Chapter 21, Problem 21.63QP

(a)

Interpretation Introduction

Interpretation:

The $\Delta G^{\circ }$ for the given reaction at ${25}^{\circ }\text{C}$ has to be calculated.

Concept introduction:

To calculate $\Delta G^{\circ }$ values for the given reaction we have to use the following formula

$\Delta G_{\text{rxn}}{}^{\circ }=\Delta G_{\text{f}}{}^{\circ }\left(products\right)-\Delta G_{\text{f}}{}^{\circ }\left(substrates\right)$

Answer to Problem 21.63QP

The $\Delta G^{\circ }$ for the given reactions at ${25}^{\circ }\text{C}$ is 1482 kJ/mol

Explanation of Solution

Record the given data.

Reactions:

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

The temperature of the reaction is ${25}^{\circ }\text{C}$

$\Delta G_{\text{f}}{}^{\circ }\left({\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\right)=\text{-741}\text{.0}\text{kJ/mol}$

The given equations, temperature and Gibbs free energies of the given iron oxide and aluminum oxides were recorded as above.

To calculate $\Delta G^{\circ }$ for ${\text{2Fe}}_{\text{2}}{\text{O}}_{\text{3(s)}}\to {\text{4Fe}}_{\text{(s)}}{\text{+ 3O}}_{\text{2(g)}}$

Substitute the corresponding values in the above formula we can get answer as follows.

$\Delta G_{\text{rxn}}{}^{\circ }=4\Delta G_{\text{f}}{}^{\circ }\left(\text{Fe}\right)+3\Delta G_{\text{f}}{}^{\circ }\left({\text{O}}_{\text{2}}\right)-2\Delta G_{\text{f}}{}^{\circ }\left({\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\right)$

$\Delta G_{\text{rxn}}{}^{\circ }=4(0)+3(0)-(\text{2)(-741}\text{.0 kJ/mol) = 1482 kJ/mol}$

Conclusion

The $\Delta G^{\circ }$ for the given reaction at ${25}^{\circ }\text{C}$ is 1482 kJ/mol

(b)

Interpretation Introduction

Interpretation:

The $\Delta G^{\circ }$ for the given reaction at ${25}^{\circ }\text{C}$ has to be calculated.

Concept introduction:

To calculate $\Delta G^{\circ }$ values for the given reaction we have to use the following formula

$\Delta G_{\text{rxn}}{}^{\circ }=\Delta G_{\text{f}}{}^{\circ }\left(products\right)-\Delta G_{\text{f}}{}^{\circ }\left(substrates\right)$

Answer to Problem 21.63QP

The $\Delta G^{\circ }$ for the given reaction at ${25}^{\circ }\text{C}$ is 3152.8 kJ/mol

Explanation of Solution

Record the given data.

Reactions:

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

The temperature of the reaction is ${25}^{\circ }\text{C}$

$\Delta G_{\text{f}}{}^{\circ }\left({\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\right)=\text{-1576}\text{.4}\text{kJ/mol}$

The given equations, temperature and Gibbs free energies of the given iron oxide and aluminum oxides were recorded as above.

To calculate $\Delta G^{\circ }$ for ${\text{2Al}}_{\text{2}}{\text{O}}_{\text{3}}\to {\text{4Al}}_{\text{(s)}}{\text{+ 3O}}_{\text{2(g)}}$

Substitute the corresponding values in the above formula we can get answer as follows.

$\Delta G_{\text{rxn}}{}^{\circ }=4\Delta G_{\text{f}}{}^{\circ }\left(\text{Al}\right)+3\Delta G_{\text{f}}{}^{\circ }\left({\text{O}}_{\text{2}}\right)-2\Delta G_{\text{f}}{}^{\circ }\left({\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\right)$

$\Delta G_{\text{rxn}}{}^{\circ }=4(0)+3(0)-(\text{2)(-1576}\text{.4 kJ/mol) = 3152}\text{.8 kJ/mol}$

Conclusion

The $\Delta G^{\circ }$ for the given reaction at ${25}^{\circ }\text{C}$ is 3152.8 kJ/mol