Question

\[\ce{∆G}\] is net energy available to do useful work and is thus a measure of “free energy”. Show mathematically that ∆G is a measure of free energy. Find the unit of \[\ce{∆G}\]. If a reaction has positive enthalpy change and positive entropy change, under what condition will the reaction be spontaneous?

Answer 1

We know, \[\ce{∆S_{total} = ∆S_{sys} + ∆S_{surr}}\]

When a system is in thermal equilibrium with its surroundings, the surroundings' temperature is the same as the system's. Furthermore, a rise in the enthalpy of the surroundings equals a decrease in the system's enthalpy. As a result of the entropy shift in the environment,

\[\ce{∆S_{surr} = ∆H_{surr}/T = - ∆H_{sys}/T}\]

\[\ce{∆S_{total} = ∆S_{sys} = (- ∆H_{sys}/T)}\]

Rearranging the above equation:

\[\ce{∆S_{total} = T∆S_{sys} - ∆H_{sys}}\]

For spontaneous process,

\[\ce{∆S_{total} > 0, so}\]

\[\ce{T∆S_{sys} - ∆H_{sys} > 0}\] 

⇒ \[\ce{(- ∆H_{sys} - T∆S_{sys}) > 0}\]

The above equation can be written as

\[\ce{- ΔG > 0}\]

\[\ce{ΔG = ΔH - TΔS < 0}\]

\[\ce{ΔH_{sys}}\] is the enthalpy change of a reaction, \[\ce{TΔS_{sys}}\] is the energy which is not available to do useful work. As a result, G is a measure of 'free energy,' as it is the net energy available to conduct beneficial work. As a result, it's also known as the reaction's free energy. At constant pressure and temperature, G gives a set of conditions for spontaneity,

(i) If \[\ce{ΔG}\] is negative (< 0), the process is spontaneous.

(ii) If \[\ce{ΔG}\] is positive (> 0), the process is nonspontaneous.

• Unit of \[\ce{ΔG}\] is Joule.

• The reaction will be spontaneous at high temperature.