Chapter 15, Problem 103A

Interpretation Introduction

Interpretation:

Temperature at which ${\text{ΔG}}_{\text{system}}=-\text{34}\text{.7kJ}$ 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 103A

Temperature at which ${\text{ΔG}}_{\text{system}}=-\text{34}\text{.7kJ}$ is 265.8K

Explanation of Solution

Given information,

$\begin{array}{l}{\text{ΔG}}_{\text{system}}=-\text{34}\text{.7kJ}\\ {\text{ΔH}}_{\text{system}}=\text{-28}\text{.8kJ}\\ {\text{ΔS}}_{\text{system}}=\text{22}\text{.2J/K}\\ {\text{ΔS}}_{\text{system}}=\text{0}\text{.0222kJ/L}\text{(Converted}\text{J}\text{to}\text{kJ)}\end{array}$

Since,

$\begin{array}{c}{\text{ΔG}}_{\text{system}}\text{=}{\text{ΔH}}_{\text{system}}{\text{-TΔS}}_{\text{system}}\\ \text{-34}\text{.7kJ}=\text{-28}\text{.8kJ-T(0}\text{.0222kJ)}\\ \text{0}\text{.0222T}=-\text{28}\text{.8+34}\text{.7kJ}\\ \text{T}=\frac{5.9}{\text{0}\text{.0222}}\\ \text{T}=\text{265}\text{.8K}\end{array}$

Conclusion

In order to calculate Gibbs free energy( $\text{ΔG}$ ), it is usually necessary to convert units because $\text{ΔS}$ is usually expressed in $\text{j/K}$, whereas $\text{ΔH}$ is expressed in $\text{kJ}$.