Gibbs free energy in chemical reactions

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 5.2.

The Gibbs free energy is defined as

\displaystyle G\equiv U-TS+PV=H-TS \ \ \ \ \ (1)

The enthalpy {H} is the total energy required to create the system from nothing, in which the environment at constant pressure {P} must be pushed back to create the volume {V} in which the new system is to be stored.

For a process occurring at constant temperature and pressure, the change in {G} is

\displaystyle \Delta G=\Delta H-T\Delta S \ \ \ \ \ (2)

We can calculate changes in {G} for a process such as a chemical reaction by considering the values for the reactants. Consider the reaction in which nitrogen and hydrogen combine to form ammonia:

\displaystyle \mbox{N}_{2}+3\mbox{H}_{2}\rightarrow2\mbox{NH}_{3} \ \ \ \ \ (3)

 

We can look up the relevant values in Schroeder’s book, where he gives values for 1 mole at {T=298\mbox{ K}} and a pressure of 1 bar. The tabulated values are

{\Delta H\mbox{ (kJ)}} {S\mbox{ (J K}^{-1}\mbox{)}}
{\mbox{N}_{2}} 0 191.61
{\mbox{H}_{2}} 0 130.68
{\mbox{NH}_{3}} {-46.11} 192.45

For the reaction 3 we combine 1 mole of {\mbox{N}_{2}} with 3 moles of {\mbox{H}_{2}} to get 2 moles of {\mbox{NH}_{3}}, so we have

\displaystyle \Delta H \displaystyle = \displaystyle -92.22\times10^{3}\mbox{ J}\ \ \ \ \ (4)
\displaystyle \Delta S \displaystyle = \displaystyle 2\times192.45-3\times130.68-191.61\ \ \ \ \ (5)
\displaystyle \displaystyle = \displaystyle -198.75\mbox{ J K}^{-1}\ \ \ \ \ (6)
\displaystyle \Delta G \displaystyle = \displaystyle \Delta H-T\Delta S\ \ \ \ \ (7)
\displaystyle \displaystyle = \displaystyle -32.99\times10^{3}\mbox{ J} \ \ \ \ \ (8)

This value is for 2 moles, so for one mole of {\mbox{NH}_{3}} we have

\displaystyle \Delta G=-16.5\mbox{ kJ} \ \ \ \ \ (9)

which is close to the value of {-16.45\mbox{ kJ}} given in Schroeder’s table. (I’m not sure if we’re supposed to be able to get closer with the given data.)

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