# Chapter 16 - Molecular substances¶

## Example E1 - Pg 385¶

In [1]:
#calculate the net dipole moment
#Initialization of variables
import math
e=1.609*math.pow(10,-19) #C
#calculations
mux=(-0.36*e*(-0.8) + 0.45*e*(2.1) )*math.pow(10,-12) /(3.33564*math.pow(10,-30))
muy=-0.96
muz=0
mux=-1.1
mu=math.sqrt(mux*mux+muy*muy+muz*muz)
#results
print '%s %.1f %s' %("Net dipole moment =",mu,"D")

Net dipole moment = 1.5 D


## Example E2 - Pg 390¶

In [2]:
#calculate the potential energy of the system
#Initialization of variables
import math
Na=6.023*math.pow(10,23) # /mol
e=1.60228*math.pow(10,-19) #C
e0=8.85419*math.pow(10,-12) #C^2/J m
#calculations
factor=Na*e*e /(4*math.pi*e0)
#Multiply by Z^2/R to get the value of potential energy. Plot the graph
#results
print '%s %.3e %s' %("Potential energy =",factor," Z*Z/R  kJ/mol")

Potential energy = 1.390e-04  Z*Z/R  kJ/mol


## Example I1 - Pg 383¶

In [3]:
#calcualte the predicted dipole moment
#Initialization of variables
EH=2.1
EBr=2.8
#calculations
diff=-EH+EBr
#results
print '%s %.1f %s' %("Prediced dipole moment =",diff,"D")

Prediced dipole moment = 0.7 D


## Example I2 - Pg 387¶

In [4]:
#calculate the potential energy
#Initialization of variables
import math
mu1=1.4 #D
mu2=1.4 #D
angle=180. #degrees
d=3 #nm
D=4.7*math.pow(10,-30) #C m
#calculations
Vmol=D*D*(1-3*math.cos(angle*math.pi/180.)*math.cos(angle*math.pi/180.))/(4*math.pi*8.854*math.pow(10,-12) *math.pow((d*math.pow(10,-9)),3))
V=Vmol*(6.023*math.pow(10,23))
#results
print '%s %.1f %s' %("Potential energy =",V," J/mol")

Potential energy = -8.9  J/mol