chapter 2 heat effects (part2)
TRANSCRIPT
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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!!