Chemical process calculations involve various computations and analyses to design, optimize, and understand chemical processes. Here are some descriptions highlighting different aspects of chemical process calculations: Material Balances: Material balances are the cornerstone of chemical process calculations, ensuring that the amounts of all components entering and leaving a system are properly accounted for. These calculations involve tracking the flow rates and compositions of substances throughout a process, often using mass or mole balances. Energy Balances: Energy balances involve quantifying the energy inputs and outputs in a chemical process. These calculations are crucial for understanding the heat transfer requirements, evaluating energy efficiency, and optimizing process conditions. Reaction Kinetics: Chemical reactions kinetics calculations focus on understanding the rates at which reactions occur and how they are influenced by various factors such as temperature, pressure, and catalysts. These calculations help in determining the optimal reaction conditions and designing reactors for desired conversion rates. Phase Equilibrium Calculations: Phase equilibrium calculations deal with determining the distribution of components between different phases in a system, such as liquid-liquid or vapor-liquid equilibrium. These calculations are essential for designing separation processes like distillation, extraction, and absorption. Thermodynamic Calculations: Thermodynamic calculations involve applying thermodynamic principles to predict the behavior of chemical systems. These calculations include determining properties such as enthalpy, entropy, Gibbs free energy, and fugacity, which are crucial for process design and optimization. Process Simulation: Process simulation involves using computer software to model and simulate chemical processes. These simulations allow engineers to predict process behavior under different operating conditions, optimize process parameters, and troubleshoot potential issues. 7 network error

1. Dr. S. K. Behera
School of Chemical Engineering
Vellore Institute of Technology, Vellore, TN, India – 632 014.
CHE1002 - Process Calculations

2. Module 7
Energy Balance

4. Heat Capacity
• Heat capacity: the amount of heat required to raise the temperature of a body by 1
degree.
– Heat capacity = Heat energy/Temperature change
• Specific heat capacity: the amount of heat required to raise the temperature of 1
gram of substance by 1 degree.
T
Q
C



T
m
Q
C




5. Heat Capacity
Empirical equation for heat capacities:
Cp and CV have units of energy per unit mass per unit temperature interval,
where the mass may be measured in mole or mass units (e.g. units of heat
capacity - J/(kg oC), J/(mol oC), etc).
Cp = a + bT + cT2 + dT3
where a, b, c and d are constants. T is in K.

6. • Cp values for mixtures: The overall heat capacity of a mixture Cp,mix
can be approximated as the sum of heat capacity contributions from the
separate components of the mixture.
Cp,mix(T) = Σ xi Cpi (T)

7. Mean Heat Capacity of Gases
1
2
1
2
1
2
2
1
)
(
ˆ
ˆ
T
T
dT
T
C
T
T
H
H
C
T
T
p
pm






)
(
3
)
(
2
)
(
2
2
2
1
2
1
2
1
1
2
2
1
T
T
T
T
c
T
T
b
a
T
T
dT
T
C
C
T
T
p
pm 




















If Cp = a + bT + cT2

8. Problem - 1
Flue gases leaving boiler stack at 250 °C have following composition by volume:
CO2 – 11.31%, H2O – 13.04%, O2 – 2.17%, and N2 – 73.48%
Calculate the heat lost in 1 kg of the gas mixture above 25 °C.
Empirical heat capacity equation for gases:
Cp = a + bT + cT2 + dT3, kJ/kmol.K
Components a b×103 c×106 d×109
CO2 21.3655 64.2841 -41.0506 9.7999
H2O 32.4921 0.0796 13.2107 -4.5474
O2 26.0257 11.7551 -2.3426 -0.5623
N2 29.5909 -5.141 13.1829 -4.968

9. Solution:

12. Problem - 2
Toluene is to be heated from 17 °C to 77 °C at the rate of 250 gm/sec. Calculate
the heat to be supplied to toluene using the heat capacity data given below:
Cp = a + bT + cT2 + dT3, kJ/kmol.K
Component a b×103 c×106 d×109
Toluene 1.8083 812.223 -1512.67 1630.01
Solution:
Molal flow rate of toluene = (0.25/92) = 2.717 ×10-3 kmol/s
)
)(
(
4
)
(
3
)
(
2
)
(
2
2
2
1
2
1
2
2
2
1
2
1
2
1
1
2
2
1
T
T
T
T
d
T
T
T
T
c
T
T
b
a
T
T
dT
T
C
C
T
T
p
pm 






























Q = (2.717 ×10-3) Cp × (60) = 26.124 kJ/s = 26.124 kW

13. 13
• The heat of formation of a chemical compound (Hf
o) is the enthalpy
change for the formation of one mole of a compound from its
component elements in its standard state.
– Note that the standard enthalpy of formation for a pure element in its
standard state is zero
Heat of Formation

14. STANDARD HEAT OF FORMATION
H2 + ½ O2  H2O Ho = -285.83 kJ
 When a mole of hydrogen and ½ mole of oxygen react to form a mole
water, 285.83 kJ of heat is evolved under standard conditions (exothermic
reaction).
 All substance involved must be in their normal physical states under
standard condition.
 The enthalpy of all elements in their most stable form and standard state
is conventionally taken as zero.
Ex: Ho
298 [O2(g)] = 0; Ho
298 [Na(s)] = 0

15. HEAT OF REACTION FROM HEAT OF FORMATION
Heat of Reaction can be computed by subtracting the sum of the Heats of
Formation of the Reactants from the sum of the Heats of Formation of the Products.
Calculating an Enthalpy Change using Standard Heats of Formation
SO2(g) + NO2(g) → SO3(g) + NO(g)
ΔH˚rcn = Σ(ΔH˚products) - Σ (ΔH˚reactants)
= (ΔH˚f, NO + ΔH˚f, SO3) - (ΔH˚f, NO2 + ΔH˚f, SO2)
ΔH ˚rcn = (______ + ______) – (_______ + ______)
)
re a c ta nts
(
)
p ro d uc ts
( 
 



 o
f
o
f
r e a c tio n H
H
H

16. STANDARD ENTHALPIES OF HEAT OF FORMATION

19. Problem - 3
Nitric acid, whose worldwide annual production is about 8 billion kg, is used to
make many products, including fertilizer, dyes, and explosives. The first step in
the industrial production process is the oxidation of ammonia:
Calculate H0
rxn from H0
f values.
4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)
NO : 90.3 kJ/mol
H2O : -241.8 kJ/mol
NH3 : -45.9 kJ/mol

20. 20
- [4(H0
f NH3(g) + 5(H0
f O2(g)]
SOLUTION: Hrxn = S mH0
f (products) - S nH0
f (reactants)
Hrxn = [4(H0
f NO(g) + 6(H0
f H2O(g)]
= (4)(90.3) + (6)(-241.8) - [(4)(-45.9) + (5)(0)]
Hrxn = -906 kJ

21. Germain Henri Hess (1802 - 1850)
“The heat evolved or absorbed in a
chemical process is the same
whether the process takes place in
one or in several steps“
It is also known as the law of
constant heat summation.

22. HESS’S LAW OF HEAT SUMMATION

23. HESS’S LAW OF HEAT SUMMATION
The enthalpy change for the oxidation of carbon (graphite) to carbon
dioxide is the same whether it is carried out Route A ( 1 stage ) or
Route B ( 2 stages)
Route A: C(s) + O2(g) » CO2(g) Ha = -393.5 kJ mol-1
Route B: C(s) + ½ O2(g) » CO (g) Hb1 = -110.5 kJ mol-1
CO(g) + ½ O2(g) » CO2 (g) Hb2 = -283 kJ mol-1
Hb1 +H b2 = -110.5 + (-283.0) = -393.5 kJ mol-1

24. Problem - 4
S(s) + O2(g)  SO2(g) H0 = -297.0 kJ
2SO3(g)  2SO2(g) + O2(g) H0 = 198.0 kJ
Find the enthalpy change of the equation:
2S(s) + 3O2(g) 2SO3(g); H0 = ?
Solution:
Multiply the first equation by 2 and reverse the second equation, they will
sum together to become the third:
2S(s) + 2O2(g) » 2SO2(g) H0 = 2(-297) kJ
2SO2(g) + O2(g) » 2SO3(g) H0 = -1(198) kJ
2S(s) + 3O2(g) » 2SO3(g) H0 = - 792 kJ

25. Problem - 5
Calculate H for the combustion of methane, CH4:
CH4 + 2O2  CO2 + 2H2O
Data: C + 2H2  CH4 Ho = -74.80 kJ
C + O2  CO2 Ho = -393.50 kJ
H2 + ½ O2  H2O Ho = -285.83 kJ
Solution:
• Reverse Rxn. 1 and change the sign on H (since CH4 must appear on reactant side)
• Let Rxn. 2 unchanged, ( CO2 is on product side)
• Multiply Rxn. 3 by 2 (Since 2 H20 required on product side)
• Sum up reaction and H
CH4  C + 2H2 Ho = +74.80 kJ
C + O2  CO2 Ho = -393.50 kJ
2H2 + O2  2 H2O Ho = -571.66 kJ
CH4 + 2O2CO2 + 2H2O Ho = -890.36 kJ

26. Problem - 6
Determine the Δ H°f for the ClF3(g)
2ClF3(g) + 2NH3(g) -> N2(g) + 6HF(g) + Cl2(g)
ΔH°rxn = -1196 kJ
H°f (NH3(g)) = -46.0 kJ/mol
H°f (HF(g)) = -271.0 Kj/mol
Solution: H°rxn = H°f (products) - H° f (reactants)
-1196KJ = 6H°f(HF) - 2H°f(NH3) - 2H°f(ClF3)
= 6mol (-271) – 2mol (-46) - 2H°f(ClF3)
= -1534 kJ - 2H°f(ClF3)
2H°f(ClF3) = -1534 kJ + 1196 kJ
= -338 kJ/mol
H°f(ClF3) = -169 kJ/mol

27. 27
HEAT OF COMBUSTION
• The heat of combustion (ΔHc
0) is the energy released as heat
when a compound undergoes complete combustion with oxygen.
• Calculating standard heat of reaction using standard heat of
combustion:
c
products
c
ts
reac
rxn H
H
H 






 
 ˆ
ˆ
ˆ
tan

28. Problem - 7
Calculate the heat of reaction for the esterification of ethyl alcohol with acetic
acid if the heats of combustion are : ethyl alcohol; -1366.91 kJ/mol, acetic acid:
-871.69 kJ/mol, ethyl acetate -2274.48 kJ/mol.
Given: C2H5OH + 3O2  2CO2 + 3H2O;  Ho = -1366.91 kJ
CH3COOH + 2O2  2CO2 + 2H2O;  Ho = -871.69 kJ
C2H5COOCH3 + 5O2  4CO2 + 4H2O;  Ho = -2274.48 kJ
Solution:
Rxn: C2H5OH + CH3COOH  C2H5COOCH3 + H2O; Hf = ?

29. C2H5OH + 3O2  2CO2 + 3H2O;  Ho = -1366.91 kJ ----(1)
CH3COOH + 2O2  2CO2 + 2H2O;  Ho = -871.69 kJ ----(2)
C2H5COOCH3 + 5O2  4CO2 + 4H2O;  Ho = -2274.48 kJ ----(3)
1. Multiply eqn. 3 with (-1):
2. Retain eqns. (1) and (2):
4CO2 + 4H2O  C2H5COOCH3 + 5O2  Ho = +2274.48 kJ
C2H5OH + 3O2  2CO2 + 3H2O  Ho = -1366.91 kJ
CH3COOH + 2O2  2CO2 + 2H2O  Ho = -871.69 kJ
C2H5OH + CH3COOH  C2H5COOCH3 + H2O; Hf = 35.88 kJ

30. 30
Enthalpy changes in reaction with different temperatures
The enthalpy change accompanying a chemical reaction can be expressed in
terms of an overall energy balance:
R
ts
reac
products
H
H
H
H
Q 







 

 ˆ
ˆ
ˆ
ˆ
tan
∑HR – Sum of enthalpies of all reactants relative to standard state
∑HP – Sum of enthalpies of all products relative to standard state
∑HR
0 – Standard heat of reactions involved

31. ADIABATIC PROCESS
Any process in which no heat can leave or enter the system.
The system does not give heat to the surroundings nor does the system
receive heat from the surroundings.

32. ADIABATIC/THEORETICAL FLAME TEMPERATURE
When a fuel is burned, a
considerable amount of
energy is released.

33. Some of this energy/heat is
transferred through the walls of
the reactor and remainder
causes a rise in temperature of
reaction products.
If less amount of heat is
transferred/lost, then higher will
be the product temperature.

34.  If the reactor is adiabatic, the highest achievable temperature is called as
adiabatic flame temperature.
 In industrial practice, though the fuel is burned in well insulated furnaces,
the heat loss is bound to be there due to radiation from the surface and
hence, the actual flame temperature is less than the adiabatic flame
temperature.

35. Problem - 8
CO at 200 °C is burned under atmospheric pressure with dry air at 500 °C in 90%
excess of that theoretically required. The products of combustion leave the
reaction chamber at 1000 °C. Calculate the heat evolved in the reaction chamber
(kcal/kmol) of CO burnt, assuming complete combustion.
Specific heat (kcal/kmol.K):
CO = 7.017, CO2 = 11.92, O2 = 7.941, air = 7.225, N2 = 7.507
Standard heat of reaction = - 67.636 ×103 kcal

38. Problem - 9
Dry methane is burnt with dry air and both are initially at 25°C. The adiabatic flame
temperature is 1350 °C. If combustion is assumed to be complete, how much
excess air is used?
Data given:
Heat of reaction, ΔHR
0 = - 0.2 × 106 cal
(Cpm)CO2 – 12.37 cal/mol. °C, (Cpm)H2O – 9.6 cal/mol. °C,
(Cpm)N2 – 7.68 cal/mol. °C, (Cpm)air – 7.74 cal/mol. °C,
Solution

41. Problem - 10
Calculate the theoretical flame temperature of a gas containing 20% CO and 80%
N2 when burnt with 150% excess air, both air and gas being at 25 °C.
Data:
Heat of formation (Cal/mol at 25°C):
CO2 = - 94052, CO = - 26412
Mean specific heat (Cal/mol.K):
CO2 = 12.1, O2 = 7.9, N2 = 7.55

42. Solution:

44. Problem – 11:
Calculate the theoretical temperature of combustion of ethane with 25% excess
air, both air and ethane being at 25 °C.
Data:
Mean specific heat (kJ/kg.K):
CO2 = 1.24, O2 = 1.1, steam = 2.41, N2 = 1.19
Standard heat of reaction = - 1541×103 kJ/kmol

45. Solution:
Basis: 1 kmol of ethane

46. T = 2180. 9 K

47. Practice Problem – 1:
Pyrites fines are roasted in a chamber plant for making sulphuric acid. The
gases leaving the roaster are at 775 K and have molar composition as
mentioned below. Calculate the heat content of 1 kmol gas mixture over 298 K,
using the heat capacity data provided below:
Component a b (x10-3) c (x10-6) d (x10-9)
SO2 24.7706 62.9481 -44.2582 11.122
O2 26.0257 11.7551 -2.3426 -0.562
SO3 22.0376 121.624 -91.8673 24.369
N2 29.5909 -5.141 13.1829 -4.968
SET A: SO2 10%, O2 20% , SO3 30% and N2 40%.
SET B: SO2 15%, O2 20% , SO3 30% and N2 35%.
SET C: SO2 20%, O2 25% , SO3 30% and N2 25%.

48. Practice Problem - 2
A solid municipal waste on combustion gave the following flue gas composition
(dry basis):
CO2 – 9.2%, CO – 1.5%, O2 – 7.3%, and N2 – 82%
What is the enthalpy difference for this gas per mole between the bottom and
the top of the stack if the temperature at the bottom of the stack is 350 °C and
the temperature at the top of the stack is 105 °C. Ignore the water vapour in the
gas. Heat capacity (Cal/mol.°C) equation for gases is given by:
Cp (N2) = 6.895 + 0.7624 ×10-3 T - 0.7009 ×10-7 T2
Cp (O2) = 7.104 + 0.7851 ×10-3 T - 0.5528 ×10-7 T2
Cp (CO2) = 8.448 + 5.757 ×10-3 T – 21.59 ×10-7 T2
Cp (CO) = 6.808 + 0.8024 ×10-3 T - 0.767 ×10-7 T2