PHYSICAL CHEMISTRY IN BRIEF

CHAP. 6: THERMODYNAMICS OF HOMOGENEOUS MIXTURES [CONTENTS] 167

Solution

According to (6.55) we have

G

E

i =Gi−Gi,id.mix,

where, according to (6.50)

Gi,id.mix=G•m,i+RTlnxi.

From (6.90) we get the relation for the partial molar Gibbs energy (chemical potential) in the

form

Gi=G•m,i+RTlnai=G•m,i+RTlnxiγi.

By subtracting we obtain the required relationG

E

i =RTlnγi.

Example

The activity coefficientsγ 1 = 2.27 andγ 2 = 1.041 were found for a methanol (1) — ethyl acetate

(2) system at 313.15 K andx 1 = 0.2. Calculate the excess Gibbs energy in this mixture.

Solution

From relation (6.104) we obtain

∆GE= 8. 314 × 313. 15 ×[0. 2 ×ln 2.27 + 0. 8 ×ln 1.041 ] = 510. 6 J mol−^1.

Thedependence of the activity coefficient on temperatureis expressed by the relation

lnγi(T 2 ) = lnγi(T 1 ) +

∫T 2

T 1

(

∂lnγi

∂T

)

p,n

dT , (6.105)

where (
∂lnγi
∂T

)

p,n

= −

Hi−Hmst,i

RT^2

. (6.106)