---
title: "The standard state of bromine at \\(298\\text{ K}\\) and \\(1\\text{ atm}\\) is \\(\\text{Br}_2\\text{(l)}\\). A student compares the standard enthalpy changes for the two reactions represented below at \\(298\\text{ K}\\):  \\[ \\text{Reaction 1: } \\text{H}_2\\text{(g)} + \\text{Br}_2\\text{(l)} \\rightarrow 2\\text{HBr(g)} \\quad \\Delta H_1^\\circ \\] \\[ \\text{Reaction 2: } \\text{H}_2\\text{(g)} + \\text{Br}_2\\text{(g)} \\rightarrow 2\\text{HBr(g)} \\quad \\Delta H_2^\\circ \\]  Which of the following correctly compares \\(\\Delta H_2^\\circ\\) to \\(\\Delta H_1^\\circ\\) and provides the correct justification?"
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url: "https://nerd-notes.com/ubq/123823/"
date_modified: "2026-09-28T12:30:27+00:00"
---

# The standard state of bromine at \(298\text{ K}\) and \(1\text{ atm}\) is \(\text{Br}_2\text{(l)}\). A student compares the standard enthalpy changes for the two reactions represented below at \(298\text{ K}\):

\[ \text{Reaction 1: } \text{H}_2\text{(g)} + \text{Br}_2\text{(l)} \rightarrow 2\text{HBr(g)} \quad \Delta H_1^\circ \]
\[ \text{Reaction 2: } \text{H}_2\text{(g)} + \text{Br}_2\text{(g)} \rightarrow 2\text{HBr(g)} \quad \Delta H_2^\circ \]

Which of the following correctly compares \(\Delta H_2^\circ\) to \(\Delta H_1^\circ\) and provides the correct justification?

The standard state of bromine at \(298\text{ K}\) and \(1\text{ atm}\) is \(\text{Br}_2\text{(l)}\). A student compares the standard enthalpy changes for the two reactions represented below at \(298\text{ K}\):

\[ \text{Reaction 1: } \text{H}_2\text{(g)} + \text{Br}_2\text{(l)} \rightarrow 2\text{HBr(g)} \quad \Delta H_1^\circ \]
\[ \text{Reaction 2: } \text{H}_2\text{(g)} + \text{Br}_2\text{(g)} \rightarrow 2\text{HBr(g)} \quad \Delta H_2^\circ \]

Which of the following correctly compares \(\Delta H_2^\circ\) to \(\Delta H_1^\circ\) and provides the correct justification?

- **A.** \(\Delta H_2^\circ\) is more negative than \(\Delta H_1^\circ\), because the covalent \(\text{Br}-\text{Br}\) bond in \(\text{Br}_2\text{(g)}\) is weaker and requires less energy to break than the covalent \(\text{Br}-\text{Br}\) bond in \(\text{Br}_2\text{(l)}\).
- **B.** \(\Delta H_2^\circ\) is more negative than \(\Delta H_1^\circ\), because energy must be absorbed to overcome intermolecular forces when \(\text{Br}_2\text{(l)}\) vaporizes, placing \(\text{Br}_2\text{(g)}\) at a higher initial enthalpy.
- **C.** \(\Delta H_2^\circ\) is less negative than \(\Delta H_1^\circ\), because the endothermic enthalpy of vaporization of \(\text{Br}_2\text{(l)}\) must be added to \(\Delta H_1^\circ\) to obtain \(\Delta H_2^\circ\).
- **D.** \(\Delta H_2^\circ\) is less negative than \(\Delta H_1^\circ\), because forming intermolecular attractions between \(\text{Br}_2\) molecules in Reaction 1 releases additional energy to the surroundings.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/123823/*