Chapter 15, Problem 102A

Interpretation Introduction

Interpretation:

The nature of the reaction needs to be determined.

${\text{CuS(s)+2O}}_{}\text{(g)}\to {\text{CuSO}}_{\text{4}}\text{(s)}$.

Concept introduction:

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

${\text{ΔG}}_{\text{rxn}}^{\text{o}}={\text{ΔH}}_{\text{rxn}}^{\text{o}}-{\text{TΔS}}_{\text{rxn}}^{\text{o}}$

Answer to Problem 102A

Since Gibbs free energy is positive, the reaction will be non-spontaneous.

Explanation of Solution

In this reaction the copper sulfide (s) converted into copper sulphate. It is exothermic reaction.

Given, change in enthalpy,

${\text{ΔH}}_{\text{rxn}}^{\text{o}}=-718.3\text{kJ}$

change in entropy,

$\begin{array}{c}{\text{ΔS}}_{\text{rxn}}^{\text{o}}=\text{-368J/K}\\ =-\text{0}\text{.368kJ/K(}\text{convert}\text{J}\text{to}\text{kJ)}\end{array}$

Since, Gibbs free energy

$\begin{array}{c}{\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{=}{\text{ΔH}}_{\text{rxn}}^{\text{o}}{\text{-TΔS}}_{\text{rxn}}^{\text{o}}\\ {\text{ΔG}}_{\text{rxn}}^{\text{o}}=-\text{718}\text{.3-298×(-0}\text{.368)}\\ {\text{ΔG}}_{\text{rxn}}^{\text{o}}=-\text{718}\text{.3+109}\text{.664}\\ {\text{ΔG}}_{\text{rxn}}^{\text{o}}=\text{-608}\text{.636}\end{array}$

The reaction is non spontaneous.

Conclusion

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