Chapter 15, Problem 101A

Interpretation Introduction

Interpretation:

${\text{ΔG}}_{\text{system}}^{\text{o}}$ is to be calculated.

Concept introduction:

Free energy change, ${\text{ΔG}}_{\text{system}}$ is the difference between the system’s change in enthalpy ( ${\text{ΔH}}_{\text{system}}$ ) and the product of the kelvin temperature and the change in entropy ( ${\text{TΔS}}_{\text{system}}$ ). That is,

${\text{ΔG}}_{\text{system}}\text{=}{\text{ΔH}}_{\text{system}}-{\text{TΔS}}_{\text{system}}$

Answer to Problem 101A

Standard free energy change, ${\text{ΔS}}_{\text{system}}^{\text{o}}$ for the change from ${\text{H}}_{\text{2}}\text{O(l)}\to {\text{H}}_{\text{2}}\text{O(g)}$ is 0.11kJ/K

Explanation of Solution

When a reaction or process occurs under standard conditions (298 K and 1 atm), the standard free energy change can be expressed as follows.

$\begin{array}{l}{\text{ΔG}}_{{}_{\text{system}}}^{\text{o}}\text{=}{\text{ΔH}}_{\text{system}}^{\text{o}}-{\text{TΔS}}_{\text{system}}^{\text{o}}\end{array}$

Given information,

$\begin{array}{c}{\text{ΔG}}_{\text{system}}^{\text{o}}=\text{8}\text{.55kJ}\\ {\text{ΔH}}_{\text{system}}^{\text{o}}=\text{44}\text{.01kJ}\\ \text{T}=\text{298K}\end{array}$

$\begin{array}{c}{\text{ΔG}}_{{}_{\text{system}}}^{\text{o}}\text{=}{\text{ΔH}}_{\text{system}}^{\text{o}}-{\text{TΔS}}_{\text{system}}^{\text{o}}\\ \text{8}\text{.557kJ}=\text{44}{\text{.01kJ-298KΔS}}_{\text{system}}^{\text{o}}\\ {\text{298KΔS}}_{\text{system}}^{\text{o}}=\text{44}\text{.01kJ-8}\text{.557kJ}\\ {\text{ΔS}}_{\text{system}}^{\text{o}}=\frac{\text{35}\text{.456kJ}}{\text{298K}}\\ {\text{ΔS}}_{\text{system}}^{\text{o}}=\text{0}\text{.11kJ/K}\end{array}$

Conclusion

If the sign of the free energy change( ${\text{ΔG}}_{\text{system}}^{\text{o}}$ ) is negative, the reaction is spontaneous and if the sign of the free energy change ( ${\text{ΔG}}_{\text{system}}^{\text{o}}$ ) is positive, the reaction is nonspontaneous.