---
title: "The phosphorylation of glucose is an essential 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}_{\\text{rxn}}\\]  In biological systems, this endergonic process is coupled to 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}_{\\text{rxn}}\\]  Which of the following statements best explains why the overall coupled process is thermodynamically favorable under standard conditions?"
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date_modified: "2026-08-21T08:17:56+00:00"
---

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

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

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

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

Which of the following statements best explains why the overall coupled process is thermodynamically favorable under standard conditions?

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

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

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

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

Which of the following statements best explains why the overall coupled process is thermodynamically favorable under standard conditions?

- **A.** The overall coupled process is thermodynamically favorable because \(\text{ATP}\) acts as a catalyst that lowers the activation energy of glucose phosphorylation.
- **B.** The overall coupled process is thermodynamically favorable because the sum of the standard Gibbs free energy changes for the coupled reactions is negative (\(\Delta G^\circ_{\text{net}} < 0\)).
- **C.** The overall coupled process is thermodynamically unfavorable because the positive \(\Delta G^\circ\) of the individual phosphorylation reaction cannot be altered by combining reactions.
- **D.** The overall coupled process is thermodynamically unfavorable because the equilibrium constant for \(\text{ATP}\) hydrolysis is less than \(1\) (\(K < 1\)).

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/119918/*