Chapter 3 - Thermochemistry

Example I1 - Pg 67

In [1]:
#Calculate the change in enthalpy
#Initialization of variables
dU=-969.6 #kJ/mol
nN2=1/2.
nCO2=2.
nO2=9./4.
T=298.15 #K
#Calculations
n=nCO2+nN2-nO2
H=dU+n*8.3145*T/1000.
#results
print '%s %.1f %s' %('Enthalpy change = ',H,'kJ/mol')

Enthalpy change =  -969.0 kJ/mol

Example E1 - Pg 58

In [2]:
#Calculate the Molar Enthalpy Change
#Initialization of variables
I=0.682 #A
V=12 #V
t=500. #s
m=4.33 #g
MW=46.07 #g/mol
#Calculations
q=I*V*t
n=m/MW
H=q/n/1000.
#Results
print '%s %.1f %s' %('Molar enthalpy change =',H,'kJ/mol')

Molar enthalpy change = 43.5 kJ/mol

Example E2 - Pg 62

In [3]:
#Calculate the Heat supplied
#Initialization of variables
m=1 #g
MW=24.31 #g/mol
H=2337 #kJ/mol
#Calculations
n=m/MW
q=n*H
#results
print '%s %.1f %s' %('Heat supplied =',q,'kJ')

Heat supplied = 96.1 kJ

Example E3 - Pg 65

In [5]:
#Calculate the sum of enthalpy changes
#Initialization of variables
HC=716.68 #kJ
HH=871.88 #kJ
HO=249.17 #kJ
Hcond=-38 #kJ
HCH=-412
HCO=-360
HOH=-463
#Calculations
H1=HC+HH+HO
H2=3*HCH+HCO+HOH
H3=Hcond
H=H1+H2+H3
#results
print '%s %d %s' %('Sum of enthalpy changes =',H,' kJ')

Sum of enthalpy changes = -259  kJ

Example E4 - Pg 68

In [6]:
#Calculate the standard enthalpy of combustion of propene
#Initialization of variables
Hf=-124 #kJ
Hoxi=-2220 #kJ
Hwater=286 #kJ
#Calculations
H=Hf+Hoxi+Hwater
#results
print '%s %d %s' %('Standard enthalpy of combustion of propene =',H,'kJ/mol')

Standard enthalpy of combustion of propene = -2058 kJ/mol

Example E5 - Pg 70

In [7]:
#Calculate the standard enthalpy of combustion of benzene
#Initialization of variables
nCO2=6 #mol
nH2O=3 #mol
nO2=15/2. #mol
nC6H6=1 #mol
HC6H6=49 #kJ/mol
HH2O=-285.83
HO2=0
HCO2=-393.51 
#Calculations
H=nCO2*HCO2+nH2O*HH2O-nC6H6*HC6H6-nO2*HO2
#results
print '%s %.1f %s' %('Standard enthalpy of combustion of benzene is',H,'kJ/mol')

Standard enthalpy of combustion of benzene is -3267.6 kJ/mol

Example E6 - Pg 74

In [8]:
#Calculate the enthalpy of formation of water
#Initialization of variables
HH2O=-241.82 #kJ/mol
T1=25 #C
T2=100 #C
CpH2O=33.58 #J/K mol
CpH2=28.84 #J/K mol
CpO2=29.37  #J/K mol
#calculations
dCp=CpH2O-CpH2-0.5*CpO2
dH=HH2O+dCp*(T2-T1)/1000.
#results
print '%s %.2f %s' %('Enthalpy of formation of water at 100 C is',dH,'kJ/mol')

Enthalpy of formation of water at 100 C is -242.57 kJ/mol