The Foundations of Chemistry

Plan

The process we must consider is

Br 2 (
)88nBr 2 (g)

By definition, the normal boiling point of a liquid is the temperature at which pure liquid and

pure gas coexist in equilibrium at 1 atm. Therefore, 
G0. We assume that 
Hrxn
H^0 rxn

and 
Srxn
S^0 rxn. We can evaluate these two quantities, substitute them in the relationship


G
HT
S,and then solve for T.

Solution

The required values (Appendix K) are as follows:

Br 2 (
)Br 2 (g)


H^0 f, kJ/mol: 0 30.91

S^0 , J/molK: 152.2 245.4

Hrxn
H^0 fBr 2 (g)
H^0 fBr 2 (
)

30.91  0 30.91 kJ/mol


SrxnS^0 Br 2 (g) S^0 Br 2 (
)

(245.4 152.2)93.2 J/molK0.0932 kJ/molK

We can now solve for the temperature at which the system is in equilibrium, that is, the boiling

point of Br 2.

Grxn
HrxnT
Srxn0so
HrxnT
Srxn

T332 K (59°C)

This is the temperature at which the system is in equilibrium, that is, the boiling point of Br 2.

The value listed in a handbook of chemistry and physics is 58.78°C.

You should now work Exercise 110.

The free energy change and spontaneity of a reaction depend on both enthalpy and

entropy changes. Both 
Hand 
Smay be either positive or negative, so we can group

reactions in four classes with respect to spontaneity (Figure 15-15).

G
HT
S (constant temperature and pressure)

1.
H(favorable) 
S(favorable) Reactions are product-

favored at all temperatures

2.
H(favorable) 
S(unfavorable) Reactions become product-

favored below a definite

temperature

3.
H(unfavorable) 
S(favorable) Reactions become product-

favored above a definite

temperature

4.
H(unfavorable) 
S(unfavorable) Reactions are reactant-

favored at all temperatures

30.91 kJ/mol



0.0932 kJ/molK

Hrxn




Srxn

632 CHAPTER 15: Chemical Thermodynamics

Actually, both
H^0 rxnand
S^0 rxnvary
with temperature, but usually not
enough to introduce signif icant errors
for modest temperature changes. The
value of
G^0 rxn, on the other hand,
is strongly dependent on the
temperature.