# Chapter 25 : Energy Balances How to Account for Chemical Reaction¶

### Example 25.1 page no. 766¶

In :
# Variables
Qa = -393.51 ;			# Heat of reaction of reaction (a) - [kJ/g mol C]
Qb = -282.99 ;			# Heat of reaction of reaction (b) - [kJ/g mol CO]
del_Ha = Qa ;			# Change in enthalpy of reaction A - [kJ/g mol C]
del_Hb = Qb ;			# Change in enthalpy of reaction B - [kJ/g mol CO]

# Calculations
del_Hfc = del_Ha - del_Hb ;			# Standard heat of formation of CO - [kJ/g mol C]

# Results
print 'Standard heat of formation of CO is %.2f kJ/g mol C.'%del_Hfc

Standard heat of formation of CO is -110.52 kJ/g mol C.


### Example 25.2 page no. 767¶

In :
# Variables
H_H2 = 0 ;			# Standard heat of formation of H2 -[kJ/ g mol H2]
H_Cl2 = 0 ;			# Standard heat of formation of Cl2 -[kJ/ g mol Cl2]
H_HCl = -92.311 ;			#  Standard heat of formation of HCl -[kJ/ g mol HCl]

# Calculations
H_f = 1*H_HCl - (1./2)*(H_H2 + H_Cl2) ; 			# Standard heat of formation of HCl by reaction - [kJ/ g mol HCl]

# Results
print 'Standard heat of formation of HCl(g) is %.3f kJ/g mol HCl.'%H_f

Standard heat of formation of HCl(g) is -92.311 kJ/g mol HCl.


### Example 25.3 page no. 771¶

In :
# Variables
H_fNH3 = -46.191 ;			# Standard heat of formation of NH3 -[kJ/ g mol]
H_fO2 = 0 ;			#Standard heat of formation of O2 -[kJ/ g mol]
H_fNO = 90.374 ;			# Standard heat of formation of NO -[kJ/ g mol]
H_fH2O = -241.826 ;			# Standard heat of formation of H2O -[kJ/ g mol]

# Calculations
H_rxn = ((4*H_fNO + 6*H_fH2O) - (4*H_fNH3 + 5*H_fO2))/4. ;			# Heat of above reaction-[kJ/ g mol NH3]

# Results
print 'Heat of above reaction is %.3f kJ/g mol NH3.'%H_rxn

Heat of above reaction is -226.174 kJ/g mol NH3.


### Example 25.4 page no. 773¶

In :
# Variables
P1 = 1. ;			# Initial pressure - [atm]
P2 = 1. ;			# Final pressure - [atm]
T1 = 500. ;			# Initial temperature -[degree C]
T2 = 500. ;			# Final temperature -[degree C]

m_CO2 = 1. ;			# Moles of CO2 - [ g mol]
m_H2 = 4. ;			# Moles of H2 - [ g mol]
m_H2O = 2. ;			# Moles of H2O - [ g mol]
m_CH4 = 1. ;			# Moles of CH4 - [ g mol]

H_fCO2 = -393.250 ;			# Heat of formation of CO2 - [kJ/g mol]
H_fH2 = 0. ;			# Heat of formation of H2 - [kJ/g mol]
H_fH2O = -241.835 ;			# Heat of formation of H2O - [kJ/g mol]
H_fCH4 = -74.848 ;			# Heat of formation of CH4 - [kJ/g mol]

H_CO2 = 21.425 ;			# Change in enthalpy during temperature change from 25 to 500 degree C  of CO2 - [kJ/g mol]
H_H2 = 13.834 ;			# Change in enthalpy during temperature change from 25 to 500 degree C of H2 - [kJ/g mol]
H_H2O = 17.010 ;			# Change in enthalpy during temperature change from 25 to 500 degree C of H2O - [kJ/g mol]
H_CH4 = 23.126 ;			# Change in enthalpy during temperature change from 25 to 500 degree C of CH4 - [kJ/g mol]

# Calculations
H_rxn_25 = (m_CH4*H_fCH4 + m_H2O*H_fH2O) - (m_CO2*H_fCO2 + m_H2*H_fH2) ;			# Heat of reaction at 25 C
sum_H_rct = m_CO2*H_CO2 + m_H2*H_H2 ;			# sum of heat of formation of reactant - [kJ]
sum_H_pdt = m_CH4*H_CH4 + m_H2O*H_H2O ;			#sum of heat of formation of product - [kJ]
# Heat of above reaction is calculated by eqn. 25.4
H_rxn_500 = sum_H_pdt - sum_H_rct + H_rxn_25  ;			# Heat of reaction at 500 C

# Results
print 'Heat of above reaction at 500 degree C and 1 atm is %.1f kJ.'%H_rxn_500

Heat of above reaction at 500 degree C and 1 atm is -184.9 kJ.


### Example 25.5 page no. 775¶

In :
# Variables
m_CO2 = 1. ;			# Moles of CO2 - [ g mol]
m_H2 = 4. ;			# Moles of H2 - [ g mol]
m_H2O = 2. ;			# Moles of H2O - [ g mol]
m_CH4 = 1. ;			# Moles of CH4 - [ g mol]
P1 = 1. ;			# Initial pressure  - [atm]
P2 = 1. ;			# Final pressure - [atm]

T1_CO2 = 800. ;			# Initial temperature of entering CO2 -[K]
T1_H2 = 298. ;			# Initial temperature of entering H2 -[K]
T2 = 1000. ;			# Temperature of exiting product - [K]

m1_CO2 = 1. ;			# Moles of entering CO2 - [ g mol]
m1_H2 = 4. ;			# Moles of entering H2 - [ g mol]
f_con = 70./100 ;			# Fractional conversion of CO2
m2_H2O = 2*f_con ;			# Moles of H2O in product - [ g mol]
m2_CH4 = 1*f_con ;			# Moles of CH4 in product - [ g mol]
m2_CO2 = m1_CO2*(1-f_con) ;			#  Moles of CO2 in product - [ g mol]
m2_H2 = m1_H2*(1-f_con) ;			#  Moles of CO2 in product - [ g mol]

H_fCO2 = -393.250 ;			# Heat of formation of CO2 - [kJ/g mol]
H_fH2 = 0 ;			# Heat of formation of H2 - [kJ/g mol]
H_fH2O = -241.835 ;			# Heat of formation of H2O - [kJ/g mol]
H_fCH4 = -74.848 ;			# Heat of formation of CH4 - [kJ/g mol]

H1_CO2 = 22.798 ;			# Change in enthalpy during temperature change from 298K to 800 K  of CO2 - [kJ/g mol]
H1_H2 = 0 ;			# Change in enthalpy during temperature change from 298K to 298 K  of H2 - [kJ/g mol]
H2_H2O = 25.986 ;			# Change in enthalpy during temperature change from 298K to 1000 K  of H2O - [kJ/g mol]
H2_CH4 = 38.325 ;			# Change in enthalpy during temperature change from 298K to 1000 K  of CH4 - [kJ/g mol]
H2_CO2 = 33.396; 			# Change in enthalpy during temperature change from 298K to 1000 K  of CO2 - [kJ/g mol]
H2_H2 = 20.620; 			# Change in enthalpy during temperature change from 298K to 1000 K  of H2 - [kJ/g mol]

# Calculations
H_rxn_25 = (m_CH4*H_fCH4 + m_H2O*H_fH2O) - (m_CO2*H_fCO2 + m_H2*H_fH2) ;			# Standard heat of reaction at 25 C-[kJ]
H_rxn_ac = f_con*H_rxn_25 ;			# Heat of reaction actual - [kJ]
sum_H_rct = m1_CO2*H1_CO2 + m1_H2*H1_H2 ;			# sum of heat of formation of reactant - [kJ]
sum_H_pdt = m2_CH4*H2_CH4 + m2_H2O*H2_H2O + m2_CO2*H2_CO2 + m2_H2*H2_H2 ;			#sum of heat of formation of product - [kJ]
H_rxn = sum_H_pdt - sum_H_rct + H_rxn_ac  ;			# Heat of reaction -[kJ/ g mol CO2]

Q = H_rxn ;			# Heat transfer to/from the reactor - [kJ]

# Results
print 'Heat transfer to/from the reactor  is %.3f kJ.Since Q is negative , the reactor losses heat.'%Q

Heat transfer to/from the reactor  is -40.515 kJ.Since Q is negative , the reactor losses heat.


### Example 25.6 page no. 776¶

In :
# Variables
H_EtOH =-1330.51  ;			# Change in enthalpy of ethanol -[kJ/g mol]
H_Ac = -887.01 ;			# Change in enthalpy of acetate -[kJ/g mol]
H_Fr = -221.75 ;			# Change in enthalpy of formate -[kJ/g mol]
H_Lc = -1330.51 ;			# Change in enthalpy of lactate -[kJ/g mol]
H_Mn = -2882.78  ;			# Change in enthalpy of mannitol -[kJ/g mol]
mol_EtOH =1.29   ;			#ethanol produced / g mol mannitol -[g mol]
mol_Ac = 0.22    ; 			#acetate produced / g mol mannitol -[g mol]
mol_Fr = 1.6    ; 			#formate produced / g mol mannitol-[g mol]
mol_Lc = 0.4    ;			#lactate produced / g mol mannitol-[g mol]
mol_Mn = 1.0    ;			#mannitol produced / g mol mannitol-[g mol]
B_growth = 40.5   ;			# Biomass growth -[g cells/g mol mannitol]

# Calculations and Results
del_H1 = H_EtOH*mol_EtOH +H_Ac*mol_Ac + H_Fr*mol_Fr + H_Lc*mol_Lc - H_Mn*mol_Mn ;			# Net enthalpy change for several products (metabolites) per g mol mannitol consumed -[kJ]
printnt ' (a) Net enthalpy change for several products (metabolites) per g mol mannitol consumed is %.2f kJ.'%del_H1

#(b)
del_H2 = del_H1 / B_growth ;			#Net enthalpy change for several products (metabolites) per g cells produced-[kJ]
print '  (b) Net enthalpy change for several products (metabolites) per g cells produced is %.2f kJ.'%del_H2

 (a) Net enthalpy change for several products (metabolites) per g mol mannitol consumed is 84.28 kJ.
(b) Net enthalpy change for several products (metabolites) per g cells produced is 2.08 kJ.


### Example 25.7 page no. 777¶

In :
# Solution

# Variables
H_Cb = -26. ;			#Standard heat of formation of carbaryl(C12H11O2N) -[kJ/ g mol]
H_HCl = -92.311 ;			#Standard heat of formation of HCl -[kJ/ g mol]
H_Ma = -20.0 ;			#Standard heat of formation of methyl amine(CH3NH2) -[kJ/ g mol]
H_Mi = -9*10**4 ;			#Standard heat of formation of methyl isocynate(C2H3NO) -[kJ/ g mol]
H_Nc = -17.9 ;			#Standard heat of formation of 1-Napthalenyl chloroformate(C11H7O2Cl) -[kJ/ g mol]
H_N = 30.9 ;			#Standard heat of formation of napthol(C10H8O) -[kJ/ g mol]
H_P = -221.85 ;			#Standard heat of formation of phosgene(COCl2) -[kJ/ g mol]

# Calculations
H_rxn_a = (2*H_HCl + 1*H_Mi) - (1*(H_Ma) + 1*H_P )  ;			# Heat of reaction (A)-[kJ]
H_rxn_b = (1*H_Cb ) - (1*(H_Mi) + 1*H_N )  ;			# Heat of reaction (B)-[kJ]
H_rxn_c = (1*H_Nc) - (1*(H_N) + 1*H_P )  ;			# Heat of reaction (C)-[kJ]
H_rxn_d = (1*H_Cb + 1*H_HCl) - (1*(H_Nc) + 1*H_Ma )  ;			# Heat of reaction (D)-[kJ]

# Results
#Bhopal Process
print ' Bhopal process .'
print '   (a) Heat of reaction (A) is %.1e kJ.'%H_rxn_a
print '   (b) Heat of reaction (B) is %.1e kJ.'%H_rxn_b

#Alternate process
print '  Alternate process .'
print '   (c) Heat of reaction (C) is %.2f kJ.'%H_rxn_c
print '   (d) Heat of reaction (D) is %.2f kJ.'%H_rxn_d
print ' The above data show that capital cost of Bhopal process could be higher than alternate process.'

 Bhopal process .
(a) Heat of reaction (A) is -9.0e+04 kJ.
(b) Heat of reaction (B) is 9.0e+04 kJ.
Alternate process .
(c) Heat of reaction (C) is 173.05 kJ.
(d) Heat of reaction (D) is -80.41 kJ.
The above data show that capital cost of Bhopal process could be higher than alternate process.


### Example 25.8 page no. 782¶

In :
# Variables
P1 = 1. ;			# Initial pressure - [atm]
P2 = 1. ;			# Final pressure - [atm]
T1 = 500. ;			# Initial temperature -[degree C]
T2 = 500. ;			# Final temperature -[degree C]

m_CO2 = 1. ;			# Moles of CO2 - [ g mol]
m_H2 = 4. ;			# Moles of H2 - [ g mol]
m_H2O = 2. ;			# Moles of H2O - [ g mol]
m_CH4 = 1. ;			# Moles of CH4 - [ g mol]

H_fCO2 = -393.250; 			# Heat of formation of CO2 - [kJ/g mol]
H_fH2 = 0 ;			# Heat of formation of H2 - [kJ/g mol]
H_fH2O = -241.835 ;			# Heat of formation of H2O - [kJ/g mol]
H_fCH4 = -74.848 ;			# Heat of formation of CH4 - [kJ/g mol]

H_CO2 = 21.425 ;			# Change in enthalpy during temperature change from 25 to 500 degree C  of CO2 - [kJ/g mol]
H_H2 = 13.834 ;			# Change in enthalpy during temperature change from 25 to 500 degree C of H2 - [kJ/g mol]
H_H2O = 17.010 ;			# Change in enthalpy during temperature change from 25 to 500 degree C of H2O - [kJ/g mol]
H_CH4 = 23.126 ;			# Change in enthalpy during temperature change from 25 to 500 degree C of CH4 - [kJ/g mol]

# Calculations
H_in  = (H_fCO2 + H_CO2)*m_CO2 + (H_fH2 + H_H2)*m_H2 ;			# Enthalpy change for inputs -[kJ]
H_out = (H_fH2O + H_H2O)*m_H2O + (H_fCH4 + H_CH4)*m_CH4 ; 			# Enthalpy change for outputs -[kJ]
del_H = H_out - H_in ;			# Net enthalpy change of process -[kJ]

# Results
print 'Heat of above reaction at 500 degree C and 1 atm is %.1f kJ.'%del_H

Heat of above reaction at 500 degree C and 1 atm is -184.9 kJ.


### Example 25.9 page no. 783¶

In :
# Solution

# Variables
m_CO = 1. ;			# Moles of CO input- [g mol]
m1_O2 = 1.5 ;			# Moles of O2 input - [g mol]
m_CO2 = 1. ;			# Moles of CO2 output - [g mol]
m2_O2 = 1. ;			# Moles of O2 output - [g mol]
T_in_CO = 298. ;			# Temperature of entering CO -[K]
T_in_O2 = 400. ;			#Temperature of entering O2 -[K]
T_out_CO2 = 300. ;			# Temperature of exiting CO2 -[K]
T_out_O2 = 300. ;			# Temperature of exiting O2 -[K]

H_fCO = -110.520 ;			# Heat of formation of CO - [kJ/g mol]
H_fO2 = 0 ;			# Heat of formation of O2 - [kJ/g mol]
H_fCO2 = -393.250 ;			# Heat of formation of CO2 - [kJ/g mol]

H_CO = 0 ;			# Change in enthalpy during temperature change from 298K to 298 K of CO - [kJ/g mol]
H1_O2 = 11.619 ;			# Change in enthalpy during temperature change from 298K to 400 K of input O2 - [kJ/g mol]
H_CO2 = 11.644 ;			# Change in enthalpy during temperature change from 298K to 300 K of CO2 - [kJ/g mol]
H2_O2 = 8.389 ;			# Change in enthalpy during temperature change from 298K to 300 K of output  O2 - [kJ/g mol]

# Calculations
H_in  = (H_fCO + H_CO)*m_CO + (H_fO2 + H1_O2)*m1_O2 ;			# Enthalpy change for inputs -[kJ]
H_out  = (H_fCO2 + H_CO2)*m_CO2 + (H_fO2 + H2_O2)*m2_O2 ;			# Enthalpy change for inputs -[kJ]
del_H = H_out - H_in ;			# Net enthalpy change of process -[kJ]

# Results
print 'Heat of above reaction  is %.1f kJ.'%del_H

Heat of above reaction  is -280.1 kJ.


### Example 25.10 page no. 788¶

In :
# Solution

# Given
Ex_hv = 29770.0 ;			# Experimental heating value of given coal - [kJ/kg]

C = 71.0/100 ;			#Fraction of C in coal
H2 = 5.6/100 ;			# Fraction of H2 in coal
N2 = 1.6/100 ;			# Fraction of N2 in coal
S = 2.7/100 ;			# Fraction of S in coal
ash = 6.1/100 ;			# Fraction of ash in coal
O2 = 13.0/100 ;			#Fraction of O2 in coal

# Calculations
HHV = 14544*C + 62028*(H2 - O2/8) + 4050*S ;			# Higher heating value (HHV) by Dulong formula -[Btu/lb]
HHV_SI = HHV *1.055/0.454 ;			# HHV in SI unt - [kJ/kg]

# Results
print 'The experimental heating value -                       %.0f kJ.'%Ex_hv
print ' Higher heating value (HHV) by Dulong formula -         %.0f kJ.'%HHV_SI
print ' The two values are quite close.'

The experimental heating value -                       29770 kJ.
Higher heating value (HHV) by Dulong formula -         29980 kJ.
The two values are quite close.


### Example 25.11 page no. 789¶

In :
# Variables
H_req = 10**6 ;			# Heat requirement - [Btu]

d_N6 = 60.2 ;			# Density of fuel no. 6-[lb/ft**3]
d_N2 = 58.7 ;			# Density of fuel no. 2-[lb/ft**3]
S_N6 = 0.72/100 ;			# Sulphur content in fuel no. 6
S_N2 = 0.62/100; 			#Sulphur content in fuel no. 2
lhv_N6 = 155000 ;			#Lower heating value of  No.6 -[Btu/gal]
lhv_N2 = 120000 ;			#Lower heating value of  No.2 -[Btu/gal]

# Calculations
S1 = H_req*d_N6*S_N6/lhv_N6 ;			# Sulphur emmited when we use fuel NO. 6-[lb]
S2 = H_req*d_N2*S_N2/lhv_N2 ;			# Sulphur emmited when we use fuel NO. 2-[lb]

# Results
print ' Sulphur emmited when we use fuel NO. 6 is %.2f lb.'%S1
print ' Sulphur emmited when we use fuel NO. 2 is %.2f lb.'%S2
print 'Clearly fuel no. 6 should be selected because of its low SO2 emmission.'

 Sulphur emmited when we use fuel NO. 6 is 2.80 lb.
Sulphur emmited when we use fuel NO. 2 is 3.03 lb.
Clearly fuel no. 6 should be selected because of its low SO2 emmission.

In :


In [ ]: