7.

1 Energy balances for reacting systems

Learning objectives
©zyBooks 11/22/23 23:26 1858759
Formulate simplified energy balances for systems including chemical reactions.
Enoch Tamale
UBCCHBE241BagherzadehFall2023
Reacting systems can become complicated when solving both material balances and energy
balances. Common assumptions, as discussed in another section, will apply for the energy balances
on reacting systems. More specifically, this section explores systems at steady state, in rigid vessels,
without significant moving parts, and no changes in kinetic or potential energy. Applying these
assumptions to the energy balance is detailed in the animation below.

PARTICIPATION
ACTIVITY
7.1.1: Simplifying the energy balance for many reacting systems.

Animation captions:

1. The general form of the energy balance is written before simplification.

2. A steady state system does not change with time, so the right hand side of the energy balance
goes to zero.
3. Kinetic and potential energy changes are negligible for many reacting systems.
4. Converting mass flow rates to component molar flow rates further simplifies the energy
balance.
5. The work term is neglected as many reactors are rigid vessels (moving boundary work). Shaft
work is negligible when a reactor has no moving parts.
6. Finally, a common simplified form of the energy balances is found for many reacting systems.

The simplified energy balance can be used in terms of mass or molar flow rates. Solving for the flow
rates is usually done using the methods discussed elsewhere, including extent of reaction, atom
balance, and others. Now, the energy balance states that an enthalpy will be needed for each
component entering or exiting the system. Other sections will focus on enthalpy calculations specific
for reacting systems.
©zyBooks 11/22/23 23:26 1858759

( )
1 →2 1 →2 Enoch Tamale dU
∑ ṁ i(h i + v +
2 i
gz i ) − ∑ ṁ e(h e + 2 v e + gz e ) + Q̇ + Ẇ =
UBCCHBE241BagherzadehFall2023
dt
inlets
⏟ ⏟ exits
⏟ ⏟ ⏟
No ΔKE No ΔPE No ΔKE No ΔPE Rigid vessel, no moving parts ⏟
Steady state
 ∑ ṁ i ⋅ h i − ∑ ṁ e ⋅ h e + Q̇ = 0
inlets exits

∑ ṅ i ⋅ h i − ∑ ṅ e ⋅ h e + Q̇ = 0
inlets exits

PARTICIPATION
ACTIVITY
7.1.2: Reacting system energy balances.
©zyBooks 11/22/23 23:26 1858759
Enoch Tamale
1) A brand new reactor has one inlet and UBCCHBE241BagherzadehFall2023
one exiting stream and is so well
insulated that an adiabatic assumption
is applied. The simplified energy
balance should be:

ṅ 1 ⋅ h 1 − ṅ 2 ⋅ h 2 = 0

True
False

2) Hydrogen and oxygen enter a reactor. Both

components react completely to form water. The
energy balance can be written as:

ṅ H ⋅ h H + ṅ O ⋅ h O − ṅ H O ⋅ h H O + Q̇ = 0
2 2 2 2 2 2

True
False

©zyBooks 11/22/23 23:26 1858759

Enoch Tamale
UBCCHBE241BagherzadehFall2023
3) Hydrogen and oxygen enter a reactor in Stream 1. Both
components react to form water, but the reaction does not
go to completion. If the exit stream is Stream 2, the energy
balance can be written as:

ṅ 1 , H ⋅ h 1 , H + ṅ 1 , O ⋅ h 1 , O − ṅ 2 , H O ⋅ h 2 , H O
2 2 2 2 2 2
©zyBooks 11/22/23 23:26 1858759
Enoch Tamale
− ṅ 2 , H ⋅ h 2 , H + ṅ 2 , O ⋅ h 2 , O + Q̇ = 0 UBCCHBE241BagherzadehFall2023
2 2 2 2

True
False

4) Two components enter a reactor where

three different reactions occur
simultaneously. Five components exit
the reactor. The system is too
complicated to write an energy
balance.
True
False

CHALLENGE
ACTIVITY
7.1.1: Simplifying the energy balance for a reactor.

495688.3717518.qx3zqy7

Start

The reactor operates at steady state. A term removed from the energy
balance is: ©zyBooks 11/22/23 23:26 1858759
Enoch Tamale
UBCCHBE241BagherzadehFall2023
a. \(\frac{dU}{dt}\)

b. \(h_e\)

c. \(g{z_i}\)
 d. \(\dot W\)

e. \(\tfrac{1}{2}\vec v_e^2\)

1 2 3 ©zyBooks 11/22/23
4 23:26 1858759
Enoch Tamale
UBCCHBE241BagherzadehFall2023
Check Next

fitness_center EXERCISE 7.1.1: Water splitting reactor. help_outline

Water (H2O) is split into hydrogen (H2) and oxygen (O2) in a reactor. While only water enters the
reactor, the conversion is incomplete. So all three components exit the reactor.

(a) Draw and label a process flow diagram. Clearly number each stream.
Solution keyboard_arrow_down
(b) Starting from the general form of the energy balance, list the assumptions that should be
used to simplify the energy balance.
Solution keyboard_arrow_down
(c) Starting from the general form of the energy balance again, apply the assumptions and
simplify the energy balance.
Solution keyboard_arrow_down
(d) Finally, write the energy balance in terms of molar flow rates for all components entering and
exiting the reactor.
Solution keyboard_arrow_down

©zyBooks 11/22/23 23:26 1858759

7.2 Heat of reaction and Hess’s law

Enoch Tamale
UBCCHBE241BagherzadehFall2023

Learning objectives

Distinguish between terms related to heat of reaction, including endothermic, exothermic, and
standard heat of reaction.
 Calculate a new heat of reaction using algebraic combination, i.e., Hess’s law.

Heat of reaction

Heat of reaction (Δhr(T, P)) is the enthalpy change for stoichiometric quantities of reactants at fixed
temperature and pressure that react completely to form products at same temperature and pressure.
Another term for heat of reaction is enthalpy of reaction. Ex: The reaction to create ammonia from
©zyBooks 11/22/23 23:26 1858759
nitrogen and hydrogen has a heat of reaction: 3H2 + 1N2 → 2NH3 and Δhr(25°C, 1 atm)
Enoch = -46.2 kJ/mol.
Tamale
UBCCHBE241BagherzadehFall2023
A heat of reaction specifies the phase (solid, liquid, or gas) of each reactant and product, so the same
stoichiometric reaction can have more than one heat of reaction. Ex: H2(g) + 0.5O2(g) → H2O(g) and
H2(g) + 0.5O2(g) → H2O(l) have different heats of reaction, since the product water is either in the gas
or liquid phase.

The heat of reaction also applies to each component in the reaction. While only one number will be
tabulated, that heat of reaction can apply to all of the components based on the stoichiometric
coefficients in the balanced reaction. Ex: For the balanced reaction 2D+E → 4F, Δhr(100°C, 1 atm) =
-50 kJ/mol. The heat of reaction can thus be represented for each component as:

− 50 kJ − 50 kJ − 50 kJ
Δh r = = =
2 mol D consumed 1 mol E consumed 4 mol F produced

Three additional new terms are tabulated below. Tables of heats of reaction and standard heats of
reaction are tabulated in reference materials and can assist with energy balance calculations.

Table 7.2.1: Heat of reaction terms.

Definition Details

Exothermic is when enthalpy change due to the reaction is negative. Δhr < 0

Endothermic is when enthalpy change due to the reaction is positive. Δhr > 0

A standard heat of reaction is the heat of reaction at a standard Δhor = Δhr

temperature and pressure, which are commonly 25°C and 1 atm. (25°C, 1 atm)

©zyBooks 11/22/23 23:26 1858759

Enoch Tamale
PARTICIPATION UBCCHBE241BagherzadehFall2023
ACTIVITY
7.2.1: Heat of reaction.
 1) 3H2 + 1N2 → 2NH3 and Δhr(25°C, 1
atm) = -46.2 kJ/mol.
The reaction is _____.
Exothermic
Endothermic

2) Reaction 1: 3H2(g) + 1N2(g) → 2NH3(g) ©zyBooks 11/22/23 23:26 1858759

Reaction 2: 3H2(g) + 1N2(g) → 2NH3(l) Enoch Tamale
UBCCHBE241BagherzadehFall2023
The heat of reaction for Reaction 1 is
_____ to the heat of reaction for
Reaction 2.
not equal
equal

3) All reactions have a heat of reaction.

True
False

4) The heat of reaction is reported as 112

kJ/mol. Therefore, the heat of reaction
is an extensive property.
True
False

5) For the balanced reaction 2D+E → 4F,

Δhr(100°C, 1 atm) = -50 kJ/mol, if 99
mol/s of F are produced at 100°C and
1 atm, what is heat produced by the
reaction?
-1200 kW
-5000 kW

Hess’s law ©zyBooks 11/22/23 23:26 1858759

Enoch Tamale
UBCCHBE241BagherzadehFall2023
If the heat of reaction is unknown for a reaction, the value can be derived by adding and subtracting
reactions with known heats of reaction. Hess’s law is the algebraic combination of known reactions
and their heats of reaction to calculate the unknown heat of reaction for another reaction.

PARTICIPATION
ACTIVITY
7.2.2: Using Hess's law to calculate a new heat of reaction.
Animation captions:

1. Hess's law combines the heat of reaction for a set of reactions to find an unknown heat of
reaction for another reaction.
2. Evaluating the algebraic equation finds the third reaction, so the heat of reaction for the third
reaction can be found.
3. Finding the unknown heat of reaction follows the same algebraic operations as the reaction.
©zyBooks 11/22/23 23:26 1858759
Enoch Tamale
UBCCHBE241BagherzadehFall2023
PARTICIPATION
ACTIVITY
7.2.3: Hess’s law calculations.

The goal is to use a set of three reactions to find the heat of reaction for the complete
combustion of liquid acetone. The four reactions of interest are:

Reaction 1: 3C(s) + 3H2(g) + 0.5O2(g) → C3H6O(l)

Reaction 2: C(s) + O2(g) → CO2(g)

Reaction 3: H2(g) + 0.5O2(g) → H2O(g)

Reaction 4: C3H6O(l) + 4O2(g) → 3CO2(g) + 3H2O(g)

1) The complete combustion of liquid

acetone is Reaction _____.
1
2
4

©zyBooks 11/22/23 23:26 1858759

Enoch Tamale
UBCCHBE241BagherzadehFall2023
2) The four reactions are balanced in this case. In order to check Reaction 4,
an atom balance on oxygen is:

2 mol O 2 mol O 1 mol O

4 mol O 2 ⋅ + _____ = 3 mol CO 2 ⋅ + 3 mol H 2O ⋅
1 mol O 2 1 mol CO 2 1 mol H 2O

©zyBooks 11/22/23 23:26 1858759

Enoch Tamale
UBCCHBE241BagherzadehFall2023

Fill in the missing term.

1 mol O
1 mol C 3H 6O ⋅ 1 mol C H O
3 6

0
2 mol O
1 mol C 3H 6O ⋅ 1 mol C H O
3 6

3) Algebraically combining Reactions 1, 2,

and 3 can be completed to create
Reaction 4. Finish the expression:

- 1∙Reaction 1 + 3∙Reaction 2 + _____ =

Reaction 4

Recalling the reactions:

Reaction 1: 3C(s) + 3 H2(g) + 0.5O2(g)
→ C3H6O(l)
Reaction 2: C(s) + O2(g) → CO2(g)
Reaction 3: H2(g) + 0.5O2(g) → H2O(g)
Reaction 4: C3H6O(l) + 4O2(g) →
3CO2(g) + 3H2O(g)
-3∙Reaction 3
3∙Reaction 2
3∙Reaction 3 ©zyBooks 11/22/23 23:26 1858759
Enoch Tamale
UBCCHBE241BagherzadehFall2023
4) Finally, the heats of reaction for the
three reactions are:
Reaction 1: Δhr1 = -248 kJ/mol

Reaction 2: Δhr2 = -394 kJ/mol

Reaction 3: Δhr3 = -242 kJ/mol

©zyBooks 11/22/23 23:26 1858759
Enoch Tamale
Reaction 4: Δhr4 = _____ UBCCHBE241BagherzadehFall2023

Fill in the heat of reaction for Reaction

4.
-2160 kJ/mol
-1660 kJ/mol
+1660 kJ/mol

CHALLENGE
ACTIVITY
7.2.1: Hess’s law calculations.

495688.3717518.qx3zqy7

Start

For the balanced stoichiometric reaction:

\(1H_2(g) \,+ \, 0.5O_2(g) \rightarrow \, 1H_2O(g) \,\,\,\Delta h_{r1} =
-241.83\tfrac{kJ}{mol}\)
The reaction is __________.

©zyBooks 11/22/23 23:26 1858759

Enoch Tamale
UBCCHBE241BagherzadehFall2023

1 2 3 4

Check Next
 fitness_center EXERCISE 7.2.1: Heat from a generic reaction. help_outline
A balance stoichiometric reaction 2D + 0.5E → 3.8F with Δhr(100°C, 1 atm)
©zyBooks = -48 kJ/mol.
11/22/23 23:26 1858759
Enoch Tamale
(a) If 62 mol/s of component F are produced at 100°C and 1 atm, how much heat (kW) is
UBCCHBE241BagherzadehFall2023
produced by the reaction?
Solution keyboard_arrow_down

7.3 Heat of formation method

Learning objectives

Generate standard heat of reaction from standard heats of formation.

Construct enthalpy paths, calculate enthalpies, and solve energy balances for a reacting system.

Heat of formation

An energy balance can still be performed if the heat of reaction is not known. This section covers
terms related to the heat of formation and using heats of formation to solve energy balance problems.

Table 7.3.1: Heat of formation terms.

Definition Nomenclature Example

The heat of formation is the enthalpy needed to For methane:

Δhf(T,P)
form a component from its atoms. Δhf = -75 kJ/mol

The standard heat of formation is commonly ©zyBooks 11/22/23 23:26 1858759

ForEnoch
oxygen in the
Tamale
tabulated at 25°C and 1 atm, and the standard Δhof (T=25oUBCCHBE241BagherzadehFall2023
C,
gas phase:
heat of formation for atoms in their native phase P=1 atm)
Δhof = 0
is zero.

Standard heats of formation are measured quantities and tabulated in many reference materials,
including the Appendix here. The heat of reaction can be calculated from the heat of formation of
each component scaled by the stoichiometric coefficients. Since standard heats of formation are
given at 25°C and 1 atm, the calculated heat of formation will also be at 25°C and 1 atm.

o
Δh r = ∑ ν i ⋅ h of, i
PARTICIPATION
ACTIVITY
7.3.1: Calculating heat of reaction from heat of formation.
©zyBooks 11/22/23 23:26 1858759
Enoch Tamale
Animation content: UBCCHBE241BagherzadehFall2023

undefined

Animation captions:

1. Finding the heat of reaction from heats of formation begins with a balanced chemical
reaction.
2. Stoichiometric coefficients are multiplied by the heat of formation for each component. Then
the terms are summed to complete the calculation.
3. Inserting the stoichiometric coefficients and tabulated heat of formation values is next.
4. Checking the units, the unknown heat of reaction can now be calculated.

PARTICIPATION
ACTIVITY
7.3.2: Heat of formation.

1) The heat of formation at 25°C and 1 atm is

equal to the standard heat of formation.

Δh f(25 oC, 1 atm) = Δh of(25 oC, 1 atm)

True
False

2) The heat of formation of carbon

dioxide is -394 kJ/mol. Three moles of ©zyBooks 11/22/23 23:26 1858759
CO2 leads to an extensive heat of Enoch Tamale
UBCCHBE241BagherzadehFall2023
formation of -394 kJ.
True
False
3) The heat of reaction is needed for the combustion of methane.

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

Δh r = ν CH ⋅ Δh f , CH + ν O ⋅ Δh f , O + ν CO ⋅ Δh f , CO
4 4 2 2 2 2
©zyBooks 11/22/23 23:26 1858759
+ ν H O ⋅ Δh f , H O Enoch Tamale
2 2
UBCCHBE241BagherzadehFall2023

Δh r = − 1 ⋅ Δh f , CH + − 2 ⋅ Δh f , O + 1 ⋅ Δh f , CO
4 2 2

+ _____ ⋅ Δh f , H O
2

Select the missing stoichiometric coefficients.

+2
-2

©zyBooks 11/22/23 23:26 1858759

Enoch Tamale
UBCCHBE241BagherzadehFall2023
4) Calculating the heat of reaction for the combustion of methane requires
looking up heat of formation values. Fill in the missing heat of formation.

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

Δh r = ν CH ⋅ Δh f , CH + ν O ⋅ Δh f , O + ν CO ⋅ Δh f , CO ©zyBooks 11/22/23 23:26 1858759

4 4 2 2 2 2 Enoch Tamale
+ ν H O ⋅ Δh f , H O UBCCHBE241BagherzadehFall2023
2 2

kJ kJ kJ
Δh r = − 1 ⋅ _____ + − 2 ⋅ 0 mol + 1 ⋅ − 394 mol + 2 ⋅ − 242 mol

-74.9 kJ/mol
0 kJ/mol

©zyBooks 11/22/23 23:26 1858759

Enoch Tamale
UBCCHBE241BagherzadehFall2023
5) Based on the information given below, the phase of the product
water is _____.

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

Δh r = ν CH ⋅ Δh f , CH + ν O ⋅ Δh f , O + ν CO ⋅ Δh f , CO ©zyBooks 11/22/23 23:26 1858759

4 4 2 2 2 2 Enoch Tamale
+ ν H O ⋅ Δh f , H O UBCCHBE241BagherzadehFall2023
2 2

kJ kJ
Δh r = − 1 ⋅ − 74.9kJ / mol + − 2 ⋅ 0 mol + 1 ⋅ − 394 mol
kJ
+ 2 ⋅ − 242 mol

Gas
Liquid

©zyBooks 11/22/23 23:26 1858759

Enoch Tamale
UBCCHBE241BagherzadehFall2023
6) The heat of reaction is calculated below. The heat of combustion is
tabulated in the Appendix also. The two values are the same.

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

Δh r = ν CH ⋅ Δh f , CH + ν O ⋅ Δh f , O + ν CO ⋅ Δh f , CO ©zyBooks 11/22/23 23:26 1858759

4 4 2 2 2 2 Enoch Tamale
+ ν H O ⋅ Δh f , H O UBCCHBE241BagherzadehFall2023
2 2

kJ kJ kJ kJ
Δh r = − 1 ⋅ − 74.9 mol + − 2 ⋅ 0 mol + 1 ⋅ − 394 mol + 2 ⋅ − 242 mol

kJ
Δh r = − 803 mol

True
False

CHALLENGE
ACTIVITY
7.3.1: Heat of formation calculations.

495688.3717518.qx3zqy7

Start

The standard heat of formation for benzene in the gas phase is:

©zyBooks 11/22/23 23:26 1858759

Enoch Tamale
UBCCHBE241BagherzadehFall2023
20.8 kJ/mol

1 2 3 4 5
 Check Next

Enthalpy path for a reacting system

©zyBooks 11/22/23 23:26 1858759
Enthalpy values are needed to solve many energy balance problems, and enthalpy Enoch is
Tamale
a strong function
UBCCHBE241BagherzadehFall2023
of temperature in many reacting systems. Pressure can also change the enthalpy, which follows the
pressure enthalpy path detailed in another section. Here, calculations of enthalpy at different
temperatures can be completed using the heat of formation and heat capacity. While the equation for
calculating temperature dependent enthalpies is given below, the animation explains the enthalpy path
that is created to derive this equation. The reference conditions for these enthalpy paths are 25°C and
1 atm, which comes from the conditions for the standard heat of formation.

T
Δh(T) = Δh of + ∫ T = 25 oCC P(T)dT
ref
 

PARTICIPATION
ACTIVITY
7.3.3: An enthalpy path from atoms in a reacting system.

Animation captions:

1. Creating an enthalpy path for a reacting system can involve components or atoms.
2. The enthalpy of methane is needed at 100°C, 1 atm, in the vapor phase starting from atoms.
3. The first enthalpy step forms methane from carbon and hydrogen.
4. The second enthalpy step changes the temperature to the desired temperature.
5. Assembling the enthalpy equation is the final step.

Heat of formation method

Solving energy balances for reacting systems can use the heat of reaction and additional enthalpy
terms. However, the heat of reaction is not usually tabulated at all of the temperatures of interest, so
calculating heats of reaction is cumbersome. More directly, the enthalpy of all inlet and outlet
components can be found using the heat of formation and a temperature change, without additional
calculations to solve the energy balance. Thus, the heat of formation ©zyBooks
method11/22/23 23:26technique
is a general 1858759 to
Enoch Tamale
solve energy balances for reacting systems. Recalling that heats ofUBCCHBE241BagherzadehFall2023
formation are readily available and
tabulated quantities, energy balance problems with reactions combine many material and energy
balance concepts in a single problem.

1. Following the 12-step method may solve the material balances. Drawing and labeling a process
flow diagram is followed by setting up material balances and extra equations. Solving material
balances may be completed if the material balances are independent from the energy balance.
2. Writing down the energy balance, any assumptions, and simplifying the energy balance.
Assumptions for reactor problems generally include steady state, no change in kinetic or potential
energy, and no work (i.e., rigid reactor). The simplified energy balance in terms of mass or molar flow
rates becomes:

∑ ṁ i ⋅ h i − ∑ ṁ e ⋅ h e + Q̇ = 0
inlets exits
©zyBooks 11/22/23 23:26 1858759
Enoch Tamale
∑ ṅ i ⋅ h i − ∑ ṅ e ⋅ h e + Q̇ = 0 UBCCHBE241BagherzadehFall2023
inlets exits

3. Choosing a reference state. Normally, 25°C and 1 atm are used, since the standard heats of
formation are tabulated at these conditions.
4. Constructing a flow rate+enthalpy table around the reactor or system of interest. Calculating all
enthalpies in the table, including all steps in the enthalpy path for each species. The heat of formation
for each inlet and outlet component is looked up and then added to temperature change or other path
steps. Reviewing the examples of enthalpy paths described elsewhere could be useful. An example
table:

Table 7.3.2: Example of a flow rate+enthalpy table.

ṅ i , j hi , j ṅ e , j he , j

Component mol kJ mol kJ

j s mol s mol

A 24 -152 11 -16.5

B 15 0 2.8 +4.60

C 0 --- 7.9 -203

5. Solving the energy balance and material balances independently - or solving the energy balance and
material balances simultaneously as a system of equations.
©zyBooks 11/22/23 23:26 1858759
Enoch Tamale
UBCCHBE241BagherzadehFall2023
Two common types of energy balances on reaction systems are: (1) Finding the heat interaction when
inlet and outlet temperatures of the reactor are known, and (2) An adiabatic reactor where the exit
temperature is unknown.

Example 7.3.1: Sulfur dioxide oxidation.

Oxidation reactions are very common. For example, sulfur dioxide is oxidized to sulfur
trioxide according to: SO2 + 0.5 O2 → SO3. Sulfur dioxide and 60.0% excess oxygen are fed to
the reactor at 25.0°C and 1 atm. The reaction proceeds to a SO2 conversion of 78.0%.
Reaction products leave the reactor at 500°C and 1 atm with a production rate of SO3 equal
to 90.0 mol/min.

The goals are to: ©zyBooks 11/22/23 23:26 1858759

Enoch Tamale
1. Calculate the feed rates of the SO2 and O2 (mol/min). UBCCHBE241BagherzadehFall2023
2. Calculate the heat interaction (kW).
3. Explore a concept question: If dry air rather than pure oxygen is used in this reaction,
will the heat interaction increase, decrease, or stay the same?

Step by step solution

Step 1. The process flow diagram involves a single process unit, a reactor.

Step 2. The basis is the production rate of SO3.

mol
ṅ 2 , SO = 90.0 min
3

Step 3. The system is the single process unit, the reactor.

Steps 4, 5, and 6. If the material balances and energy balance are independent, the material
balances can be solved first. With a reacting system, atom balances are selected. Only two
atoms exist in the three components, S and O.

1 mol S 1 mol S ©zyBooks

1 mol S 11/22/23 23:26 1858759
S: ṅ 1 , SO ⋅ 1 mol SO = ṅ 2 , SO ⋅ 1 mol SO + ṅ 2 , SO ⋅ 1 mol SO Enoch Tamale
2 2 2 2 3 3
UBCCHBE241BagherzadehFall2023

2 mol O 2 mol O
O: ṅ 1 , SO ⋅ 1 mol SO + ṅ 1 , O ⋅ 1 mol O =
2 2 2 2

2 mol O 3 mol O 2 mol O

ṅ 2 , SO ⋅ 1 mol SO + ṅ 2 , SO ⋅ 1 mol SO + ṅ 2 , O ⋅ 1 mol O
2 2 3 3 2 2
Step 7. The conversion and excess provide two extra equations.

ṅ 1 , SO − ṅ 2 , SO
2 2
Conversion: f SO =
2 ṅ 1 , SO
2

ṅ 1 , O − ṅ STOICH , O
2 2
Excess: f Ex , O = ©zyBooks 11/22/23 23:26 1858759
2 ṅ STOICH , O Enoch Tamale
2
UBCCHBE241BagherzadehFall2023
Step 8. Listing the 4 unknowns:

ṅ 1 , SO ṅ 1 , O ṅ 2 , SO ṅ 2 , O
2 2 2 2

Step 9. The 4 unknowns match the 4 independent equations, so the system can be solved.
No additional systems are needed (Step 10) to solve the material balance portion of the
problem.

Step 11 involves solving for all of the unknown flow rates. The S balance and conversion
have the same two unknown flow rates. Combining these equations finds the entering and
exiting SO2.

mol
Simplified S balance: ṅ 1 , SO − ṅ 2 , SO = 90 min
2 2

ṅ 1 , SO − ṅ 2 , SO
2 2
f SO =
2 ṅ 1 , SO
2

mol
90 min
0.78 =
ṅ 1 , SO
2

mol
ṅ 1 , SO = 115 min
2

Plugging back into the S balance:

mol mol ©zyBooks 11/22/23 23:26 1858759

115 min − ṅ 2 , SO = 90 min Enoch Tamale
2
UBCCHBE241BagherzadehFall2023
mol
ṅ 2 , SO = 25 min
2

Now, the reaction stoichiometry is needed to use the extra equation for excess. The reaction
should be SO2 + 0.5O2 → SO3. Thus, the stoichiometric amount of oxygen in Stream 1 is half
of the flow rate of SO2 in Stream 1. Solving the excess and O atom balance finds the final two
component flow rates.

ṅ 1 , O − ṅ STOICH , O
2 2
f Ex , O =
2 ṅ STOICH , O
2

mol
ṅ 1 , O − 57.5 min ©zyBooks 11/22/23 23:26 1858759
2 Enoch Tamale
0.600 = mol UBCCHBE241BagherzadehFall2023
57.5 min

mol
ṅ 1 , O = 92 min
2

mol mol mol mol

115 min ⋅ 2 + 92 min ⋅ 2 = 25 min ⋅ 2 + 90 min ⋅ 3 + ṅ 2 , O ⋅ 2
2

mol
ṅ 2 , O = 47 min
2

Step 12. The calculations can be checked using either atom balance.

Now, the energy balance can be solved using the flow rates. First, the energy balance is
simplified using appropriate assumptions. In this case, a steady state system, no work, and
no changes in kinetic or potential energy are appropriate. Solving for the heat interaction
gives the energy balance as:

Q̇ = (ṅ 2 , SO ⋅ h 2 , SO + ṅ 2 , SO ⋅ h 2 , SO + ṅ 2 , O ⋅ h 2 , O ) − (ṅ 1 , SO ⋅ h 1 , SO + ṅ 1 , O ⋅ h 1 , O )
2 2 3 3 2 2 2 2 2 2

Next, a flow rate+enthalpy table organizes the calculated enthalpies. The inlet enthalpies are
at 25°C, so the enthalpies are simply the standard heats of formation. The exit enthalpies
need to account to heat of formation and the temperature changes to 500°C. Heat capacity
and standard heat of formation values are found in the Appendix or reference materials.

o
h 1 , j(T = 25 oC) = Δh f , j

500 oC
h 2 , j(T = 500 oC) = Δh of, j + ∫ 25 oC C P , jdT©zyBooks 11/22/23 23:26 1858759
Enoch Tamale
UBCCHBE241BagherzadehFall2023
ṅ 1 , j h1 , j ṅ 2 , j h2 , j
mol kJ mol kJ
Component
min mol min mol

SO2 115 -297 25 -275

 SO3 0 --- 90 -364

O2 92 0 47 +15.0

Plugging the values in the table into the energy balance finds the heat interaction after a unit
conversion.
©zyBooks 11/22/23 23:26 1858759
kJ Enoch Tamale
Q̇ = − 4770 min UBCCHBE241BagherzadehFall2023

Q̇ = − 79.5 kW

Finally, the concept question asks: If dry air rather than pure oxygen is used in this reaction,
will the heat interaction increase, decrease, or stay the same?

Adding nitrogen at 25°C would not change the enthalpy of Stream 1, since the standard heat
of formation of nitrogen gas is 0. However, the enthalpy of stream 2 should increase as the
enthalpy of nitrogen at 500°C is a positive number, similar to the oxygen value calculated
above. So the overall heat interaction would increase.

PARTICIPATION
ACTIVITY 7.3.4: Sulfur dioxide oxidation details.

Based on the example detailed above, answer the following questions.

1) Atom balances and extra equations

were used to solve the material
balances. An alternative method to
solve this reacting system is _____.
Multiplication
Mole balances
Extent of reaction

2) The standard heat of formation is

needed to calculate the enthalpy for
©zyBooks 11/22/23 23:26 1858759
both components in Stream 1. Find the Enoch Tamale
standard heat of formation for SO2. UBCCHBE241BagherzadehFall2023

-296.8 kJ/mol
-395.2 kJ/mol
0 kJ/mol
3) The enthalpies for all three
components in Stream 2 are needed to
solve the energy balance. When using
the heat of formation method to solve
energy balance problems, the standard
heat of formation for SO2 should _____
be included when calculating the
©zyBooks 11/22/23 23:26 1858759
enthalpy in Stream 2. Enoch Tamale
UBCCHBE241BagherzadehFall2023
Always
Sometimes
Never

4) Find the c coefficient for SO2 that is used to

calculate the enthalpy change.

kJ b
h 2 = − 296.8 mol + [a ⋅ (T 2 − T 1) + ⋅ (T 22 − T 11)
2

c 3 3 d 4 4
+ ⋅ (T 2 − T 1) + ⋅ (T 2 − T 1)]
3 4

3.891x10-2
-3.10x10-8
-8.54x10-8

5) Robert needs to calculate the enthalpy

of SO2 in Stream 2 quickly, so he
ignores the b, c, and d terms in the heat
capacity. The enthalpy of SO2 in Stream
2 is now:
-275 kJ/mol
-278 kJ/mol
-297 kJ/mol

©zyBooks 11/22/23 23:26 1858759

Enoch Tamale
UBCCHBE241BagherzadehFall2023
 6) Finally, the concept question
qualitatively examined the addition of
nitrogen on the heat interaction.
Quantifying the specific enthalpy for
nitrogen exiting the reactor at 500°C
gives:
15.0 kJ/mol ©zyBooks 11/22/23 23:26 1858759
Enoch Tamale
0 kJ/mol UBCCHBE241BagherzadehFall2023
14.2 kJ/mol

Flow rate+enthalpy tables in a spreadsheet

A flow rate+enthalpy table helps organize problem solving, however, calculations are repetitive.
Therefore, spreadsheets are commonly used to quickly complete energy balance calculations, which
are shown in the animations below.

PARTICIPATION
ACTIVITY 7.3.5: Flow rate+enthalpy table in a spreadsheet.

Animation captions:

1. Calculating enthalpy as a function of temperature starts by listing each component in the

system.
2. Standard heat of formation and heat capacity coefficients are copied from the Appendix for
each component.
3. After solving a system of equations, such as material balances and extra equations, the inlet
and outlet molar flow rates are entered.
4. The inlet, outlet, and reference temperatures are entered.
5. Using the enthalpy equation, h₁ and h₂ for each component are calculated.

PARTICIPATION
ACTIVITY
7.3.6: Finding heat interaction from a flow rate+enthalpy table.

Animation content: ©zyBooks 11/22/23 23:26 1858759

Enoch Tamale
UBCCHBE241BagherzadehFall2023
undefined

Animation captions:

1. The table starts with listing each component. After solving material balances in many cases,
inlet and outlet molar flow rates can also be entered.
2. Inlet, outlet, and reference temperatures are entered next.
 3. Enthalpies are calculated for h₁ and h₂ for each component.
4. The inlet and outlet enthalpy rates are calculated in column A. Finally, the heat transfer rate
can be calculated as the difference between Stream 1 and 2's enthalpies.

CHALLENGE
ACTIVITY
7.3.2: Acetic acid series reaction energy balance.

©zyBooks 11/22/23 23:26 1858759

495688.3717518.qx3zqy7 Enoch Tamale
UBCCHBE241BagherzadehFall2023
Start

Engineers Yolanda and Charles operate a reactor where two unbalanced, gas phase reaction
CH_3COOH + H_2 \rightarrow C_2H_5OH + H_2O \) and \( C_2H_5OH \rightarrow (C_2H_5)_
\). In the first reaction, acetic acid (AA or \(CH_3COOH\)) and hydrogen (H or \(H_2\)) react t
ethanol (E or \(C_2H_5OH\)) and water (W or \(H_2O\)); and ethanol decomposes to diethyl e
or \((C_2H_5)_2O\)) and water in the second reaction. A stream of 60.3 mol/s hydrogen and
balance acetic acid at 147\(^o\)C is fed to the reactor, which operates at steady state. The st
exiting the reactor contains acetic acid, hydrogen, ethanol, water, and diethyl ether at 297\(^o
component flow rate of acetic acid exiting the reactor is 3.81 mol/s. The fraction conversion
acid is 85.2%, and the selectivity, which is the ratio of the amount of ethanol to diethyl ether i
exiting stream, is 3.14. Assume all components are in the gas phase and no phase changes
the reactor.
After drawing and labeling a process flow diagram including numbering of each stream and
components present in each stream, answer the following question.

How many components enter the reactor?

Ex:8

1 2 3 4 5 6 7

Check Next

 

©zyBooks 11/22/23 23:26 1858759

Enoch Tamale
fitness_center EXERCISE 7.3.1: Methanol formation energy balance. help_outline
UBCCHBE241BagherzadehFall2023

A reversible reaction between carbon monoxide (CO) and hydrogen (H2) can be used to form
methanol (CH3OH). The equilibrium constant for the reversible reaction varies with temperature
as Keq = 3.14 + 0.0247∙ T(°C). The feed to the poorly insulated reactor enters at 67.3°C and 345
mol/min. The feed gas mixture contains CO and H2 with a carbon monoxide mole fraction of
 0.327. The flow rate of the product gas stream at equilibrium is 181 mol/min. Assume that
pressure effects are negligible.
(a) Draw and label a process flow diagram. Number each stream and label the components
(M=Methanol; CO=Carbon Monoxide; H=Hydrogen).
Solution keyboard_arrow_down
(b) Calculate the temperature of the product stream (°C). ©zyBooks 11/22/23 23:26 1858759
Enoch Tamale
Solution keyboard_arrow_down UBCCHBE241BagherzadehFall2023

(c) Calculate the heat interaction (kW) between the reactor and its surroundings.
Solution keyboard_arrow_down
(d) If the temperature of the feed increases by 7°F, will Keq increase, decrease, or stay the same?
Solution keyboard_arrow_down

7.4 Combustion reactions and the energy

balance

Learning objectives

Apply definitions related to combustion, including heating value, heat of combustion, adiabatic
flame temperature, etc., to formulate and solve energy balance problems.

As discussed in another section, the burning of fuel or combustion is a common type of chemical
reaction. Additional definitions are needed to convey new terms related to energy balances and
combustion. These terms can lead to extra equations or information needed to solve material and
energy balance problems.

Heat of combustion and heating value

©zyBooks
The heat of combustion (Δhc) is the heat of reaction for a combustion 11/22/23
reaction. 23:26 1858759
The heating value
Enoch Tamale
(HV) is the negative of the heat of combustion. The symbolic relationship between these terms is:
UBCCHBE241BagherzadehFall2023

Δhr = Δhc = -HV

Analogous to the heat of reaction definitions, a temperature and pressure is normally specified for a
heat of combustion. Also, the standard heat of combustion (Δhoc) is the heat of combustion at 25°C
and 1 atm (Δhoc = Δhc(T=25°C, P=1 atm)).
The higher heating value (HHV) is the energy per mole of the complete combustion reaction of a fu

reacting with oxygen to form carbon dioxide and water in the liquid phase.

aF + bO2 → cCO2 + dH2O(l) Δhr = Δhc = -HHV

The lower heating value (LHV) is the energy per mole of the complete combustion reaction of a fue
reacting with oxygen to form carbon dioxide and water in the gas phase.

aF + bO2 → cCO2 + dH2O(g) Δhr = Δhc = -LHV ©zyBooks 11/22/23 23:26 1858759
Enoch Tamale
Combustion reactions release significant amounts of energy (e.g., UBCCHBE241BagherzadehFall2023
car engines or power plants). If a
reactor is well insulated and all of the heat from the combustion process remains in the reactor, the
products will increase in temperature. The exit temperature of the product gases is known as the
adiabatic flame temperature.

Several other common terms are defined to help identify safe operating conditions. Autoignition
temperature is the temperature where no external heat source is needed to ignite or burn the fuel. T
explosive range or lower and upper flammability limits are also important. The lower flammability
limit is the autoignition temperature. And the upper flammability limit is the highest temperature
where self-sustaining combustion can occur.

PARTICIPATION
ACTIVITY
7.4.1: Combustion and energy.

1) The autoignition temperature is always

equal to the lower flammability limit.
True
False

2) The complete combustion of propane

is:

C3H8(l) + 5 O2(g) → 3 CO2(g) + 4

H2O(g)

Δhr = _____
-LHV
-HHV ©zyBooks 11/22/23 23:26 1858759
Enoch Tamale
UBCCHBE241BagherzadehFall2023
3) The complete combustion of n-butane

is:

C4H10(l) + 6.5O2(g) → 4CO2(g) +

5H2O(g)

Δhc = -2860 kJ/mol

©zyBooks 11/22/23 23:26 1858759
The higher heating value can be found Enoch Tamale
using only the given information. UBCCHBE241BagherzadehFall2023

True
False

4) The complete combustion of n-butane

is:

C4H10(l) + 6.5O2(g) → 4CO2(g) +

5H2O(g)

Δhc1 = -2660 kJ/mol

The phase change of water is:

H2O(g) → H2O(l)

Δhr2 = -44.0 kJ/mol

Both reactions occur at 25°C in this

case, so the higher heating value can
be found using Hess’s law. The higher
heating value is _____ kJ/mol.
2880
2700

CHALLENGE
ACTIVITY
7.4.1: Heat of combustion calculations. ©zyBooks 11/22/23 23:26 1858759
Enoch Tamale
UBCCHBE241BagherzadehFall2023
495688.3717518.qx3zqy7

Start

Select the correct term from the list for the definition: The energy per mole of
the complete combustion reaction of a fuel reacting with oxygen to form carbon
 dioxide and water in the liquid phase. 

1 2 3 4 5

©zyBooks 11/22/23 23:26 1858759

Check Next Enoch Tamale
UBCCHBE241BagherzadehFall2023

Example 7.4.1: Mixed hydrocarbon combustion.

A mixed hydrocarbon fuel of butane (C4H10) and propane (C3H8) is burned in air (molar ratio 
of 3.76 N2: 1 O2). The feed to the reactor (fuel + air) is at 200°C and 1 atm combines a fresh
feed stream and a recycle stream. The hydrocarbon fuel entering the reactor at 25 mol/s and
is composed of 62 mol% butane and the rest propane while the air enters the reactor at 238
mol/s. The single pass conversions of the butane and propane are the same value. The
product stream contains 4.6 moles of carbon dioxide (CO2) for every mole of carbon
monoxide (CO). Additionally, 3 moles of water (H2O) for every mole of hydrogen (H2) are
produced. The product gas contains 3.5 mol/s CO. The products leave the reactor as an ideal
gas at 600°C and 3 atm. A separator splits the exit stream from the reactor into two streams:
a product stream containing all of the CO, CO2, H2, and water, and a stream containing the
unreacted hydrocarbons and air. The stream containing the unreacted components is split
with 80% of the total molar flow rate being recycled and the rest being purged. Additional
data are given to simplify calculations: The stream exiting the reactor (includes all
components) has an enthalpy of -9.3 kJ/mol and an internal energy of -13.5 kJ/mol
(reference temperature is 25°C).

The goals are to:

1. Calculate the component flow rates (mol/s) of all species exiting the reactor.
2. Calculate the overall conversion (%) of oxygen.
©zyBooks 11/22/23 23:26 1858759
3. Calculate the heat interaction of the reactor (kW). Enoch Tamale
4. Explore a concept question: The temperature of the feed toUBCCHBE241BagherzadehFall2023
the reactor increases by
100°C. Will the magnitude of the heat interaction of the reactor increase, decrease, or
stay the same?

Step by step solution

Step 1. The process flow diagram involves three process units, a reactor, a separator, and a

valve (labeled split in the PFD). The reactor leads into the separator whose exit stream is split
to create a recycle and a purge stream.

©zyBooks 11/22/23 23:26 1858759

Enoch Tamale
UBCCHBE241BagherzadehFall2023

Step 2. The basis is the feed to the reactor (Stream 2) as this stream is fully defined.

Step 3. With the feed to the reactor serving as the basis, the subsystem around reactor is a
logical place to start.

Steps 4, 5, and 6. With multiple fuels, complete combustion, partial combustion, and other
reactions creating hydrogen, the reaction stoichiometry is quite complicated. Therefore, atom
balances should be the preferred choice. Four atom balances can be written: C, H, O, and N.
The subscript P is for propane, and B is for butane.

©zyBooks 11/22/23 23:26 1858759

Enoch Tamale
UBCCHBE241BagherzadehFall2023
 3 atom C 4 atom C 3 atom C 4 atom C
C: ṅ 2 , P ⋅ 1 mol P + ṅ 2 , B ⋅ 1 mol B = ṅ 3 , P ⋅ 1 mol P + ṅ 3 , B ⋅ 1 mol B + 

1 atom C 1 atom C
ṅ 3 , CO ⋅ 1 mol CO + ṅ 3 , CO ⋅ 1 mol CO
2 2

8 atom H 10 atom H 8 atom H 10 atom H

H: ṅ 2 , P ⋅ 1 mol P + ṅ 2 , B ⋅ 1 mol B = ṅ 3 , P ⋅ 1 mol P + ṅ©zyBooks
3 , B ⋅ 1 mol + 23:26 1858759
11/22/23
B
Enoch Tamale
2 atom H 2 atom H UBCCHBE241BagherzadehFall2023
ṅ 3 , H ⋅ 1 mol H + ṅ 3 , H O ⋅ 1 mol H O
2 2 2 2

2 atom O 2 atom O 1 atom O

O: ṅ 2 , O ⋅ 1 mol O = ṅ 3 , O ⋅ 1 mol O + ṅ 3 , CO ⋅ 1 mol CO +
2 2 2 2

2 atom O 1 atom O
ṅ 3 , CO ⋅ 1 mol CO + ṅ 3 , H O ⋅ 1 mol H O
2 2 2 2

N: ṅ 2 , N = ṅ 3 , N
2 2

Step 7. Three additional extra equations are included in the problem statement. First, the
conversion of propane and butane are equal. Next, two relationships between flow rates of
components exiting the reactor are given.

ṅ 2 , B − ṅ 3 , B ṅ 2 , P − ṅ 3 , P
Conversion: f B = = fP = = f conv
ṅ 2 , B ṅ 2 , P

Flow rate relationship: ṅ 3 , CO = 4.6 ⋅ ṅ 3 , CO

2

Flow rate relationship: ṅ 3 , H O = 3 ⋅ ṅ 3 , H

2 2

Step 8. The 7 unknowns are the component flow rates exiting the reactor:

ṅ 3 , Bṅ 3 , Pṅ 3 , H ṅ 3 , H Oṅ 3 , CO ṅ 3 , O ṅ 3 , N

2 2 2 2 2

Step 9. With 4 balances and 3 extra equations, the 7 unknowns can be solved. Additional
systems are needed (Step 10), but the reactor subsystem will be solved first.
©zyBooks 11/22/23 23:26 1858759
Enoch Tamale
Solving the equations in Step 11 involves a number of steps. First, the nitrogen balance can
UBCCHBE241BagherzadehFall2023
be solved. Next, the CO flow rate was given in the problem statement and one extra equation
can then be solved for the CO2 flow rate.
 ṅ 2 , N = ṅ 3 , N
2 2 

mol
ṅ 3 , N = 188 s
2

mol
ṅ 3 , CO = 3.5 s
©zyBooks 11/22/23 23:26 1858759
Enoch Tamale
ṅ 3 , CO = 4.6 ⋅ ṅ 3 , CO UBCCHBE241BagherzadehFall2023
2

mol
ṅ 3 , CO = 16.1 s
2

Solving for the five additional unknowns uses the 5 remaining equations. The C balance
contains two unknowns: the exiting butane and propane flow rates. The conversion equation
contains the same two unknowns. Solving for the exiting butane and propane flow rates
follows:

mol 3 atom C mol 4 atom C mol 3 atom C

9.5 s ⋅ 1 mol P + 15.5 s ⋅ 1 mol B = 9.5 s ⋅ (1 − f conv) ⋅ 1 mol P +

mol 4 atom C mol 1 atom C mol 1 atom C

15.5 s ⋅ (1 − f conv) ⋅ 1 molB + 3.5 s ⋅ 1 mol CO + 16.1 s ⋅ 1 mol CO
2

f conv = 0.217

Next, use conversion to find flow rates of exiting B and P.

mol
ṅ 3 , B = 12.1 s

mol
ṅ 3 , P = 7.44 s

Now, the hydrogen balance has two unknowns as does the remaining extra equation. Solving
for the hydrogen and water flow rates should then be completed.

mol 8 atom H mol 10 atom H mol 8 atom H

9.5 s ⋅ 1 mol P + 15.5 s ⋅ 1 mol B = 7.44 s ⋅ 1 mol P +

mol 10 atom H 2 atom H ©zyBooks 11/22/23 23:26 1858759

2 atom H
Enoch
12.1 s ⋅ 1 mol B + ṅ 3 , H ⋅ 1 mol H + 3 ⋅ ṅ 3 , H ⋅ 1 mol H O Tamale
2 2 2
UBCCHBE241BagherzadehFall2023
2

mol
ṅ 3 , H = 6.31 s
2

mol
ṅ 3 , H O = 18.9 s
2
Finally, the oxygen balance has only one remaining unknown and can be solved.

mol 2 atom O 2 atom O mol 1 atom O
50.0 s ⋅ 1 mol O = ṅ 3 , O ⋅ 1 mol O + 3.50 s ⋅ 1 mol CO +
2 2 2

mol 2 atom O mol 1 atom O

16.1 s ⋅ 1 mol CO + 18.9 s ⋅ 1 mol H O
2 2

mol ©zyBooks 11/22/23 23:26 1858759

ṅ 3 , O = 22.7 s Enoch Tamale
2 UBCCHBE241BagherzadehFall2023

Checking the reactor subsystem (Step 12) can be completed by using C or H atom balances
without the extra equations inserted. The first goal has now been completed.

Returning from Step 10 back to Step 3 will help address the next goal of finding the overall
conversion of oxygen. First, the overall conversion is defined. Then the separator subsystem
is selected and solved next. Since oxygen is the only component of interest, the oxygen
component flow rate can be found for the separator subsystem.

ṅ 1 , O − ṅ 6 , O
2 2
f Overall , O =
2 ṅ 1 , O
2

ṅ 3 , O = ṅ 5 , O
2 2

mol
ṅ 5 , O = 22.7 s
2

Returning to Step 3 again identifies the split point as another subsystem that can be solved
for the oxygen component flow rate. Here, the purge ratio serves as an extra equation.

ṅ 5 , O = ṅ 6 , O + ṅ 7 , O
2 2 2

ṅ 7 , O = 0.8 ⋅ ṅ 5 , O
2 2

ṅ 5 , O = ṅ 6 , O + 0.8 ⋅ ṅ 5 , O
2 2 2

©zyBooks 11/22/23 23:26 1858759

mol Enoch Tamale
ṅ 6 , O = 4.54 s UBCCHBE241BagherzadehFall2023
2

mol
ṅ 7 , O = 18.2 s
2

One additional system is needed to complete to find the overall conversion of oxygen. The
mix point between Streams 1, 2, and 7 can now be solved for the fresh feed flow rate of
oxygen. Finally, the overall conversion is found to complete the second goal.


ṅ 1 , O + ṅ 7 , O = ṅ 2 , O
2 2 2

mol
ṅ 1 , O = 31.8 s
2

©zyBooks 11/22/23 23:26 1858759

ṅ 1 , O − ṅ 6 , O Enoch Tamale
2 2
f Overall , O = UBCCHBE241BagherzadehFall2023
2 ṅ 1 , O
2

f Overall , O = 0.857 or 85.7%

2

Finding the heat interaction of the reactor is the next goal. The reactor operates at steady
state, with no work, and changes in potential and kinetic energy are negligible. Thus, the
simplified energy balance includes an enthalpy term for each component k in Stream 2 and a
grouped component for Stream 3 as the problem statement provided the enthalpy of Stream
3.

Q̇ = ṅ 3 ⋅ h 3 − ∑ ṅ 2 , k ⋅ h 2 , k
k

Q̇ = ṅ 3 ⋅ h 3 − (ṅ 2 , B ⋅ h 2 , B + ṅ 2 , P ⋅ h 2 , P + ṅ 2 , O ⋅ h 2 , O + ṅ 2 , N ⋅ h 2 , N )
2 2 2 2

Making a table with the component flow rates and enthalpy values helps organize each
calculation. Enthalpy calculations for Stream 2 require a path that includes the standard heat
of formation and a temperature change step. Data for the heat of formation and heat
capacity are included in the Appendix.

200 oC
h 2 , k(T = 200 oC) = Δh of, k + ∫ 25 oC C p , k ⋅ dT

mol kJ
Component (k) ṅ 2 , i( ) h 2 , i( )
s mol

P 9.5 -87.8

B 15.5 -103.5
©zyBooks 11/22/23 23:26 1858759
Enoch Tamale
O2 50 5.3 UBCCHBE241BagherzadehFall2023

N2 188 5.1

Completing the heat interaction calculation accomplishes the fourth goal.

 Q̇ = ṅ 3 ⋅ h 3 − ∑ ṅ 2 , k ⋅ h 2 , k 
k

Q̇ = − 2560 kW − ( − 1210 kW)

Q̇ = − 1350 kW

Finally, the concept question asks: The temperature of the feed to©zyBooks
the reactor increases
11/22/23 23:26by
1858759
Enoch Tamale
100°C. Will the magnitude of the heat interaction of the reactor increase, decrease, or stay
UBCCHBE241BagherzadehFall2023
the same?

The enthalpies of the inlet components will increase with increasing temperature. So the heat
interaction will become a smaller negative number. Therefore, the magnitude, or absolute
value, of the heat interaction will decrease. The reactor has a smaller heat loss when the
temperature difference between the inlet and exit streams gets smaller.

PARTICIPATION
ACTIVITY 7.4.2: Mixed hydrocarbon details.

Using the example worked above as reference, answer the following questions.

1) In steps 4, 5, and 6, atom balances

were selected when solving the reactor
subsystem. Identify one reason for
selecting atom balances over other
types of balances.
Moles are always conserved.
Energy is conserved.
Multiple reactions occur in the
reactor.

2) The overall conversion of oxygen was

found after solving the separator and
mix point subsystems. Now, calculate
the single pass conversion of oxygen.
©zyBooks 11/22/23 23:26 1858759
85.7% Enoch Tamale
UBCCHBE241BagherzadehFall2023
54.6%
50.0%
 3) Complete the enthalpy calculation for

the stream exiting the reactor.

kJ
ṅ 3 ⋅ h 3 = _____ ⋅ − 9.3 mol

263 mol/s
©zyBooks 11/22/23 23:26 1858759
275 mol/s Enoch Tamale
UBCCHBE241BagherzadehFall2023
188 mol/s

4) The heat interaction for the reactor was

found to be -1350 kW. Therefore, the
set of reactions should be considered
_____.
Adiabatic
Endothermic
Exothermic

fitness_center EXERCISE 7.4.1: Ethane combustion energy balance. help_outline

Ethane (C2H6) is combusted in 18.1% excess oxygen (O2) in an adiabatic reactor. The feed of
ethane enters the reactor at 25.8 mol/hr and 93.7°C. The stream exiting the reactor contains
ethane, oxygen, carbon dioxide (CO2), carbon monoxide (CO) and water (H2O), so ethane reacts
by both complete and partial combustion reactions. The conversion of ethane is 15.8% and the
conversion of oxygen is 12.3%. Assume that pressure effects are negligible.

(a) Draw and label a process flow diagram. Number each stream and label the components
(E=ethane; O2=oxygen; CO2=carbon dioxide, CO=carbon monoxide; W=water).
Solution keyboard_arrow_down
(b) Calculate the component molar flow rates (mol/hr) for all of the exiting components.
Solution keyboard_arrow_down
©zyBooks 11/22/23 23:26 1858759
(c) Calculate the adiabatic flame temperature (°C) for the reactor. Enoch Tamale
UBCCHBE241BagherzadehFall2023
Solution keyboard_arrow_down
(d) If the reactor’s insulation is damaged and the reactor could lose heat to the surroundings,
would the exit temperature increase, decrease, or stay the same?
Solution keyboard_arrow_down
7.5 Simultaneous material and energy balances

Learning objectives
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Generate both material and energy balances for an engineering problem.
Enoch Tamale
Solve, or simultaneously solve, systems of material and energy balances.
UBCCHBE241BagherzadehFall2023

Many real world chemical engineering problems are more difficult than the examples provided here.
While material balances and energy balances can be solved independently in many cases,
simultaneous solutions of the balances may be required for more complicated systems.
Computations may be completed by hand in some cases or require software tools. A list of commonly
used software tools for solving set of equations and balances is included at the end of the section.
One example will demonstrate a difficult example where solving a set of simultaneous equations and
balances is an efficient problem solving technique.

Example 7.5.1: Unknown hydrocarbon combustion.

An unknown molecule CxHyOz and O2 are fed to a reactor at 100°C and 1 atm. The flow rate
of O2 (in units of mol/s) is related to the heat interaction (in units of kW) as: − 486 ⋅ ṅ O = Q̇.
2
The product stream emerges from the reactor at 1 atm and 25°C contains 15 mol/s of CO, 30
mol/s of CO2, and 60 mol/s of H2O. The sum of x, y, and z in the unknown molecule is 17.
o kJ kJ
Some properties of the unknown molecule are available: Δh f = − 104 mol and C P = 0.07
mol ⋅ oC
.

Assuming that all reactants and products are in the gas phase, the goals are to:

1. Determine the composition of the hydrocarbon (Find x, y, and z).

2. Determine the component molar flow rates of the reactants (mol/s).
3. Calculate the heating or cooling rate (in kW) of the reactor.
4. Explore a concept question: The product stream now leaves the process at 5 atm.
Assuming the product gases behave ideally, will the heating©zyBooks
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reactor increase, decrease, or stay the same?
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Step 1. The process flow diagram involves one process unit, a reactor.
Step 2. The basis is the stream exiting the reactor (Stream 3), as all three component flow
rates are given. ©zyBooks 11/22/23 23:26 1858759
Enoch Tamale
UBCCHBE241BagherzadehFall2023
Step 3. The single process unit serves as the only system of interest.

Steps 4, 5, and 6. Atom balances should be the preferred choice as the number of atoms in
the fuel are an unknown. The subscript U is used for the unknown fuel.

1C 1C
C: x ⋅ ṅ U = mol CO ⋅ ṅ CO + mol CO ⋅ ṅ CO
2 2

2H
H: y ⋅ ṅ U = mol H O ⋅ ṅ H O
2 2

2O 1O 2O 1O
O: z ⋅ ṅ U + mol O ⋅ ṅ O = mol CO ⋅ ṅ CO + mol CO ⋅ ṅ CO + mol H O ⋅ ṅ H O
2 2 2 2 2 2

Step 7. Two extra equations are included in the problem statement. The first extra equation is
the relationship between the heat interaction and the oxygen flow rate, and the second
equation is the sum of the coefficients in the unknown hydrocarbon.

− 486 ⋅ ṅ O = Q̇
2

x + y + z = 17

Step 8. The 6 unknowns are x, y, and z; the component flow rate of oxygen, the component
flow rate of unknown molecule, and the heat transfer rate:
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xyzṅ 1 , Uṅ 1 , O Q̇ Enoch Tamale
2
UBCCHBE241BagherzadehFall2023
Step 9. The C balance has 2 unknowns, the H balance has 2 unknowns, and the O balance
has 3 unknowns. The feed flow rate-heat interaction relationship has 2 unknowns, and the
sum of coefficient for unknown hydrocarbon has 3 unknowns. Therefore, with 5 equations
and 6 unknowns, the material balances cannot be solved. The energy balance provides the
6th independent equation for completing this problem.
The energy balance is simplified by a number of assumptions: steady state, no changes in
kinetic or potential energy, and no work. Simplifying the energy balance gives:

∑
inlets
⏟
1 →2
ṁ i(h i + 2 v i + gz i
⏟
)−
exits
⏟
1
∑ ṁ e(h e + 2 →v 2e + gz i
⏟
) + Q̇ + Ẇ
⏟
= ( ) dU
dt Sys

No ΔKE No ΔPE No ΔKE No ΔPE Rigid Vessel ⏟

Steady State
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Enoch Tamale
∑ ṁ ih i − ∑ ṁ eh e + Q̇ = 0 UBCCHBE241BagherzadehFall2023
inlets exits

ṅ U ⋅ h U(T in) + ṅ O ⋅ h O (T in) − ṅ CO ⋅ h CO(T out) − ṅ CO ⋅ h CO (T out) − ṅ H O ⋅ h H O(T out) + Q̇ = 0

2 2 2 2 2 2

Next, all enthalpies can be calculated from heat of formation and heat capacity data in the
Appendix.

T kJ kJ kJ
h U(T in) = h U(100 oC) = Δh 0f + ∫ T in C PdT = − 104 mol + 0.07 o
oC ⋅ (100 − 25) C = − 99 mol
ref mol ⋅

T kJ kJ
h O (T in) = h O (100 oC) = Δh 0f + ∫ T in C PdT = 0 + 0.03 ⋅ (100 − 25) oC = 2 mol
2 2 ref mol ⋅ oC

kJ
h CO(T out) = h CO(25 oC) = Δh 0f = − 111 mol

0 kJ
h CO (T out) = h CO (25 oC) = Δh f = − 393 mol
2 2

0 kJ
h H O(T out) = h H O(25 oC) = Δh f = − 242 mol
2 2

Inserting the enthalpy values into the energy balance, with each term in kW, gives the final,
simplified equation. Additional systems are not needed (Step 10) as the reactor is the only
system. Now, a set of 6 equations and 6 unknowns can be solved using a computational
solver (Step 11).

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ṅ U ⋅ − 99 + ṅ O ⋅ 2 + 1665 + 11790 + 14520 + Q̇ = 0
2 Enoch Tamale
UBCCHBE241BagherzadehFall2023

mol mol
x = 4.07, y = 10.8, z = 2.09, Q̇ = − 27.2 MW, ṅ U = 11.1 s , ṅ O = 55.9 s
2

Step 12. Check using any single balance or extra equation.

Finally, the concept question asks: The product stream now leaves the process at 5 atm.
Assuming the product gases behave ideally, will the heating or cooling rate of the reactor
increase, decrease, or stay the same?

Recalling that enthalpy changes of an ideal gas are independent of pressure changes, the
previously derived equation was: Δh = Δu + RΔT. Both terms are only functions of
temperature, and independent of pressure, so the heat interaction should stay the same.
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Enoch Tamale
UBCCHBE241BagherzadehFall2023
PARTICIPATION
ACTIVITY 7.5.1: Unknown hydrocarbon details.

Based on the example above, answer the following questions.

1) The quantities x, y, and z cannot be any

value. Selecting the correct limitation, x,
y, and z must be _____.
An integer
Less than five
A positive number

2) Mass balances would be very difficult

to write, as the molecular weight of the
fuel adds an extra unknown to the set
of equations. Based on the final
answer, the molecular weight of the
fuel is:
93
12
29

3) The reactor is in a facility that uses

English units. Convert the heat
interaction from -27,200 kW to
_____BTU/hr.
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-25,800 Enoch Tamale
UBCCHBE241BagherzadehFall2023
-92,800,000
-97,900,000

Exploring further:
 Common tools for solving sets of equations:

Wolfram Alpha
MATLAB
Mathematica
Polymath
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Solver in Microsoft Excel Enoch Tamale
UBCCHBE241BagherzadehFall2023

7.6 Reaction + energy balances problems

Additional problems

CHALLENGE
ACTIVITY
7.6.1: Reaction + energy balance vocabulary.

Select the term that best matches the following:

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is a general technique to solve energy balances for reacting systems.

1 2 3 4

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CHALLENGE
ACTIVITY 7.6.2: Fluidized bed biomass reactor.

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 Engineers Angeline and Matt operate a biomass fluidized bed reactor. The composition of th
is unknown (B or C\(_X\)H\(_{Y}\)O\(_{Z}\)).The feed to the reactor contains biomass and 20
water (W or H\(_2\)O) and a total flow rate of 9.63 mol/s. The reaction stoichiometry is unkn
gas product stream is produced. The product stream exits at 23.2 mol/s and contains mole
0.536 for CO\(_2\), 0.142 for acetic acid (AA or CH\(_3\)COOH), and the balance water. The f
at 167\(^o\)C and the products exit at 331\(^o\)C. Assume all components are in the gas pha
phase changes occur in the reactor.
After drawing and labeling a process flow diagram including numbering
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each stream and
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Enoch
components (AA, B, CO\(_2\), W) in each stream, answer the following question. Tamale
UBCCHBE241BagherzadehFall2023

How many components exit in the vapor phase?

Ex:8

1 2 3 4 5 6 7

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CHALLENGE
ACTIVITY 7.6.3: Acetic acid and hydrogen reactor.
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Engineers Lucy and Urban operate a reactor where two unbalanced, gas phase reactions occ
CH\(_3\)COOH + H\(_2\) \(\rightarrow\) C\(_2\)H\(_5\)OH + H\(_2\)O and C\(_2\)H\(_5\)OH
(\rightarrow\) (C\(_2\)H\(_5\))\(_2\)O + H\(_2\)O
In the first reaction, acetic acid (CH\(_3\)COOH) and hydrogen (H\(_2\)) react to produce eth
(_2\)H\(_5\)OH) and water (H\(_2\)O); and ethanol decomposes to diethyl ether ((C\(_2\)H\(_
(_2\)O) and water in the second reaction. A stream of 54.6 mol/s hydrogen and the balance a
at 173\(^o\)C is fed to the reactor, which operates at steady state. The stream exiting the rea
contains acetic acid, hydrogen, ethanol, water, and diethyl ether at 311\(^o\)C. The compone
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rate of acetic acid entering the reactor is 14.1 mol/s. The fraction conversion
Enochof Tamale
acetic acid is
and the selectivity, which is the ratio of the amount of ethanol toUBCCHBE241BagherzadehFall2023
diethyl ether in the exiting s
8.27. Assume all components are in the gas phase and no phase changes occur in the react

Determine the unknown component molar flow rates in the reactor effluent.

\(\dot n_{2,H} =\,\) \(\dot n_{2,E} =\,\) \(\dot n_{2,W} =\,\) \(\dot n_{2,DE}
 Ex: 30.3 mol/s Ex: 6.37 mol/s Ex: 16.0 mol/s Ex: 1.28 m

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Enoch Tamale
UBCCHBE241BagherzadehFall2023
CHALLENGE
ACTIVITY
7.6.4: Acetic acid parallel reactions with energy balance.

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Parallel gas phase reactions occur between acetic acid (AA, CH\(_3\)COOH) and hydrogen (H
(_2\)). Engineering students Hildegard and Stephen identify the reaction products as ethanol
(_2\)H\(_5\)OH) and water (W, H\(_2\)O) for the first reaction, ethyl acetate (EA, C\(_4\)H\(_8
and water for the second reaction. The reactor operates at steady state with acetic acid and
entering at 137\(^o\)C.The reactor effluent contains acetic acid, ethanol, water, and ethyl ace
281\(^o\)C. Three component flow rates exiting the reactor are known: 14.4 mol/min acetic a
mol/min ethanol, and 0.912 mol/min ethyl acetate. Assume all components are in the gas ph
no phase changes occur in the reactor.

Determine the unknown component molar flow rates.

 

\(\dot n_{1,AA} =\,\) \(\dot n_{1,H} =\,\) \(\dot n_{2,W} =\,\)

Ex: 30.7 mol/min Ex: 33.5 mol/min Ex: 34.9 mol/min

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UBCCHBE241BagherzadehFall2023

CHALLENGE
ACTIVITY
7.6.5: Butanol combustion energy balance.

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 Engineers Mike and Sue alternate working the day and night shifts operating a combustion r
from the air-conditioned control room where the temperature is a comfortable 68.5\(^o\)F. B
C\(_4\)H\(_9\)OH) and dry air enter the continuous steady-state reactor as a single stream. T
inlet flow rate to the reactor at 151\(^o\)C is 338 mol/s with 0.0582 mol fraction butanol, 0.2
fraction oxygen (O\(_2\)) and the balance nitrogen (N\(_2\)). Both complete and partial comb
reactions occur, and the exit stream contains butanol, oxygen, nitrogen, water (W, H\(_2\)O),
monoxide (CO), and carbon dioxide (CO\(_2\)). Based on the composition of the exit
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engineers have measured the conversion of butanol to be 54.7%, the ratio of CO\(_2\):CO is 7
UBCCHBE241BagherzadehFall2023
and the temperature of the exiting stream to be 294\(^o\)C. Assume all components are in th
phase and no phase changes occur in the reactor.

Determine some unknown component molar flow rates.

\(\dot n_{2,O_2} =\,\) \(\dot n_{2,CO_2} =\,\) \(\dot n_{2,W} =\,\)

Ex: 15.4 mol/s Ex: 51.9 mol/s Ex: 36.7 mol/s

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CHALLENGE 
ACTIVITY 7.6.6: Pentane combustion energy balance.

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Combustion reactors produce a signifcant amount of heat that can serve many uses in a lar
so engineers Ann and Sylvester are completing some calculations. n-Pentane (P, C\(_5\)H\(_
and dry air enter a continuous, steady-state reactor as a single stream. Flowing into the reac
123\(^o\)C is 2090 mol/hr n-pentane, some oxygen (O\(_2\)), and the balance nitrogen (N\(_
Oxygen is fed at 45.8% excess. For every mole of oxygen in the feed, 3.75 moles nitrogen are
complete combustion occurs, so the reactor effluent contains all©zyBooks
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water (W, H\(_2\)O), and carbon dioxide (CO\(_2\)). The reactors is specified to maintain a co
UBCCHBE241BagherzadehFall2023
of n-pentane as 81.5%. The stream exiting the reactor is measured as 237\(^o\)C. Assume a
components are in the gas phase and no phase changes occur in the reactor.

Determine some unknown component molar flow rates.

\(\dot n_{2,O_2} =\,\) \(\dot n_{2,CO_2} =\,\) \(\dot n_{2,W} =\,\)

 Ex: 7340 mol/hr Ex: 6.30E+ mol/hr Ex: 14500 mol/hr

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UBCCHBE241BagherzadehFall2023
CHALLENGE
ACTIVITY 7.6.7: Sugars fermentation adiabatic reactor.

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William and Angela are college students discussing how ethanol (E, C\(_2\)H\(_5\)OH) can b
produced by the fermentation of sugars (S, C\(_{7.2}\)H\(_{12.3}\)O\(_{8.25}\)) in the presenc
yeast according to the following unbalanced chemical reaction: C\(_{7.2}\)H\(_{12.3}\)O\(_{8
(\rightarrow\) C\(_2\)H\(_5\)OH(l) + CO\(_2\)(g). Flowing into the reactor at 25.0\(^o\)C and
mol/day are water (W, H\(_2\)O) with a mole fraction of 0.952, sugars with a mole fraction of
and the balance yeast (Y). Water is an inert component, and yeast can be consider a non-con
catalyst. The reactor effluent contains water, yeast, unreacted sugars, ethanol, and carbon di
(CO\(_2\)). At 55.0\(^o\)C, the stream exiting the reactor contains 12.9 mol/day carbon dioxi
 

Determine some unknown component molar flow rates.

\(\dot n_{2,S} =\,\) \(\dot n_{2,Y} =\,\) \(\dot n_{2,E} =\,\)

Ex: 1.11 mol/day Ex: 0.715 mol/day Ex: 5.80 mol/day

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