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
V dV dT U
C V dV
dT U T
dU U
T V
T V
• Constant volume process
• Ideal gases
• Approximately zero for low-pressure gases
dT C
dU
V U
TT12C
VdT
4.1 Sensible Heat Effects
P dP dT H
C P dP
dT H T
dH H
T P
T P
• Constant pressure process
• Ideal gases
• Approximately zero for low-pressure gases
dT C
dH
P H
TT12C
PdT
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
2
2
A BT CT DT R
C
PigR C
C
V
P
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
Mean Heat Capacity
2
1 2
1 2
1
1 3
2
2 2
3 2
) (
T
T T
T T
T P
CT DT AT BT
dT DT
CT BT
A dT
C H
1 2 1
2
1 2
2 1
2
)
1( T T
dT C T
T
H T H
T C
T
T P
p
) ( T
2T
1C
H
p
Always two temperature values are required !
Mean Heat Capacity
Tabulation
Based on a reference temperature T
refMean Heat Capacity
Calculation of Enthalpy
) ( T
2T
1C
H
p
) (
) ( ) (
)
(
2 11
2 ref p T ref
T
p
T T C T T
C
H
) (
) (
)
( T
1T
2H T T
2H T T
1H
ref
ref
Enthalpy Calculation
1. Heat Capacity
2. Table
3. Mean heat capacity
Heat capacity for mixtures
( )
)
( T y C T C
pm i pi
21
)
T
(
T
C
pmT dT H
y
i mole fraction of i
21 2
1
T T
2 T 2
T
C
PdT ( A BT CT DT ) dT
H
) T T
( C
H
p 2
1
H n
outH ˆ
outn
inH ˆ
inH ˆ
: Specific enthalpy (from table)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.
) T .
T .
. ( R
C
p,CH4 1 702 9 081 10
3 2 164 10
6 2From Table C.1
J , dT
) T .
T .
. ( . H
Q
.
.
778 19 10
164 2 10
081 9 702 1 314 8
15 873
15 533
2 6
3
Energy balance for steady-state flow
Q
H
s fs
W Q m
zg u
H
2 2
14.2 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 - Latent heat of fusion - Latent heat of sublimation
Hˆ
v Hˆ
m Hˆ
s
Estimation and Correlation of Latent Heat
Clapeyron Equation
Derivation will be given in Chapter 6
slope of vapor pressure curve, volume change heat of vaporization
dT V dP T H
sat
Estimation and Correlation of Latent Heat
Standard Heat of Vaporization
Trouton’s Rule : Rough estimates (30 % accuracy for hydrocarbons)
Chen’s equation (2 % accuracy for hydrocarbons)
Standard Heat of Fusion
(K) 109
0
(K) 088
0 (kJ/mol)
b b
v
. T
T H ˆ .
) / ( 07 . 1
) log
0297 .
0 0327 .
0 ) / ( 0331 .
0 (kJ/mol) (
ˆ
10c b
c c
b b
v
T T
P T
T H T
(K) 050
0
(K) 025 0
(K) 0092 0
(kJ/mol)
m m m m
T .
T .
T .
H ˆ
(Non-polar liquids) (water, alcohols)
(metallic elements) (inorganic compounds)
(organic compounds)
Heat of Vaporization at other temperature
Watson Correlation
38 . 0
1 2 1
2
) ( )
(
T T
T T T
H T
H
c c v
v
38 . 0
1 2 1
2
1
) 1 ( )
(
r r v
v
T
T T H T
H
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)
Reactants : stoichiometric quantities
Complete Reaction
Reactants are fed at T, P
Products are emerging at T, P
ex) CaC
2(s) + 2 H
2O (l) Ca(OH)
2(s) + C
2H
2(g)
reactants products
r
( T , P ) H H
H ˆ
mol /
kJ 4 . 125 )
atm 1
, C 25 ( H ˆ
or
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
Extent of Reaction, x
Amount of reactants with stoichiometric coefficient of “1” that have been reacted.
produced C
mol 3
50 reacted
B mol 1
50 reacted
A mol 2
ˆ 50 kJ kJ kJ
H
r
A A
r
n H H
n
A: moles of A consumed or produced
A: the stoichiometric coefficient of A kJ
H ˆ
r
50
x
A rA
r
n H
H H
Properties of Heat of Reaction
“Standard” heat of reaction
at reference T and P (25
oC, 1 atm)
Exothermic : Endothermic :
Value depends on stoichiometric eqn.
CH
4(g) + 2 O
2(g) CO
2(g) + 2H
2O (l) H
r(25
oC) = -890.3 kJ/mol 2CH
4(g) + 4 O
2(g) 2CO
2(g) + 4H
2O (l) H
r(25
oC) = -1780.6 kJ/mol
Value depends on the state (gas, liquid, solid)
CH
4(g) + 2 O
2(g) CO
2(g) + 2H
2O (l) H
r(25
oC) = -890.3 kJ/mol CH
4(g) + 2 O
2(g) CO
2(g) + 2H
2O (g) H
r(25
oC) = -802.6 kJ/mol
0
H
r 0
H
r) C (
H or
H
ro
r25
o
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 + O
2(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)
C + ½ O
2(+ ½ O
2) CO (+ ½ O
2)
CO
23
ˆ
0r H
0 1
ˆ
rH H
H H ˆ
0r2kJ mol
H H
H
r r r/ 52 . 110
99 . 282 51
. 393
ˆ ) ˆ (
ˆ
0 3 0 1 0 2
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, H
r, 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 (O
2instead of O)
Heat of formation
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 )
Benzene :
- 6C (s) + 3H2 (g) C6H6 (l)
mol / kJ .
H ˆ
of 365 14
mol / kJ .
H ˆ
of 48 66
H ˆ
f
Determination of Heats of Reaction using Heats of Formation
Heats of reactions can be determined from heats of formation using Hess’s Law
Example
C
5H
12(l) + 8O
2(g) 5CO
2(g) + 6H
2O (l)
reactants
0 products
0
0
( ˆ ) ( ˆ )
ˆ
r iH
f i iH
f iH
) l ( H f C o )
l ( O f H
o )
g ( f CO r o
o
( C H ) ( H ˆ ) ( H ˆ ) ( H ˆ )
H ˆ
12 5 2
2
6
12
5
5
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 : CO
2(g), H
2O (l), SO
2(g), NO
2(g)
Using Hess’s Law heat of reaction can be calculated from heats of combustion
Example
- C2H6 C2H4+ H2
products reactants
c i o i
c i o r i
o
( H ˆ ) ( H ˆ )
H ˆ
2 4
2 6
2 oc H
H c C o H
c C r o
o
( H ˆ ) ( H ˆ ) ( H ˆ )
H ˆ
4.6 Temperature Dependence of H o
General Procedure
Choice of reference conditions 1. Heat of Reactionmethod
– Usually for single reaction, Hr is known
– Reactants and products : T0 where Hr is known – Non-reactive species (e.g., N2) : any convenient T
2. Heat of Formationmethod
– Usually for multiple reactions, Hris unknown – Reactants and Products :elements at 25 oC
» Use sum of heats of formation
– Non-reactive species (e.g., N2) : any convenient T
reactant product
i i i
i A
or
AR
H ˆ n H n H
H n
reactant product
i i i
i
H n H
n
H
nA: moles of A consumed or produced
A: the stoichiometric coefficient of A
Method 1 vs. method 2
Heat of reaction method
Heat of formation method
Reactants T
inProducts T
outH
Reactants 25
oC
Products 25
oC
H
ro
reactant product
i i i
i A
r o
AR
H ˆ n H n H
H n
Reactants T
inProducts T
outH
Elements 25
oC
H H
ofC
pdT
reactant product
i i i
i
H n H
n
H
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
H2O 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
Reactants Tin = 25 oC
Products Tout = ?
H
Reactants 25 oC
Products 25 oC
H
roReactants 1 mol CH4 2.4 mol O2 9.03 mol N2
Products 1 mol CO2 2 mol H2O 0.4 mol O2 9.03 mol N2
20 % excess air?
reactant product
i i i
i A
or
AR Hˆ nH nH
H n
Example 4.7 - solution
Using the mean heat capacity,
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)
H2O - - 2 41.42(T-298.15)
CO2 - - 1 52.77(T-298.15)
K mol / J . C
K mol / J . C
K mol / J . C
K mol / J . C
CO , p
O H , p
N , p
O , p
77 52
42 41
58 32
42 34
2 2 2 2
) ( T
2T
1C
H
p
K . T
T . .
H n H
H ˆ n
H n
i i i iA or AR
8 2106 0
6 443 6
934585
0
reactant product
* Repeat this problem yourself using the heat capacity (use the value A, B, C, and D from Table C.1)
2
2
A BT CT DT R
CPig
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.
Reactants T
in= 600K
Products T
out= 1300 K
H
Elements 25
oC
H C dT H
of p
reactant product
i i i
i
H n H
n
H
Reactants 1 mol CH4 2 mol H2O
Products
CH4 = 1 – x1 = 0
H2O = 2 – x1 – x2 = 1 – x2 CO = x1 – x2 = 1 – x2 CO2 = x2
H2 = 3 x1 + x2 = 3 + x2 Total = 5 mol
CH4(g) + H2O(g) CO (g) + 3H2 (g) x1
CO(g) + H2O(g) CO2 (g) + H2 (g) x2
To find the amount of product, use the extent of reaction that you have learned from “Intro. to Chem & Bio Eng”
CH4 is completely converted x1 = 1 mol
Example 4.8 - solution
Products
CH4 = 1 – x1 = 0
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 Since the product stream contains 17.4 mol % CO,
(1 – x2)/5 = 0.174 x2 = 0.13 mol
substance nin Hin nout Hout
CH4 1 H1 0 0
H2O 2 H2 0.87 H3
CO - - 0.87 H4
CO2 - - 0.13 H5
H2 - - 3.13 H6
H C dT H
of p
K mol / J . C
K mol / J . C
K mol / J . C
K mol / J . C
H , p
CO , p
CO , p
O H , p
78 29
85 49
70 31
66 38
2 2 2
) T .
T .
. ( R
Cp,CH4 17029081103 2164106 2
Example 4.8 - Quiz
Perform the rest of calculation to obtain Q = H.
0
509 393
525 110
830 285
520 74
2 2 2
4
o H , f o
CO , f o
CO , f o
O H , f o
CH , f
H
mol / J , H
mol / J , H
mol / J , H
mol / J , H
Heat of formation at 298 K is
(from Table C.4)
(naturally occurring substance)
Example 4.8 - solution
0
509 393
525 110
830 285
520 74
2 2 2
4
o H , f o
CO , f o
CO , f o
O H , f o
CH , f
H
mol / J , H
mol / J , H
mol / J , H
mol / J , H
Heat of formation at 298 K is
(naturally occurring substance)
mol / J , ) (
. H
H
mol / J , )
( . H
H
mol / J , )
( . H
H
mol / J , )
( . H
H
mol / J , )
( . H
H
mol / J , dT
) T .
T .
. ( . H
H
o H , f o
CO , f o
CO , f o
O H , f o
O H , f o
CH , f
588 47 298
1300 78 29
559 343 298
1300 85 49
850 115 298
1300 70 31
092 247 298
1300 66 38
154 274 298
600 66 38
145 61 10
164 2 10
081 9 702 1 314 8
2 6
2 5
4
2 3
2 2
600
298
2 6 3
4 1
J ,
H n H
n Q
H
i i i i980 397
reactant product