Question

Consider the reaction

${\mathbf{Fe}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}\mathbf{\left(}\mathbf{s}\mathbf{\right)}}\mathbf{+}\mathbf{3}{\mathbf{H}}_{\mathbf{2}}\to \mathbf{2}{\mathbf{Fe}}_{\mathbf{\left(}\mathbf{s}\mathbf{\right)}}\mathbf{+}\mathbf{3}{\mathbf{H}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{\left(}\mathbf{g}\mathbf{\right)}}$

1. Use $\mathit{\Delta }{\mathit{G}}_{\mathbf{r}}^{\mathbf{o}}$values in appendix 4 to calculate $\mathit{\Delta }{\mathit{G}}^{\mathbf{o}}$for third reaction.
2. Is this reaction spontaneous under standard conditions at 298K.
3. The value of $\mathit{\Delta }{\mathit{H}}^{\mathbf{o}}$for this reaction is 100KJ. At what temperatures is this reaction spontaneous at standard conditions? Assume that $\mathit{\Delta }{\mathit{H}}^{\mathbf{o}}$ and $\mathit{\Delta }{\mathit{S}}^{\mathbf{o}}$do not depend on temperature.

Short answer

$\Delta G_R^0=53kJ$is the value for the equation a. And $\Delta H_{}^0=634K$for equation c.

Step-by-step solution

Explanation for  in this equation

On the Gibbs free energy,

$$\begin{gathered} \mathit{}\Delta G_R^0=\left[2\Delta G_r^0\left(F{e}_{\left(s\right)}\right)+3\Delta G_r^0\left(H_2{O}_{\left(g\right)}\right)\right]-\left[1\Delta G_r^0(F{e}_2{O}_{3\left(s\right)}+3\Delta G_r^0{H}_{2\left(g\right)})\right] \\ =\left[2\text{mol}\times 0kJ/\text{mol}+3\text{mol}\times -229kJ/\text{mol}\right]-\left[1\text{mol}\times -740kJ/\text{mol + 3mol}\times \text{0kJ/mol}\right] \\ \Delta G_R^0=53kJ \end{gathered}$$

Explanation for b.

The value of delta G of the reaction has a positive sign. Hence, the reaction is non-spontaneous.

Explanation for C.

$\Delta T° \text{For} \text{the} \text{reaction} T=25C$

From part a $\Delta G_R^0=53kJ\mathit{,}\mathit{}\Delta H_{}^{\mathit{0}}=100kJ$

$$\begin{gathered} T=(25+273)K \\ =298K \end{gathered}$$

In the calculation,

$$\begin{gathered} \Delta G_{}^0=\Delta H°-T\Delta S° \\ \Delta S° =\frac{\Delta H°-\Delta G_{}^0}{T} \\ =\frac{(100-53)kJ}{298K} \\ =157.7J/K \\ \Delta H_{}^0=T\Delta S_{}^0 \\ =\frac{\Delta H_{}^0}{\Delta S_{}^0} \\ =\frac{100}{0.1577} \\ \Delta H_{}^0=634K \end{gathered}$$

Conclusion

Therefore, $∆{\mathrm{GR}}_{\mathrm{R}}^0=53\mathrm{kJ}$is the value for the equation, standard condition, but it is a non-spontaneous reaction and $\Delta H_{}^0=634K$are the solution for a,b, and c question, respectively.