---
title: "The standard enthalpies of formation, \\(\\Delta H_f^\\circ\\), for gaseous and liquid water at \\(298\\text{ K}\\) and \\(1\\text{ atm}\\) are given in the table below.  | Substance | State | \\(\\Delta H_f^\\circ\\text{ (kJ/mol)}\\) | | :— | :— | :— | | \\(\\text{H}_2\\text{O}(g)\\) | Gas | \\(-242\\) | | \\(\\text{H}_2\\text{O}(l)\\) | Liquid | \\(-286\\) |  Which of the following best explains why the standard enthalpy of formation of \\(\\text{H}_2\\text{O}(l)\\) is more negative than that of \\(\\text{H}_2\\text{O}(g)\\)?"
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url: "https://nerd-notes.com/ubq/123820/"
date_modified: "2026-09-28T12:30:26+00:00"
---

# The standard enthalpies of formation, \(\Delta H_f^\circ\), for gaseous and liquid water at \(298\text{ K}\) and \(1\text{ atm}\) are given in the table below.

| Substance | State | \(\Delta H_f^\circ\text{ (kJ/mol)}\) |
| :— | :— | :— |
| \(\text{H}_2\text{O}(g)\) | Gas | \(-242\) |
| \(\text{H}_2\text{O}(l)\) | Liquid | \(-286\) |

Which of the following best explains why the standard enthalpy of formation of \(\text{H}_2\text{O}(l)\) is more negative than that of \(\text{H}_2\text{O}(g)\)?

The standard enthalpies of formation, \(\Delta H_f^\circ\), for gaseous and liquid water at \(298\text{ K}\) and \(1\text{ atm}\) are given in the table below.

| Substance | State | \(\Delta H_f^\circ\text{ (kJ/mol)}\) |
| :--- | :--- | :--- |
| \(\text{H}_2\text{O}(g)\) | Gas | \(-242\) |
| \(\text{H}_2\text{O}(l)\) | Liquid | \(-286\) |

Which of the following best explains why the standard enthalpy of formation of \(\text{H}_2\text{O}(l)\) is more negative than that of \(\text{H}_2\text{O}(g)\)?

- **A.** The value of \(\Delta H_f^\circ\) is more negative for \(\text{H}_2\text{O}(l)\) because the covalent \(\text{O}-\text{H}\) bonds in the liquid state are significantly shorter and have higher bond dissociation enthalpies than those in the gaseous state.
- **B.** The value of \(\Delta H_f^\circ\) is more negative for \(\text{H}_2\text{O}(l)\) because the formation of intermolecular attractions, predominantly hydrogen bonding, releases energy as the gas condenses to the liquid state.
- **C.** The value of \(\Delta H_f^\circ\) is more negative for \(\text{H}_2\text{O}(l)\) because breaking the \(\text{H}-\text{H}\) and \(\text{O}=\text{O}\) reactant bonds requires less energy when the synthesized product is in the liquid phase than when it is in the gas phase.
- **D.** The value of \(\Delta H_f^\circ\) is more negative for \(\text{H}_2\text{O}(l)\) because liquid water has lower standard entropy than gaseous water, which directly decreases the enthalpy of formation by \(T\Delta S^\circ\).

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/123820/*