Thermodynamics and Chemistry

CHAPTER 11 REACTIONS AND OTHER CHEMICAL PROCESSES

11.9 EFFECTS OFTEMPERATURE ANDPRESSURE ONEQUILIBRIUMPOSITION 357

which shows how infinitesimal changes inT,p, andeqare related.
Now we are ready to see howeqis affected by changes inTorp. Solving Eq.11.9.6
for deqgives

deqD

ÅrH

T

dTÅrVdp



@^2 G

@^2



T;p

(11.9.7)

(closed system)

The right side of Eq.11.9.7is the expression for the total differential ofin a closed
system at reaction equilibrium, withTandpas the independent variables. Thus, at constant
pressure the equilibrium shifts with temperature according to

@eq
@T



p

D

ÅrH

T



@^2 G

@^2



T;p

(11.9.8)

(closed system)

and at constant temperature the equilibrium shifts with pressure according to

@eq
@p



T

D

ÅrV



@^2 G

@^2



T;p

(11.9.9)

(closed system)

Because the partial second derivative.@^2 G=@^2 /T;pis positive, Eqs.11.9.8and11.9.9
show that.@eq=@T /pandÅrHhave the same sign, whereas.@eq=@p/TandÅrV have
opposite signs.
These statements express the application to temperature and pressure changes of what
is known asLe Chatelier’s principleˆ : When a change is made to a closed system at equilib-
rium, the equilibrium shifts in the direction that tends to oppose the change. Here are two
examples.

1.SupposeÅrH is negative—the reaction is exothermic. Since.@eq=@T /phas the
same sign asÅrH, an increase in temperature causeseqto decrease: the equilibrium
shifts to the left. This is the shift that would reduce the temperature if the reaction
were adiabatic.
2.IfÅrV is positive, the volume increases as the reaction proceeds to the right at con-
stantT andp. .@eq=@p/T has the opposite sign, so if we increase the pressure
isothermally by reducing the volume, the equilibrium shifts to the left. This is the
shift that would reduce the pressure if the reaction occurred at constantTandV.
It is easy to misuse or to be misled by Le Chatelier’s principle. Consider the solutionˆ
process B(s)!B(sln) for which.@eq=@T /p, the rate of change of solubility withT, has
the same sign as the molar differential enthalpy of solutionÅsolHat saturation. The sign
ofÅsolHat saturation may be different from the sign of the molarintegralenthalpy of
solution,ÅHm(sol). This is the situation for the dissolution of sodium acetate shown in Fig.
11.9on page 326. The equilibrium position (saturation) with one kilogram of water is at
sol 15 mol, indicated in the figure by an open circle. At this position,ÅsolHis positive