Concise Physical Chemistry

c18 JWBS043-Rogers September 13, 2010 11:29 Printer Name: Yet to Come

PROBLEMS AND EXAMPLES 301

The spectral input to this calculation is the line separation

2 B= 3 .8604 cm−^1 = 386 .04 m−^1

soB= 193 .02 m−^1. This is multiplied bycto obtain the frequency in s−^1 and

bond lengthrin pm:

I=

h

4 π^2 ( 2 Bc)

=

2. 799 × 10 −^44

B

=

2. 799 × 10 −^44

193. 02

= 1. 450 × 10 −^46

r^2 =

1. 450 × 10 −^46

1. 138 × 10 −^26

= 1. 27 × 10 −^20

r= 1. 129 × 10 −^10 m= 112 .9pm

Example 18.2 The Dipole Moment of Sulfur Dioxide

Sulfur dioxide, SO 2 , has a total molar polarization of 68.2 cm^3 mol−^1 at 298 K and

56.0 cm^3 mol−^1 at 398 K. What is the dipole moment of SO 2?

Solution 18.2 The inverse of the lower temperature is 0.00336 and the in-

verse of the upper temperature is 0.00251. The molar polarizations are 68.2 and

56.0 cm^3 mol−^1. The slope of the curve of molar polarization vs. the inverse of

Tis 1. 44 × 104 cm^3 Kmol−^1. (Probably for historical reasons, the nonstandard unit

system of erg cm is often used.) The experimental slope is related to the dipole

moment as

slope=

4 πNAμ^2

9 kB

= 1. 44 × 104

Solving forμ^2 , we get

μ^2 =

9 kB

4 πNA

slope= 1. 641 × 10 −^47

(

1. 44 × 104

)

It is convenient at this point to change the units ofkBfrom J K−^1 to erg K−^1 .This

requires multiplication by 10^7 to give

μ^2 = 1. 641 × 10 −^40

(

1. 44 × 104

)

= 2. 363 × 10 −^36

μ=

√

2. 363 × 10 −^36 = 1. 54 × 10 −^18 = 1 .54 D

Problem 18.1

The molecule^1 H^35 Cl absorbs light atλ= 2. 991 × 10 −^4 cm= 2. 991 × 10 −^6 m. Tak-

ing^1 H^35 Cl to be a harmonic oscillator, what are the differences in energy between