# Chapter 04: Thermochemistry¶

## Example Problem 4.1, Page Number 68¶

In :
#Varialble Declaration
DH0_H2O = 241.8          #Std Enthalpy of reaxtion of Water Fomation backward rxn, kJ/mol
DH0_2H = 2*218.0         #Std Enthalpy of formation of Hydrogen atom, kJ/mol
DH0_O = 249.2            #Std Enthalpy of formation of Oxygen atom, kJ/mol
R = 8.314                #Ideal gas constant, J/(mol.K)
Dn = 2.0
T = 298.15               #Std. Temperature, K
#Calculation
DH0_2HO = DH0_H2O + DH0_2H + DH0_O
DU0 = (DH0_2HO - Dn*R*T*1e-3)/2

#Results
print 'Avergae Enthalpy change required for breaking both OH bonds %4.1f kJ/mol'%DH0_2HO
print 'Average bond energy required for breaking both OH bonds %4.1f kJ/mol'%DU0

Avergae Enthalpy change required for breaking both OH bonds 927.0 kJ/mol
Average bond energy required for breaking both OH bonds 461.0 kJ/mol


## Example 4.2, Page Number 70¶

In :
import numpy as np
from sympy import symbols, integrate

#Variable Declaration
a = ([29.064, 31.695, 28.165])             #Constant 'a' in Heat capacity equation, J/(mol.K)
b = ([-0.8363e-3, 10.143e-3, 1.809e-3])    #Constant 'b' in Heat capacity equation, J/(mol.K)
c = ([20.111e-7, -40.373e-7, 15.464e-7])   #Constant 'a' in Heat capacity equation, J/(mol.K)
delHf0HCl = -92.3                          #Std. Heat of formation of HCl, kJ/mol
T1, T2 = 298.15, 1450                      #Std and final temperature, K

#Calculations
T = symbols('T')
DA = a-(a+a)/2
DB = b-(b+b)/2
DC = c-(c+c)/2

expr = integrate( DA + DB*T + DC*T**2, (T,T1,T2))
DHR1450= expr/1000 + delHf0HCl

#Results
print 'Heat of reaction for HCl formation is %4.1f kJ/mol'%DHR1450

Heat of reaction for HCl formation is -95.1 kJ/mol


## Example 4.3, Page Number 72¶

In :
#Varialble Declaration
ms1 = 0.972          #Mass of cyclohexane, g
DT1 = 2.98           #Change in temperature for bath, °C
DUR1 = -3913e3       #Std Internal energy change, J/mol
mw = 1.812e3         #Mass of water, g
ms2 = 0.857          #Mass of benzene, g
Ms1 = 84.16
Ms2 = 78.12
DT2 = 2.36           #Change in temperature for bath, °C
Mw = 18.02
Cpw = 75.3

#Calculation

Ccal = ((-ms1/Ms1)*DUR1-(mw/Mw)*Cpw*DT1)/DT1
DUR2 = (-Ms2/ms2)*((mw/Mw)*Cpw*DT2+Ccal*DT2)

#Results
print 'Calorimeter constant %4.2e J/°C'%Ccal
print 'Enthalpy of rection for benzene %4.2e J/mol'%DUR2

Calorimeter constant 7.59e+03 J/°C
Enthalpy of rection for benzene -3.26e+06 J/mol


## Example 4.4, Page Number 73¶

In :
#Varialble Declaration
ms = 1.423           #Mass of Na2SO4, g
mw = 100.34          #Mass of Na2SO4, g
DT = 0.037           #Change in temperature for solution, K
Mw = 18.02           #Molecular wt of Water
Ms = 142.04          #Molecular wt of ms Na2SO4
Ccal = 342.5         #Calorimeter constant, J/K
#Data
DHfNa = -240.1
DHfSO4 = -909.3
DHfNa2SO4 = -1387.1

#Calculation
DHs = (-Ms/ms)*((mw/Mw)*Cpw*DT+Ccal*DT)
DHsolD = 2*DHfNa + DHfSO4 - DHfNa2SO4

#Results
print 'Enthalpy of solution for Na2SO4 %4.2e J/mol'%DHs
print 'Enthalpy of solution for Na2SO4 from Data %4.2e J/mol'%DHsolD

Enthalpy of solution for Na2SO4 -2.81e+03 J/mol
Enthalpy of solution for Na2SO4 from Data -2.40e+00 J/mol