Physical Chemistry Third Edition

1104 26 Equilibrium Statistical Mechanics. II. Statistical Thermodynamics

approximately equal to 1.000. The translational partition function of the molecule is given

by the same expression as that of the atom except for the mass. Its value is

z′tr,H◦

2

NAv

 4. 152 × 105

The dissociation energy,De, is equal to 4.75 eV for the H 2 molecule. The value of∆ε 0 is

equal toD 0 , which is equal toDeminus the zero-point vibrational energy of the molecule:

∆ε 0  4 .75 eV−

1

2

hc ̃νe 7. 61 × 10 −^19 J−

1

2

hc ̃νe

 7. 61 × 10 −^19 J−

1

2

(6. 6261 × 10 −^34 J s)(2. 9979 × 1010 cm s−^1 )(4401.2cm−^1 )

 7. 17 × 10 −^19 J

e−∆ε^0 /kBTexp

[

− 7. 17 × 10 −^19 J

(1. 3807 × 10 −^23 JK−^1 )(500 K)

]

 7. 8 × 10 −^46

Ke−∆ε^0 /kBT(z◦el,H)^2 (z′tr,H◦/NAv)^2 (z′tr,H 2 ◦/NAv)−^1 (zrot, H′◦ 2 )−^1



(7. 8 × 10 −^46 )(2.00)^2 (1. 468 × 105 )^2

(4. 152 × 105 )(2.85)

 5. 7 × 10 −^41

We can calculate the thermodynamic value ofKby using techniques discussed in Part I of

this textbook. The standard-state Gibbs energy change of the reaction is 385.914 kJ mol−^1

and the thermodynamic value ofKis

Ke−∆G

◦/RT

exp

(

−385914 J mol−^1

(8.3145 J K−^1 mol−^1 )(500.0K)

)

 4. 84 × 10 −^41

Exercise 26.10

Without making detailed calculations, estimate the equilibrium constant for the reaction at

298.15 K:

H 2 +D 2 
2HD

Assume that the electronic and vibrational partition functions equal unity and that the rotational

partition functions cancel except for the symmetry numbers. Also assume that∆ε 0 ≈0.

We can express the equilibrium constant for a dilute gas reaction in terms of molar

concentrations. This equilibrium constant is proportional toKand is denoted byKc.

Kc

∏c

a 1

(ca

c◦

)νa

(26.3-16)

where we denote the molar concentration of substanceabyca, expressed in mol m−^3.

ca

na

V



Na

NAvV

(26.3-17)

so that

z′a

Na



z′a

NAvna



z′a

NAvVca



z′a

NAvVc◦

c◦

ca

(26.3-18)