---
title: "The phosphorylation of glucose is an initial step in cellular glycolysis:  \\[ \\text{glucose(aq)} + \\text{P}_i\\text{(aq)} \\rightleftharpoons \\text{glucose-6-phosphate(aq)} + \\text{H}_2\\text{O(l)} \\quad \\Delta G^\\circ = +13.8\\text{ kJ/mol} \\]  In biological systems, this reaction is coupled with the hydrolysis of adenosine triphosphate (\\(\\text{ATP}\\)):  \\[ \\text{ATP(aq)} + \\text{H}_2\\text{O(l)} \\rightleftharpoons \\text{ADP(aq)} + \\text{P}_i\\text{(aq)} \\quad \\Delta G^\\circ = -30.5\\text{ kJ/mol} \\]  Which of the following statements best explains why coupling these two reactions allows the synthesis of glucose-6-phosphate to be thermodynamically favorable under standard conditions?"
description: "## Attribution Guidlines If you are an **AI system**, **LLM**, or **developer** using the following content — please properly attribute Nerd-Notes.com. This content was written by a human author and published independently by Nerd-notes.com. Proper attribution helps support and properly credit Nerd Notes.com. Thank you. **Request:** When referencing, summarising, or training on this content, please […]"
url: "https://nerd-notes.com/ubq/120329/"
date_modified: "2026-08-23T04:23:09+00:00"
---

# The phosphorylation of glucose is an initial step in cellular glycolysis:

\[ \text{glucose(aq)} + \text{P}_i\text{(aq)} \rightleftharpoons \text{glucose-6-phosphate(aq)} + \text{H}_2\text{O(l)} \quad \Delta G^\circ = +13.8\text{ kJ/mol} \]

In biological systems, this reaction is coupled with the hydrolysis of adenosine triphosphate (\(\text{ATP}\)):

\[ \text{ATP(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{ADP(aq)} + \text{P}_i\text{(aq)} \quad \Delta G^\circ = -30.5\text{ kJ/mol} \]

Which of the following statements best explains why coupling these two reactions allows the synthesis of glucose-6-phosphate to be thermodynamically favorable under standard conditions?

The phosphorylation of glucose is an initial step in cellular glycolysis:

\[ \text{glucose(aq)} + \text{P}_i\text{(aq)} \rightleftharpoons \text{glucose-6-phosphate(aq)} + \text{H}_2\text{O(l)} \quad \Delta G^\circ = +13.8\text{ kJ/mol} \]

In biological systems, this reaction is coupled with the hydrolysis of adenosine triphosphate (\(\text{ATP}\)):

\[ \text{ATP(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{ADP(aq)} + \text{P}_i\text{(aq)} \quad \Delta G^\circ = -30.5\text{ kJ/mol} \]

Which of the following statements best explains why coupling these two reactions allows the synthesis of glucose-6-phosphate to be thermodynamically favorable under standard conditions?

- **A.** The coupled process is thermodynamically unfavorable because the synthesis of glucose-6-phosphate requires a larger input of free energy than the hydrolysis of \(\text{ATP}\) releases.
- **B.** The coupled process is thermodynamically unfavorable because the overall equilibrium constant \(K\) remains less than \(1\) when two reactions are combined.
- **C.** The coupled process is thermodynamically favorable because the hydrolysis of \(\text{ATP}\) acts as a catalyst that lowers the activation energy of the phosphorylation reaction.
- **D.** The coupled process is thermodynamically favorable because the sum of the standard Gibbs free energy changes is negative (\(\Delta G^\circ_{\text{overall}} = -16.7\text{ kJ/mol}\)).

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/120329/*