Concise Physical Chemistry

c07 JWBS043-Rogers September 13, 2010 11:25 Printer Name: Yet to Come

98 EQUILIBRIUM

so

⎡

⎢

⎢

⎣

∂

(

RlnKeq

)

∂

(

1

T

)

⎤

⎥

⎥

⎦

p

=−
rH◦

where
rH◦ is the standard enthalpy of reaction. Now,dT−^1 /dT=−T−^2 and

d(1/T)=−T−^2 dT,so

d

(

RlnKeq

)

=−
H◦d

(

1

T

)

=

H◦

T^2

dT

This equation can be integrated as

∫

dlnKeq=−

rH◦

R

∫

1

T^2

dT

to give thevan’t Hoff equation

lnKeq=−

rH◦

R

(

−

1

T

)

+const.=

rH◦

R

(

1

T

)

+const.

This is the equation of a straight line of lnKeqplotted against (1/T) with
rH◦/R

as the slope. It applies so long as
rH◦remains constant. The equation as written

implies that the slope will be positive, but this does not follow because
rH◦may

be positive, negative, or zero for endothermic, exothermic, or null-thermal reactions.

Common examples are melting, freezing, or mixing of two isomers of a liquid alkane

which are endothermic, exothermic, and null-thermal.

The van’t Hoff equation can be integrated between limits to give an expression for

a new equilibrium constant at a new temperature from known values of
rH◦,Keq,

and the initial and revised temperatures,TandT′:

∫Keq′

Keq

dlnKeq=−

rH◦

R

∫T′

T

1

T^2

dT

ln

Keq′

Keq

=−

rH◦

R

(

1

T′

−

1

T

)

=

rH◦

R

(

1

T

−

1

T′

)

Conversely, the integrated form is a way of determining
rH◦from two experimental

determinations ofKeqat different temperatures.