial reactant and the highest energy barrier along the entire pathway. Let's map out the energy landscape. Starting with A at 0 kJ/mol, intermediate B sits at +8 kJ/mol. The transition state for B→C requires 75 kJ/mol above B, placing it at 8 + 75 = 83 kJ/mol above A. Intermediate C is at +15 kJ/mol, and the transition state for C→D needs 85 kJ/mol above C, putting it at 15 + 85 = 100 kJ/mol above A. The overall activation energy is determined by the highest energy barrier from start to finish. Comparing the two transition states (83 kJ/mol and 100 kJ/mol above A), the C→D transition state is highest at 100 kJ/mol above the starting material A. Choice A (75 kJ/mol) incorrectly uses just the activation energy of the B→C step without accounting for B's elevated position. Choice B (83 kJ/mol) represents the energy of the first transition state above A, missing that there's a higher barrier later. Choice C (85 kJ/mol) gives only the activation energy for the final step, ignoring that C is already elevated above A. Study tip: For multi-step mechanisms, always sketch an energy diagram showing all intermediates and transition states relative to the starting material. The overall activation energy is the height of the tallest peak above your starting point.

Question 8

A proposed enzyme mechanism involves substrate binding followed by two chemical transformation steps:

Step 1: $E + S \rightleftharpoons ES$ (rapid equilibrium) Step 2: $ES \rightarrow EI$ (slow) Step 3: $EI \rightarrow E + P$ (fast)

If the energy profile shows the ES complex is 15 kJ/mol below the free enzyme and substrate, the first chemical step has an activation energy of 65 kJ/mol, and the second chemical step has an activation energy of 25 kJ/mol, what is the overall activation energy for product formation?

25 kJ/mol
50 kJ/mol (correct answer)
65 kJ/mol
80 kJ/mol
90 kJ/mol

Explanation: When analyzing enzyme mechanisms with multiple steps, the overall activation energy is determined by the highest energy barrier the reaction must overcome, not the sum of individual barriers. To find the overall activation energy, you need to trace the energy path from reactants to the highest transition state. Starting from free enzyme and substrate, the ES complex forms at -15 kJ/mol (below the starting point). From this ES complex, the first chemical step has an activation energy of 65 kJ/mol, meaning the first transition state sits at -15 + 65 = 50 kJ/mol above the initial state. The second step's activation energy of 25 kJ/mol is measured from the EI intermediate, but since this barrier is lower and occurs after the rate-determining step, it doesn't affect the overall activation energy. The overall activation energy is 50 kJ/mol, making answer B correct. Answer A (25 kJ/mol) incorrectly uses only the activation energy of the second step, ignoring that you must consider the path from the initial reactants. Answer C (65 kJ/mol) makes the common error of using the activation energy of the slow step directly, without accounting for the ES complex being 15 kJ/mol below the starting materials. Answer D (80 kJ/mol) incorrectly adds the two activation energies together (65 + 15), treating them as cumulative barriers. Remember: in multi-step mechanisms, draw an energy diagram and identify the highest point above your starting materials—that's your overall activation energy.

Question 9

A proposed mechanism for the reaction $2A + B \rightarrow C + D$ involves three elementary steps:

Step 1: $A + B \rightleftharpoons I_1$ (fast equilibrium) Step 2: $I_1 + A \rightarrow I_2$ (slow) Step 3: $I_2 \rightarrow C + D$ (fast)

If the energy profile shows that $I_1$ is 20 kJ/mol below the reactants and $I_2$ is 15 kJ/mol above the reactants, and the activation energy for step 2 is 85 kJ/mol, what is the energy of the rate-determining transition state relative to the original reactants?

65 kJ/mol (correct answer)
70 kJ/mol
85 kJ/mol
100 kJ/mol
105 kJ/mol

Explanation: When analyzing multi-step reaction mechanisms, you need to identify the rate-determining step and carefully track energy levels relative to the starting materials. The rate-determining step controls the overall reaction rate and determines which transition state has the highest energy barrier. Since step 2 is labeled as "slow," it's the rate-determining step. To find the energy of its transition state, you start from the energy of

I_1

(the reactant for step 2) and add the activation energy. The problem states that

I_1

is 20 kJ/mol below the original reactants, so

I_1

sits at -20 kJ/mol relative to the starting point. The activation energy for step 2 is 85 kJ/mol, which means the transition state is 85 kJ/mol above

I_1

. Therefore: -20 + 85 = 65 kJ/mol above the original reactants. Looking at the wrong answers: B (70 kJ/mol) might result from incorrectly using

I_2

's energy as a reference point instead of

I_1

. C (85 kJ/mol) represents the activation energy itself, ignoring that we need the energy relative to the original reactants, not relative to

I_1

. D (100 kJ/mol) could come from mistakenly adding the activation energy to the energy of

I_2

instead of

I_1

. The correct answer is A (65 kJ/mol). Study tip: Always identify which intermediate serves as the reactant for the rate-determining step, then add the activation energy to that intermediate's energy level—not to the original reactants or products.

Question 10

For a three-step reaction mechanism, the energy profile shows that the first step has an activation energy of 80 kJ/mol, the second step has an activation energy of 65 kJ/mol, and the third step has an activation energy of 95 kJ/mol. If the first intermediate is 20 kJ/mol above the reactants and the second intermediate is 35 kJ/mol above the reactants, which statement about this mechanism is correct?

The first step is rate-determining because it has the highest energy intermediate formed in the reaction pathway
The second step is rate-determining because it has the lowest activation energy among all three steps
The third step is rate-determining because it has the highest activation energy relative to its starting point (correct answer)
The rate-determining step cannot be identified without knowing the energy of the final products in this mechanism
All three steps proceed at equal rates because the energy differences between intermediates are relatively small

Explanation: When analyzing multi-step reaction mechanisms, the rate-determining step is the slowest step in the overall process. This step controls the overall reaction rate because the entire mechanism cannot proceed faster than its slowest component. To identify it, you need to compare the activation energies from each step's starting point, not just the absolute activation energy values. Let's trace through each step's energy barrier. The first step starts from the reactants (0 kJ/mol baseline) and has an 80 kJ/mol activation energy. The second step starts from the first intermediate (20 kJ/mol above reactants) with a 65 kJ/mol activation energy, giving a total energy barrier of 20 + 65 = 85 kJ/mol from the baseline. The third step starts from the second intermediate (35 kJ/mol above reactants) with a 95 kJ/mol activation energy, requiring 35 + 95 = 130 kJ/mol total energy from the baseline. This makes the third step rate-determining with the highest overall energy barrier. Answer A incorrectly focuses on intermediate stability rather than activation barriers. Answer B makes the common error of comparing only the relative activation energies (65 kJ/mol being lowest) without considering where each step begins. Answer D is wrong because product energy doesn't affect which step is rate-determining—only the energy barriers matter. Remember: for multi-step mechanisms, always calculate the total energy required to reach each transition state from your reference point (usually the reactants) to identify the rate-determining step.

Question 11

A proposed mechanism for the decomposition of ozone involves two elementary steps:

Step 1: $O_3(g) + Cl(g) \rightarrow ClO(g) + O_2(g)$ (fast equilibrium) Step 2: $ClO(g) + O(g) \rightarrow Cl(g) + O_2(g)$ (slow)

Based on the energy profile shown, what is the activation energy for the rate-determining step?

45 kJ/mol
75 kJ/mol
120 kJ/mol (correct answer)
165 kJ/mol
210 kJ/mol

Explanation: The rate-determining step is Step 2 (the slow step). On the energy profile, this corresponds to the transition from the intermediate (ClO + O) to the second transition state. The activation energy is the difference between the energy of the intermediate (at 45 kJ/mol) and the second transition state (at 165 kJ/mol): 165 - 45 = 120 kJ/mol. Choice A is the energy of the intermediate above reactants. Choice B is the activation energy for the reverse of step 1. Choice D is the energy of the second transition state. Choice E is the overall energy change.

Question 12

The energy profile below represents a reaction that can proceed through two different pathways. Pathway 1 (solid line) involves a single step, while Pathway 2 (dashed line) involves two steps with an intermediate. Under conditions where both pathways are possible, which statement best describes the kinetic behavior?

Pathway 1 will be faster because it involves fewer molecular collisions and has a lower entropy requirement
Pathway 2 will be faster because the rate-determining step has a lower activation energy than Pathway 1 (correct answer)
Both pathways will proceed at equal rates because they have the same overall thermodynamic favorability
Pathway 1 will be faster because its single transition state is lower in energy than either transition state in Pathway 2
The relative rates depend on temperature, with Pathway 1 favored at high temperature and Pathway 2 favored at low temperature

Explanation: The rate of a reaction is determined by the highest activation energy barrier. In Pathway 1, the activation energy is 120 kJ/mol. In Pathway 2, the rate-determining step has an activation energy of 95 kJ/mol (from reactants to the higher of the two transition states). Since 95 < 120, Pathway 2 will be faster. Choice A incorrectly focuses on molecular collisions rather than activation energy. Choice C confuses thermodynamics with kinetics. Choice D incorrectly compares individual transition states rather than the rate-determining barrier. Choice E incorrectly suggests temperature dependence when the activation energy difference is clear.

Question 13

The energy profile below shows a reversible reaction mechanism with two steps. At equilibrium, the forward and reverse rates of each elementary step are equal. If the temperature is increased, which statement best describes the effect on the energy profile and reaction behavior?

The activation energies decrease proportionally, making both forward and reverse reactions faster while maintaining the same equilibrium position
The energy differences between reactants, intermediates, and products remain unchanged, but reaction rates increase due to more molecules having sufficient kinetic energy (correct answer)
The activation energies increase due to enhanced molecular motion, but the equilibrium shifts toward the endothermic direction
The intermediate becomes more stable relative to reactants and products, changing the overall mechanism pathway
The rate-determining step changes because different activation energies have different temperature dependencies according to the Arrhenius equation

Explanation: Temperature affects reaction rates through the Arrhenius equation (k = Ae^(-Ea/RT)) but does not change the intrinsic energy profile of the reaction. The activation energies and energy differences remain constant, but higher temperature means more molecules have kinetic energy ≥ Ea, increasing reaction rates. The equilibrium position may shift according to Le Châtelier's principle, but the energy profile itself is unchanged. Choice A incorrectly states activation energies decrease. Choice C incorrectly states activation energies increase. Choice D incorrectly suggests the intermediate stability changes. Choice E incorrectly suggests the rate-determining step changes due to temperature.

Question 14

The diagram compares the energy profiles for the same reaction under three different conditions: uncatalyzed (solid line), with catalyst A (dashed line), and with catalyst B (dotted line). Both catalysts provide two-step pathways with different intermediates. Based on the energy barriers shown, which statement about the relative reaction rates is correct?

Uncatalyzed reaction is fastest because it involves the fewest number of elementary steps in the mechanism
Catalyst A provides the fastest rate because its first step has the lowest activation energy among all pathways
Catalyst B provides the fastest rate because both of its steps have lower activation energies than the uncatalyzed reaction (correct answer)
Catalyst A and B provide equal rates because they both lower the overall activation energy by the same amount
The relative rates depend on reaction conditions since both catalysts have rate-determining steps with similar activation energies

Explanation: The reaction rate is determined by the highest activation energy barrier in each pathway. Uncatalyzed: 130 kJ/mol. Catalyst A: rate-determining step is 85 kJ/mol (higher of the two barriers). Catalyst B: rate-determining step is 70 kJ/mol (higher of its two barriers). Since 70 < 85 < 130, catalyst B provides the fastest rate. Choice A incorrectly focuses on number of steps. Choice B looks at the wrong barrier. Choice D incorrectly states equal rates. Choice E incorrectly suggests similar activation energies when there's a clear 15 kJ/mol difference.

Question 15

The diagram shows the energy profile for a catalyzed reaction (solid line) and the same reaction without a catalyst (dashed line). If the uncatalyzed reaction has an activation energy of 150 kJ/mol and the catalyzed pathway involves two steps with activation energies of 60 kJ/mol and 75 kJ/mol respectively, what is the energy of the intermediate in the catalyzed pathway relative to the reactants?

15 kJ/mol above reactants
35 kJ/mol above reactants
45 kJ/mol above reactants
60 kJ/mol above reactants

Explanation: C

Question 16

The decomposition of $N_2O_5$ follows a two-step mechanism. The energy profile below shows both steps, where the first step is endothermic and the second step is exothermic. If a catalyst is introduced that lowers the activation energy of the rate-determining step by 40 kJ/mol while leaving the other step unchanged, what would be the new overall activation energy?

45 kJ/mol
65 kJ/mol
85 kJ/mol (correct answer)
105 kJ/mol
125 kJ/mol

Explanation: From the energy profile, the first step has an activation energy of 85 kJ/mol (from 0 to 85 kJ/mol), and the second step has an activation energy of 125 kJ/mol (from 30 kJ/mol to 155 kJ/mol above reactants). The rate-determining step is the second step (125 kJ/mol barrier). When the catalyst lowers this by 40 kJ/mol, the new activation energy for the second step becomes 125 - 40 = 85 kJ/mol above reactants. The first step remains at 85 kJ/mol above reactants. Since both barriers are now equal, the overall activation energy is 85 kJ/mol. Choice A is too low. Choice B assumes the wrong step was catalyzed. Choice D is the original first step barrier. Choice E is the original second step barrier.

Question 17

The energy profile diagram shows a reaction that can proceed through two parallel pathways from the same starting intermediate. Pathway X leads to product P, while pathway Y leads to product Q. Under kinetic control conditions, which product will be formed preferentially and why?

Product P, because it has a lower final energy and is thermodynamically more stable than product Q
Product Q, because its pathway has a lower activation energy barrier from the intermediate to the transition state (correct answer)
Product P, because the overall energy change from reactants to P is more favorable than to Q
Both products form in equal amounts, because they start from the same intermediate with equal concentrations
Product Q, because it forms through a mechanism with fewer bond-breaking steps based on the energy profile shape

Explanation: Under kinetic control, the product that forms faster is favored, which depends on activation energy, not thermodynamic stability. Pathway X (to P) has an activation energy of 95 - 30 = 65 kJ/mol from the intermediate. Pathway Y (to Q) has an activation energy of 85 - 30 = 55 kJ/mol from the intermediate. Since pathway Y has a lower activation energy (55 < 65 kJ/mol), product Q will form faster and be the kinetic product. Choice A focuses on thermodynamic stability. Choice C considers overall energy change rather than kinetics. Choice D ignores activation energy differences. Choice E makes unfounded assumptions about bond-breaking.

Question 18

The diagram shows energy profiles for the same reaction catalyzed by two different enzymes. Enzyme A provides a single-step pathway, while Enzyme B provides a two-step pathway with an intermediate. If both enzymes are present in equal concentrations, which statement best predicts the reaction outcome?

Only Enzyme A will be active because single-step mechanisms are always preferred over multi-step mechanisms
Only Enzyme B will be active because it provides the lower overall activation energy for the reaction
Both enzymes will be equally active because they both lower the activation energy compared to the uncatalyzed reaction
Enzyme A will dominate because its rate-determining step has a lower activation energy than Enzyme B's rate-determining step

Explanation: D