PHYSICAL CHEMISTRY IN BRIEF

CHAP. 8: CHEMICAL EQUILIBRIUM [CONTENTS] 254

Solution

K•=K◦

(

pst

p

)∆ν

= 1. 794

1

√

101. 3

= 0. 178.

If we choose the standard state of a pure component at the temperature and pressure of
the system, even if under the given conditions the pure substances would be in another state
of matter or structural modification than in a mixture, the following conversions are necessary:

• Conversion ofG◦m,i(g) toG•m,i(l)

G•m,i(l) =G◦m,i(g) +RTln

fis

f◦

+

∫p

psi

Vm,i(l)dp , (8.37)

wherepsidenotes the saturated vapour pressure of substanceiat a standard temperature,fis
denotes the fugacity of the same substance at a standard temperature and a pressure equal to
the saturated vapour pressure, andf◦= 101.325 kPa.

Example

Calculate∆fG•H 2 O(l) at 25◦C and pressure 0.3 MPa. The value ∆fG◦H 2 O(g) = − 228. 590

k J mol−^1 for this temperature, and the saturated vapour pressure of water at 25◦C isps= 3.168

kPa.

Solution

Substituting in the above relation while assuming ideal behaviour of water vapour yields

∆fG•H 2 O(l) = −228590 +R· 298 .15 ln

3. 168

101. 25

+ 18· 10 −^3 (300− 3 .168)

= − 237174 J mol−^1.

We assumed that the density of liquid water (1 g cm−^3 ) did not depend on pressure.

• Conversion ofG•m,itoG[mc],i

G[mc],i=G•m,i−RTln

γ[ic]ci

csst

. (8.38)