Question

The values in Appendix D, calculate the values of$\mathbf{∆}\mathit{G}\mathbf{°}$and$\mathbf{∆}\mathit{H}\mathbf{°}$for the reaction.

$\mathbf{3}{\mathbf{\text{Fe}}}_{\mathbf{2}}{\mathbf{\text{O}}}_{\mathbf{3}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{\to }\mathbf{2}{\mathbf{\text{Fe}}}_{\mathbf{3}}{\mathbf{\text{O}}}_{\mathbf{4}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{\text{O}}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}$

at$\mathbf{25}\mathbf{°}\mathit{C}$. Which of the two oxides is more stable at$\mathbf{25}\mathbf{°}\mathit{C}$androle="math" localid="1663661928430" ${\mathit{P}}_{{\mathbf{\text{O}}}_{\mathbf{2}}}\mathbf{=}\mathbf{1}\mathbf{}\mathbf{\text{atm}}$
?

Short answer

$\Delta H°=+235.8\text{kJ}$- Positive value denotes an endothermic reaction.

$\Delta G°=+195.6\text{kJ}$-Positive value indicates that the reaction is not spontaneous.

Oxide ${\text{Fe}}_2{\text{O}}_3\left(s\right)$is more stable than oxide${\text{Fe}}_3{\text{O}}_4\left(s\right)$

Step-by-step solution

 Step 1: Given data

The given chemical equation is: $3{\text{Fe}}_2{\text{O}}_3\left(s\right)\to 2{\text{Fe}}_3{\text{O}}_4\left(s\right)+\frac{1}{2}{\text{O}}_2\left(g\right)$.

$P_{{\text{O}}_2}=1\text{atm}$nd Temperature$25°\text{C}$ .

$\Delta H_f^{°}$of ${\text{Fe}}_2{\text{O}}_3\left(s\right)$is $-1118.4{\text{kJmol}}^{-1}$

$\Delta H_f^{°}$of ${\text{O}}_2\left(g\right)$is $0$.

$\Delta H_f^{°}$of${\text{Fe}}_2{\text{O}}_3\left(s\right)$ is $-824.2{\text{kJmol}}^{-1}$.

$\Delta H_f^{°}$represents standard enthalpy of the molecules.

$\Delta G_f^{°}$of${\text{Fe}}_2{\text{O}}_3\left(s\right)$ is$-1015.5{\text{kJmol}}^{-1}$ .

$\Delta G_f^{°}$of${\text{O}}_2\left(g\right)$ is$0$

$\Delta G_f^{°}$of ${\text{Fe}}_2{\text{O}}_3\left(s\right)$is$-742.2{\text{kJmolmol}}^{-1}$ .

$\Delta G_f^{°}$denotes Standard Gibbs free energy of formation.

Concept used

Enthalpy is a thermodynamic quantity equivalent to the total heat content of a system. It is equal to the internal energy of the system plus the product of pressure and volume and whenever a reaction has a positive change in enthalpy, which by definition means that heat is being absorbed, we call it an endothermic reaction.

$\mathbf{∆}\mathit{G}\mathbf{°}$is the change in free energy. Reactions with a positive ∆G need an input of energy in order to take place (are non-spontaneous).

Calculate the enthalpy change

The given chemical equation is:$3{\text{Fe}}_2{\text{O}}_3\left(s\right)\to 2{\text{Fe}}_3{\text{O}}_4\left(s\right)+\frac{1}{2}{\text{O}}_2\left(g\right)$

By the use of given data, the enthalpy change has been calculated as follows:

$\Delta H°=n\Delta H_F^{°}($Products$)-n\Delta H_f^{°}($Reactants)

$$\begin{gathered} \Delta H°=\left[2\Delta H_f^{°}\left[{\text{Fe}}_3{\text{O}}_4\left(s\right)\right]+\frac{1}{2}\Delta H_f^{°}\left[{\text{O}}_2\left(g\right)\right]\right]-3\Delta H_f^{°}\left[{\text{Fe}}_2{\text{O}}_3\left(s\right)\right] \\ =2\text{mol}\times \left(-1118.4{\text{kJmol}}^{-1}\right)+\frac{1}{2}\times 0-3\text{mol}\times \left(-824.2{\text{kJmol}}^{-1}\right) \\ =-2236.8\text{kJ}+2472.6\text{kJ} \\ =+235.8\text{kJ} \end{gathered}$$

Calculate the free energy

The given chemical equation is:$3{\text{Fe}}_2{\text{O}}_3\left(s\right)\to 2{\text{Fe}}_3{\text{O}}_4\left(s\right)+\frac{1}{2}{\text{O}}_2\left(g\right)$

$\Delta G°={\left(\Delta G_f^{°}\right)}_{\text{produts}}-{\left(\Delta G_f^{°}\right)}_{\text{reactants}}$

By the use of given data, the free energy of the system has been calculated as follows:

role="math" localid="1663662807361" $$\begin{gathered} \Delta G°=\left[2\Delta G_f^{°}\left[{\text{Fe}}_3{\text{O}}_4\left(s\right)\right]+\frac{1}{2}\Delta G_f^{°}\left[{\text{O}}_2\left(g\right)\right]\right]-3\Delta G_f^{°}\left[{\text{Fe}}_2{\text{O}}_3\left(s\right)\right] \\ =2\text{mol}\times \left(-1015.5{\text{kJmol}}^{-1}\right)+\frac{1}{2}\times 0-3\text{mol}\times \left(-742.2{\text{kJmolmol}}^{-1}\right) \\ =-2031.0\text{kJ}+2226.6\text{kJ} \\ =+195.6\text{kJ} \end{gathered}$$