Reaction Gibbs energy

With reference to eq14 of the previous article, where \Delta G_{sys}^o=\Delta H_{sys}^o-T\Delta S_{sys}^o, we can define the standard reaction Gibbs energy as:

\Delta G_r^o=\Delta H_r^o-T\Delta S_r^o\; \; \; \; \; \; \; \; 15

or in general, the reaction Gibbs energy as: \Delta G_r=\Delta H_r-T\Delta S_r.

Furthermore, just as the standard reaction enthalpy is \Delta H_r^o=\sum_Pv_P(\Delta H_f^o)_P-\sum_Rv_R(\Delta H_f^o)_R, and the standard reaction entropy of a system is \Delta S_r^o=\sum_Pv_P(\Delta S_m^o)_P-\sum_Rv_R(\Delta S_m^o)_R, the standard reaction Gibbs energy of a system is:

\Delta G_r^o=\sum_Pv_P(\Delta G_f^o)_P- \sum_Rv_R(\Delta G_f^o)_R\; \; \; \; \; \; \; \; 16

where ΔGfo is the standard reaction Gibbs energy for the formation of a compound from its elements in their reference states, and vP and vR are the stoichiometric coefficients of the products and reactants respectively. ΔGro, like ΔHro, is computed using energies of formation, as we cannot determine the absolute Gibbs energy of a substance (unlike entropy). Even so, the terms ΔGro, ΔHro and ΔSro in eq15 are comparable, as the reference states in \sum_Pv_P(\Delta G_f^o)_P and \sum_Rv_R(\Delta G_f^o)_R as well as those in \sum_Pv_P(\Delta H_f^o)_P and \sum_Rv_R(\Delta H_f^o)_R cancel out when we compute ΔGro and ΔHro respectively (see sections on standard enthalpy change of formation and Hess Law for details).

Combining eq15 and eq16,

\sum_Pv_P(\Delta G_f^o)_P- \sum_Rv_R(\Delta G_f^o)_R=\Delta H_r^o-T\Delta S_r^o\; \; \; \; \; \; \; \; 17

Finally, ΔGro is used in the same way as ΔGsyso to predict the spontaneity of reactions, where ΔGro < 0 for a spontaneous reaction and ΔGro = 0 when the reaction attains equilibrium.



Calculate the standard combustion Gibbs energy for the reaction C_2H_6(g)+\frac{7}{2}O_2(g)\rightarrow 2CO_2(g)+3H_2O(l) given ΔGfo[C2H2(g)] = -32.9 kJmol-1, ΔGfo[CO2(g)] = -394.4 kJmol-1 and ΔGfo[H2O(l)] = -237.2 kJmol-1.


Using eq16 and noting that ΔGfo[O2(g)] = 0,

\Delta G_r^o=(2)(-394.4)+(3)(-237.2)-(-32.9)=-1467.5\, kJmol^{-1}

Since ΔGro < 0, the reaction is spontaneous.



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