---
title: "A student determines the standard enthalpy change, \\(\\Delta H^\\circ_{\\text{rxn}}\\), for the synthesis of liquid ethanol represented by the following equation:  \\[ \\text{C}_2\\text{H}_4(g) + \\text{H}_2\\text{O}(g) \\rightarrow \\text{C}_2\\text{H}_5\\text{OH}(l) \\]  The student calculates \\(\\Delta H^\\circ_{\\text{rxn}}\\) using standard enthalpies of formation, \\(\\Delta H_f^\\circ\\), obtaining a value of \\(-88 \\text{ kJ/mol}_{\\text{rxn}}\\). When estimating \\(\\Delta H^\\circ_{\\text{rxn}}\\) using average bond enthalpies, the student obtains a value of \\(-42 \\text{ kJ/mol}_{\\text{rxn}}\\). Which of the following statements best accounts for why the bond enthalpy method yields a significantly less exothermic value?"
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/123802/"
date_modified: "2026-09-28T12:30:22+00:00"
---

# A student determines the standard enthalpy change, \(\Delta H^\circ_{\text{rxn}}\), for the synthesis of liquid ethanol represented by the following equation:

\[ \text{C}_2\text{H}_4(g) + \text{H}_2\text{O}(g) \rightarrow \text{C}_2\text{H}_5\text{OH}(l) \]

The student calculates \(\Delta H^\circ_{\text{rxn}}\) using standard enthalpies of formation, \(\Delta H_f^\circ\), obtaining a value of \(-88 \text{ kJ/mol}_{\text{rxn}}\). When estimating \(\Delta H^\circ_{\text{rxn}}\) using average bond enthalpies, the student obtains a value of \(-42 \text{ kJ/mol}_{\text{rxn}}\). Which of the following statements best accounts for why the bond enthalpy method yields a significantly less exothermic value?

A student determines the standard enthalpy change, \(\Delta H^\circ_{\text{rxn}}\), for the synthesis of liquid ethanol represented by the following equation:

\[ \text{C}_2\text{H}_4(g) + \text{H}_2\text{O}(g) \rightarrow \text{C}_2\text{H}_5\text{OH}(l) \]

The student calculates \(\Delta H^\circ_{\text{rxn}}\) using standard enthalpies of formation, \(\Delta H_f^\circ\), obtaining a value of \(-88 \text{ kJ/mol}_{\text{rxn}}\). When estimating \(\Delta H^\circ_{\text{rxn}}\) using average bond enthalpies, the student obtains a value of \(-42 \text{ kJ/mol}_{\text{rxn}}\). Which of the following statements best accounts for why the bond enthalpy method yields a significantly less exothermic value?

- **A.** The bond enthalpy calculation is an estimate that assumes all species are in the gas phase and fails to include the exothermic release of energy associated with forming intermolecular attractions as \(\text{C}_2\text{H}_5\text{OH}(l)\) condenses.
- **B.** The bond enthalpy calculation overestimates the energy required to break bonds because gaseous reactants possess significant intermolecular attractions that must be overcome before chemical bonds can break.
- **C.** The bond enthalpy calculation yields an inaccurate result because standard enthalpies of formation are general averages across many molecular environments, whereas bond enthalpies are exact values for specific molecules.
- **D.** The bond enthalpy calculation yields a less exothermic value because the breaking of chemical bonds in the reactants is an exothermic process, whereas forming chemical bonds in the product is an endothermic process.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/123802/*