Lect 14. Enthalpy Calculations

Calculation of Enthalpy Changes

Manolito E. Bambase JrAssistant Professor, Department of Chemical Engineering

CEAT, University of the Philippines, Los Banos, Laguna, Philippines

14

hg

14.1 Heat Capacity Equations

As shown previously, the change in enthalpy can be calculated using the heat capacity CP.

1

PH C dT

To give the heat capacity some physical meaning, CP represents the amount of energy required to increase the temperature of a given amount of substance by 1 degree.

Common units for CP:

0

kJ Btu

kmol K lbmol F


Heat Capacities of Ideal Gases

2

Type of Molecule

Monoatomic

Polyatomic, linear

Polyatomic, nonlinear

High Temperature Low Temperature

5R/2

(3n 3/2)R

(3n 2)R

n = number of atoms per molecule ; R = ideal gas constant ; CV = CP R

5R/2

7R/2

4R

For gas mixtures, the CP of the mixture is the mole weighted average of the heat capacities of the components:

3

N

P avgi 1

C

i Pix C

If CP is not constant for the given temperature range, it may be expressed as a function of temperature in a power series such as:

CP = a + bT + gT2

where a, b, and g are constants specific to a particular substance(Appendix E.1, Basic Principles and Calculations in Chemical Engineering, 6th edition, by David M. Himmelblau)


Example 141. Calculation of H for a Gas Mixture

An economic feasibility study indicates that solid municipal waste can be burned to a gas of the following composition (on a dry basis):

4

CO2

CO

O2

N2

9.2%

1.5%

7.3%

82.0%

What is the enthalpy difference for this gas per lbmol between the bottom and the top of the stack if the temperature at the bottom of the stack is 5500F and the temperature at the top is 2000F.

The heat capacity equations for the gases are:[T in 0F; CP in Btu/(lbmol)(

0F)]

N2: CP = 6.895 + 0.7624 10-3T 0.7009 10-7T2

O2: CP = 7.104 + 0.7851 10-3T 0.5528 10-7T2

CO: CP = 6.865 + 0.8024 10-3T 0.7367 10-7T2

CO2: CP = 8.448 + 5.757 10-3T 21.59 10-7T2

+ 3.059 10-10T3

5


Basis: 1.00 lbmol of gas mixture

The enthalpy difference (H) is calculated as:

6


PH C dT

The CP of the gas mixture is determined from the equation:

N 2 O 2 CO 2 CO

N

P avgi 1

P N2 P O2 P CO2 P CO Pavg

C

C x C x C x C x C

i Pix C

7Example 141. Calculation of H for a Gas Mixture

N 2

O 2

CO 2

CO

N2 P

O2 P

CO2 P

CO P

x C 0.82 6.895 ....

x C 0.073 7.104 ....

x C 0.092 8.448 ....

x C 0.015 6.865 ....

Obtaining SxiCPi:

3 7 2P avg10 3

C 7.053 1.2242 10 T 2.6124 10 T

0.2814 10 T

Solving for H:

8

3 7 2200

10 3550

32 2

7 103 3 4 4

7.053 1.2242 10 T 2.6124 10 TH

0.2814 10 T

1.2242 10H 7.053 200 550 200 550

2

2.6124 10 0.2814 10200 550 200 550

3 4

H 2616Btu / lbmol

dT


14.2 Tables of Enthalpy Values

Enthalpies at various temperatures can also be obtained from tables.

From Basic Principles and Calculations in Chemical Engineering by David M. Himmelblau (6th edition):

Table D.2 Enthalpies of Paraffinic Hydrocarbons (C1 C6)

Table D.3 Enthalpies of Other Hydrocarbons (ethylene, propylene, butene, acetylene, benzene)

Table D.4 Enthalpies of Nitrogen and Some of its Oxides

Table D.5 Enthalpies of Sulfur Compounds

Table D.6 Enthalpies of Combustion Gases

9

14.3 Enthalpy Calculations from Standard Heat of Formation

The enthalpy from a standard reference state is given by

10

R

T

fT

H H PC dT

where Hfo is the standard heat of formation and TR is the

reference temperature.

For a mixture,

R

N T

fT

i 1

H Hmixture PC dT

11

The standard heat of formation (Hfo) is the special enthalpy for

the formation of 1 mole of a compound from its constituents elements, for example

C(s) + 0.5O2(g) ======> CO(g)

The reactants and products must be at 250C and 1 atm.

The reaction may not represent a real reaction. By definition, the heat of formation for the elements is zero in the standard state.

From Basic Principles and Calculations in Chemical Engineering by David M. Himmelblau (6th edition):

Table F.1 Heats of Formation and Combustion


12

Consider an open system with no chemical reaction:


Open System(No Reaction)

Input, T1 Output, T2

A: HA1B: HB1

A: HA2B: HB2

The enthalpy difference between inlet and outlet will be

H = Hout Hin = (HA2 + HB2) (HA1 + HB1)

13

Calculating the enthalpies from standard heat of formation


2 2

R R

1 1

R R

T T

fA fBT T

T T

fA fBT T

H H H

H H

PA PB

PA PB

C dT C dT

C dT C dT

Simplifying,

2 2

1 1

T T

T TH PA PBC dT C dT

14

Consider an open system with chemical reaction:


Open System

A+B ===> C+D

Input, T1 Output, T2

A: HAB: HB

C: HCD: HD

The enthalpy difference between inlet and outlet will be

H = Hout Hin = (HC + HD) (HA + HB)

15

Calculating the enthalpies from standard heat of formation


2 2

R R

1 1

R R

T T

fC fDT T

T T

fA fBT T

H H H

H H

PC PD

PA PB

C dT C dT

C dT C dT

Rearranging the terms,

2R

2 1 1

R R R

T

fC fD fA fBT

T T T

T T T

H H H H H PC

PD PA PB

C dT

C dT C dT C dT

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The group of terms involving the standard heats of formation is called the standard heat of reaction, HoR.

In general,


R f fH nH nHproducts reactants ii

The n in the equation is the stoichiometric coefficient of species i in the chemical reaction.

The standard heat of reaction is the difference between the heats of formation of the products and that of the reactants.

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Calculate HRo for the following reaction:

4 NH3(g) + 5 O2(g) =====> 4 NO(g) + 6 H2O(g)

The standard heats of formation of the products and reactants are:

NH3: 46.191 kJ/mol NO = +90.374 kJ/molO2: 0 H2O = 241.826 kJ/mol

And the standard heat of reaction:

HRo = [4(90.374) + 6( 241.826)] [4( 46.191) + 5(0)]

HRo = 904.696 kJ/4 mol NH3

Example 14-2. Calculation of Standard Heat of Reaction

LECT 14. Calculation of Enthalpy Changes.vsd123456789101112131415161718

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Lect 14. Enthalpy Calculations