Physical Chemistry Third Edition

308 7 Chemical Equilibrium

c.

T∆

⎛

⎝

−

(

G◦m−Hm,298◦ ◦

)

T

⎞

⎠(298.15 K)2(213.795JK−^1 mol−^1 )

− 2

(

197 .653JK−^1 mol−^1

)

− 205 .147JK−^1 mol−^1

(298.15 K)(− 0 .172863 kJ K−^1 mol−^1 )

− 51 .539 kJ mol−^1

∆G◦∆H◦ 298 −T∆

⎛

⎝

−

(

G◦m−H◦m,298◦

)

T

⎞

⎠

− 565 .990 kJ mol−^1 + 51 .539 kJ mol−^1

− 514 .451 kJ mol−^1

Exercise 7.1

a.Using Gibbs energy changes of formation from Table A.8, calculate∆G◦at 298.15 K for the

reaction

PCl 5 (g)
PCl 3 (g)+Cl 2 (g)

b.Calculate∆H◦and∆S◦at 298.15 K for the same reaction.

c.Calculate∆G◦at 298.15 K for the same reaction using Eq. (7.1-15). Compare with the answer

of part a.

d.Calculate∆G◦at 298.15 K for the same reaction using values of−(G◦m−Hm,298◦ )/Tand

the value of∆H◦.

The Gibbs Energy Change at Fixed Composition

Using the identity that a sum of logarithms is equal to the logarithm of a product, we

write Eq. (7.1-12) in the form

(

∂G

∂ξ

)

T,P

∆G◦+RTln(Q) (7.1-16)

where

Qav 11 av 22 ···avcc

∏c

i 1

avii (7.1-17)

The notationΠdenotes a product of factors, just asΣdenotes a sum of terms. The

quantityQis called theactivity quotient. It is called a quotient because the factors for

reactants have negative exponents.