Chapter 2 Heat Effects (Part2)

8/8/2019 Chapter 2 Heat Effects (Part2)

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Chapter 2

Heat Effects

Chemical Engineering

Thermodynamics


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2.3 Standard Heat of Reaction,

Formation and Combustion

2.4 Temperature Dependence of

Delta H

Chapter Outline


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2.3 Standard Heat of Reaction,

Formation and CombustionWe have discussed the heateffects for physical processes.

Most of heat effects for chemical reactionsare calculated in standard way.

The heat of specific reaction depends on

temperature of both reactants and products.

Chemical processes alsoaccompanied by transfer ofheat and temperature changes.


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Standard Heat of Reaction

Enthalpy change of reaction is called theheat of reaction.

HQ (!

Given the reaction:

mMlLbBaA p

The standard heat of reaction is defined asthe enthalpy change when a moles ofA and

b moles of react to form lmoles of and

moles of in their standard states at temp. T.


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A standard state is a particular state of aspecies at temperature Tand at specified

conditions of pressure, composition and

physical conditions (gas, liquid, solid).

What is standard state?

Standard state pressure: 1 bar (105 Pa).Standard state composition: the states of

pure species:-Gases: ideal gas state at 1 bar.

Liquids and solids: the real liquid/solid at 1 bar.


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Standard Heat of Formation

Standard heat of any reaction can becalculated if the standard heats of formation

of the compounds taking part in the reaction

are known.A formation reaction is a reaction which

forms a single compound form its constituent

elements.

OHCH2HO2

1C 322 p

4232

SOHSOOH p

from elements

from other compounds


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Heat of formation is based on 1 mol of thecompound formed.

Heat of reaction at any temperature can be

calculated for heat capacity data (Table C.1).

At temperature 298.15K (25C), heat of

reaction can be calculated using standard heat

of formation, for compounds fromTable C.4 in Appendix C.

Q

298fH(


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Table C.4


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Example 4.5

Calculate the standard heat at 25C for the

following reaction:

222 l2O2(g)Ol(g)4p

Standard heats of formation at 298.15K from

Table C.4 are:

Solution:

HCl (g):-92,307 J H2O (g):-241,818 J


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(g)2Cl(g)2H4HCl(g) 22 pQ

298H( = (4)(92,307)

O(g)2H(g)

O(g)2H 222

pQ

298H(

= (2)(-241,818)

(g)2ClO(g)2H(g)O4HCl(g) 222 pQ

298H( = -114,408 J

Solution- continue:


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Standard Heat of Combustion

A combustion reaction is a reaction between an

element or compound and oxygen to form

specified combustion products.

Data are always based on 1 mole of the

substance burned.


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(g)CO4(g)4O4C(s) 22 p

O(l)5H(g)O2

12(g)5H

222

p

(g)O2

16(g)HCO(l)5H(g)4CO 210422 p

Q

298H( = (4)(-393,509)

Q

298

H( =(5)(-285,830)Q

298H( = 2,877,296

(g)HC(g)5H4C(s) 1042 pQ

298H( = -125,790 J

Example: The formation ofn-butane is from

combination of combustion reactions.


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2.4 Temperature Dependence ofQH(

General chemical reaction:

.......... 44332211 p AAAA RRRR

where are stoichiometric numbers and

stands for a chemical formula.iR

iA

iRT

he sign for is positive (+) for products andnegative (-) for reactants.

The sign allows the definition of a standard

heat of reaction.


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Standard heat of reaction using stoichiometric

numbers:|(i

iiHHQQ

R

where is the enthalpy of compound i in itsstandard state. The summation, is over all

products and reactants.

Q

iH

This enthalpy of compound i in its standard stateis equal to its heat of formation, plus

standard state enthalpies of its constituents

elements.

Q

iHQ

ifH(


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As the standard-state of enthalpies of elements

are set to zero as basic of calculation,

The enthalpy of compound i in its standard state,

is equal to its heat of formation.Qi

H

(!(

ifi iHH

QQR

Hence, the standard of heat of reaction;


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Example 4.5

222 Cl2O2H(g)OHCl(g)4 p

(!(i

fi iHHQQ

Ruse

QQQ

HClO2H42 ff HHH ((!(

J408,114)307,92)(4()818,241)(2( !!( QH

The answer is equal to previous solution !!

Calculate the standard heat at 25C for the

following reaction:


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We have discussed the standard heat of

reaction for reference temperature 298 K.Let us calculate the standard heat ofreaction at other temperature.

By using the value at referencetemperature.

HOW???


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For standard reaction, products and reactants

are at standard pressure (1 bar).

The standard-state enthalpies are function of

temperature only.

dTCdHiPi

QQ!

Multiplying by and summing over products

and reactants:

iR

dTCdHiP

i

i

i

ii

QQ ! RR

remember this?


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Since is a constant, it may be placed inside

the differential:

iR

dTCHdiP

i

i

i

ii

QQ ! RR

We already knew before that:

Similarly, we can also define the standard heatcapacity as:

|(i

iiHHQQ

R

QQ

iP

i

iP CC !( R

dTCdHiP

i

i

i

ii

QQ ! RR


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Finally, we can get the fundamental equation

relating heats of reaction to temperature;

dTCHdP

QQ(!(

Integrating equation above;

((!(

T

T

P dT

R

CRHH

0

0

Q

QQ

Where and are heats of reaction at

temperature and at reference temperature, .

QH(

Q

0H(

T 0T

H

eat of reactionat 298 K.


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(T

T

P dTR

C

0

Q

Solving the would give:

(

(

((!

(

XX

XXX1

13

12

10

33

0

22

000

T

DT

CTTA

R

dTCT

T

P

Q

|(i

iiAA R

where

similar with , and .B( C( D(

remember0T

T|X


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For mean heat capacity of reaction,

2

0

22

001

31

2 T

DT

CT

BA

R

CHP

XXXX

(

(

((!

(Q

Hence, for heat reaction with mean heat

capacity,

00 TTCHH HP ((!(QQQ


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Guide: (!(i

fi iHHQQ

R

|(i

iiAA R

0T

T

|X

(

(

((

X

XXXX

11

31

21

0

33

0

22

00 T

DT

CT

BTA

((!(

T

T

P dT

R

CRHH

0

0

Q

QQ

Which equationshould I use?


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What have you learned


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Heats effects in Industry

Sharpen your knowledge with

Example 4.7, Example 4.8 and Example 4.9.

The applications of heat effects elementin industry are not as easy as you think!!

Documents

Chapter 2 Heat Effects (Part2)