2.6 Gibbs Free Energy

According to the second law, \(\Delta _{universe} S^\circ > 0\), which was expressed in another form in Eq. (16).

\[\Delta _{universe} S^\circ = \Delta _{sys} S^\circ - \frac{\Delta _{sys} H^\circ }{T} >0\]

Multiply both sides of the above equation by \(T\),

\[T\Delta _{universe} S^\circ = T \Delta _{sys} S^\circ - \Delta _{sys} H^\circ >0\]

Now reverse signs on both sides of the above equation,

(18)\[-T \Delta _{universe} S^\circ = \Delta _{sys} H^\circ -T \Delta _{sys} S^\circ <0\]

According to Eq. (18), processes that occur at constant \(P\) and \(T\) are spontaneous if \(\Delta _{sys} H^\circ\) and \(\Delta _{sys} S^\circ\) are such that \(\Delta _{sys} H^\circ -T \Delta _{sys} S^\circ <0\). To simplify this expression, we introduce a new thermodynamic function called Gibbs free energy (\(G\)) or free energy. Gibbs free energy, \(G\), is the energy available for work.

(19)\[\Delta _{sys} G^\circ = \Delta _{sys} H^\circ -T \Delta _{sys} S^\circ \]

Free energy is a state function and, as the name implies, has the same units as and \(T \Delta _{sys} S^\circ\) (\(\pu{kJ}\) or \(\pu{kJ mol−1}\)) If a reaction is accompanied by the release of usable energy (\(\Delta _{sys} G^\circ <0\)), the reaction is guaranteed to be spontaneous.

Table 7 \(\Delta _{sys} G^\circ\) and spontaneity of reactions.

\(\Delta _{sys} G^\circ\) condition	Reaction condition
\(\Delta _{sys} G^\circ <0\)	Reaction is spontaneous in forward direction
\(\Delta _{sys} G^\circ =0\)	Reaction is at equilibrium
\(\Delta _{sys} G^\circ >0\)	Reaction is nonspontaneous as written

Standard Gibbs free energy, \(\Delta _{f} G^\circ\)

The standard free energy of the reaction is the free-energy change for a reaction when it occurs under standard-state conditions - i.e. when reactants in their standard states are converted to products in their standard states.

The CRC Handbook of Physics and Chemistry is a good reference for obtaining \(\Delta_f G^\circ\) for most chemicals. The online version is available at Standard Thermodynamic Properties of Chemical Substances (chemnetbase.com)

\(\Delta_{rxn}G^\circ\) is calculated as follows:

(20)\[\Delta _{rxn} G^\circ = \sum \Delta _{products} G^\circ - \sum \Delta _{reactants} G^\circ\]

\(\Delta _{rxn} G^\circ\) can be calculated by looking up standard free energy of formation (\(\Delta _{f} G^\circ\)) from thermodynamic data tables.

Example: Calculating \(\Delta_{rxn}G^\circ\)

Example 1

Let’s calculate \(\Delta_{rxn}G^\circ\) for the following reaction at \(\pu{25 ^\circ C}\):

\[ \ce{ C (s) + O2 (g) -> CO2 (g) }\]

\(\Delta_f G^\circ _\ce{C(s)} = \pu{0 kJ mol-1}\), \(\Delta_f G^\circ _\ce{O2(g)} = \pu{0 kJ mol-1}\), and \(\Delta_f G^\circ _\ce{CO2(g)} = \pu{-394.4 kJ mol-1}\) at \(\pu{25 ^\circ C}\).

\[\begin{split} \begin{aligned} \Delta _{rxn} G^\circ &= [\Delta _f G^\circ _\ce{CO2 (g)}] - [\Delta _f G^\circ _\ce{C (s)} + \Delta _f G_\ce{O2 (g)}]\\ &= [(\pu{-394.4 kJ mol-1})] - [(\pu{0 kJ mol-1}) + (\pu{0 kJ mol-1})]\\ &= \pu{-394.4 kJ mol-1} \end{aligned} >\end{split}\]

Example 2

Let’s calculate \(\Delta_{rxn}G^\circ\) for the following reaction at \(\pu{25 ^\circ C}\):

\[ \ce{ CH4(g) + 2 O2(g) -> CO2(g) + 2 H2O(l) } \]

\(\Delta_f G^\circ _\ce{CH4(g)} = \pu{-50.8 kJ mol-1}\), \(\Delta_f G^\circ _\ce{CO2(g)} = \pu{-394.4 kJ mol-1}\), \(\Delta_f G^\circ _\ce{O2(g)} = \pu{0 kJ mol-1}\), and \(\Delta_f G^\circ _\ce{H2O(l)} = \pu{-237.2 kJ mol-1}\).

\[\begin{split} \begin{aligned} \Delta _{rxn} G^\circ &= [\Delta_f G^\circ _\ce{CO2 (g)} + 2 \cdot\Delta_f G^\circ _\ce{H2O (l)}] - [\Delta_f G^\circ _\ce{CH4 (g)} + 2\cdot \Delta_f G^\circ _\ce{O2 (g)}]\\ &= [(\pu{-394.4 kJ mol-1}) + 2(\pu{-237.2 kJ mol-1})] - [(\pu{-50.8 kJ mol-1}) + 2(\pu{0 kJ mol-1})]\\ &= \pu{-818.0 kJ mol-1} \end{aligned} >\end{split}\]

Predicting sign of \(\Delta_{rxn}G^\circ\)

\(\Delta_{rxn}G^\circ\) is a function of \(T\) as can be seen in Eq. (20). Also, \(\Delta _{rxn} G^\circ = 0\) at equilibrium (see Table 7). Eq. (20) can be rearranged to determine the exact \(T\) where equilibrium is reached.

\[\begin{split}\Delta _{rxn} G^\circ = \Delta _{sys} H^\circ -T \Delta _{sys} S^\circ = 0\\\end{split}\]

Rearranging the above expression, we get

(21)\[T = \frac{\Delta _{sys} H^\circ}{\Delta _{sys} S^\circ }\]

Equations (19) and (21) can be used to predict the sign of \(\Delta _{sys} G^\circ\) as shown in Table 8.

Table 8 Predicting sign of \(\Delta _{sys} G^\circ\) using signs of \(\Delta _{sys}S^\circ\) and \(\Delta _{sys} H^\circ\).

When \(\Delta _{sys} H^\circ\) is	and \(\Delta _{sys}S^\circ\) is	then \(\Delta _{sys} G^\circ\) will be	and the process is
\(-\)	\(+\)	\(-\)	Spontaneous
\(+\)	\(-\)	\(+\)	Nonspontaneous
\(-\)	\(-\)	\(-\) when \(T{\Delta _{sys}S^\circ} < {\Delta _{sys} H^\circ}\)	Spontaneous at low \(T\)
		\(+\) when \(T {\Delta _{sys} S^\circ} > {\Delta _{sys} H^\circ}\)	Nonspontaneous at high \(T\)
\(+\)	\(+\)	\(-\) when \(T{\Delta _{sys}S^\circ} > {\Delta _{sys} H^\circ}\)	Spontaneous at high \(T\)
		\(+\) when \(T{\Delta _{sys}S^\circ} < {\Delta _{sys} H^\circ}\)	Nonspontaneous at low \(T\)

Calculating \(T\) at equilibrium

In earlier example, we saw how \(\Delta _{rxn} G^\circ\) was calculated at \(\pu{25 ^\circ C}\) using Eq. (20) - however, \(\Delta _{rxn} G^\circ\) is a function of \(T\). How do we calculate (\(\Delta _{rxn} G^\circ\) at other \(T\)? In such cases, \(\Delta _{rxn} S^\circ\) and \(\Delta _{rxn} H^\circ\) are determined separately and applied directly in Eq. (21) at the specified \(T\).

Using Eq. (21), the exact \(T\) where a system reaches equilibrium could be calculated. The system will no longer be in equilibrium at a \(T\) above or below this value. The exact definition of equilibrium will be explored in the next chapter.

Example: Calculating \(\Delta _{rxn} G^\circ\) at a specified \(T\)

Let’s calculate \(\Delta _{rxn} G^\circ\) for decomposition of limestone to quick lime and \(\ce{CO2(g)}\) as per the following reaction at \(\pu{835 ^\circ C}\):

\[ \ce{CaCO3(s) -> CaO(s) + CO2(g)} \]

Thermodynamic data for these substances at \(\pu{25 ^\circ C}\) are as follows:

Chemical	\(\Delta_f H^\circ\), \(\pu{kJ mol-1}\)	\(S^\circ\), \(\pu{J K-1 mol−1}\)
\(\ce{CaCO3(s)}\)	\(-1206.9\)	\(92.9\)
\(\ce{CaO(s)}\)	\(-635.6\)	\(39.8\)
\(\ce{CO2(g)}\)	\(-393.5\)	\(213.6\)

From these data we can determine \(\Delta_{rxn}H^\circ = \pu{177.8 kJ mol-1}\) and \(\Delta_{rxn}S^\circ = \pu{160.5 J K-1 mol-1}\). From these data, let’s calculate \(\Delta _{rxn} G^\circ\) at \(\pu{835 ^\circ C}\) (= \(\pu{1108 K}\)).

\[\begin{split} \begin{aligned} \Delta _{rxn} G^\circ &= \Delta_{rxn}H^\circ -T \Delta_{rxn}S^\circ\\ &=\pu{177.8 kJ mol-1} - (\pu{1108 K})(\pu{160.5 J K-1 mol-1}) \left(\frac{\pu{1 kJ}}{\pu{e3 J}}\right)\\ & =\pu{0 kJ mol-1} \end{aligned} >\end{split}\]

In this example, the system reached equilibrium at \(\pu{835 ^\circ C}\)!