Chapter 17.4, Problem 35SP

Interpretation Introduction

Interpretation: The standard enthalpy change for the reaction is to be found.

Concept introduction: The standard enthalpy of formation is the amount of energy released or consumed when one mole of a substance is formed under standard conditions from its pure elements.

Standard temperature and pressure conditions are 298 K and 1 atm.

The enthalpy of formation for an element in its elemental state or naturally occurring state will always be zero because it takes no energy to form a naturally occurring compound.

Answer to Problem 35SP

  $\text{ΔH°}$ for the given reaction ${\text{Br}}_{\text{2}}\text{(g)}\to {\text{Br}}_{\text{2}}\text{(l)}$ is $\text{-31kJ/mol}$

Explanation of Solution

At 298 K and 1 atm, the standard state of bromine is liquid, thus standard enthalpy of formation of ${\text{Br}}_{\text{2}}\text{(l)}$ is zero.

At 298 K and 1 atm, the standard enthalpy of formation of ${\text{Br}}_{\text{2}}\text{(g)}$ is $31\text{kJ/mol}$ .

The standard enthalpy changes of the reaction ${\text{Br}}_{\text{2}}\text{(g)}\to {\text{Br}}_{\text{2}}\text{(l)}$ is

  $\text{ΔH°=H}{°}_f{\text{Br}}_{\text{2}}\text{(l)-H}{°}_f{\text{Br}}_{\text{2}}\text{(g)}$

  $\begin{array}{l}\text{ΔH°= 0kJ/mol-(+31kJ/mol)}\\ \text{ΔH°= -31kJ/mol}\end{array}$