Chapter4 Thermochemistry

Example 4.1, Page no.14

In [2]:
import math
#initialisation of variables
E= -1148.93  #in kcal moleˆ−1
R= 1.987     #cal moleˆ−1 Kˆ−1
T= 25        #in Celsius
n=4 
#CALCULATIONS
E1= (E*1000-R*n*(273.1+T))/1000 
#RESULTS
E1=round(E1,2)
print 'heat absorbed=',E1,'kcal moleˆ−1'

heat absorbed= -1151.3 kcal moleˆ−1

Example 4.2, Page no.14

In [3]:
import math
#given
Hr1= -71.03  #in kcal
Hr2= 70.96   #in kcal
#CALCULATIONS
H= Hr1+Hr2 
#RESULTS
print 'Enthalpy of transition=',H,'kcal'

Enthalpy of transition= -0.07 kcal

Example 4.3,Page no.15

In [4]:
import math
#initialisation of variables
Hr1= -70.96   #in kcal
Hr2= -23.49   #in kcal
Hr3= -31.14   #in kcal
Hr4= -68.32   #in kcal 
#CALCULATIONS
H= Hr1+Hr2+Hr3+Hr4 
#RESULTS
print 'Enthalpy of formation=',H,'kcal'

Enthalpy of formation= -193.91 kcal

Example 4.4,Page no.15

In [5]:
import math
#intialisation of variables
dH= -310.615  #in kcal
HfCO2= -94.52  #in kcal
HfH2O= -68.3174  #kcal 
#CALCULATIONS
HfCH2= -dH+2*HfCO2+HfH2O 
#RESULTS
HfCH2=round(HfCH2,2)
print 'Enthalpy of formation of acetylene=',HfCH2,'kcal moleˆ−1'

Enthalpy of formation of acetylene= 53.26 kcal moleˆ−1

Example 4.5,Page no.16

In [6]:
import math
#initialisation of variables
dH= -687.982  #in kcal
HCO2= -94.0518  #in kcal
#CALCULATIONS
H= -dH+4*HCO2+5*HCO2
#RESULTS
H=round(H,3)
print 'Enthalpy of formation of n butane=',H,'kcal moleˆ−1'

Enthalpy of formation of n butane= -158.484 kcal moleˆ−1

Example 4.6,Page no.16

In [7]:
import math
#initialisation of variables
HfAlO2= -399.1  #in kcal
HfFe2O2= -196.5  #in kcal
#CALCULATIONS
dH= HfAlO2 -HfFe2O2
#RESULTS
print 'Enthalpy change=',dH,'kcal moleˆ−1'

Enthalpy change= -202.6 kcal moleˆ−1

Example 4.7,Page no.16

In [8]:
import math
#initialisation of variables
Hr= -17.74  #in kcal
Hr1= 15.31  #in kcal
#CALCULATIONS
dH= Hr+Hr1
#RESULTS
print 'integral heat of dilution=',dH,'kcal'

integral heat of dilution= -2.43 kcal

Example 4.8,Page no.17

In [9]:
import math
#initialisation of variables
dHr= -0.56   #in kcal
dHr1= -18.85 #in kcal
#CALCULATIONS
dH= dHr+dHr1
#RESULTS
print 'integral heat of hydration=',dH,'kcal'

integral heat of hydration= -19.41 kcal

Example 4.9,Page no.17

In [10]:
import math
#initialisation of variables
HfHcl= -22.063 #in kcal
H298= -17.74  #in kcal
#CALCULATIONS
HfHcl200H2O= HfHcl+H298 
#RESULTS
print 'enthalpy of formation=',HfHcl200H2O,'kcal moleˆ−1'

enthalpy of formation= -39.803 kcal moleˆ−1

Example 4.10,Page no.17

In [11]:
import math
#initialisation of variables
HNaCl= -97.219   #in kcal
HH2O= -68.3174   #in kcal
HHcl= -39.713    #in kcal
HNaOH= -112.108  #in kcal
#CALCULATIONS
H298= HNaCl+HH2O -HHcl -HNaOH
#RESULTS
H298=round(H298,2)
print 'dH298=',H298,'kcal'

dH298= -13.72 kcal

Example 4.11,Page no.18

In [2]:
import math
#initialisation of variables
T1=1000    #in K
T2=300     #in K
k1=6.0954         #in cal degˆ−1 moleˆ−1 
k2=3.2533*10**-3  #in cal degˆ−2 moleˆ−1
k3=-1.071*10**-6   #in cal degˆ−3 moleˆ−1
#CALCULATIONS
dH=k1*(T1-T2)+(k2*(T1**2-T2**2)/2)+(k3*(T1**3-T2**3)/3)
#RESULTS
dH=round(dH,2)
print 'dH=',dH,'cal moleˆ−1'

dH= 5399.67 cal moleˆ−1

Example 4.12,Page no.18

In [13]:
import math
#initialisation of variables
dH273=-79.7    #in cal gˆ−1
T1=263         #in K
T2=273         #in K
dCp=-0.51      #in cal moleˆ−1 degˆ−1
#CALCULATIONS
H263=dH273+dCp*(T1-T2) 
#RESULTS
print 'H263=',H263,'cal gˆ−1'

H263= -74.6 cal gˆ−1

Example 4.13,Page no.19

In [14]:
import math
#initialisation of variables
dH293=-115595.8   #in cal
T1=1500           #in K
T2=298            #in K
k1=-5.6146        #cal degˆ−1 moleˆ−1
k2=1.8931*10**-3  #cal degˆ−2 moleˆ−1
k3=4.723*10**-7   #cal degˆ−3 moleˆ−1 
#CALCULATIONS
dH=dH293+ k1*(T1-T2)+(k2*(T1**2-T2**2)/2)+(k3*(T1**3-T2**3)/3)
#RESULTS
dH=round(dH,2)
print 'dH=',dH,'cal'

dH= -119771.7 cal