# Chapter 12 : Energy Balance Thermochemistry¶

## Example 12.1 pageno : 475¶

In [1]:
# variables
N = 100.                            #mol gas mixture burned

#CO(g) + 1/2 O2(g) = CO2 -                    Hr1 = - 282.91kJ/mol
#H2(g) + 1/2 O2(g) = H2O -                    Hr2 = - 241.83kJ/mol
Hr1 = - 282.91                      #kJ/mol
Hr2 = - 241.83                      #kJ/mol
Nco1 = 20.
Nh21 = 30.
Nn21 = 50.

# Calculation
Htotal = Nco1*Hr1 + Nh21*Hr2

# Result
print "the amount of heat liberated on the complete combustion of 100mol of the gas mixture = ",-Htotal,"kJ"
Ncoreac = Nco1 * 0.9
Nh2reac = Nh21 * 0.8
Htotal1 = Ncoreac*Hr1 + Nh2reac*Hr2
print "the amount of heat liberated if only 90% of CO and 80% of H2 react of 100mol of the gas mixture = ",-Htotal1,"kJ"

the amount of heat liberated on the complete combustion of 100mol of the gas mixture =  12913.1 kJ
the amount of heat liberated if only 90% of CO and 80% of H2 react of 100mol of the gas mixture =  10896.3 kJ


## Example 12.2 pageno : 477¶

In [2]:
# variables
#C(s) + 2H2(g) = CH4(g)          Hf = ?
Hc = -393.51                    #kJ/mol
Hh2 = -285.84                   #kJ/mol
Hch4 = - 890.4                  #kJ/mol

# Calculation
#heat of reaction can be calculated from the heat of combustion data
# using following equation, the heat of reaction is the sum of the heat of combustion of all the reactants in the desired reaction minus the sum of the heat of combustion of all the products of the desired reaction. Here the reactants are one mole of Carbon and two moles hydrogen, and the product is one mole of methane,there heat of reaction is
Hf = 1 * Hc + 2 * Hh2 - 1 * Hch4

# Result
print "Heat of formation of methane = ",Hf,"kJ"

Heat of formation of methane =  -74.79 kJ


## Example 12.3 pageno : 478¶

In [3]:
# variables
m = 1.                      #kg of coal burned
xc = 0.7
xh2 = 0.055
xn2 = 0.015
xs = 0.03
xo = 0.13
xash = 0.07
Hvap = 2370.                #kJ/kg
C = 29000.                  #kJ/kg

# Calculation
Nh2 = xh2 * m / 2.016
Nwater = Nh2                # ( amount of water formed )
mwater = Nwater * 18.016
Hreq = mwater * Hvap
Hnet = C - Hreq

# Result
print "Net heating value of coal = %.1f"%Hnet,"kJ/kg"

Net heating value of coal = 27835.1 kJ/kg


## Example 12.4 pageno : 479¶

In [4]:
# variables
#C2H5OH(l) + CH3COOH(l) = C2H5COOCH3(l) + H2O(l) H = ?
Hc2h5oh = -1366.91                      #kJ/mol
Hch3cooh = -871.69                      #kJ/mol
Hc2h5cooch3 = -2274.48                  #kJ/mol

# Calculation
#to calculate heat of reaction from the heat of combustion data ,
#Hreac = Hreac - Hprod
Hreac = Hc2h5oh + Hch3cooh - Hc2h5cooch3

# Result
print "Heat of reaction for the esterification of ethyl alcohol with acetic acid = ",Hreac,"kJ"

Heat of reaction for the esterification of ethyl alcohol with acetic acid =  35.88 kJ


## Example 12.5 pageno : 479¶

In [5]:
# variables
#C2H4(g) + H2O(g) = C2H5OH(g)
#2CO2(g) + 3H2O(l) = C2H5OH(l) + 3O2(g)     H = 1366.91kJ  (A)
Hc2h4 = -1410.99                    #kJ/mol
Hvap = 44.04                        #kJ/mol
Hc2h5oh = 42.37                     #kJ/mol
#C2H4(g) + 3H2O(l) = C2H5OH(l) + 3O2(g)     H = -1410.99kJ (B)
#H2O(l) = H2O(g)                            H = 44.04kJ    (C)
#C2H5OH(l) = C2H5OH(g)                      H = 42.37kJ    (D)
#A + B + D - C gives the required reaction
Ha = 1366.91                        #kJ
Hb = -1410.99                       #kJ
Hc = 44.04                          #kJ
Hd = 42.37                          #kJ

# Calculation
Hreac = Ha + Hb + Hd - Hc

# Result
print "The heat of reaction = ",Hreac,"kJ"

The heat of reaction =  -45.75 kJ


## Example 12.6 pageno : 480¶

In [6]:
# variables
#C2H5(g) + 5/2O2(g) = 2CO2(g) + H2O(l)             H1 = -1299.6kJ    (A)
#C(s) + O2(g) = CO2(g)                             H2 = -393.51kJ    (B)
#H2(g) + 1/2O2(g) = H2O(l)                         H3 = -285.84kJ    (C)
#2C(s) + H2(g) = C2H2(g)                           H = ?
H1 = -1299.6                        #kJ
H2 = -393.51                        #kJ
H3 = -285.84                        #kJ

# Calculation
Hreac = 2 * H2 + H3 - H1

# Result
print "Heat of formation of acetylene = ",Hreac,"kJ"

Heat of formation of acetylene =  226.74 kJ


## Example 12.7 pageno : 480¶

In [7]:
# variables
m = 100.                    #kg of pyrites charged
xfes2in = 0.8
xganguein = 0.2
xfes2out = 0.05

# Calculation
#let x be the FeS2 in the feed, then, Fe2O3 = (80 - x)*159.69 / (119.98*2)
#and gangue = 20, total = 73.24 + 0.3345, be FeS2 is only 5 % in the product, hence
x = 0.05 * 73.24 / (1 - 0.05*0.3345)
mfes2reacted = m*xfes2in - x
#4FeS2 + 11O2 = 2Fe2O3 + 8SO2
Hfes2 = -178.02             #kJ/mol
Hfe2o3 = -822.71            #kJ/mol
Hso2 = -296.9               #kJ/mol
Hreac = 2 * Hfe2o3 + 8 * Hso2 - 4 * Hfes2
N = mfes2reacted *1000/ 119.98
H = Hreac * N / 4
H1 = H/m                    #(heat of reaction per kg of coal burnt)

# Result
print "Heat of reaction per 1 kg of coal burned = %.3e"%H1,"kJ"

Heat of reaction per 1 kg of coal burned = -5.258e+03 kJ


## Example 12.8 pageno : 481¶

In [8]:
# variables
#CH3OH(l) + 3/2O2(g) = CO2(g) + 2H2O(l)         H = -726.55kJ
H1 = -726.55                    #kJ
Hco2 = -393.51                  #kJ/mol
Hh2o = -285.84                  #kJ/mol

# Calculation
Hch3oh = Hco2 + 2 * Hh2o - H1

# Result
print "Heat of formation of liquid methanol = ",Hch3oh,"kJ"

Heat of formation of liquid methanol =  -238.64 kJ


## Example 12.9 pageno : 482¶

In [6]:
# variables
N = 100.                        #mol fuel gas
Nco = 21.
Nh2 = 15.6
Nco2 = 9.0
Nch4 = 2.
Nc2h4 = 0.4
Nn2 = 52.
Hco = 282.99                    #kJ/mol ( heat of combustion )
Hh2 = 285.84                    #kJ/mol ( heat of combustion )
Hch4 = 890.4                    #kJ/mol ( heat of combustion )
Hc2h4 = 1410.99                 #kJ/mol ( heat of combustion )
Hvap = 44.04                    #kJ/mol

# Calculation
H = round(Nco * Hco + Nh2 * Hh2 + Nch4*Hch4 + Nc2h4*Hc2h4)         #kJ
V = round(N * 22.4143/1000,2)
H1 = round(H / V)                      #kJ/m**3

#on combustion, 1 mol hydrogen gives 1 mol of water, 1 mol of
# methane gives 2 mol of water and 1 mol of ethylene gives 2 moles of water
Nwater = Nh2 + 2 * Nch4 + 2 * Nc2h4
Hvap1 = round(Hvap * Nwater,2)
Hnet = H1 - Hvap1

# Result
print "Net heating value of the fuel = %.2f"%Hnet,"kJ"

Net heating value of the fuel = 4792.58 kJ


## Example 12.10 pageno : 483¶

In [10]:
# variables
# C5H12(g) + 8O2(g) = 5CO2(g) + 6H20(l)
Hfco2 = -393.51                         #kJ
Hfh2o = - 241.826                       #kJ
Hfc5h12 = -146.4                        #kJ
Hvap = 43.967                           #kJ/mol

# Calculation
H1 = 6*Hfh2o +5*Hfco2 - Hfc5h12
H2 = 6 * (-Hvap)
Hreac = H1 + H2
# Result
print "heat of reaction = ",Hreac,"kJ"

heat of reaction =  -3535.908 kJ


## Example 12.11 pageno : 484¶

In [11]:
# variables
m = 1.                          #kg of oil burned
mc = 0.9                        #kg
mh2 = 0.1                       #kg

# Calculation
Mc = mc / 12                    #kmol
#C(s) + O2(g) = CO2(g)
Nh2 = mh2 / 2.016               #kmol

#change in the no. of gaseous components accompanying the
#combustion of 1 mole of hydrogen in liquid state is -1/2 mol, therefore for Nh2 mol
R = 8.314
T = 298.                        #K
x = Nh2 * R * T / (-2)
Qv = -43000                     #kJ/kg
Qp = Qv + x

# Result
print "the constant pressure heat of combustion = %.1f"%Qp,"kJ/kg"

the constant pressure heat of combustion = -43061.4 kJ/kg


## Example 12.12 pageno : 487¶

In [1]:
# variables
#1 - N2, 2 - H2, 3 - NH3
a1 = 27.31
a2 = 29.09
a3 = 25.48
b1 = 5.2335*10**-3
b2 = -8.374*10**-4
b3 = 36.89 * 10**-3
c1 = -4.1868 * 10**-9
c2 = 2.0139*10**-6
c3 = -6.305*10**-6
H1 = -46191.                    #J
T1 = 298.                       #K

# Calculation
#1/2 N2 + 3/2 H2 = NH3               H = -46.191kJ
#Ht = H + a*T + b*T**2 / 2+ c*T**3 / 3
#at 298,
a = a3 - a1 / 2 - 3 * a2 / 2
b = b3 - b1 / 2 - 3 * b2 / 2
c = c3 - c1 / 2 - 3 * c2 / 2
H = H1 -a * T1 - b * (T1**2) / 2 - c * (T1**3) / 3
T2 = 700.                       #K
H2 = H + a * T2 + b * (T2**2) / 2 + c * (T2**3) / 3

# Result

print "Heat of reaction at 700K = %d"%H2,"kJ"

Heat of reaction at 700K = -52835 kJ


## Example 12.13 pageno : 488¶

In [13]:
#CO(g) + 2H2(g) = CH3OH(g)
import math

# Variables
T1 = 298.                   #K
T2 = 1073.                  #K
#Cp(CH3OH) = 18.382 + 101.564 * 10**-3 * T - 28.683 * 10**-6 * T**2
#Cp(CO) = 28.068 + 4.631 * 10**-3 * T - 2.5773 * 10**4 * T**-2
#Cp(H2)  = 27.012 + 3.509 * 10**-3 * T + 6.9006 * 10**4 * T**-2
#for reactants,

# Calculation
def f1(T):
return 28.068 + 4.631 * 10**-3 * T - 2.5773 * 10**4 * T**-2

def f2(T):
return 27.012 + 3.509 * 10**-3 * T + 6.9006 * 10**4 * T**-2

#for product,

def f2(T):
return 18.382 + 101.564 * 10**-3 * T - 28.683 * 10**-6 * T**2

#H298 = Hproducts - Hreactants
#CO + 2H2 = CH3OH            Ha1 = -238.64kJ
Ha1 = -238.64               #kJ

#CH3OH(l) = CH3OH(g)         Hvap = 37.98kJ
Hvap = 37.98                #kJ

#CO(g) + 2H2(g) = CH3OH(g)   Ha2 = -200.66kJ
Ha2 = Ha1 + Hvap            #kJ
Hco = -110.6                #kJ/mol
H298 = Ha2 - (Hco)
Htotal = H1/1000 + H298 + H2/1000

# Result
print "The heat of reaction at 773K = %.3f"%Htotal,"kJ/mol"

The heat of reaction at 773K = -103.497 kJ/mol


## Example 12.14 pageno : 489¶

In [14]:
# variables
Nco = 1.                            #mol CO reacted
#CO + 1/2 O2 = CO2
No2 = Nco / 2
Pexcess = 100.

# Calculation
Nosupp = No2 * ( 1 + Pexcess / 100 ) #oxygen supplied
Nn2 = Nosupp * 79. / 21
Nco2 = Nco
Noremain = Nosupp - No2
T1 = 298.                           #K
T2 = 400.                           #K
Hr1 = -282.99                       #kJ
T3 = 600.                           #K
SHco = 29.1                         #J/molK
SHo2 = 29.7                         #J/molK
SHn2 = 29.10                        #J/molK
SHco2 = 41.45                       #J/molK
H1 = (Nosupp * SHo2 + Nn2 * SHn2 + Nco * SHco) * (T1 - T2) #enthalpy of cooling of reactants from 298 to 400 K
H2 = (Nco2 * SHco2 + Nn2 * SHn2 + Noremain * SHo2) * (T3 - T1) #enthalpy of heating the products from 298K to 600K
H = H1/1000 + Hr1 + H2/1000

# Result
print "Heat change at 600K = %.1f"%H,"kJ"

Heat change at 600K = -250.1 kJ


## Example 12.15 pageno : 490¶

In [15]:
# variables
#CO(g) + H2O(g) = CO2(g) + H2(g)            H298 = -41.190
T1 = 298.                       #K
Pconv = 75.                     #%
T2 = 800.                       #K
H298 = -41.190                  #kJ
Hco = 30.35                     #J/molK
Hco2 = 45.64                    #J/molK
Hwater = 36.                    #J/molK
Hh2 = 29.3                      #J/molK
Nco = 1.                        #mol
Nh2o = 1.                       #mol

# Calculation
Ncofinal = Nco * (1 - Pconv/100)
Nwaterf = Ncofinal
Nco2final = Nco - Ncofinal
Nh2final = Nco2final
H2 = (Nco2final * Hco2 + Nh2final * Hh2 + Ncofinal * Hco + Nwaterf * Hwater) * (T2 - T1)
Hr1 = H298 * (Nco - Ncofinal)
Hr2 = Hr1 * 1000 + H2
mh2 = Nh2final * 2.016 * 10**-3 #kg
#therfore for 1000k H2,
Hr = Hr2 * 1000 / (mh2 * 1000)  #kJ

# Result
print "Amount of heat change for 1000kg of hydrogen produced = %.3e"%Hr,"kJ"

Amount of heat change for 1000kg of hydrogen produced = 3.736e+06 kJ


## Example 12.16 pageno : 490¶

In [16]:
# variables
#CO2(g) + C(s) = 2CO(g)                H1298 = 170kJ/mol
#O2(g) + 2C(s) = 2CO(g)                H2298 = -221.2kJ/mol
T2 = 1298.                      #K
T1 = 298.                       #K
Hc = 0.02                       #kJ/molK
Ho = 0.03                       #kJ/molK
Hco = 0.03                      #kJ/molK
Hco2 = 0.05                     #kJ/molK

# Calculation
#let the flue gas contain x mol CO2 per mole of oxygen, product contains 2(1+x)mol CO. Nitrogen in reactant and product remain the same
#enthalpy of cooling xmol CO2, 1 mol O2 and 2 + xmol carbon from 1298 to 298K is given as, H1 = (Hco2 * x + Ho * 1 + Hc * (2 + x)) * (298 - 1298)
#H1 = (-70x - 70)kJ
#enthalpy of heating the product, H2 = 2 * ( 1 + x )* Hco * (1298 - 298)
#H2 = 60 + 60x kJ
#Hr = 170x - 221.2
#Htotal = 0 = H1 + H2 + Hr
x = (221.2 + 70 - 60)/(170 + 60 - 70)

# Result
print "moles of CO2 present per mol of oxygen in feed stream = ",x,"mol"

moles of CO2 present per mol of oxygen in feed stream =  1.445 mol


## Example 12.17 pageno : 491¶

In [2]:
# variables
N = 100.                        #mol flue gas
#Carbon balance,
#x is the feed of methane, w is water in flue ga, y is the oxygen supplied
xco2 = 0.019
xch2o = 0.117
xo2 = 0.038
xch4 = 0.826

# Calculation
xc = xco2 + xch2o + xch4
Nc = xc * N
Nch4i = Nc
#Hydrogen balance,
xh2 = xch2o + xch4*2
w = 2 * (Nch4i) - xh2*N
#oxygen balance
No2s = (xco2 + xch2o/2 + xo2)*N + w/2
y = No2s
T1 = 298.                       #K
T2 = 573.                       #K
T3 = 673.                       #K

#oxygen cooled from 573K and methane from 673 to 298K
Ho573 = 30.5                    #J/molK
Hch4673 = 45.9                  #J/molK
H1 = y * Ho573 * (T1 - T2) + Nch4i * Hch4673 * (T1 - T3)
#CH4 + O2 = CH2O + H2O        Hr1 = -282.926kJ
#CH4 + 2O2 = CO2 + 2H2O       Hr2 = -802.372kJ
Hr1 = -282.926 #kJ
Hr2 = -802.372 #kJ
H2 = xch2o*N*Hr1 + xco2*N*Hr2
T4 = 873.                       #K
Ho = 31.9
Hch4 = 51.4
Hco2 = 46.3
Hch2o = 47.1
Hh2o = 36.3
H3 = ((xco2 * Hco2 + xo2 * Ho + xch4 * Hch4 + Hch2o*xch2o)*N + w * Hh2o)*(T4 - T1)
Htotal = H1/1000 + H2 + H3/1000
Nch2o = xch2o * N
mch2o = Nch2o * 30.016/1000     #kg

#for 1000 kg of formaldehyde produced,
H = Htotal * 1000 / mch2o
# Result
print "The amount of heat to be removed per 1000kg of formaldehyde produced = %.1e"%H,"kJ"

# Note: answers are slightly different because of rounding error.

The amount of heat to be removed per 1000kg of formaldehyde produced = -9.8e+06 kJ


## Example 12.18 pageno : 494¶

In [18]:
# variables
Nn2 = 1.            #kmol/s ( basis - feed conisting of 1 kmol of N2  and 3 kmol of H2 )
Nh2 = 3.            #kmol/s

#let x be the fraction converted
T1 = 700.           #K
Hr1 = -94.2         #kJ/mol

# Calculation
#heat liberated = Hr1 * x
#Product consists of 2x kmol NH3, (1-x)kmol N2, and 3(1-x)kmol Hydrogen
T2 = 800.           #K
Hn2 = 0.03          #kJ/molK
Hh2 = 0.0289        #kJ/molK
Hnh3 = 0.0492       #kJ/molK

#H2 = (1-x)*0.03*10**3 * 100 + 3*(1-x)*0.0289*1000*100 + 2*x*0.0492*1000*100
#H2 = 11.67*1000 - 1.83*10**3*x kJ
#reaction is adiabatic, hence, H1 = H2
#solving this we get,
x = 0.1215
Convmax = x * 100

# Result
print "The maximum conversion for nitrogen should be ",Convmax,"%"

The maximum conversion for nitrogen should be  12.15 %


## Example 12.19 pageno : 494¶

In [19]:
# variables
Nco = 1.                        #mol CO

# CO + 1/2 O2 = CO2
O2r = 1.                        #mol
N2r = 3.76                      #mol
COr = 1.                        #mol
O2p = 0.5                       #mol
N2p = 3.76                      #mol
CO2p = 1.                       #mol
Hco = 29.23                     #J/molK
Ho2 = 34.83                     #J/molK
Hn2 = 33.03                     #J/molK
Hco2 = 53.59                    #J/molK
Hcomb1 = -282.99                #kJ/mol
T1 = 298.                       #K
T2 = 373.                       #K

# Calculation
H1 = (O2r * Ho2 + N2r * Hn2 + COr * Hco) * (T1 - T2)
#For product at temp T, H2 = (O2p * Ho2 + N2p * Hn2 + CO2p * Hco2) * (T - T1)
#For adiabatic condition, -(H1 + Hcomb1) = H2
T = -(H1 + Hcomb1 * 1000) / (O2p * Ho2 + N2p * Hn2 + CO2p * Hco2) + T1

# Result
print "Theoretical flame temperature = %.1f"%T,"K"

Theoretical flame temperature = 1820.1 K


## Example 12.20 pageno : 495¶

In [20]:
# variables
N = 1.                                  #kmol hydrogen burned
No = N/2

# Calculation
Nosupplied = 2 * No
Nair = Nosupplied * 100 / 21.
Nn2 = Nair - Nosupplied
#Reactants, H2 = 1kmol, Air = 4.762 kmol
#Product, Water vapour = 1kmol, Oxygen = 0.5kmol, N2 = 3.762kmol
#Cp(water) = 30.475 + (9.652*10**-3)*T + 1.189 * 10**-6 * T**2
#Cp(nitrogen) = 27.034 + 5.815 * 10**-3 *T - 0.2889 * 10**-6 * T**2
#Cp(oxygen) = 25.611 + 13.260 * 10**-3 * T - 4.2077 * 10**-6 * T**2
#H2 = integration(298 to T of (1 * Cp(water) + 0.5 * Cp(oxygen) + 3.762 * Cp(nitrogen)))
#therefore, H2 = 140.34 * T + 31.222 * 10**-3 * T**2 - 4.928 * 10**-6 * T**2 - 44463.54 kJ
H298 = -241.826 * 10**3                 #kJ

#H2 = -H1 - H298
#H1 = 0
#therefore using equation H2, the value of T is obtained to be
T = 1609.8                              #K

# Result
print "Temperature of the reaction products = ",T,"K"

Temperature of the reaction products =  1609.8 K
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