KWAME NKRUMAH UNIVERSITY OF SCIENCE AND TECHNOLOGY

CHEMICAL ENGINEERING DEPARTMENT

CHE 252: CHEMICAL PROCESS CALCULATIONS II
INSTRUCTOR: Dr. (Mrs.) Mizpah A. D. Rockson

LECTURE 7: ENERGY BALANCE ON REACTIVE PROCESSES

Learning Objectives

At the end of this lecture the students are expected to do the following:

1. Explain the concepts of heat of reaction; heat of formation; combustion; heat of

combustion; standard heats of formation, combustion, and reaction;
2. Calculate the total enthalpy change, given (a) the amount of any reactant consumed or any
product generated in a reaction at a given temperature and pressure and (b) the heat of the
reaction at that temperature and pressure.
3. Determine a heat of reaction from heats of other reactions using Hess's law. Determine
standard enthalpies and internal energies of reaction from known standard heats of
formation and heats of combustion.
4. Solve reactive-system energy balance problems for (a) the heat transfer required for
specified inlet and outlet conditions, (b) the outlet temperature corresponding to a specified
heat input (e.g., for an adiabatic reactor), and (c) the product composition corresponding to
a specified heat input and a specified outlet temperature.

8.1 HEATS OF REACTION

The heat of reaction (or enthalpy of reaction), ∆𝐻̂𝑟 (T, P), is the enthalpy change for a process in
which stoichiometric quantities of reactants at temperature T and pressure P react completely in a
single reaction to form products at the same temperature and pressure.

Consider the reaction between solid calcium carbide and liquid water to form solid calcium
hydroxide and gaseous acetylene,

𝐶𝑎𝐶2 (𝑠) + 2𝐻2 𝑂(𝑙) → 𝐶𝑎(𝑂𝐻)2 (𝑠) + 𝐶2 𝐻2 (𝑔)

̂𝑟 (25°C, 1 atm) = -125.4 kJ/mol

the heat of reaction for this at 25°C and 1 atm is ∆𝐻

From the above it means that if 1 mol of solid calcium carbide reacts completely with 2 mol of
liquid water to form 1 mol of solid calcium hydroxide and 1 mol of gaseous acetylene, and the
initial and final temperatures are both 25°C and the initial and final pressures are both 1 atm, then

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Hproducts - Hreactants = -125.4 kJ. If the reaction is run under conditions such that the energy balance
reduces to 𝑄 = ∆𝐻, then 125.4 kJ of heat must be transferred from the reactor in the course of the
reaction.

If a reaction takes place at a temperature To and pressure Po and the extent of reaction is ξ, the
associated enthalpy change is
∆𝐻 = 𝜉∆𝐻 ̂𝑟 (𝑇𝑜 , 𝑃𝑜 )

For a continuous process, nA,r (mol) would be replaced by 𝑛̇ 𝐴,𝑟 (mol/s) in this expression, ξ(mol)
would be replaced by ξ̇(mol/s), and ∆𝐻 (kJ) would be replaced by ∆𝐻̇ (kJ/s).

Following are several important terms and observations related to heats of reaction:

1. If ∆𝐻̂𝑟 (T, P) is negative the reaction is exothermic at temperature T and pressure P, and if
∆𝐻 ̂𝑟 (T, P) is positive the reaction is endothermic at T and P. These definitions of exothermic
and endothermic are equivalent to the ones given earlier in terms of chemical bond strengths.

2. At low and moderate pressures, ∆H ̂ r (T, P) is nearly independent of pressure. We will presume
̂ r (T).
this independence in the balance of in this lecture and write the heat of reaction as ∆H

3. The value of the heat of a reaction depends on how the stoichiometric equation is written.

For example,
̂𝑟1(25℃) = −890.3 𝑘𝐽⁄𝑚𝑜𝑙
𝐶𝐻4 (𝑔) + 2𝑂2 (𝑔) → 𝐶𝑂2 (𝑔) + 2𝐻2 𝑂(𝑙): ∆𝐻

̂𝑟2 (25℃) = −1780.6 𝑘𝐽⁄𝑚𝑜𝑙

2𝐶𝐻4 (𝑔) + 4𝑂2 (𝑔) → 2𝐶𝑂2 (𝑔) + 4𝐻2 𝑂(𝑙): ∆𝐻

The first line states that the combined enthalpy of 1 gram-mole of CO2 plus 2 gram-moles of liquid
water is 890.3 kJ lower than the combined enthalpy of 1 gram-mole of methane plus 2 gram-moles
of oxygen at 25°C. Doubling the quantity of reactants at a given condition doubles the total
enthalpy of the reactants at that condition, and similarly for the products.

The difference between the product and reactant enthalpies in the second reaction (by definition,
̂𝑟2 must therefore be double the enthalpy difference in the first reaction ∆𝐻
∆𝐻 ̂𝑟1.

4. The value of a heat of reaction depends on the states of aggregation (gas, liquid, or solid) of the
reactants and products. For example,

̂𝑟1(25℃) = −890.3 𝑘𝐽⁄𝑚𝑜𝑙

𝐶𝐻4 (𝑔) + 2𝑂2 (𝑔) → 𝐶𝑂2 (𝑔) + 2𝐻2 𝑂(𝑙): ∆𝐻

̂𝑟1 (25℃) = −802.3 𝑘𝐽⁄𝑚𝑜𝑙

𝐶𝐻4 (𝑔) + 2𝑂2 (𝑔) → 𝐶𝑂2 (𝑔) + 2𝐻2 𝑂(𝑔): ∆𝐻

The only difference between the reactions is that the water formed is a liquid in the first one and a
vapor in the second. Since enthalpy is a state function, the difference between the two heats of

2
reaction must be the enthalpy change associated with the vaporization of 2 mol of water at 25°C-
̂𝑣 (25°C).
that is, 2∆𝐻

5. The standard heat of reaction, ∆𝐻̂𝑟𝑜 is the heat of reaction when both the reactants and products
are at a specified reference temperature and pressure, usually 25°C and 1 atm.

Example 7.1 Calculation of Heats of Reaction

The standard heat of the combustion of n-butane vapor is

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C4 H10 (g) + ̂ ro (25℃) = −2878 kJ⁄mol
O2 (g) → 4CO2 (g) + 5H2 O(l): ∆H
2

Calculate the rate of enthalpy change, ∆Ḣ (kJ/s), if 2400 mol/s of CO2 is produced in this reaction
and the reactants and products are all at 25°C.

2. What is the standard heat of the reaction

2C4 H10 (g) + 13O2 (g) → 8CO2 (g) + 10H2 O(l)

Calculate ∆Ḣ if 2400 mol/s of CO2 is produced in this reaction and the reactants and products are
all at 25OC.

3. The heats of vaporization of n-butane and water at 25°C are 19.2 kJ/mol and 44.0 kJ/mol
respectively.
What is the standard heat of the reaction

13
C4 H10 (l) +O (g) → 4CO2 (g) + 5H2 O(v)
2 2
Calculate ∆Ḣ if 2400 mol/s of CO2 is produced in this reaction and the reactants and products are
all at 25°C.

Solution
1.
(𝑛̇ 𝐶𝑂2 )𝑟 2400 𝑚𝑜𝑙 ⁄𝑠
𝜉= = = 600 𝑚𝑜𝑙 ⁄𝑠
|𝑣𝐶𝑂2 | 4

𝑚𝑜𝑙 𝑘𝐽
∆𝐻̇ = 𝜉̇ ∆𝐻
̂𝑟𝑜 = (600 ) (−2878 ) = −1.73 × 106 𝑘𝐽/𝑠
𝑠 𝑚𝑜𝑙

2. Since doubling the stoichiometric coefficients of a reaction must double the heat of reaction
∆𝐻̂𝑟2
𝑜
= 2∆𝐻̂𝑟1
𝑜
= 2(−2878 𝑘𝐽⁄𝑚𝑜𝑙) = −5756 𝑘𝐽⁄𝑚𝑜𝑙

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 The enthalpy change associated with the production of 2400 mol/s of CO2 at 25°C cannot
depend on how the stoichiometric equation is written (the same quantities of reactants and
products at the same temperatures must have the same enthalpies), and so ∆𝐻̇ must be the
value calculated in question 1 (a). Let us do the calculation and prove it, however.
(𝑛̇ 𝐶𝑂2 )𝑜𝑢𝑡 2400 𝑚𝑜𝑙 ⁄𝑠
𝜉= = = 300 𝑚𝑜𝑙 ⁄𝑠
|𝑣𝐶𝑂2 | 8

𝑚𝑜𝑙 𝑘𝐽
∆𝐻̇ = 𝜉̇ ∆𝐻
̂𝑟𝑜 = (300 ) (−5756 ) = −1.73 × 106 𝑘𝐽/𝑠
𝑠 𝑚𝑜𝑙

3. Compare the two reactions:

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̂ r1
C4 H10 (g) + O2 (g) → 4CO2 (g) + 5H2 O(l): ∆H o (25℃)
= −2878 kJ⁄mol
2
13
C4 H10 (l) + ̂ r2
O (g) → 4CO2 (g) + 5H2 O(v): ∆H o (25℃)
=?
2 2
The total enthalpy of the products in the second reaction [4 mol CO2(g) + 5 mol H2O(g) at 25°C]
is greater than that of the products in the first reaction [4 mol CO2(g) + 5 mol H2O(l) at 25°C] by
five times the heat of vaporization of water. Similarly, the total enthalpy of the reactants in the
second reaction is lower than that of the reactants in the first reaction by the heat of vaporization
of butane.
Since∆𝐻 ̂𝑟 = Hproducts - Hreactants, it follows that

̂ r2
∆H o ̂ r1
= ∆H o ̂𝑣 )H + (∆𝐻
+ 5(∆𝐻 ̂𝑣 )C = [−2878 + 5(44.0) + 19.2] kJ⁄mol
2O 4H10

= −2639 kJ⁄mol

𝑚𝑜𝑙 𝑘𝐽
∆𝐻̇ = 𝜉̇ ∆𝐻
̂𝑟2
𝑜
= (600 ) (−2639 ) = −1.58 × 106 𝑘𝐽/𝑠
𝑠 𝑚𝑜𝑙

For a closed System

If a reaction takes place in a closed reactor at constant volume, the heat released or absorbed is
determined by the change in internal energy between reactants and products. The internal energy
of reaction, ∆𝑈 ̂𝑟 (T), is the difference Uproducts - Ureactants if stoichiometric quantities of reactants
react completely at temperature T.

Suppose a reaction occurs, and vi is the stoichiometric coefficient of the ith gaseous reactant or
product. If ideal gas behavior can be assumed and specific volumes of liquid and solid reactants
and products are negligible compared with those of the gases, the internal energy of reaction is
related to the heat of reaction bv

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 ̂𝑟 (𝑇) = ∆𝐻
∆𝑈 ̂𝑟 (𝑇) − 𝑅𝑇 ( ∑ |𝑣𝑖 | − ∑ |𝑣𝑖 |)
𝑔𝑎𝑠𝑒𝑜𝑢𝑠 𝑔𝑎𝑠𝑒𝑜𝑢𝑠
𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠
For example
19
C6 H14 (l) + O (g) → 6CO(g) + 7H2 O(v)
2 2
The internal energy of reaction is
19 7
̂𝑟 (𝑇) = ∆𝐻
∆𝑈 ̂𝑟 (𝑇) − 𝑅𝑇(6 + 7 − ̂𝑟 (𝑇) − 𝑅𝑇
) = ∆𝐻
2 2
̂𝑟 = ∆𝐻
If there are no gaseous reactants or products then to a good approximation ∆𝑈 ̂𝑟 .

̂𝒓
Example 7.2 Evaluation of ∆𝑼

The standard heat of reaction

C2 H4 (g) + 2Cl2 (g) → C2 HCl3 (l) + H2 (g) + HCl(g)

̂ ro = −420.8 kJ⁄mol. Calculate ∆U

is ∆H ̂ ro for this reaction.

Solution
From stoichiometric equation
∑ 𝑣𝑖 (𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑔𝑎𝑠𝑒𝑠) = 1 + 1 = 2

∑ 𝑣𝑖 (𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡 𝑔𝑎𝑠𝑒𝑠) = 1 + 2 = 3

̂𝑟 (𝑇) = ∆𝐻
∆𝑈 ̂𝑟 (𝑇) − 𝑅𝑇 ( ∑ |𝑣𝑖 | − ∑ ̂𝑟 − 𝑅𝑇(2 − 3)
|𝑣𝑖 |) = ∆𝐻
𝑔𝑎𝑠𝑒𝑜𝑢𝑠 𝑔𝑎𝑠𝑒𝑜𝑢𝑠
𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠
8.314 𝐽 1 𝑘𝐽
= −420.8 𝑘𝐽⁄𝑚𝑜𝑙 − ( ) × (298 𝐾) × (−1) × ( 3 )
𝑚𝑜𝑙 ∙ 𝐾 10 𝐽
= −418.3 𝑘𝐽/𝑚𝑜𝑙

7.2 MEASUREMENT AND CALCULATION OF HEATS OF REACTION: HESS' LAW

Hess's law states that if the stoichiometric equation for reaction 1 can be obtained by algebraic
operations (multiplication by constants, addition, and subtraction) on stoichiometric equations for
̂𝑟1
reactions 2,3, ... , then the heat of reaction ∆𝐻 𝑜
can be obtained by performing the same operations
on the heats of reactions ∆𝐻 ̂𝑟2 , ∆𝐻
𝑜 ̂𝑟3...
𝑜

Example 7.3 Hess’ Law

The standard heats of the following combustion reactions have been determined experimentally:

5
 7
̂𝑟1
(1) 𝐶2 𝐻6 + 2 𝑂2 → 2𝐶𝑂2 + 3𝐻2 𝑂: ∆𝐻 𝑜
= −1559.8 𝑘𝐽⁄𝑚𝑜𝑙
(2) 𝐶 + 𝑂2 → 𝐶𝑂2 : ∆𝐻̂𝑟2
𝑜
= −393.5 𝑘𝐽⁄𝑚𝑜𝑙
1
̂𝑟3
(3) 𝐻2 + 2 𝑂2 → 𝐻2 𝑂: ∆𝐻 𝑜
= −285.8 𝑘𝐽⁄𝑚𝑜𝑙
Use Hess’s law and the given heats of reaction to determine the standard heat of the reaction
(4) 2𝐶 + 3𝐻2 → 𝐶2 𝐻6 : ∆𝐻̂𝑟4
𝑜
=?

Since form equations above (1-4) it can be seen that (4) = 2 × (2) + 3 × (3) − (1)

From Hess’s law

̂𝑟4
∆𝐻 𝑜 ̂𝑟2
= 2∆𝐻 𝑜 ̂𝑟3
+ 3∆𝐻 𝑜 ̂𝑟1
− ∆𝐻 𝑜
= 2(−393.5) + 3(−285.8) −(−1559.8) = −84.6 𝑘𝐽⁄𝑚𝑜𝑙

This heat of reaction could not have been measured directly, since you cannot react carbon and
hydrogen in such a way that ethane is the only reaction product.

Example 7.4
Calculate the heat of combustion for C2H6 from the following reactions:
(1) 𝐶2 𝐻4 + 3𝑂2 → 2𝐶𝑂2 + 2𝐻2 𝑂: ∆𝐻 ̂𝑟1
𝑜
= −1409.5 𝑘𝐽⁄𝑚𝑜𝑙
(2) 𝐶2 𝐻4 + 𝐻2 → 𝐶2 𝐻6 : ∆𝐻̂𝑟2 = −136.7 𝑘𝐽⁄𝑚𝑜𝑙
𝑜
1
(3) 𝐻2 + 𝑂2 → 𝐻2 𝑂: ∆𝐻 ̂𝑟3
𝑜
= −285.8 𝑘𝐽⁄𝑚𝑜𝑙
2
The reaction for combustion of ethane is as follows:
𝐶2 𝐻6 + 3.5𝑂2 → 2𝐶𝑂2 + 3𝐻2 𝑂: ∆𝐻 ̂𝑟1
𝑜
=?

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7.3 FORMATION REACTIONS AND HEATS OF FORMATION

A formation reaction of a compound is the reaction in which the compound is formed from its
elemental constituents as they normally occur in nature (e.g., O2 rather than O). The enthalpy
change associated with the formation of 1 mole of the compound at a reference temperature and
̂𝑓𝑜 .
pressure (usually 25°C and 1 atm) is the standard heat of formation of the compound, ∆𝐻

Standard heats of formation for many compounds are listed in Table B.1 (Felder and Rosseau) and
on pp. 2-187 through 2-198 of Perry's Chemical Engineers' Handbook. For example, ∆𝐻 ̂𝑓𝑜 for
crystalline ammonium nitrate is given in Table B.1 as - 365.14 kJ/mol, signifying
3
𝑁2 (𝑔) + 2𝐻2 (𝑔) + 𝑂2 (𝑔) → 𝑁𝐻4 𝑁𝑂3 (𝑐): ∆𝐻 ̂𝑓𝑜 = −365.14 𝑘𝐽⁄𝑚𝑜𝑙
2
̂𝑓𝑜 = 48.66 kJ/mol or
Similarly, for liquid benzene ∆𝐻
̂𝑓𝑜 = 48.66 𝑘𝐽/𝑚𝑜𝑙
6𝐶(𝑠) + 3𝐻2 (𝑔) → 𝐶6 𝐻6 (𝑙): ∆𝐻
The standard heat of formation of an elemental species (e.g., O2) is zero.

It may be shown using Hess's law that if vi is the stoichiometric coefficient of the ith species
̂𝑓𝑖
participating in a reaction (+ for products, - for reactants) and : ∆𝐻 𝑜
is the standard heat of
formation of this species, then the standard heat of the reaction is
̂𝑟𝑜 = ∑ 𝑣𝑖 ∆𝐻
∆𝐻 ̂𝑓𝑖
𝑜
= ̂𝑓𝑖
∑ |𝑣𝑖 | ∆𝐻 𝑜
− ∑ ̂𝑓𝑖
|𝑣𝑖 | ∆𝐻 𝑜

𝑖 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠
Example 7.5
Consider the combustion of liquid ethanol as shown in the following reaction scheme:

𝐶2 𝐻5 𝑂𝐻(𝑙) + 3𝑂2 (𝑔) → 2𝐶𝑂2 (𝑔) + 3𝐻2 𝑂(𝑙)

Use heat of formation to determine the standard heat of reaction.

̂𝑟𝑜 = 3∆𝐻
̂𝑓,𝐻
𝑜 ̂𝑓,𝐶𝑂
𝑜 𝑜
̂𝑓,𝐶
∆𝐻 2 𝑂(𝑙)
+ 2∆𝐻 2
− 0 − ∆𝐻 2 𝐻5 𝑂𝐻(𝑙)

Substitute the values of the standard heat of formation:

̂𝑟𝑜 (𝑘𝐽⁄𝑚𝑜𝑙 ) = 3(−285.84) + 2(−393.51) − 0— (277.63) = −1366.9 𝑘𝐽⁄𝑚𝑜𝑙

∆𝐻

7.4 HEATS OF COMBUSTION

The standard heat of combustion of a substance, ∆𝐻 ̂𝑐𝑜 , is the heat of the combustion of that
substance with oxygen to yield specified products [e.g., CO2(g) and H2O(l)], with both reactants
and products at 25oC and 1 atm (the arbitrary but conventional reference state).

Table B.l (Felder and Rosseau) lists standard heats of combustion for a number of substances. The
given values are based on the following assumptions: (a) all carbon in the fuel forms CO 2(g), (b)
all hydrogen forms H2O(I), (c) all sulfur forms SO2(g), and (d) all nitrogen forms NO2(g). The

7
 ̂𝑐𝑜 = 1366.9
standard heat of combustion of liquid ethanol, for example, is given in Table B.l as ∆𝐻
kJ/mol, which signifies

̂𝑐𝑜 (25°C, 1 atm) = -1366.9 kJ/mol

𝐶2 𝐻5 𝑂𝐻(𝑙) + 3𝑂2 (𝑔) → 2𝐶𝑂2 (𝑔) + 3𝐻2 𝑂(𝑙): ∆𝐻

Additional heats of combustion are given on pp. 2-195 through 2-199 of Perry's Chemical
Engineers' Handbook.
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. A hypothetical reaction path may be constructed in which (a) all combustible reactants are
burned with O2 at 25°C and (b) CO2 and H2O combine to form the reaction products plus O2. Step
(b) involves the reverse of the combustion reactions of the reaction products. Since both steps
involve only combustion reactions, the total enthalpy change-which equals the desired heat of
reaction-can be determined entirely from heats of combustion as
̂𝑟𝑜 = − ∑ 𝑣𝑖 (∆𝐻
∆𝐻 ̂𝑐𝑜 ) = ∑ ̂𝑐𝑜 ) −
|𝑣𝑖 | (∆𝐻 ̂𝑐𝑜 )
∑ |𝑣𝑖 | (∆𝐻
𝑖 𝑖 𝑖
𝑖 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠

If any reactants or products are combustion products (i.e., CO2, H2O, SO2), their heats of
combustion are equal to zero.

Consider the formation of pentane:

5𝐶(𝑠) + 6𝐻2 (𝑔) → 𝐶5 𝐻12 (𝑙) ̂𝑟𝑜 =?
∆𝐻

Carbon, hydrogen, and pentane can all be burned, and their standard heats of combustion can be
determined experimentally. Therefore,

̂𝑟𝑜 = 5∆𝐻
̂𝑐,𝐶(𝑠)
𝑜 ̂𝑐,𝐻
𝑜 𝑜
̂𝑐,𝐶
∆𝐻 + 6∆𝐻 2 (𝑔)
− ∆𝐻 5 𝐻12 (𝑙)
Example 7.6

Consider the combustion of liquid ethanol as shown in the following reaction scheme:

𝐶2 𝐻5 𝑂𝐻(𝑙) + 3𝑂2 (𝑔) → 2𝐶𝑂2 (𝑔) + 3𝐻2 𝑂(𝑙)

Use heat of formation to determine the standard heat of combustion.

The standard heat of reaction is calculated from the standard heat of combustion as

̂𝑟𝑜 = ∆𝐻 𝑜
̂𝑐,𝐶 ̂𝑐,𝑂
𝑜 ̂𝑐,𝐻
𝑜 ̂𝑐,𝐶𝑂
𝑜
∆𝐻 2 𝐻5 𝑂𝐻(𝑙)
+ 3∆𝐻 2 (𝑔)
− 3∆𝐻 2 𝑂(𝑙)
− 2∆𝐻 2 (𝑔)

Substitute the values of the standard heat of combustion, knowing that the magnitudes of the
standard heat of combustion of oxygen, water, and carbon dioxide are zero:
∆𝐻̂𝑟𝑜 = −1366.91 + 0 − 0 − 0 = −1366.9 𝑘𝐽/𝑚𝑜𝑙

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7.5 Energy Balance for Reactive Processes
For energy balances with reaction, we have two methods for solving these types of problems: the
heat of reaction method (extent of reaction) and the heat of formation method (element balance).
These two methods differ in the choice of the reference state.

7.5.1Heat of Reaction Method

The heat of reaction method is ideal when there is a single reaction for which ∆𝐻 ̂𝑟𝑜 is known. This
method requires calculation of the extent of reaction, ξ. The extent of reaction can be obtained by
performing material balance for any reactant or product for which the feed and product flow rates
are known. The reference state is such that all reactant and product species are at 25°C and 1 atm
in the states for which the heat of reaction is known (Figure 7.1).

Figure 7.1: Rate of change in enthalpy for a reactive process.

For a single reaction at a reference state of 25°C and 1 atm while reactant and product are at
different inlet and exit temperatures
∆𝐻̇ = 𝜉∆𝐻
̂𝑟𝑜 + ∑ 𝑛̇ 𝑖 𝐻
̂𝑖 − ∑ 𝑛̇ 𝑖 𝐻
̂𝑖
𝑜𝑢𝑡 𝑖𝑛
For multiple reactions where the reference state is 25°C and 1 atm and the inlet and exit streams
are at temperatures other than the reference states,
∆𝐻̇ = ∑ ̂𝑟,𝑗
𝜉𝑗 ∆𝐻 𝑜 ̂𝑜𝑢𝑡 − ∑ 𝑛̇ 𝑖𝑛 𝐻
+ ∑ 𝑛̇ 𝑜𝑢𝑡 𝐻 ̂𝑖𝑛
𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛𝑠

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7.5.2 Heat of Formation or Element Balance Method
In the heat of formation method, the heats of reaction terms (∆𝐻 ̂𝑟𝑜 ) are not required as they are
implicitly included when heats of formation of the reactants are subtracted from the products. For
single and multiple reactions

∆𝐻̇ = ∑ 𝑛̇ 𝑜𝑢𝑡 𝐻
̂𝑜𝑢𝑡 − ∑ 𝑛̇ 𝑖𝑛 𝐻
̂𝑖𝑛

Figure 7.2 Heat of formation or element balance method.

where the specific molar enthalpy of component i in the inlet streams is

𝑇1𝑛
̂𝑖,𝑖𝑛 = ∫
𝐻 ̂𝑓,𝑖
𝐶𝑝,𝑖 𝑑𝑇 + ∆𝐻 𝑜
25
Specific molar enthalpy of component i in the exit streams is
𝑇𝑜𝑢𝑡
̂𝑖,𝑜𝑢𝑡 = ∫
𝐻 ̂𝑓,𝑖
𝐶𝑝,𝑖 𝑑𝑇 + ∆𝐻 𝑜
25

Example 7.7 Energy Balance About an Ammonia Oxidizer

The standard heat of reaction for the oxidation of ammonia is given below:

̂ 𝑜𝑟 = −904.7 𝑘𝐽⁄𝑚𝑜𝑙
4𝑁𝐻3 (𝑔) + 5𝑂2 (𝑔) → 4𝑁𝑂(𝑔) + 6𝐻2 𝑂(𝑣): ∆𝐻

One hundred mol NH3/s and 200 mol O2/s at 25°C are fed into a reactor in which the ammonia is
completely consumed. The product gas emerges at 300°C. Calculate the rate at which heat must
be transferred to or from the reactor, assuming operation at approximately 1 atm.

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Solution

Since only one reaction takes place and ∆𝐻̂𝑟𝑜 is known, we will use the heat of reaction method for the energy
balance, choosing as references the reactant and product species in the states for which the heat of reaction is
given. The enthalpy table appears as follows:

References: 𝑁𝐻3 (𝑔), 𝑂2 (𝑔), 𝑁𝑂(𝑔), 𝐻2 𝑂(𝑣) at 25oC and 1 atm

Substance 𝒏̇ 𝒊𝒏 𝑯̂ 𝒊𝒏 𝒏̇ 𝒐𝒖𝒕 ̂ 𝒐𝒖𝒕
𝑯
(mol/s) (kJ/mol) (mol/s) (kJ/mol)
𝑁𝐻3 100 0 - -
𝑂2 200 0 75 ̂𝟏
𝑯
𝑁𝑂 - - 100 ̂𝟐
𝑯
𝐻2 𝑂 - - 150 ̂𝟑
𝑯

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Thus, 19,700 kW of heat must be transferred from the reactor to maintain the product temperature
at 300°C. If less heat were transferred, more of the heat of reaction would go into the reaction
mixture and the outlet temperature would increase.

Example 7.8 Energy Balance on a Methane Oxidation Reactor

Methane is oxidized with air to produce formaldehyde in a continuous reactor. A competing
reaction is the combustion of methane to form CO2.
A flowchart of the process for an assumed basis of 100 mol of methane fed to the reactor is shown
below:

Basis: 100 mol CH4 Fed

Since the component amounts of all streams are known, we may proceed directly to the energy
balance. We choose as references the elemental species that form the reactants and products at
25°C and I atm (the state for which heats of formation are known) and the nonreactive species N2
(g) also at 25°C and I atm (the reference state for Table B.8). The inlet-outlet enthalpy table is
shown below.

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Calculate Unknown Enthalpies
In the following calculations, values of ∆𝐻̂𝑓𝑜 come from Table B.1, formulas for Cp(T) come from
Table B.2, and values of H(T) for O2 and N2 are specific enthalpies relative to the gaseous species
at 25°C taken from Table B.8. Effects of any pressure changes on enthalpies are neglected and the
details of the calculations are not shown.

As each of these values is calculated, it should be substituted in the inlet-outlet enthalpy table. The
table finally appears as follows:

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Evaluate ∆𝐻
̂𝑜𝑢𝑡 − ∑ 𝑛𝑖𝑛 𝐻
∆𝐻 = ∑ 𝑛𝑜𝑢𝑡 𝐻 ̂𝑖𝑛 = −15,300 kJ
Energy Balance
Remember that we are dealing with a continuous process and hence an open system. [The reason
we use n (mol) and not 𝑛̇ (mol/s) is that we took 100 mol CH4 as a basis of calculation.] With ∆𝐸𝑘 ,
∆𝐸𝑝 and Ws neglected, the open system energy balance yields
Q = ∆𝐻 = -15,300 kJ

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