4.

2 Heat effects for chemical reactions and industrial

applications

Introduction
Chemical reactions are accompanied by the transfer of heat or by
temperature changes during the reaction or by both. These heat
effects occurring in chemical reactions are manifestation of the
differences in molecular structure and energy of the products and
reactants. In this topic, a brief discussion on standard heat of
reaction, standard heat of formation and standard heat of combustion
serve a review before the topic on temperature dependence of heat
of reaction and its application in industrial reaction are discussed.

Topic learning objectives

At the end of the topic, the students should be able to

1)calculate the standard heat of formation, combustion and reaction
2)solve problems on heat effects involving industrial reactions.

Lesson Proper
In the flow-calorimetric determination of heat of combustion of fuel
gases, the fuel is mixed with air at ambient temperature and mixture
flows into the combustion chamber. The combustion products enter a
water jacketed section where they are cooled to the temperature of
the reactant. There is no shaft work and the calorimeter is designed so
that changes in potential and kinetic energy are negligible. The
energy balance equation becomes,
Q=∆H
Where Q is the energy absorbed by the water and ∆H is enthalpy of
reaction or heat of reaction.
 Consider the reaction: aA + bB → cC + dD

Enthalpy changes= (summation of enthalpy)products-

(summation of enthalpy)reactants

Standard heat of reaction is the enthalpy change when a moles of A

and b moles of B in their standard states at temperature T react to
form c moles of C and d moles of D in their standard state at the same
temperature T (25oC/298K).

Standard state is particular state of a species at temperature T and at

specified conditions of pressure, composition and physical conditions
as, gas, liquid, or solid.
*standard-state pressure: 1 bar( 105 Pa)
Older data: 1 standard
atmosphere(101,325 Pa)
*standard composition: pure species
For liquids and solids: The real pure liquid or solid at 1 bar.
For gases: The pure substance in the ideal-gas state at 1
bar.
o
C p is the standard-state heat capacity. Since the standard state for
gases is the ideal gas state, Cop is similar to Cigp and table C- may apply.

Except for temperature, all conditions for a standard state are fixed,
the temperature is always the temperature of the system. The
symbol, ∆Ho 298 indicates that the heat of reaction is the standard
value for a temperature of 298K.

Standard heat of formation

A formation reaction is a reaction which forms one mole of a single
compound from its constituent elements. The enthalpy of all
elements in the reaction is zero.

∆H = H of products- H of the reactants.

Standard Heat of combustion

A combustion reaction is a reaction between an element or
compound and oxygen to form a specified combustion products. For
organic substances composed of carbon, hydrogen and oxygen, the
products are carbon dioxide and water which must be specified either
liquid or vapour. Many standard heat of formation are calculated
from the heat of combustion. The enthalpy of a new reaction may be
determined without the actual experiment through the use of heat of
combustion. Hess Law on heat summation is applied.

Calculation of heat of reaction at other temperatures.

A general chemical equation may be written:
V1A1 + v2 A2 +… → v3 A3 + v4 A4
Where v1, v2, v3 and v4 ate the stoichiometric coefficient: (+) for
products, (-) for reactants

The fundamental equation relating heat of reaction to temperature;

d∆Ho = ∆Co p dT
T
o o ⟨ Cop ⟩
Integration gives ∆H = ∆H 0 + R ∫ R
dT
¿

where Cop similar to C ig p

∆Ho is the heat of reaction at temperature T and
∆Ho0 is the standard heat of reaction at reference
temperature To(298K).
An analogue of integral showing the temperature dependence of the
heat capacity of each reactant and product is presented below,
T
∆B ∆C ∆ D (τ −1 ¿ ¿ )
∫ ∆ Cop
R
dT = (∆A)To(τ −1 ¿+
2
To2(τ 2 -1) + 3
To3(τ 3-1) + ¿ τ
¿

Where parameters A, B,C and D are heat capacities of the

products and reactants are obtained in Appendix C.

∆A= ∑ viAi summation is overall the products and reactants and the
same analogy is applied for ∆B, ∆C and ∆D.
Sample calculation of these is illustrated in the learning activity
below.

Alternative solution makes use of the mean heat change of reaction,

⟨ Cp ⟩ H ∆B ∆C ∆D 1
=∆A +¿ 2
To(τ +1) + 3
To2 (τ 2 +τ +1) + ( )]
¿2 τ
R

The heat of reaction at any temperature T,

∆Ho= ∆Ho0 + ⟨ Cp ⟩H (T-To)

Learning Activity on Heat of reaction at To=298K, Heat of reaction at

any temperature.
4.21 Determine the standard heat of reaction at 25 oC
4.22Determine the standard heat of reaction at the temperature T
4.23Develop a general equation for standard heat of reaction as
function of temperature
a)N2 (g) + 3H2 (g) → 2 NH3 (g) (for 4.22; T=
600 oC)
b)4NH3 (g) + 5 O2 (g) → 4NO (g) + 6 H2O (g) (for 4.22; T=
500 oC)
d)CaC2 (s) + H2O (l) → C2H2 (g) + CaO (s) (for 4,22;
T= 600 oC)
e) 2 Na (s) + 2 H2O (g) → 2 NaOH (s) + H2 (g) (for 4.22; T=
600 oC)
f) 6NO2 (g) + 3 NH3 (g) → 7 N2 (g) + 12 H2O (g) ( for 4.22;
T=650 oC)
Let us solve case d, where the reaction is
CaC2 (s) + H2O (l) → C2H2 (g) + CaO (s) (for 4,22; T=
600 oC)
The chemical equation is balanced.
The standard heat of reaction is calculated using enthalpies from
any physical data.
∆Hof= ∑ nH product -∑ nH reactants
=[H C2H2 + H CaO]- [H CaC2 + H H2O]
=[227480-635,090] – [-59,800-285,830] = -61,980 Joules
600+273
D#4.22 use T= 600, τ =¿ 298 = 2.929

Let us solve for the heat of reaction at T= 600 oC. Heat capacities
A,B,C and D for the reactants and products are obtained from
appendix C
Reactant/ ni Paramet Parameter B Paramete Paramet
product er A rC er D
CaO(s) + 6.104 0.443E-3 -1.047E5
1
C2H2(g) + 6.132 1.952E-3 -1.299E5
1
H2O (l) -1 8.712 1.25E-3 -0.18E-6
CaC2 (s) -1 8.254 1.429E-3 -1.042E5
∆ A=∑ niAi = - ∆ B=∑ niBi=0. ∆C ∆ D=∑ niDi
4.73 284E-3 =niCi=0.18 = -
E-6 1.304E5
Substitute these values into the mean heat capacity change
 ⟨ ∆Cp ⟩ H ∆B ∆C ∆D 1
R
=∆A +¿ 2 To(τ +1) + 3 To2 (τ 2 +τ +1) + ¿ 2 ( τ )] where τ =¿
600+273
298
= 2.929

0.284 E−3 0.18 E−6

=-4.73 + 2 (298)(2.929+1) + 3
−1.304 E 5 1
(298)2[(2.929)2+2.929+1)]- ( 298 ) 2 2.929 = -4.99
Simplying the¿ side∧cross multiplying with Rgives the mean heat capcity

mean heat capcity ⟨ ∆ Cp ⟩ H =

-4.99 R
Heat of reaction at 600 oC using the formula
∆Ho= ∆Ho0 + ⟨ Cp ⟩H (T-To)

∆Ho = -61,980 + (-4.99)(8.314)(873-298)= -85,835 J

This is the heat of reaction liberated at 600 K. The
standard heat of reaction at 25 oC(298K) is -61,980 J.

D#4.23. expression for enthalpy change at any temperature T

1.468
∆Ho=-61,980 +R[-4.73 +0.042(τ +1 ¿+5.328E-3(τ 2+ τ+ 1¿- τ ](T-298)

a) #4.21
N2 (g) + 3H2 (g) → 2 NH3 (g)
∆Hof= ∑ nH product -∑ nH reactants
= 2(-46,110) – [0] = -92,220 Joules Enthalpy of formation
of elements are zero

a) #4.22 Summation of niVi are reflected in the excel derived

table below.
 ∆A= -5.871 ; ∆B= 4.18E-3; ∆C=0 and ∆D= 6.61 E+4
Substitute these values into the mean heat capacity change
⟨ Cp ⟩ H ∆B ∆C ∆D 1
R
=∆A +¿ 2 To(τ +1) + 3 To2 (τ 2 +τ +1) + ¿ 2 ( τ )] where τ =¿
600+273
298
= 2.929
4.18 E−3 6.61 E 4 1
= -5.871 +¿ 2 (298) (2.92+1) +0+ (292)2 ( 2.929 )]
The mean heat capacity , ⟨ ∆ Cp o ⟩ = R(-3.169)
The heat of reaction at 873K
∆Ho= ∆Ho0 + ⟨ Cp ⟩H (T-To)
=-92,220 + (-3.169 R)(873-298) = 107,369 J

Expression of heat of reaction at any temperature T

0.7443
∆Ho= -92,220 + R[-5.871+0.62282(τ +1 ¿+ τ ] (T-298)

Mean heat capacities for a) N2 (g) + 3H2 (g) → 2 NH3 (g)

|v| A |v| B |v| C D |v|*D

*A *B
NH3 2 3.57 7.15 3.02E 6.04E - -
8 6 -03 -03 1.86E+0 3.72E+0
4 4
H2 -3 3.24 - 4.22E - 8.30E+0 -
9 9.74 -04 1.27E 3 2.49E+0
7 -03 4
N2 -1 3.28 -3.28 5.93E - 4.00E+0 -
-04 5.93E 3 4.00E+0
-04 3
∆A= ∆B= ∆D=
-5.871 4.18E-03 -6.61E+04

Heat Effects of Industrial reactions

Industrial reactions are rarely carried out under standard state
conditions. Other factors which differentiate industrial reaction from
standard state reactions includes;(a) reaction may not go to
completion, (b) reactants may be in excess or may require preheating,
(c) the presents of inerts and (d) the final temperature may differ from
the initial temperature.

For combustion reaction, a maximum attainable temperature (also

called theoretical flame temperature) is attained when the reaction is
assumed to go to completion adiabatically. Other assumptions like
negligible kinetic and potential energy and the absence of shaft work
may reduce the overall energy balance to ∆H=0.

In calculating the final temperature, any convenient path between the

initial and final state may be used. Consider the right triangle below.
Suppose the products are at a temperature T higher than 298 K, the
heat of reaction is represented by the hypothenuse and is equal to the
sum of enthalpy change involve in the two paths shown. The first
being the standard heat of reaction(represented by the horizontal
side of the triangle) and the other is by raising the products from 298
K to the final state at higher temperature (vertical side of the triangle)
State 2; Products at higher T

State 1 Reactants 298 Products at 298

T

∆H1-2= standard heat of reaction + ∫ Cp products dT

298

If the reactants are initially preheated at state 1 prior to the leading to

the product at state 2, the single step from state 1 to state 2 is
replaced by 3 steps as shown in the diagram and the total enthalpy
change is,
T1

∆H1-2= ∫ Cpreactants dT +¿ standard heat of reaction +

298
T2

∫ Cp products dT
298

The phase change in single step is represented by the inclined line.

The enthalpy change in this single step is equivalent to the sum of the
following:
(a) The standard heat of reaction is represented by the horizontal line.
(b) The enthalpy change involved in preheating the reactants is
represented by the shorter vertical line.
(c) The enthalpy changed involved in raising the temperature of the
products from 298K to the final temperature is represented by the
longer vertical line.
 State 2, Products at T2

State 1, Reactants at T1

Reactant at 298 Products at 298

Learning activity on flame temperature and excess reactant supplied.

Dry methane is burned with dry air. Both are at 25 oC initially. The
flame temperature is 1300 oC. If combustion is complete,
determine the % excess air supplied.
Given; Flame temperature= 1300 oC
1 mol CH4 at 25 oC

Dry air at 25 oC
Required: % excess air supplied
Solution:
∆Ho rxn= ∑ nH products - ∑ nH reactants
CH4(g) +2O2 (g) → CO2 (g) + 2H2O (g) ∆Ho 298= -
191,758 cal/gmol
Apply ICC: Initial concentration, Change in concentration and
Concentration upon completion.
 CH4(g) +2O2 (g) → CO2 (g) + 2H2O (g)
1 2+X 0 0 X represent
oxygen from excess air
-1 -2 +1 +2
Flue gases moles Cp at 2372 oF n Cp
CO2 1 12.3 12.3
H2O 2 9.65 19.3
O2 X 8.1 8.1X
N2 (2+X)79/21 7.65 57.759 + 28.979
X
∑ nCpm=36.879 X + 89.359
25 Tf

Hess equation: ∆H= ∫ nCpm reactants dT + ∆H o

rxn 298 ∫ nCpm products dT
25 25

0 = 0 (no preheating) -191,758 + ( 36.879X + 89.359)

(1300-25)
Adiabatic/flame temperature, ∆H=0
X= 1.655 mole
1.655
% excess Oxygen= 2 x 100= 82.76%

Learning Activity on enthalpy change and heat transfer

#4.31 Smith and Van Ness
Ethylene gas and steam at 320 oC and atmospheric pressure are fed
to a reaction process as equimolar mixture. The process produces
ethanol by the reaction
C2H2 (g) + H2O (g) → C2H5OH (l)
The liquid ethanol exits the process at 25 oC. What is the heat
transfer associated with this overall process per mole of ethanol
produced?
Solution;
The overall process is broken into two steps: (a) Cooling the
reactants from 320 oC to 25 oC and (b) standard reaction which
takes place at 298 K. The product liquid ethanol exit at 25 oC, hence,
there is no further heating
From the heat of formation in Appendix C, the standard heat of
reaction is computed,
∆Ho rxn = ∑ n x enthalphy of products - ∑ n x enthalpy of reactants
= -277,690 – [52,510 +2( -241,818)] = - 153,436
joules
Separately evaluate the heat capacity temperature dependency of
the reactants using the formula,
⟨ Cp ⟩
= A + B2 To(τ +1) + C3 To 2(τ 2 + τ +1) + ¿D2 τ where
R
320+ 273
τ=
298
= 1.99

Values of the parameters and the computer value of ⟨ Cp ⟩ is

presented;
Parameters/Cp A B C
D ⟨ Cp ⟩
Ethylene 1.424 14.394E-3 4.392E
7.428
Water 3.47 1.45E-3
0.121E6 3.667

The enthalpy change for cooling the reactants from 320 oC to 25

oC,
∆Ho cooling of reactants = 8.314{1mole(7.428)+ 2 moles(3.667)]
(298-593)= -36,205.7 J
Total Enthalpy change from state 1, reactants at 593 k to products at
298 K
= ∆Ho cooling of reactants + ∆Ho rxn = -36,205.7 + (-153,436) = -
189,641 j

Learning Activity involving side reaction, cooling of reactants and

heating of products.
#4.32 Smith and Van Ness
A gas mixture of methane and steam at atmospheric pressure and
500 oC is fed to a reactor, where the following reactions occur:
CH4 + H2O →CO + 3H2 and CO + H2O →CO2 + H2
The product streams leaves the reactor at 850 oC. Its
composition(mole fraction) is:
Y co2= 0.0275 Y CO= 0.1725 Y H2O=0.1725
YH2= 0.6275
Determine the quantity of heat added to the reactor per mole
product gas
Solution;
This problem is similar to Problem 4.8 in the book where the mole
ratio of the reactants are given and the fraction of CH4 consumed in
the two reactions are computed.

Compute the heat of reactions;

CH4 + H2O →CO + 3H2 ∆H rxn = 205,813
CO + H2O →CO2 + H2 ∆H rxn= -41,166 J
CH4 + 2H20→CO2 + 4H2 ∆H rxn= 164,647 J the two
reactions were added

Assume that water to methane ratio is 2. This was not stated in this
problem. This maybe verified from the solution on the fraction of
methane used in the two reactions in Problem 4.8. Computation of
the mole fractions show that they are the same with the mole
fractions of the gases in this problem. Therefore, the assumption is
correct.

The total enthalpy in the whole process is the sum of enthalpy in

cooling the reactant, the heat of reactions and the enthalpy for
heating the products to 850 oC

Process of cooling the reactants from 500 oC to 25 oC

⟨ Cp ⟩ B C D
=A + 2
To(τ +1) + 3
To 2(τ 2 + τ +1) + ¿2τ
where
R
500+ 273
τ=
298
= 2.59

Values of the parameters and the computer value of ⟨ Cp ⟩ is

presented;
Parameters/Cp A B C
D ⟨ Cp ⟩
methane 1.702 9.081E-3 -2.14E-6
5.907
Water 3.47 1.45E-3
0.121E6 4.772
∆Ho cooling of reactants =8.314{1(5.096)+ 2(4.772)](298-500)= -
25,945.8 J

Process of heating the Products from 298 to 1123 K where τ =

1123/298= 3.77
Parameters/Cp A B C
D ⟨ Cp ⟩
H2 3.249 0.442E-3
0.083E5 3.574
CO2 5.457 1.045E-3 -
1.57 E5 5.731
CO 3.376 0.557E-3 -
0.031E5 3.763
Water 3.47 1.45E-3
0.121E6 4.536
Enthalpy change for heating products from 25 oC to 850 oC
∆Ho heating = R∑ nCp product (850−25)
=8.314[0.1725(4.536+3.763)
+0.0275(5.731)+ 0.6275(3.574)](850-25)
= 18,382.7 J
Total enthalpy change= H cooling + H from reaction + heating of
product
= -25,945.8 + 164,647 + 18,381.7 J= 157,083 J
One mole of CO2 and 4 moles of H2 are produced.
157083
Quantity of heat added per mole of product gas, Q= 5 moles =
31,416.6

Assignment
1.solve for the remaining reactions e,f in the sample problem above.
2. 4.19
3.# 4.27
4. Methane is burned with the theoretical amount of air but not all of
the methane is burned. However, all of the carbon that burns goes to
CO2. If the methane and air enter at 25 oC and the flame
temperature is 400 oC, what fraction of methane is burned?
4.28
4.33
4,37
4.42
4.45 4.46 letters a,b and c only
4.50