Gibbs energy

Gibbs energy is a thermodynamic state function that is used to measure the spontaneity of a process at constant temperature and pressure.

As explained in an earlier article, the change in total entropy can help determine if a process in an isolated adiabatic system is spontaneous. However, most reactions proceed at constant pressure and temperature, which typically happens when the system is contained in a vessel with a movable piston and is in thermal contact with a temperature-controlled bath. In such cases, the system’s temperature is kept constant through the transfer of thermal energy from the bath, making the system no longer adiabatic.

Therefore, we need another state function to predict if such reactions are spontaneous or not. Consider a closed system that is in thermal equilibrium with a constant temperature bath. From eq139, $dS\geq\frac{dq}{T_{sur}}\;\;\Rightarrow\;\;dS\geq\frac{dq}{T_{sys}}$ . Let $T_{sys}=T$ and substitute $dS\geq\frac{dq}{T}$ in the differential form of eq24 to give

$dU\leq TdS+dw$

If the system only does pV work,

$dU\leq TdS-pdV$

Since $p$ and $T$ are constant and the change in enthalpy at constant pressure is defined as $dH=dU+pdV$

$d(H-TS)\leq 0\;\;\;\;\;\;\;\;(142)$

Since $H$ , $T$ and $S$ are all state functions, we can define a new state function:

$G=H-TS\;\;\;\;\;\;\;\;(143)$

where $G$ is called the Gibbs energy.

Substituting eq143 in eq142,

$dG\leq 0\;\;\;\;\;\;\;\;(144)$

Since we have derived eq144 using eq139, where the $>$ equality identifies with irreversible processes and the $=$ relation, with reversible processes, the $<$ equality and $=$ relation in eq144 must also identify with the respective processes, where

$dG<0\;\;\;\;\;irreversible\;(spontaneous)\;process\;at\;constant\;p\;and\;T\;\;\;(145)$

$dG=0\;\;\;\;\;reversible\;process\;at\;constant\;p\;and\;T\;\;\;(146)$

Question

Why is an irreversible process a spontaneous process?

Answer

See this article for explanation.

As mentioned in the article on reversible and irreversible processes, a reversible process is an idealisation, which makes all real processes irreversible or spontaneous. Hence, the Gibbs energy of a system undergoing a real process at constant $p$ and $T$ decreases until such time when the process attains equilibrium.

Although all real processes are spontaneous, they may not occur within a finite timeframe. Whether a real process occurs is also dependent on chemical kinetics.

The integrated form of eq144 is:

$\Delta G\leq 0$

where

$\Delta G=\Delta H-T\Delta S\;\;\;\;\;\;\;\;(147)$

A useful equation can be derived from eq143 as follows:

$dG=dH-d(TS)$

$dG=dU+pdV+Vdp-TdS-SdT$

If the closed system only involves pV work, we can substitute eq121 in the above equation to give:

$dG=Vdp-SdT\;\;\;\;\;\;\;\;(148)$

Eq148 becomes $dG=0$ for a reversible process carried out at constant pressure and temperature, which is consistent with eq144. Eq148 also shows that the measure of the change in Gibbs energy of a process in general is not restricted to the process being carried out at constant pressure and constant temperature. However, becomes a reflection of whether a process is reversible or irreversible if the process occurs at constant pressure and constant temperature.

Finally, Gibbs energy is previously called Gibbs free energy, where the term ‘free’ refers to energy that is “free to do non-pV work”. However in 1988, IUPAC dropped the term to avoid confusion.

Question

Show that $\biggr$\frac{\partial G}{\partial p}\biggr$_T=\biggr$\frac{\partial G}{\partial p}\biggr$_{T,n}=V$

Answer

At constant , eq148 becomes $\biggr$\frac{\partial G}{\partial p}\biggr$_T=V$ . We can further impose the condition that the components $n$ of the system remain constant without any loss of generality, i.e.