Chapter 4.

Heat Effects
 Introduction

 Heat Transfer – Common operation in chemical

industry
 Sensible Heat effect
 Latent Heat
 Heat of Reaction
 Heat of Mixing, Heat of Solution Covered later
(Thermodynamics II)
 4.1 Sensible Heat Effects

 Heat transfer to system with no phase transition, no

reaction, no change in composition  T changes

 U   U   U 
dU    dT    dV  CV dT    dV
 T V  V T  V T

• Constant volume process

• Ideal gases
• Approximately zero for low-pressure gases

T2
dU  CV dT U   CV dT
T1
 4.1 Sensible Heat Effects

 H   H   H 
dH    dT    dP  CP dT    dP
 T  P  P T  P T

• Constant pressure process

• Ideal gases
• Approximately zero for low-pressure gases

T2
dH  CP dT H   CP dT
T1
 4.1 Sensible Heat Effect

 Expressions for constant pressure heat capacities

 Heat capacity for a substance is a function of T and P
 Calculation of Fluid properties
- Based on ideal gas : Ideal gas properties
- Departure from ideal gas : residual properties

 Ideal Gas Heat Capacity

- Heat capacity at P  0
- Valid for low pressure gases
CV  CP  R

CPig
 A  BT  CT 2  DT 2
R

Values A, B, C, and D for many chemical substances  Table C.1

 Mean Heat Capacity

 Calculation of enthalpy value

 evaluation of integrals are required
T2 T2
H   CP dT   ( A  BT  CT 2  DT  2 )dT
T1 T1
T2
 BT 2 CT 3 1 
  AT    DT 
 2 3  T1

 Mean Heat Capacity

T2

H 2  H1 T1 CP dT
C p (T1  T2 )  
T2  T1 T2  T1

H  C p (T2  T1 )

Always two temperature values are required !

 Mean Heat Capacity

 Tabulation
 Based on a reference temperature Tref
 Mean Heat Capacity

 Calculation of Enthalpy

H  C p (T2  T1 )

H (T1  T2 )  H (Tref  T2 )  H (Tref  T1 )

H  (C p )T2 (T2  Tref )  (C p )T1 (T1  Tref )

 Enthalpy Calculation

T2 T2
1. Heat Capacity H   CP dT   (A  BT  CT 2  DT2 )dT
T1 T1

2. Table H   n outĤout   n in Ĥin Ĥ : Specific enthalpy (from table)

3. Mean heat capacity

H  Cp (T2  T1 )

 Heat capacity for mixtures

C pm (T )   yi C pi (T ) yi  mole fraction of i

T2
H   C pm (T )dT
T1
 Example 4.2

 Calculate the heat required to raise the temperature of 1 mol of

methane from 260 oC to 600 oC in a steady-flow process at a pressure
sufficiently low that methane may be considered as an ideal gas.

Energy balance for steady-state flow

 1  
  H  u 2  zg m   Q  W s H  Q
 2   fs

From Table C.1

C p ,CH 4  R( 1.702  9.081103 T  2.164 106 T 2 )

873.15
Q  H   8.314( 1.702  9.08110
3
T  2.164 106 T 2 )dT 19,778 J
533.15
 4.2 Latent Heat of Pure Substances

 Heat Effects of Phase Change

 Latent Heat
 Phase changes are usually accompanied by large changes in U and H
 fusion

 vaporization

 sublimation

 Latent Heat
 the enthalpy changes associated with phase transition of a unit amount of a
substances applied heat goes to phase changes

 at constant T and P

 no temperature change
- Latent heat of vaporization Ĥ v
- Latent heat of fusion Ĥ m
- Latent heat of sublimation Ĥ s
 Estimation and Correlation of Latent Heat

 Clapeyron Equation

dP sat
H  TV
dT
slope of vapor pressure curve, volume change  heat of vaporization

 Derivation will be given in Chapter 6

 Estimation and Correlation of Latent Heat
 Standard Heat of Vaporization
 Trouton’s Rule : Rough estimates (30 % accuracy for hydrocarbons)

 0.088Tb (K) (Non-polar liquids)

Ĥ v (kJ/mol)
 0.109Tb (K) (water, alcohols)

 Chen’s equation (2 % accuracy for hydrocarbons)

Tb (0.0331(Tb / Tc )  0.0327  0.0297 log10 Pc )

Hˆ v (kJ/mol) 
1.07  (Tb / Tc )

 Standard Heat of Fusion

 0.0092Tm (K) (metallic elements)

Ĥ m (kJ/mol)  0.025Tm (K) (inorganic compounds)

 0.050Tm (K) (organic compounds)

 Heat of Vaporization at other temperature

 Watson Correlation

0.38
 Tc  T2 
H v (T2 )  H v (T1 ) 
 Tc  T1 

0.38
 1  Tr 2 
H v (T2 )  H v (T1 ) 
 1  Tr1 
 4.3 Standard Heat of Reaction

 Heat effects accompanying when reaction occur

 Reaction may be carried out in many different ways

 Tabulation of all possible reactions is impossible

 Convenient way
- “Standard way”  Q = H, and H is the state function

- The required data to be a minimum

- Use Hess’s Law

 Heat of Reaction

 Heat of Reaction (Enthalpy of Reaction)

Ĥ r (T, P)  H products  H reactants

 Reactants : stoichiometric quantities

 Complete Reaction
 Reactants are fed at T, P
 Products are emerging at T, P

ex) CaC2 (s) + 2 H2O (l)  Ca(OH)2 (s) + C2H2 (g)

Ĥ r (25o C,1 atm)  125.4kJ / mol

 Standard Condition

 A standard state is a particular state of a species at temperature T

and at specified conditions of pressure, composition and physical

condition as, e.g., gas, liquid or solid.

• Standard state pressure : 1 atm (101,325 Pa)

• Standard state temperature : 25 oC (298.15 K)

• Gas : The pure substance in ideal gas state at 1 bar

• Liquid and solid : The real pure liquid or solid at 1 bar
 Heat of Reaction : Per mole of what ?

 Example
2A + B  3C Ĥ r  50kJ
 50kJ  50kJ  50kJ
Hˆ r   
2 mol A reacted 1 mol B reacted 3 mol C produced

H r
H  nA nA : moles of A consumed or produced
A
A : the stoichiometric coefficient of A

 Extent of Reaction, x
 Amount of reactants with stoichiometric coefficient of “1” that
have been reacted.
H r
H  n A  H r  x
A
 Properties of Heat of Reaction
 “Standard” heat of reaction H r o or H r ( 25o C )
 at reference T and P (25 oC, 1 atm)

 Exothermic : H r  0
Endothermic : H r  0

 Value depends on stoichiometric eqn.

CH4(g) + 2 O2(g)  CO2 (g) + 2H2O (l) Hr(25 oC) = -890.3 kJ/mol
2CH4(g) + 4 O2(g)  2CO2 (g) + 4H2O (l) Hr(25 oC) = -1780.6 kJ/mol

 Value depends on the state (gas, liquid, solid)

CH4(g) + 2 O2(g)  CO2 (g) + 2H2O (l) Hr(25 oC) = -890.3 kJ/mol
CH4(g) + 2 O2(g)  CO2 (g) + 2H2O (g) Hr(25 oC) = -802.6 kJ/mol
 Measurement of Heats of Reaction

 Measurements of heats of reaction : Calorimeter

 Temperature rise or fall of the fluid can be measured and heat of
reactions are determined.

A: Small funnel
B: Styrofoam lid
C: Two styrofoam cups glued together
D: Magnetic stir bar (inside the cups)
E: Magnetic stir plate
F: Temperature probe
G: Thermometer
 Difficulties for measuring heat of reaction

 Some reactions cannot be accomplished.

 Ex) 2C + O2(g)  2 CO (incomplete combustion)
 It is impossible to determine the heat of the incomplete combustion only.
 The reaction cannot proceed at such low temperature.

 Alternative method
 C + O2  CO2 Hr = -393.51 kJ/mol (Easily measured)
 CO + ½ O2  CO2 Hr = -282.99 kJ/mol (Easily measured)

Hˆ 0 r 3
C + ½ O2 (+ ½ O2) CO (+ ½ O2) Hˆ 0 r 3  Hˆ 0 r1  (Hˆ 0 r 2 )
 393.51  282.99
H  Hˆ 0 r1 H  Hˆ 0 r 2  110.52kJ / mol
CO2
 Hess’s Law

If the stoichiometric equation for a reaction can be obtained by algebraic

operations (+, –, ×, / ) on the other stoichometric equations (2,3,…), then

the heat of reaction can be obtained by performing the same operations

on the heats of reactions (2,3,…).

 H is state property

 the enthalpy change for a reaction, Hr, is the same whether it occurs by

one step or by a series of steps.

 Only depends on the initial and final state.

 4.4 Heats of Formation
 Formation reaction :
 Reaction which the compound is formed from its atomic constituents.
 Normally occur in nature (O2 instead of O)
 Heat of formation Ĥ f
 H associated with the formation of 1 mole of the compound.
 Reference state: 25 oC and 1 atm

 Examples
 Ammonium Nitrate :
- N2(g) + 2H2(g) + 3/2 O2(g)  NH4NO3 (c ) Ĥ o f  365.14kJ / mol
 Benzene :
- 6C (s) + 3H2 (g)  C6H6 (l) Ĥ o f  48.66kJ / mol
 Determination of Heats of Reaction using Heats of
Formation
 Heats of reactions can be determined from heats of formation using

Hess’s Law

Hˆ 0 r  i
 ( H
products
ˆ 0
f )i  i
 ( H
reactants
ˆ 0
f )i

 Example

 C5H12(l) + 8O2 (g)  5CO2(g) + 6H2O (l)

Ĥ o r ( C5 H12 )  5( Ĥ o f )CO ( g )  6( Ĥ o f )H O( l )  ( Ĥ o f )C H

2 2 5 12 ( l )
 4.5 Standard Heats of Combustion
 Standard Heat of Combustion
 Heat of reaction of the substance with oxygen to yield specified products
 Condition : 25 oC, 1 atm
 Products : CO2 (g), H2O (l), SO2 (g), NO2 (g)

 Using Hess’s Law heat of reaction can be calculated from heats of

combustion

Ĥ o r  i
 (
reactants
Ĥ o
c )i  i
 ( 
products
Ĥ o
c )i

 Example
- C2H6  C2H4 + H2

Ĥo r  (Ĥo c )C2H6  (Ĥo c )C2H4  (Ĥo c ) H2

 4.6 Temperature Dependence of Ho
 General Procedure
 Choice of reference conditions
1. Heat of Reaction method
– Usually for single reaction, Hr is known
– Reactants and products : T0 where Hr is known
– Non-reactive species (e.g., N2) : any convenient T

n AR Ĥ o r
n H n H
nA : moles of A consumed or produced
H   
A
i i i i
product reactant A : the stoichiometric coefficient of A

2. Heat of Formation method

– Usually for multiple reactions, Hr is unknown
– Reactants and Products :elements at 25 oC
» Use sum of heats of formation
– Non-reactive species (e.g., N2) : any convenient T

H  n H
product
i i  n H
reactant
i i
 Method 1 vs. method 2
 Heat of reaction method
H
Reactants Products
Tin Tout
n AR Ĥ o r
H 
A
 n H
product
i i  n H
reactant
i i

Reactants Hro Products

25 oC 25 oC

 Heat of formation method

H
Products
n H n H
Reactants
Tin Tout H  i i  i i
product reactant

H  H of   Cp dT
Elements
25 oC
 H Calculations
 In the textbook of Thermodynamics
 Table C.1 – C3 : Heat capacity of gases, solids, and liquids (constants – A, B, C, D)
 Table C4 : Heat of formation
 Table F.1 – F.4 : Steam table
 In the textbook of Introduction to Chemical & Biological Engineering
 Table B.1 : Latent heat, heat of formation, heat of combustion
 Table B.2 : Heat capacity (constants – A, B, C, D)
 Table B.5 – B.7 : Steam table
 Table B.8 & B.9 : Specific enthalpy for simple gases (Air, O2, N2, H2, CO, CO2, H2O)
 Table B.10 : Heat capacity for Kopp’s rule
 It is convenient to prepare the enthalpy table.

substance nreactant Hreactant nproduct Hproduct

NH3 n1 H1
O2 n2 H2 n3 H3
NO n4 H4
H2 O n5 H5
 Example 4.7

 What is the maximum temperature that can be reached by the

combustion of methane with 20 % excess air? Both the methane

and the air enter the burner at 25 oC.

 Processes with unknown outlet conditions : adiabatic reactors

 Adiabatic reactor
- No heat exchange with surrounding  maximum temperature

 Q = H = 0, then solve for unknown T

 Example 4.7 - solution
The standard heat of reaction is
CH4(g) + 2 O2(g)  CO2 (g) + 2H2O (g) Hr(25 oC) = -802,256 J/mol
Since Hr(25 oC) is known, use the heat of reaction method

H n AR Ĥ o r
Reactants Products
H 
A
 n H
product
i i  n H
reactant
i i

Tin = 25 oC Tout = ?
20 % excess air?

Reactants Hro Products

25 oC 25 oC

Reactants Products
1 mol CH4 1 mol CO2
2.4 mol O2 2 mol H2O
9.03 mol N2 0.4 mol O2
9.03 mol N2
 Example 4.7 - solution
Using the mean heat capacity, C p ,O 2  34.42 J / mol  K
C p ,N 2  32.58 J / mol  K
H  C p (T2  T1 )
C p ,H 2O  41.42 J / mol  K
C p ,CO 2  52.77 J / mol  K
Enthalpy table

substance nin Hin nout Hout

CH4 1 0 - -
O2 2.4 0 0.4 34.42(T-298.15)
N2 9.03 0 9.03 32.58(T-298.15)
H2 O - - 2 41.42(T-298.15)
CO2 - - 1 52.77(T-298.15)

n AR Ĥ o r
H 
A
 n H
product
i i  n H
reactant
i i 0

 934585.6  443.6T  0 T  2106.8K

* Repeat this problem yourself using the heat capacity CPig

 A  BT  CT 2  DT 2
(use the value A, B, C, and D from Table C.1) R
 Example 4.8

One method for the manufacture of “synthesis gas” (a mixture of CO and H2) is the
catalytic reforming of CH4 with steam at high temperature and atmospheric pressure:

CH4(g) + H2O(g)  CO (g) + 3H2 (g)

The only other reaction to be considered is the water-gas-shift reaction:

CO(g) + H2O(g)  CO2 (g) + H2 (g)

If the reactants are supplied in the ratio, 2 mol steam (H2O) to 1 mol CH4, and if heat is
supplied to the reactor so that the products reach a temperature of 1300K, the CH4 is
completely converted and the product stream contains 17.4 mol % CO. Assuming the
reactants to be preheated to 600K, calculate the heat requirement for the reactor.
 Example 4.8 - solution
Since there are two reactions, it is convenient to use the heat of formation method.
H
n H n H
Reactants Products
Tin = 600K Tout = 1300 K H  i i  i i
product reactant

H  H of   C p dT
Elements To find the amount of product, use the extent of reaction
25 oC that you have learned from “Intro. to Chem & Bio Eng”
CH4(g) + H2O(g)  CO (g) + 3H2 (g) x1
CH4 is completely converted  x1 = 1 mol

Reactants Products
CO(g) + H2O(g)  CO2 (g) + H2 (g) x2
1 mol CH4 CH4 = 1 – x1 = 0
2 mol H2O H2O = 2 – x1 – x2 = 1 – x2
CO = x1 – x2 = 1 – x2
CO2 = x2
H2 = 3 x1 + x2 = 3 + x2
Total = 5 mol
 Example 4.8 - solution
Since the product stream contains 17.4 mol % CO, Products
(1 – x2)/5 = 0.174 CH4 = 1 – x1 = 0
x2 = 0.13 mol H2O = 2 – x1 – x2 = 1 – x2 = 0.87 mol
CO = x1 – x2 = 1 – x2 = 0.87 mol
CO2 = x2 = 0.13 mol
H2 = 3 x1 + x2 = 3 + x2 = 3.13 mol

substance nin Hin nout Hout

CH4 1 H1 0 0
H2 O 2 H2 0.87 H3
CO - - 0.87 H4
CO2 - - 0.13 H5
H2 - - 3.13 H6

H  H of   C p dT C p ,H 2O  38.66 J / mol  K
C p ,CO  31.70 J / mol  K
C p ,CH 4  R( 1.702  9.081103 T  2.164 106 T 2 )
C p ,CO 2  49.85 J / mol  K
C p ,H 2  29.78 J / mol  K
 Example 4.8 - Quiz
Perform the rest of calculation to obtain Q = H.
Heat of formation at 298 K is H of ,CH 4  74,520 J / mol
(from Table C.4)
H of ,H 2O  285,830 J / mol
H of ,CO  110,525 J / mol
H of ,CO2  393,509 J / mol
H of ,H 2  0 (naturally occurring substance)
 Example 4.8 - solution
Heat of formation at 298 K is H of ,CH 4  74,520 J / mol
H of ,H 2O  285,830 J / mol
H of ,CO  110,525 J / mol
H of ,CO2  393,509 J / mol
H of ,H 2  0 (naturally occurring substance)
600
H1  H o
f ,CH 4   8.314( 1.702  9.08110 3 T  2.164 10 6 T 2 )dT   61,145 J / mol
298

H 2  H of ,H 2O  38.66( 600  298 )  274,154 J / mol

H 3  H of ,H 2O  38.66( 1300  298 )  247 ,092 J / mol
H 4  H of ,CO  31.70( 1300  298 )  115,850 J / mol
H 5  H of ,CO2  49.85( 1300  298 )  343,559 J / mol
H 6  H of ,H 2  29.78( 1300  298 )  47 ,588 J / mol

H  Q  n H
product
i i  n H
reactant
i i

 397 ,980 J
 Homework

 Problems
 4.2, 4.5, 4.21 [(a),(b),(c)], 4.32, 4.38, 4.41
 Due:

 Other Recommend Problems

 4.35, 4.39, 4.46 [(a), (b), (c), (d)]