Page 132 - Physical Chemistry

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Therefore, for a material change at constant T and P in a closed system in mechanical Section 4.3
and thermal equilibrium and capable of doing only P-V work, we have The Gibbs and Helmholtz Energies

d1H TS2 0 const. T, P (4.16)
where the equality sign holds at material equilibrium. G
Thus, the state function H TS continually decreases during material changes at Const. T, P
constant T and P until equilibrium is reached. The condition for material equilibrium
at constant T and P in a closed system doing P-V work only is minimization of the sys-
tem’s state function H TS. This state function is called the Gibbs function, the
Gibbs energy, or the Gibbs free energy and is symbolized by G:
G H TS U PV TS (4.17)*
Equilibrium
G decreases during the approach to equilibrium at constant T and P, reaching a mini- reached
mum at equilibrium (Fig. 4.2). As G of the system decreases at constant T and P, S univ
increases [see Eq. (4.21)]. Since U, V, and S are extensive, G is extensive.
Both A and G have units of energy (J or cal). However, they are not energies in the Time
sense of being conserved. G syst G surr need not be constant in a process, nor need A syst
A surr remain constant. Note that A and G are defined for any system to which meaning- Figure 4.2
ful values of U, T, S, P, V can be assigned, not just for systems held at constant T and For a closed system with P-V work
V or constant T and P. only, the Gibbs energy is
Summarizing, we have shown that: minimized if equilibrium is
reached under conditions of
In a closed system capable of doing only P-V work, the constant-T-and-V material- constant T and P.
equilibrium condition is the minimization of the Helmholtz energy A, and the
constant-T-and-P material-equilibrium condition is the minimization of the Gibbs
energy G:
dA 0 at equilib., const. T, V (4.18)*
dG 0 at equilib., const. T, P (4.19)*

where dG is the infinitesimal change in G due to an infinitesimal amount of chemical
reaction or phase change at constant T and P.

EXAMPLE 4.1 G and A for a phase change
Calculate G and A for the vaporization of 1.00 mol of H O at 1.00 atm and
2
100°C. Use data from Prob. 2.49.
We have G H TS. For this process, T is constant and G G G
1
2
H TS (H TS ) H T S:
1
1
2
2
¢G ¢H T ¢S const. T (4.20)
The process is reversible and isothermal, so dS dq/T and S q/T [Eq. (3.24)].
Since P is constant and only P-V work is done, we have H q q. Therefore
P
(4.20) gives G q T(q/T ) 0. The result G 0 makes sense because
areversible (equilibrium) process in a system at constant T and P has dG 0
[Eq. (4.19)].
From A U TS, we get A U T S at constant T. Use of U
q w and S q/T gives A q w q w. The work is reversible P-V
2
work at constant pressure, so w PdV P V. From the 100°C density
1
3
in Prob. 2.49, the molar volume of H O(l) at 100°C is 18.8 cm /mol. We can
2