---
title: "An aqueous reaction \\(\\text{A(aq)} + \\text{B(aq)} \\rightarrow \\text{P(aq)}\\) is catalyzed by an enzyme \\(\\text{E}\\) according to the following proposed three-step mechanism:  \\[ \\text{Step 1: } \\text{E(aq)} + \\text{A(aq)} \\rightleftharpoons \\text{I}_1\\text{(aq)} \\quad \\text{(fast)} \\] \\[ \\text{Step 2: } \\text{I}_1\\text{(aq)} + \\text{B(aq)} \\rightarrow \\text{I}_2\\text{(aq)} \\quad \\text{(slow, rate-determining)} \\] \\[ \\text{Step 3: } \\text{I}_2\\text{(aq)} \\rightarrow \\text{P(aq)} + \\text{E(aq)} \\quad \\text{(fast)} \\]  The reaction energy profile for the original enzyme-catalyzed reaction is shown by the solid curve. A genetically modified variant of the enzyme selectively stabilizes intermediate \\(\\text{I}_1\\), lowering its potential energy as shown by the dashed curve, while the energies of the reactants, products, and all transition states (\\(\\text{TS}_1\\), \\(\\text{TS}_2\\), and \\(\\text{TS}_3\\)) remain unchanged.  Which of the following correctly predicts the effect of this modification on the activation energy of Step 2 (\\(E_{a,2}\\)) and the overall rate of product formation?"
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url: "https://nerd-notes.com/ubq/123695/"
date_modified: "2026-09-28T12:02:00+00:00"
---

# An aqueous reaction \(\text{A(aq)} + \text{B(aq)} \rightarrow \text{P(aq)}\) is catalyzed by an enzyme \(\text{E}\) according to the following proposed three-step mechanism:

\[ \text{Step 1: } \text{E(aq)} + \text{A(aq)} \rightleftharpoons \text{I}_1\text{(aq)} \quad \text{(fast)} \]
\[ \text{Step 2: } \text{I}_1\text{(aq)} + \text{B(aq)} \rightarrow \text{I}_2\text{(aq)} \quad \text{(slow, rate-determining)} \]
\[ \text{Step 3: } \text{I}_2\text{(aq)} \rightarrow \text{P(aq)} + \text{E(aq)} \quad \text{(fast)} \]

The reaction energy profile for the original enzyme-catalyzed reaction is shown by the solid curve. A genetically modified variant of the enzyme selectively stabilizes intermediate \(\text{I}_1\), lowering its potential energy as shown by the dashed curve, while the energies of the reactants, products, and all transition states (\(\text{TS}_1\), \(\text{TS}_2\), and \(\text{TS}_3\)) remain unchanged.

Which of the following correctly predicts the effect of this modification on the activation energy of Step 2 (\(E_{a,2}\)) and the overall rate of product formation?

An aqueous reaction \(\text{A(aq)} + \text{B(aq)} \rightarrow \text{P(aq)}\) is catalyzed by an enzyme \(\text{E}\) according to the following proposed three-step mechanism:

\[ \text{Step 1: } \text{E(aq)} + \text{A(aq)} \rightleftharpoons \text{I}_1\text{(aq)} \quad \text{(fast)} \]
\[ \text{Step 2: } \text{I}_1\text{(aq)} + \text{B(aq)} \rightarrow \text{I}_2\text{(aq)} \quad \text{(slow, rate-determining)} \]
\[ \text{Step 3: } \text{I}_2\text{(aq)} \rightarrow \text{P(aq)} + \text{E(aq)} \quad \text{(fast)} \]

The reaction energy profile for the original enzyme-catalyzed reaction is shown by the solid curve. A genetically modified variant of the enzyme selectively stabilizes intermediate \(\text{I}_1\), lowering its potential energy as shown by the dashed curve, while the energies of the reactants, products, and all transition states (\(\text{TS}_1\), \(\text{TS}_2\), and \(\text{TS}_3\)) remain unchanged.

Which of the following correctly predicts the effect of this modification on the activation energy of Step 2 (\(E_{a,2}\)) and the overall rate of product formation?

![Grayscale reaction energy profile plot showing Potential Energy on the vertical axis versus Reaction Progress on the horizontal axis. Axes are bare lines with arrowheads. A solid black curve starts at the initial reactant baseline labeled E + A + B, rises to a first peak labeled TS1, dips to a first local minimum labeled I1 + B, rises to a higher second peak labeled TS2, dips to a second local minimum labeled I2, rises to a lower third peak labeled TS3, and descends to a final product level labeled P + E below the reactant baseline. A dashed gray curve follows the solid curve to TS1, dips to a noticeably deeper local minimum labeled I1' + B, and then rises back up to reach the exact same peak height at TS2 before rejoining the solid curve for the remainder of the pathway. No other lines, gridlines, labels, or annotations appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1790596920-BSBy7I.jpg)

- **A.** \(E_{a,2}\) increases, and the overall rate of product formation decreases because a larger energy barrier must be overcome from \(\text{I}_1\) to reach \(\text{TS}_2\).
- **B.** \(E_{a,2}\) decreases, and the overall rate of product formation increases because lowering the potential energy of \(\text{I}_1\) stabilizes the enzyme-substrate complex.
- **C.** \(E_{a,2}\) remains unchanged, and the overall rate of product formation decreases because the lower potential energy of \(\text{I}_1\) reduces the frequency of collisions with \(\text{B}\).
- **D.** \(E_{a,2}\) increases, but the overall rate of product formation increases because the formation of \(\text{I}_1\) from reactants becomes more thermodynamically favorable.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/123695/*