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Balances on Reactive Processes
Shamsuzzoha, PhD
Department of Chemical Engineering, KFUPM
Energy Balances on Reactive Processes
Objectives:
 Explain the concepts of heat of reaction (exothermic & endothermic), Hess’s
Law, heat of formation, heat of combustion
 Explain the concepts of combustion related knowledge; heating value, adiabatic
flame temperature, flammability limit, flash point and flame properties
 Calculate the enthalpy (and internal energy) change due to reaction using Hess’s
and from tabulated data of the standard heat of formation and the standard
heat of combustion.
 Write and solve an energy balance on a chemical reactor using either the heat of
reaction method or the heat of formation method
 Write and solve a reactive-energy system for the heat transfer for specified
inlet / outlet conditions, the outlet temperature for a specified heat input and
product composition for a specified outlet temperature
 Solve energy problems for processes involving solutions for which heats of
solution are significant
 Convert a higher heating value of a fuel to a lower heating value vice versa
Energy Balance involving Chemical Reaction
DHr = -ve (exothermic)
DHr = +ve (endothermic)
Tin, Pin, Hreac
Tin, Pin = Tout, Pout
Hprod  Hreac
Tout, Pout, Hprod
Qin/Qout
2A + B
3C
(heat/enthalpy of reaction) : DĤr = ∑Ĥproduct – ∑Ĥreactant
  ξ ΔH
ˆ   n H
ˆ   n H
ˆ
Q
i
r
i
i
i
i
out
in
9.1
Heat of Reaction (or Enthalpy of Reaction)

ˆ (T, P)
Heat of reaction, ΔH
, is the enthalpy change when stoichiometric
r
quantities of reactants at temperature T and pressure P are consumed
completely to form products at the same temperature and pressure

ˆ o , is the heat of reaction at a
The standard heat of reaction, ΔH
r
specified reference temperature and pressure (e.g. 25oC & 1 atm)
Example 1
CH 4 (g)  2O 2 (g)  CO 2 (g)  2H 2O(l)
ˆ    890.3
ΔH
r
kJ
mol
1 g-mole of gaseous methane and 2 g-moles of gaseous oxygen at 25oC and 1 atm
react completely to form 1 g-mole of gaseous carbon dioxide and 2 g-moles of liquid
water and the products are brought back to 25oC and 1 atm and the net enthalpy
would be – 890.3 kJ. Q = DH , 890.3 KJ of heat would have to be transferred away
from the reactor to keep the products at 25oC.
Example 2
6C(s)  3H 2 (g)  C6 H 6 (l)
kJ

ˆ
ΔH r   48.66
mol
6 g-moles of solid carbon and 3 g-moles of gaseous hydrogen at 25oC and 1 atm
react completely to form 1 g-mole of liquid benzene the product is brought
back to 25oC and 1 atm the specific enthalpy would be + 44.86 kJ. Q = DH ,
44.86 KJ of heat would have to be added into the reactor to keep the product
at 25oC.
Note :
ˆ (T, P) is nearly independent
At low and moderate pressures ΔH
r
ˆ (T)
of pressure. Hence the heat of reaction may be written as ΔH
r
Types of Heat of Reaction
ˆ  (T,P) < 0 => the reaction is exothermic .. Example 1
 If ΔH
r
 less energy is required to break the reactant bonds than is released
when the product bonds form, resulting in a net heat release as the
reaction proceeds (i.e. Enthalpy of Reactants > Enthalpy of Products).
 This energy may be transferred from the reactor or else may serve to
raise the temperature of the reaction mixture
 Recall Example 1;
ˆ
ˆ
ˆ
ΔHr (P, T)  Hprod (P, T)  Hreact (P, T)  - 890.3 kJ/mol
ˆ (T,P) > 0 => the reaction endothermic .. Example 2
 If ΔH
r
 more energy is required to break the reactant bonds than is released
when the product bonds form, resulting in a net absorption of energy as
the reaction proceeds (i.e. Enthalpy of Reactants < Enthalpy of
Products)
 Unless this energy is supplied to the reactor as heat, the temperature
decreases
 Recall Example 2; ΔH
ˆ (P, T)  H
ˆ
ˆ
(P, T)  H
(P, T)   48.66 kJ/mol

r
prod
react
Total enthalpy change due to reaction
 Recall Example 1;
CH 4 (g)  2O 2 (g)  CO 2 (g)  2H 2O(l)
ˆ    890.3
ΔH
r
kJ
mol
The standard (T=25oC, P=1 atm) heat of reaction is - 890.3 kJ per ? mol of
WHAT?
 890.3 kJ
 890.3 kJ
ˆ  
ΔH

r
1 mol of CH 4 reacted
2 mols of O 2 reacted

 890.3 kJ
 890.3 kJ

1 mol of CO 2 generated
2 mols of liquid H 2O generated
and if for example, 2 mols /s of CH4 was reacted, the associated enthalpy
change is
ΔH  890.3
kJ
mols
kJ
x2
of CH 4 reacted  - 1780.6
1 mol of CH 4 reacted
s
s
and if for example, 4 mols /s of O2 was reacted the associated enthalpy
change is
ΔH  890.3
kJ
mols
kJ
x4
of O 2 reacted  - 1780.6
2 mols of O 2 reacted
s
s
and if for example, 2 mols /s of CO2 was generated, the associated enthalpy
change is
ΔH  890.3
kJ
mols
kJ
x2
of CO2 generated  - 1780.6
1 mols of CO2 generated
s
s
and if for example, 4 mols /s of H2O was generated, the associated enthalpy
change is
ΔH  890.3
kJ
mols
kJ
x4
of H 2 O generated  - 1780.6
2 mols of H 2 O generated
s
s
 In general, if vA is the stoichiometric coefficient of a reactant or reaction
product A ( negative if it is a reactant, positive if it is a product) and nA,r moles
of A are consumed / reacted or generated at T=To and P=Po, the associated
enthalpy change is
ˆ T , P 
ΔH
r
o
o
ΔH 
n A,r
νA
 Recall Chapter 4, the extent of reaction, ξ ;
 Then the associated enthalpy change is
ξ
n A,out - n A,in
νA

ˆ (T , P )
ΔH  ξΔH
r
o
o
n A,r
νA
The value of the heat of a reaction
depends on how stoichiometric equation is written
CH 4 (g)  2O 2 (g)  CO 2 (g)  2H 2O(l)
ˆ    890.3
ΔH
r
kJ
mol
1 g-mole of gaseous methane and 2 g-moles of gaseous oxygen at 25oC and 1 atm
react completely to form 1 g-mole of gaseous carbon dioxide and 2 g-moles of liquid
water and the products are brought back to 25oC and 1 atm and the net enthalpy
would be – 890.3 kJ. Q = DH , 890.3 KJ of heat would have to be transferred away
from the reactor to keep the products at 25oC.
2CH 4 (g)  4O 2 (g)  2CO 2 (g)  4H 2O(l)
ˆ    1780.6
ΔH
r
kJ
mol
2g-moles of gaseous methane and 4 g-moles of gaseous oxygen at 25oC and 1 atm
react completely to form 2 g-moles of gaseous carbon dioxide and 4 g-moles of liquid
water and the products are brought back to 25oC and 1 atm and the net enthalpy
would be – 1780.6 kJ. Q = DH , 1780.6 KJ of heat would have to be transferred away
from the reactor to keep the products at 25oC.
The value of the heat of a reaction
depends the states of aggregation (solid, liquid or gas)
of the reactants and products
kJ
mol
kJ
ΔĤ r2   802.3
mol
ΔĤ r1   890.3
CH 4 (g)  2O 2 (g)  CO 2 (g)  2H 2 O(l)
CH 4 (g)  2O 2 (g)  CO 2 (g)  2H 2 O(g)
The only difference between the reactions is that the water formed as a liquid in
the first reaction and as a vapor in the second reaction
hence the difference between the two heats of reaction must be the enthalpy
change associated with the vaporization of 2 mols of water at 25oC
i.e.
kJ
r2 - r1 : - 2H 2 O (g)  2H 2 O (l) ; ΔĤ  -802.3 - (-890.3)  88
mol
kJ
ΔĤ  ΔĤ or2 - ΔĤ or1  2ΔĤ vap H 2 O, 25o C  88
mol


kJ
ΔĤ   2204
mol
kJ

ΔĤ r2   2220
mol

r1
C3 H 8 (l)  5O 2 (g)  3CO 2 (g)  4H 2 O(l)
C3 H 8 (g)  5O 2 (g)  3CO 2 (g)  4H 2 O(l)
The only difference between the reactions is that propane exists as a liquid in the
first reaction and as a vapor in the second reaction
hence the difference between the two heats of reaction must be the enthalpy
change associated with the vaporization of 1 mol of propane at 25oC
i.e.
kJ
r2  r1 ; C3H 8 (l)  C3H 8 (g) ; ΔĤ  - 2204 - (-2220)  16
mol
ΔĤ  ΔĤ - ΔĤ  ΔĤ vap
o
r2
o
r1


kJ
C3H 8 , 25 C  - 2204 - (-2220)  16
mol
o
C 3 H 8 (l)  5O 2 (g)  3CO 2 (g)  4H 2O(l)
C 3 H 8 (l)  5O 2 (g)  3CO 2 (g)  4H 2O(g)
kJ
mol
ˆ    2028 kJ
ΔH
r
mol
ˆ    2204
ΔH
r
The only difference between the reactions is that the water formed as a liquid in
the first reaction and as a vapor in the second reaction
hence the difference between the two heats of reaction must be the enthalpy
change associated with the vaporization of 4 mols of water at 25oC
i.e.


kJ
o
ˆ
ˆ
ΔH  4ΔH vap H 2O, 25 C  - 2028 - (-2204)  176
mol
C 3 H 8 (g)  5O 2 (g)  3CO 2 (g)  4H 2O(l)
C 3 H 8 (g)  5O 2 (g)  3CO 2 (g)  4H 2O(g)
kJ
mol
ˆ    2044 kJ
ΔH
r
mol
ˆ    2220
ΔH
r
The only difference between the reactions is that the water formed as a liquid in
the first reaction and as a vapor in the second reaction
hence the difference between the two heats of reaction must be the enthalpy
change associated with the vaporization of 4 mols of water at 25oC
i.e.


kJ
o
ˆ
ˆ
ΔH  4ΔH vap H 2O, 25 C  - 2044 - (-2220)  176
mol
Working session 1
The standard heat of the combustion reaction of liquid n-Hexane to form
CO2(g) and H2O(l) with all reactants and products at 25oC and 1 atm, is -4163.1
kJ/mol. The heat of vaporization of hexane and water at 25oC is 31.49 kJ/mol
and 44 kJ/mol, respectively.
a.
Use the given data to calculate the standard heat of reaction at 25oC for
the combustion of n-hexane vapor to form CO2(g) and H2O(g).
b.
At what rate in kJ/s is heat absorbed or released (state which) if 54.5
kg/s of O2 is consumed in the combustion of n-hexane vapor , water vapor
is the products, and the reactants and products are all at 25oC.
Working session 1 - Problem 9.3
a.
Use the given data to calculate the standard heat of reaction at 25oC for
the combustion of n-hexane vapor to form CO2(g) and H2O(g).
C6 H 14 (l)  9.5O 2 (g)  6CO 2 (g)  7H 2O(l)
C6 H 14 (g)  9.5O 2 (g)  6CO 2 (g)  7H 2O(g)
ˆ    4163.1
ΔH
r1
ˆ  ?
ΔH
r2
kJ
mol
kJ
mol
(Hypothetical reaction steps)
C6H14 (g), 25oC
O2 (g), 25oC
ˆ  -ΔH
ˆ
DH
1
vap., C H
6
o
ˆo
ΔH
r2
H2 O (g), 25oC,
CO2 (g), 25oC
ˆ  7ΔH
ˆ
DH
3
vap., H O
2
14
C6H14 (l), 25 C
O2 (g), 25oC
ˆ  ΔH
ˆo
DH
2
r1
H2 O (l), 25oC
CO2 (g), 25oC
Working session 1 - Problem 9.3
o
ˆ
ΔH
(25
C)  31.49
vap, C 6 H 6
kJ
mol
o
ˆ
ΔH
vap, H 2O (25 C)  44
ˆ   ΔH
ˆ  ΔH
ˆ  ΔH
ˆ
ΔH
r2
1
2
3
ˆ
ˆ
ˆ
 - ΔH
vap., C6 H 1 4  ΔH r1  7ΔH vap., H 2O

 

kJ
kJ




  - 31.49
  - 4163.1

mol C6 H 14  
mol C6 H 14 


  7 mol H 2O formed 
kJ

 
  44
mol
H
O
mol
C
H
reacted
2
6
14



 - 3886.6
kJ
mol
kJ
mol
Working session 1 - Problem 9.3
b.
If Q=DH, at what rate in kJ/s is heat absorbed or released (state which)
if 54.5 kg/s of O2 is consumed in the combustion of n-hexane vapor , water
vapor is the products, and the reactants and products are all at 77oF.
  ΔH
  ξ ΔH
ˆ (25 o C)
Q
r2
ξ 
n O 2 ,out - n O 2 ,in
ν O2
- 54.5

kg kmol 1000 mol
x
x
s 32 kg
kmol
- 9.5
 179.3
mol
s
   179.3 mol   - 3886.6 kJ   - 696867 kJ
Q
s 
mol 
s

Negative enthalpy indicates
an exothermic process
(i.e. heat is released )
The heat of a reaction at constant volume
 Determined by the change on internal energy (DU) between reactants
and products

ˆ (T)  ΔH
ˆ (T)  Δ PV
ˆ  ΔH
ˆ (T)  RT
ΔU
r
r
r
r
  νi   νi
gaseous
 gaseous
reactants
 products
 





Note: This last relation assumes that the gases behave as ideal gases, and that
the molar volumes of the liquid or solid components are small. If there are no
ˆ (T)  ΔH
ˆ (T) .
gaseous species involved in the reaction, ΔU
r
r
Example :
C6 H14 (g)  9.5O2 (g)  6CO 2 (g)  7H 2O(g)

ˆ (T)  ΔH
ˆ (T)  RT
ΔU
r
r
  νi   νi
gaseous
 gaseous
reactants
 products


ˆ
  ΔH r (T)  RT7  6  9.5 - 1


9.2 Measurement and Calculation of
Heats of Reaction: Hess’s Law
20
9.2 Measurement and Calculation of Heats of Reaction:
Hess’s Law
9.2 Hess’s Law
 Just as the species in a chemical reaction can be treated as
algebraic quantities, so can the heats of reaction (and other
thermodynamic properties). Example:
A + B + E -->D + F



Rxn 1:
Rxn 2:
Rxn 3 :
A + B ---> C + D
C + E ---> F
Rxn 1 + Rxn 2
∆Horxn,1
∆Horxn,2
∆Horxn,3
∆Horxn,3= ∆Horxn,1 + ∆Horxn,2
 HESS’s law :- If the stoichiometric equation for a reaction can be
derived from other reaction equations (by multiplication by constants,
addition and subtraction) then the heat of reaction for the first
reaction can be derived by performing the same algebraic operations on
the heats of reaction for the other reactions.
Example :
kJ
o
ΔH
o  393.5 kJ
r1
ΔH r1  393.5 mol
mol
11
kJ
o
kJ
Rxn2
...
CO

O

CO
ΔH


282.99
o
Rxn2 ... CO  2 O22  CO22
ΔH r2r2  282.99 mol
2
mol
11
kJ
o
kJ
Rxn3
ΔH
Rxn3 ...
... C
C
 2O
O22 
 CO
CO
ΔH r3or3 
 ?? mol
2
mol
__________
__________
__________
__________
__________________________________________________
________________
______
According
According to
to Hess'
Hess' ss law
law ::
11
Rxn3

Rxn1
Rxn2
C

Rxn3  Rxn1 - Rxn2
C 2O
O22 
 CO
CO
2
kJ
o
o
o
o
and
ΔH
ΔH
o  ΔH r1
o - ΔH r2
o
o  110.52 kJ
r3
r3
and ΔH r3  ΔH r1 - ΔH r2
ΔH r3  110.52 mol
mol
__________
__________
__________
__________
__________
________________________________________________________
______
Rxn1
Rxn1 ...
... C
C
O
O22 
 CO
CO22
Example 3 – Problem 9.6
 Formaldehyde may be produced in the reaction between methanol and
oxygen
2CH3OH (l) + O2(g)  2HCHO (g) + 2H2O(l)
DHor = -326.2 kJ/mol
The standard heat of combustion of hydrogen is
H2(g) + ½ O2(g)  H2O (l)
DHor = -285.8 kJ/mol
Use these heats of reaction and Hess’s law to estimate the standard heat
of direct decomposition of methanol to form formaldehyde
CH3OH (l)  HCHO (g) + H2 (g)
Example 3 – Problem 9.6
Rxn 1 : 2CH3OH (l) + O2(g)  2HCHO (g) + 2H2O(l)
Rxn 2 : H2(g) + ½ O2(g)  H2O (l)
o
DH rxn1 = -326.2 kJ/mol
o
DH rxn2 = -285.8 kJ/mol
o
Rxn 3 : CH3OH (l)  HCHO (g) + H2 (g)
DH rxn3 = ? kJ/mol
________________________________________________________________
According to Hess’s law
Rxn 3 = ½ Rxn 1 – Rxn2
o
DH
o
DH
o
o
rxn3
= ½ DH
rxn3
= ½ (-326.2)- (-285.8) = 122.7 kJ/mol
rxn1
- DH
rxn2
Working session 2
a) Calculate DĤ (kJ/mol) for the following reaction using the listed
standard enthalpy of reaction data
R4 : 2N2(g) + 5O2(g)  2N2O5(s)
R1 : N2(g) + 3O2(g) + H2(g)  2HNO3(aq)
DĤo= -414 kJ/mol
R2: N2O5(s) + H2O(l)  2HNO3(aq)
DĤo= -86 kJ/mol
R3 :O2(g) + 2H2(g)  2 H2O(l)
DĤo= -571.6 kJ/mol
R4 = 2R1 – 2R2 – R1
9.3 Heats of Formation
ˆ , for a chemical species is equal to the heat of
 The heat of formation, ΔH
fi
reaction for the reaction forming 1 mol of this species from its “elements” at
standard conditions.
Note:
Elements may be monatomic (e.g., C) or diatomic (e.g., H2),
and the standard state may be any phase (c, l, or g). In
this notation, “c” means crystalline or solid phase.
ˆ o ) of the compound the enthalpy change
 The standard heat of formation ( ΔH
fi
associated with the formation of 1 mole of the compound at a reference
temperature and pressure (usually 25oC and 1 atm)




Do not depend on pressure for low and moderate pressure (i.e. < 10 atm)
Depend on the phase (gas, liquid, solid)
The standard heat of formation an elemental species (e.g. O2, N2, H2) is
zero.
The standard heat of formation of some substances is given in Table B-1
Table B-1 :- Standard Heat of Formation
 An example - heat of formation for liquid propane:
3Cc   4H 2 (g)  C3 H 8 (l)
ˆ o  119.8kJ/mol 25 oC,1atm
ΔH

f

Note: This value is from Table B.1. Values for the heats of formation
for C(c) and H2(g) are listed in Table B.1 as zero, indicating that
these are elements. The heat of formation for propane as a gas
is different (-103.8 kJ/mol).
 Heats of reaction from heats of formation (by Hess’s law):
 
o
o
ˆ
ˆ
ΔHr   ν i ΔHf i 
i
 
o
ˆ
 νi ΔHf i 
products
 
o
ˆ
 νi ΔHf
reactants
i
Calculation of Heat of Reaction from Heats
of Formation
 An example - consider the reaction
CH 4  2O2  CO 2  2H2O(v)
ˆ

Write the formula for ΔH r in terms of the standard heats of
formation of the reactants and products

ˆ   ΔH
ˆ
ΔH
r2
f
C(s)  O 2 (g)  CO 2 (g)
H 2 (g) 

1
O 2 (g)  H 2O(g)
2
 
ˆ   2 ΔH
ˆ
ΔH
r
f


CH 4 (g)
CO 2 (g)
ˆ   ΔH
ˆ
ΔH
r3
f
 
ˆ

Δ
H
f
H 2O(g)

ˆ   ΔH
ˆ
ΔH
r1
f
C(s)  2H 2 (g)  CH 4 (g)
CO 2 (g)

H 2O(g)
 ΔHˆ  

f CH4 (g)
Example 4
Calculate the standard heat of reaction for methane using the heats of
formation if the water is formed as a liquid:
CH 4 (g)  2O2 (g)  CO 2 (g)  2H2O(l)
 
ˆ   2 ΔH
ˆ
ΔH
r
f
 
ˆ

2
Δ
H
f
H 2O(l)
CO2 (g)
 ΔHˆ 

f CH4 (g)
  
ˆ
 2 ΔH
f
O 2 (g)
Read the standard heat of formation of respective species from Table B.1.
 
kJ  
kJ  
kJ   kJ 
ΔĤ r  2 - 285.84
   - 393.5
   - 74.85
  2 0

mol  
mol  
mol   mol 
 
kJ
ΔĤ r  890.33
mol
Example 5
 Calculate the standard heat of reaction for methane the heats of
formation if the water is formed as a vapor:
CH 4 (g)  2O2 (g)  CO2 (g)  2H2O(g)
 
ΔĤ r  2 ΔĤ f
   ΔĤ 


Δ
Ĥ
f
H 2 O(g)
CO 2 (g)

f CH 4 (g)
  
 2 ΔĤ f
O 2 (g)
Read the standard heat of formation of respective species from Table
B.1.
ˆ   2 - 241.83 kJ    - 393.5 kJ
ΔH
r

mol  
mol
 
ˆ   802.31 kJ
ΔH
r
mol
kJ   kJ
 
   - 74.85
  2 0
mol   mol
 



Rxn 1 : CH 4 (g)  2O 2 (g)  CO 2 (g)  2H 2O(l)
Rxn 2 : CH 4 (g)  2O 2 (g)  CO 2 (g)  2H 2O(g)
 Rxn1  Rxn2  2H 2O(l)  2H 2O(g)
ΔHˆ 
ΔHˆ 
vap

ˆo
 ΔH
r
o
r Rxn 2



ˆo

Δ
H
r
Rxn 2
ˆo
 ΔH
r

Rxn1
ˆ
where, ΔH
vap  44

Rxn1
 88
ΔHˆ 
kJ
mol
kJ
ˆo
ΔH


802.3
r Rxn 2
mol
ˆ
ΔH
vap  ?
o
r Rxn1
 890.3
 
 
kJ
mol
ˆ
 2ΔH
vap
kJ
is the heat of vaporizati on of water at 25o C and 1 atm
mol
Working session 3 - Problem 9.8
 Trichloroethylene, C2HCl3 , may be produced in a two-step reaction sequence as follows;
C2H4 (g) + 2Cl2(g)  C2H2Cl4 (l) + H2(g)
o
DH r = -387.56 kJ/mol
C2H2Cl4(l)  C2HCl3(l) + HCl (g)
The standard heat of formation of liquid trichloroethylene is -276.2 kJ/mol
a)
b)
Calculate the standard heat of formation of tetrachloroethane, C2H2Cl4 (l)
and the standard heat of the second reaction
Use Hess’s law to calculate the standard heat of reaction of
C2H4 (g) + 2Cl2(g)  C2HCl3 (l) + H2(g) + HCl(g)
c)
If 300 mol/h of C2HCl3 (l) is produced in the reaction of part (b) and the
reactants and products are all at 25oC and 1 atm, how much heat is evolved or
absorbed in the process?
Working session 3 - Problem 9.8
a)
Calculate the standard heat of formation of tetrachloroethane, C2H2Cl4 (l)
o
C2H4 (g) + 2Cl2(g)  C2H2Cl4 (l) + H2(g)
o
o
o
DH r = (DH f) H2(g) + (DH f)
C2H2Cl4(l)
DH r = -385.76 kJ/mol
o
o
- (DH f) C2H4(g) - 2(DH f)
Cl2(g)
read the standard heat of formation of ethylene, chlorine and hydrogen
from Table B-1
o
-385.76 = [0 + (DH f) C2H2Cl4(l) ] - [(52.28 - 2(0)]
o
(DH f) C2H2Cl4(l) = - 333.48 kJ/mol
Working session 3 - Problem 9.8
Calculate the standard heat of the second reaction
C2H2Cl4(l)  C2HCl3(l) + HCl (g)
o
o
DH r = (DH f)
HCl(g)
o
o
+ (DH f) C2HCl3(l) - (DH f) C2H4Cl4(l)
Read the standard heat of formation of hydrochloric acid from Table B-1 & the
standard heat of formation of liquid trichloroethylene and tetrachloroethane are 276.2 kJ/mol and -333.48 kJ/mol, respectively.
o
DH r = -92.31 - 276.2 – (- 333.48) = - 35.03 kJ/mol
Working session 3 - Problem 9.8
b)
Use Hess’s law to calculate the standard heat of reaction of
C2H4 (g) + 2Cl2(g)  C2HCl3 (l) + H2(g) + HCl(g)
According to Hess’s law
o
Rxn 1 : C2H4 (g) + 2Cl2(g)  C2H2Cl4 (l) + H2(g)
DH
Rxn 2 : C2H2Cl4(l)  C2HCl3(l) + HCl (g)
DH
o
rxn1
= -385.76 kJ/mol
rxn2
= -35.08 kJ/mol
o
Rxn 3 : C2H4 (g) + 2Cl2(g)  C2HCl3 (l) + H2(g) + HCl(g)
DH rxn3 = ? kJ/mol
_________________________________________________________________
Rxn 3 = Rxn 1 + Rxn 2
o
DH
rxn3
= - 35.03 - 385.76 = - 420.76 kJ/mol
Working session 3 - Problem 9.8
c)
If 300 mol/h of C2HCl3 (l) is produced in the reaction of part (b) and the
reactants and products are all at 25oC and 1 atm, how much heat is evolved or
absorbed in the process?
C2H4 (g) + 2Cl2(g)  C2HCl3 (l) + H2(g) + HCl(g)
o
DH r = -420.76 kJ/mol
ˆo
Δ
H
r
  ΔH
 
Q
n A,r
νA
- 420.76

kJ
mol C 2 HCl 3 formed
mol C 2 HCl 3 formed
hr
kJ
x 300
x
 - 35.06
1
hr
3600 s
s
9.4 Heats of Combustion
 Combustion involves a chemical reaction between combustible species and oxygen
(usually from air) to produce heat and combustion products (mainly water and
carbon dioxide plus some other by-products such as carbon monoxide, oxides of
nitrogen / sulphur , unburned fuels and radicals)
 The standard heat of combustion, DH°c, is the heat released when one mole of a
substance reacts with oxygen to yield specified products in a specified phase:


CO2 is always gas phase.
H2O can be liquid or vapor with the reactants and products at 25°C, 1 atm.
The standard heat of combustion of some substances is given in Table B-1
Important note : Please read carefully the footnotes in page 629
regarding the use of these and other values provided by the
table )
Table B-1:- Standard Heat of Combustion
Foot Note (pg. 629): standard states of products CO2(g), H2O(l) SO2(g), HCl(aq) & N2(g)
Higher and Lower Heats of Combustion


Heating value (HV) or calorific value (CV) is other common terms used to describe
the heat of combustion; However, othey are expressed as a negative value of
standard heat of combustion (-DH c) …..hence HV or CV is always positive
Often two heats of combustion are listed for hydrogen containing species: lower and
higher.

Lower heating or calorific value (LHV or LCV ): product water is in the gas phase.

Higher heating or calorific value (HHV or HCV): product water is in the liquid
phase.

These two metrics differ by the molar heat of vaporization of water times the
moles of water produced.
LHV (or LCV) = HHV (or HCV) - nwDHvap
…. where nw = mole of water formed / mole of fuel burned
DHvap = 44 kJ/mol ( at T = 25oC and P = 1 atm)

Lower heating value is usually more relevant for engineering purposes because
water leaves practical combustion devices as a vapor.
Heat of Reactions from Heat of Combustion


The standard heat of combustion in Table B-1 (Please take note of
the assumptions applied)
The standard heats of reactions that involve only combustible
substances and combustion products can be calculated from tabulated
standard heats of combustion, in another application of Hess’s law
   ν ΔHˆ    ν ΔHˆ 
ˆ o   ν ΔH
ˆo 
ΔH
r
i
c i
i
i
reactants
o
c i
i
o
c i
products
If any of reactants or products are themselves combustion products
ˆ o terms in the above equation should
(CO2, H2O(l), SO2(g)…) their ΔH
c
be set equal to zero
Calculation of Heat of Reaction from Heats
of Combustion
 An example - consider the reaction
C2 H 2 (g)  2H 2 (g)  C2 H 6 (g)
ˆ  in terms of the standard heats of
Write the formula for ΔH
r
combustion of the reactants and products
C 2 H 2 (g) 
5
O 2 (g)  2CO 2 (g)  H 2O(g)
2
1
O 2 (g)  H 2O(g)
2
7
C 2 H 6 (g)  O 2 (g)  2CO 2 (g)  3H 2O(g)
2
 
H 2(g)




ˆ   ΔH
ˆ
ΔH
r3
c
   ΔHˆ 
 2 ΔHˆ c
C2 H 2(g)

ˆ   ΔH
ˆ
ΔH
r2
c
H 2 (g) 
ΔHˆ r  ΔHˆ c

ˆ   ΔH
ˆ
ΔH
r1
c

c C2 H 6(g)
kJ
C 2 H 2 (g)
mol
kJ


285.84
H 2O(g)
mol
kJ


1559.9
C 2 H 6 (g)
mol
 1299.6
kJ
  311.38 mol
Calculation of Heat of Reaction from Heat
Formation
C2 H 2 (g)  2H 2 (g)  C2 H 6 (g)


 - 84.67


0
2C(s)  3H 2 (g)  C 2 H 6 (g)
ˆ   ΔH
ˆ
ΔH
r,1
f
H 2 (g)
ˆ   ΔH
ˆ
ΔH
r,2
f

ˆ   ΔH
ˆ
ΔH
r,3
f
2C(s)  H 2 (g)  C 2 H 2 (g)

ΔHˆ r  ΔHˆ f

C2 H 6( g)
 ΔHˆ 

f C H ( g)
2 2

 2 ΔHˆ f

g
g
g
kJ
mol
kJ
mol
  226.75
kJ
mol
kJ
   311.42 mol
H 2( g)
Working session 4
 Calculate the standard heat of the acetylene hydrogenation reaction
DHor = ? kJ/mol
C2H2 (g) + 2H2(g)  C2H6 (g)
a)
using the standard heat of combustion (Table B-1)

ˆ o  ΔH
ˆo
ΔH
r
c

C2 H 2

ˆo
 2 ΔH
c

H2

ˆo
 ΔH
c

C2H 6
 - 1299.6  2(-285.84) - (-1559.9)
kJ
 - 311.38
mol
Working session 4
b)
using the standard heat of formation (Table B-1)

ˆ o  ΔH
ˆo
ΔH
r
f

C2 H 6

ˆo
 ΔH
f

C2H 2
 - 84.67  226.75 - 2(0)
kJ
 - 311.42
mol

ˆo
 2 ΔH
f

H2
Working session 4
c) Standard Heat of Combustion + Hess’s law
Rxn 1 : C2H2 (g) + 2.5O2(g)  2CO2 (g) + H2O (l)
DHoc = -1299.6 kJ/mol
Rxn 2 : H2(g) + 0.5O2(g)  H2O (l)
DHoc = -285.84 kJ/mol
Rnx 3 : C2H6 (g) + 3.5O2(g)  2CO2 (g) + 3H2O (l)
DHoc = -1559.9 kJ/mol
_________________________________________________________________
Rxn 4 : C2H2 (g) + 2H2(g)  C2H6 (g)
DHoc = ? kJ/mol
Rxn 4 = Rxn 1 + 2Rxn2 – Rxn3
o
DH c = -1299.6 + 2(-285.84) – (-1559.9) kJ/mol = -311.38 kJ/mol
Calculate the standard heat of the acetylene hydrogenation reaction using the standard
heat of formation and Hess’s law
Working session 4
c) Standard Heat of Formation + Hess’s law
Rxn 1 : 2C(s) + H2 (g)  C2H2(g)
DHof = +226.75 kJ/mol
Rxn 2 : H2(g)
DHof = 0 kJ/mol
Rnx 3 : 2C2(g) + 3H2(g)  C2H6 (g)
DHof = -84.67 kJ/mol
_________________________________________________________________
Rxn 4 : C2H2 (g) + 2H2(g)  C2H6 (g)
DHof = ? kJ/mol
Rxn 4 = Rxn 3 - Rxn2 – Rxn1
o
DH f = (-84.67 + 0 – 226.75) kJ/mol = -311.42 kJ/mol