Question #44233

A closed reaction vessel at $298 \text{ K}$ contains a standard acidic aqueous solution ($[\ce{H^+}] = 1.0 \text{ M}$). To this, $1.0 \text{ mole}$ of $\ce{KMnO4}$ and $4.0 \text{ moles}$ of $\ce{MnSO4}$ are added. The system is allowed to reach complete thermodynamic equilibrium.

Given the standard reduction potentials at $298 \text{ K}$: $E^\circ(\ce{MnO4^- / Mn^{2+}}) = +1.51 \text{ V}$ $E^\circ(\ce{MnO2 / Mn^{2+}}) = +1.23 \text{ V}$ $E^\circ(\ce{Mn^{3+} / Mn^{2+}}) = +1.51 \text{ V}$

Calculate the total standard Gibbs free energy change ($\Delta G^\circ_{total}$) for the complete conversion of the given amounts of reactants to their most thermodynamically stable products. (Assume ideal behavior, standard states for all species except the initial stoichiometric amounts provided, and $F = 96500 \text{ C mol}^{-1}$)