The Concept of Free Energy
283
2. At 25.0°C for the reaction (p 374)
AgCl <=* Ag
+ + Cl~
AG° = -(4.57)(298)log^ e = -(4.57)(298)log(1.6 x KT
10
)
= +13,350cal/mole
3. At 355.0°C for the reaction (p 257)
Hacg) + 1-2(9) <=* 2HI (0)
AG° = -(4.57)(628)log/i: e = -(4.57)(628)log(54.4)
= -4994 cal per 2 moles of HI
= -2497 cal/mole
Standard Free Energies of Formation
Equation 17-13 may be used to calculate standard free energies of reaction from
£°ceii values derived from Table 17-1, and Equation 17-14 may be used if values
of AT e are known. There is also a third way by which AG° may be calculated. Just
as for enthalpy (H, see pp 215-219), it is impossible to know the actual free
energy content (G) of any substance; only changes can be measured. Also, as
for enthalpy, it is possible to construct a table of standard free energies of
formation for each substance, based on the arbitrary assumption that the elements in their standard states (physical forms stable at 25.0°C and 1 atm) have
zero free energy of formation. You recall that these "formation" reactions
correspond to the formation of compounds directly from the elements. Table
17-2 shows a few selected values of AG f °. Once such a table has been constructed, it is possible to calculate the AG° or K e for any reaction for which data
are available—even though the reaction might not actually occur because the
activation energy is too high or because other reactions occur instead. Just as
with enthalpy, we can say
AG° = 2(AG f °) products - 2(AG?) reactants
(17-16)
and apply it in the following typical manner.
PROBLEM:
What is the change in standard free energy, and what is the value of K e , at 25.0°C
for the gaseous dissociation of PC1 5 into PC1 3 and C1 2 ?
SOLUTION:
Take the needed standard free energies of formation from Table 17-2 and use them
with the balanced chemical equation.
283
2. At 25.0°C for the reaction (p 374)
AgCl <=* Ag
+ + Cl~
AG° = -(4.57)(298)log^ e = -(4.57)(298)log(1.6 x KT
10
)
= +13,350cal/mole
3. At 355.0°C for the reaction (p 257)
Hacg) + 1-2(9) <=* 2HI (0)
AG° = -(4.57)(628)log/i: e = -(4.57)(628)log(54.4)
= -4994 cal per 2 moles of HI
= -2497 cal/mole
Standard Free Energies of Formation
Equation 17-13 may be used to calculate standard free energies of reaction from
£°ceii values derived from Table 17-1, and Equation 17-14 may be used if values
of AT e are known. There is also a third way by which AG° may be calculated. Just
as for enthalpy (H, see pp 215-219), it is impossible to know the actual free
energy content (G) of any substance; only changes can be measured. Also, as
for enthalpy, it is possible to construct a table of standard free energies of
formation for each substance, based on the arbitrary assumption that the elements in their standard states (physical forms stable at 25.0°C and 1 atm) have
zero free energy of formation. You recall that these "formation" reactions
correspond to the formation of compounds directly from the elements. Table
17-2 shows a few selected values of AG f °. Once such a table has been constructed, it is possible to calculate the AG° or K e for any reaction for which data
are available—even though the reaction might not actually occur because the
activation energy is too high or because other reactions occur instead. Just as
with enthalpy, we can say
AG° = 2(AG f °) products - 2(AG?) reactants
(17-16)
and apply it in the following typical manner.
PROBLEM:
What is the change in standard free energy, and what is the value of K e , at 25.0°C
for the gaseous dissociation of PC1 5 into PC1 3 and C1 2 ?
SOLUTION:
Take the needed standard free energies of formation from Table 17-2 and use them
with the balanced chemical equation.