Physical Chemistry Third Edition

4.4 Gibbs Energy Calculations 179

We cannot use this relation, because a constant can be added to the value of the entropy

without any physical effect. The assignment of “absolute” entropies is based on a

conventional assignment of zero entropy for elements at 0 K and does not provide

unique values of entropies. Adding a constant toSwould change the value of the

integrand in Eq. (4.4-16), changing the value of the integral.

The most we can do is to consider an isothermal process that can be carried out

once at temperatureT 1 and a fixed pressurePand again at temperatureT 2 and pressure

P. We write Eq. (4.4-16) once for the initial state and once for the final state. The

difference of these equations gives

∆G(T 2 ,P)−∆G(T 1 ,P)−

∫T 2

T 1

∆S(T,P)dT (closed system) (4.4-17)

where∆G(T,P) and∆S(T,P) pertain to a process at a constant temperatureTand

a constant pressureP. Equation (4.4-17) does not mean that we are considering a

nonisothermal process. It gives the difference between∆Gfor an isothermal process

carried out atT 2 and∆Gfor the same isothermal process carried out atT 1.

If∆Sis nearly independent of temperature betweenT 1 andT 2 , Eq. (4.4-17) becomes

∆G(T 2 ,P)−∆G(T 1 ,P)≈−∆S(T 2 −T 1 ) (∆Sindependent of temperature)

(4.4-18)

This equation should be a usable approximation if the difference betweenT 2 andT 1 is

not very large. An alternate to Eq. (4.4-17) is known as theGibbs–Helmholtz equation:

∆G(T 2 ,P)

T 2

−

∆G(T 1 ,P)

T 1

−

∫T 2

T 1

∆H(T,P)

T^2

dT (4.4-19)

If∆His nearly independent of temperature,

∆G(T 2 ,P)

T 2

−

∆G(T 1 ,P)

T 1

∆H

(

1

T 2

−

1

T 1

)

(4.4-20)

Thermodynamics applies equally to chemical reactions and to physical processes such

as fusion or vaporization. We can apply Eqs. (4.4-17) and (4.4-19) to physical processes

as well as to chemical reactions.

EXAMPLE4.16

Derive Eq. (4.4-19).

Solution

(

∂∆G/T

∂T

)

P



1

T

(

∂∆G

∂T

)

P

−

∆G

T^2

−

∆S

T

−

∆G

T^2

−

∆H

T^2

∆G(T 2 ,P)

T 2

−

∆G(T 1 ,P)

T 1



∫T 2

T 1

(

∂(∆G/T)

∂T

)

P

dT

−

∫T 2

T 1

∆H

T^2

dT (4.4-21)