Chapter 23, Problem 63AP

Given that $\Delta G({\text{Fe}}_{\text{2}}{\text{O}}_3)=-741.0\text{ kJ/mol}$ and that $\Delta G({\text{Al}}_{\text{2}}{\text{O}}_3)=-1576.4\text{ kJ/mol}$, calculate $\Delta G°$ for the following reactions at: $25°\text{C}$

(a)

$2{\text{Fe}}_2{\text{O}}_3(s)\to 4\text{Fe}(s)+3{\text{O}}_2(g)$

(b)

$2{\text{Al}}_2{\text{O}}_3(s)\to 4\text{Al}(s)+3{\text{O}}_2(g)$

Interpretation Introduction

Interpretation:

The $\Delta {\text{G}}^{\circ }$ value for the given reactions is to be determined with given $\Delta {\text{G}}^{\circ }$ value for ${\text{Fe}}_2{\text{O}}_3$ and ${\text{Al}}_2{\text{O}}_3$.

Concept Introduction:

The $\Delta {\text{G}}^{\circ }$ of areaction is calculated by the expression as

$\Delta {\text{G}}^{\circ }=\sum \Delta {\text{G}}_{\text{f}}{}^{\circ }\left(\text{products}\right)-\sum \Delta {\text{G}}_{\text{f}}{}^{\circ }\left(\text{reactant}\right)$.

Here, $\Delta {\text{G}}^{\circ }$ is the change in standard Gibbs free energy, $\sum \Delta {\text{G}}_{\text{f}}{}^{\circ }\left(\text{products}\right)$ is the sum of standard Gibbs free energy of products, and $\sum \Delta {\text{G}}_{\text{f}}{}^{\circ }\left(\text{reactant}\right)$ is the sum of standard Gibbs free energy of reactants.

Answer to Problem 63AP

Solution:

The $\Delta {\text{G}}^{\circ }$ valuefor the reaction is equal to $1482\text{ kJ/mol}$.

The $\Delta {\text{G}}^{\circ }$ value for the reaction is equal to $3152.8\text{ kJ/mol}$.

Explanation of Solution

a) ${\text{2Fe}}_2{\text{O}}_3\left(s\right)\to 4\text{Fe}\left(s\right)+3{\text{O}}_2\left(g\right)$.

The value of $\Delta {\text{G}}^{\circ }{}_{\text{f}}\left({\text{Fe}}_2{\text{O}}_3\right)$ is $-741.00\text{ kJ/mol}$.

Thevalue of $\Delta {\text{G}}^{\circ }$ for the given reaction is calculated by using the relation given below:

$\Delta {\text{G}}^{\circ }=4\Delta {\text{G}}_{\text{f}}{}^{\circ }\left(\text{Fe}\right)+3\Delta {\text{G}}_{\text{f}}{}^{\circ }\left({\text{O}}_2\right)-2\Delta {\text{G}}_{\text{f}}{}^{\circ }\left({\text{Fe}}_2{\text{O}}_3\right)$.

Here, $\Delta {\text{G}}^{\circ }$ is the change in standard Gibbs free energy, $\Delta {\text{G}}_{\text{f}}{}^{\circ }\left(\text{Fe}\right)$ is the change in standard Gibbs free energy of Fe, $\Delta {\text{G}}_{\text{f}}{}^{\circ }\left({\text{O}}_2\right)$ is the change in standard Gibbs free energy of ${\text{O}}_2$ and $\Delta {\text{G}}_{\text{f}}{}^{\circ }\left({\text{Fe}}_2{\text{O}}_3\right)$ is the change in standard Gibbs free energy of ${\text{Fe}}_2{\text{O}}_3$.

The value of change in standard Gibbs free energy for atoms in their standard state is zero. In the reaction, Fe and ${\text{O}}_2$ are present in their standard state. Thus, the value of $\Delta {\text{G}}_{\text{f}}{}^{\circ }\left({\text{O}}_2\right)$ and $\Delta {\text{G}}_{\text{f}}{}^{\circ }\left(\text{Fe}\right)$ is zero.

Substitute $0$ for $\Delta {\text{G}}_{\text{f}}{}^{\circ }\left({\text{O}}_2\right)$, $0$ for $\Delta {\text{G}}_{\text{f}}{}^{\circ }\left(\text{Fe}\right)$, and $-741.00\text{ kJ/mol}$ for $\Delta {\text{G}}^{\circ }{}_{\text{f}}\left({\text{Fe}}_2{\text{O}}_3\right)$ in the equation as:

$\begin{array}{c}\Delta {\text{G}}^{\circ }=4\Delta {\text{G}}_{\text{f}}{}^{\circ }\left(\text{Fe}\right)+3\Delta {\text{G}}_{\text{f}}{}^{\circ }\left({\text{O}}_2\right)-2\Delta {\text{G}}_{\text{f}}{}^{\circ }\left({\text{Fe}}_2{\text{O}}_3\right)\\ =4\left(0\right)+3\left(0\right)-2\left(-741.0\text{ kJ/mol}\right)\\ =1482\text{ kJ/mol}\text{.}\end{array}$

b) ${\text{2Al}}_2{\text{O}}_3\left(s\right)\to 4\text{Al}\left(s\right)+3{\text{O}}_2\left(g\right)$

The value of $\Delta {\text{G}}^{\circ }{}_{\text{f}}\left({\text{Al}}_2{\text{O}}_3\right)$ is $\text{-1576}\text{.4}\text{kJ/mol}$.

The value of $\Delta {\text{G}}^{\circ }$ for the given reaction is calculated by using the relation given below:

$\Delta {\text{G}}^{\circ }=4\Delta {\text{G}}_{\text{f}}{}^{\circ }\left(\text{Al}\right)+3\Delta {\text{G}}_{\text{f}}{}^{\circ }\left({\text{O}}_2\right)-2\Delta {\text{G}}_{\text{f}}{}^{\circ }\left({\text{Al}}_2{\text{O}}_3\right)$.

Here, $\Delta {\text{G}}^{\circ }$ is the change in standard Gibbs free energy, $\Delta {\text{G}}_{\text{f}}{}^{\circ }\left(\text{Al}\right)$ is the change in standard Gibbs free energy of Al, $\Delta {\text{G}}_{\text{f}}{}^{\circ }\left({\text{O}}_2\right)$ is the change in standard Gibbs free energy of ${\text{O}}_2$, and $\Delta {\text{G}}_{\text{f}}{}^{\circ }\left({\text{Al}}_2{\text{O}}_3\right)$ is the change in standard Gibbs free energy of ${\text{Al}}_2{\text{O}}_3$.

The value of change in standard Gibbs free energy for atoms in their standard state is zero. In the reaction, Al and ${\text{O}}_2$ are present in their standard state. Thus, the value of $\Delta {\text{G}}_{\text{f}}{}^{\circ }\left({\text{O}}_2\right)$ and $\Delta {\text{G}}_{\text{f}}{}^{\circ }\left(\text{Al}\right)$ is zero.

Substitute $0$ for $\Delta {\text{G}}_{\text{f}}{}^{\circ }\left({\text{O}}_2\right)$, $0$ for $\Delta {\text{G}}_{\text{f}}{}^{\circ }\left(\text{Al}\right)$ and $\text{-1576}\text{.4}\text{kJ/mol}$ for $\Delta {\text{G}}^{\circ }{}_{\text{f}}\left({\text{Al}}_2{\text{O}}_3\right)$ in the equation as:

$\begin{array}{c}\Delta {\text{G}}^{\circ }=4\Delta {\text{G}}_{\text{f}}{}^{\circ }\left(\text{Al}\right)+3\Delta {\text{G}}_{\text{f}}{}^{\circ }\left({\text{O}}_2\right)-2\Delta {\text{G}}_{\text{f}}{}^{\circ }\left({\text{Al}}_2{\text{O}}_3\right)\\ =4\left(0\right)+3\left(0\right)-2\left(-1576.4\text{ kJ/mol}\right)\\ =3152.8\text{ kJ/mol}\text{.}\end{array}$