Heat Effects of Chemical Reactions
Enthalpy change for reactions involving compounds
Enthalpy of formation of a compound at standard conditions is obtained
from the literature as standard enthalpy of formation
ΔHo298(CO2(g)) = -393690 J/mole at 298 K
𝑎𝐶 + 𝑏𝑂2
298 K, 1 atm
𝑐𝐶𝑂2
Elements are assigned 0 enthalpy of formation
Enthalpy of a reaction involving compounds is calculated using Hess’ law:
∆𝐻𝑅𝑥𝑛 =
∆𝐻𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 −
∆𝐻𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠
The enthalpy changes are calculated from standard enthalpies of formation
if the reaction occurs at standard conditions
∆𝐻𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 =
𝐻 𝑜 298 (𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠)
∆𝐻𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 =
𝐻𝑜 298 (𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠)
Consider the chemical reaction:
298 K, 1 atm
𝐶𝐻4 + 2𝑂2
𝐶𝑂2 + 2𝐻2 𝑂
∆𝐻𝑅𝑥𝑛 = ∆𝐻 𝑜 298
= (𝐻 𝑜 298 𝐶𝑂2 + 2𝐻 𝑜 298 𝐻2 𝑂 ) − (𝐻 𝑜 298 𝐶𝐻4 + 2𝐻 𝑜 298 𝑂2 )
𝐻 𝑜 298 𝐶𝑂2 = ∆𝐻 𝑜 298 𝐶𝑂2 + 𝐻 𝑜 298 𝐶 + 𝐻 𝑜 298 𝑂2
since ∆𝐻 𝑜 298 𝐶𝑂2 = 𝐻 𝑜 298 𝐶𝑂2 − 𝐻 𝑜 298 𝐶 + 𝐻 𝑜 298 𝑂2
Similarly,
𝐻𝑜
298
𝐻2 𝑂 =
∆𝐻𝑜
298
𝐻 𝑜 298 𝐶𝐻4 = ∆𝐻 𝑜 298
1 𝑜
𝐻2 𝑂 + 298 𝐻2 + 𝐻 298 𝑂2
2
𝑜
𝐶𝐻4 + 𝐻 298 𝐶 + 2𝐻𝑜 298 𝐻2
𝐻𝑜
So ∆𝐻𝑅𝑥𝑛 = ∆𝐻𝑜 298 = ∆𝐻𝑜 298 𝐶𝑂2 + 2∆𝐻 𝑜 298 𝐻2 𝑂 − ∆𝐻 𝑜 298 𝐶𝐻4
∆𝐻 𝑜 298 =
∆𝐻𝑜 298 (𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠) −
∆𝐻𝑜 298 (𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠)
Molecular interpretation of enthalpy change
Enthalpy change of systems involving gaseous compounds may be explained
using bond enthalpies
Consider the reaction
CH4 ( g )  2O2 ( g )  CO2 ( g )  2H 2O( g )
Bond configuration for above reaction:
H
H
C
H + 2 (O==O)
O==C==O + 2(H—O—H)
H
Each bond has certain H*
O==O -414 kJ, C==O -498 kJ, O—H -741 kJ, C—H -464 kJ
ΔH˚= ΔH*(bonds made) – ΔH*(bonds broken)
ΔH˚= 2*(-498)+4(-741)-2(-414)-4(-464) = -686 kJ
Hess’ Law and its applications
Enthalpy change accompanying a chemical reaction is the same whether it
takes place in one or several stages since enthalpy is a state function
A
ΔH
B
1
4
X
2
Y
3
Z
ΔH˚= ΔH˚(1)+ΔH˚(2)+ΔH˚(3)+ΔH˚(4)
Reaction
AX
XY
YZ
ZB
AB
Enthalpy change
ΔH˚(1)
ΔH˚(2)
ΔH˚(3)
ΔH˚(4)
ΔH˚
Hess’ law is useful for calculating the unknown enthalpy change of a
reaction using known reactions
Hess’ Law and its applications
Example – Calculate the standard enthalpy of formation of solid Fe3O4 from
the following enthalpy data
ΔH˚298 = -264500 J
Fe  1 / 2O2  FeO
ΔH˚298 = -292500 J
2 FeO  1 / 2O2  Fe2O3
ΔH˚298 = -230650 J
2 Fe3O4  1 / 2O2  3Fe2O3
Main Reaction
3Fe  2O2  Fe3O4
ΔH˚298 = ?
Add the reactions in the following order to obtain the net reaction
6 Fe  3O2  6 FeO
6 FeO  3 / 2O2  3Fe2O3
3Fe2O3  2 Fe3O4  1 / 2O2
Net reaction:
6 Fe  4O2  2 Fe3O4
ΔH= 6 ΔH1+3 ΔH2- ΔH3
3Fe  2O2  Fe3O4
ΔH/2= -1117240 J/mole
Example –
Calculate ΔH for the reaction C2H4 (g) + H2 (g) = C2H6 (g), from the following
data
C2H4 (g) + 3 O2 (g) 2 CO2 (g) + 2 H2O (l) DH = -1411. kJ/mole
C2H6 (g) + 7/2 O2 (g) 2 CO2 (g) + 3 H2O (l) DH = -1560. kJ/mole
H2 (g) + 1/2 O2 (g) H2O (l) DH = -285.8 kJ/mole
Example –
Calculate ΔH for the reaction 4 NH3 (g) + 5 O2 (g) = 4 NO (g) + 6 H2O (g),
from the following data
N2 (g) + O2 (g) = 2 NO (g) ΔH = -180.5 kJ
N2 (g) + 3 H2 (g) = 2 NH3 (g) ΔH = -91.8 kJ
2 H2 (g) + O2 (g) = 2 H2O (g) ΔH = -483.6 kJ
Isothermal enthalpy change for reactions not occuring at 298 K
Consider a general reaction with components at the same temperature
different than 298 K:
HT
aA(T )  bB(T ) D

 cC (T )  dD(T )
ΔH˚298 are tabulated for components
ΔH= cΔHC + dΔHD - aΔHA - bΔHB
 H 

  CP
 T  P
dH C
dDH
dH D
dH A
dH B
c
d
a
b
dT
dT
dT
dT
dT
 DH 

  DCP cC PC  dC PD  aC PA  bC PB
 T  P
DCP   CP( prod.)  CP( react.)
Isothermal enthalpy change for reactions not occuring at 298 K
ΔH values for chemical reactions can be determined from ΔH˚298 values:

DH T
T
dDH   DC P dT
DH 298
DH T  DH
298
o
298
T
  DC P dT
Kirchoff equation
298
ΔHT can also be calculated from the enthalpy increment equations:
DHT  DH
o
298
  ( HT  H 298 ) prod.  ( HT  H 298 )react.
Example – Find the net heat available or required when the following
reaction takes place at 800 K
CaO( s)  CO2 ( g )  CaCO3 ( s)
Substance
CaO(s)
CO2(g)
CaCO3(s)
ΔHo298 (kJ/mole)
CP (J/mole K)
-634.3
49.62+4.52*10-3 *T-6.95*105*T-2
-393.5
44.14+9.04*10-3 *T-8.54*105*T-2
-1206.7
104.52+21.92*10-3 *T-25.94*105*T-2
o
o
o
o
DH 298
 DH 298
(CaCO3 )  DH 298
(CaO)  DH 298
(CO2 )
= -1206.7 - 634.3 + 393.5 = -178.9 kJ
DC P  C P (CaCO3 )  C P (CaO )  C P (CO2 )
= 10.76 + 8.36*10-3*T-10.45*105*T-2 J/K
DH
o
800
800
 178900   (10.76  8.36 *103 T  10.45 *105 T 2 )dT
298
o
DH 800
 178900  5505  173395
J
Alternatively ΔHT can be calculated from Hess’ law
H 800
CaO(800)  CO2 (800) D
 CaCO3 (800)
1
2
4
H 298
CaO(298)  CO2 (298) D
 CaCO3 (298)
3
298
DH1   CP (CaO) dT
800
298
DH 2   CP (CO2 ) dT
800
o
DH 3  DH 298
800
DH 4   CP (CaCO3 ) dT
298
DH T  DH1  DH 2  DH 3  DH 4
Isothermal enthalpy change involving phase change
Consider a general reaction with components at the same temperature
different than 298 K where liquid B and C phases are solid at 298 K:
HT
aA( s)(T )  bB(l )(T ) D

 cC (l )(T )  dD( s)(T )
Kirschoff equation can be used to calculate ΔH˚T :
DHT  DH
298
ΔCP(1)
Tm(C) ΔCP(2)
Tm(B) ΔCP(3)
o
298
T
  DCP dT
298
T
Component C transforms to liquid at Tm(C)
Component B transforms to liquid at Tm(B)
ΔCP seperated into 3 ranges
Isothermal enthalpy change involving phase change
DH T
aA( s)(T )  bB(l )(T ) 
 cC (l )(T )  dD( s)(T )
298
ΔCP(1)
Tm(C) ΔCP(2)
Tm(B) ΔCP(3)
DC  cCP (C ( s ))  dCP ( D ( s ))  aCP ( A( s ))  bCP ( B ( s ))
I
P
DCPII  cCP (C (l ))  dCP ( D ( s ))  aCP ( A( s ))  bCP ( B ( s ))
DC
III
P
 cCP (C (l ))  dCP ( D ( s ))  aCP ( A( s ))  bCP ( B (l ))
Therefore
DH T  DH
o
298

Tm ( C )
298
DC dT  cDH m (C )  
I
P
Tm ( B )
Tm ( C )
DC dT  bDH m ( B )  
II
P
T
Tm ( B )
DC PIII dT
Since HT-H298 enthalpy increment equations incorporate phase changes,
o
DHT  DH 298
  ( HT  H 298 ) prod.   ( HT  H 298 ) react.
Isothermal enthalpy change involving phase change
Σ(HT-H298)prod.
HT-H298
cΔHm(C)
bΔHm(B)
ΔH
Σ(HT-H298)react.
ΔHT
Tm(C)
Tm(B)
T
Another alternative method that can be used in enthalpy computations for
chemical reactions that involve phase transformations is the application of
Hess’ law
DH T
aA( s)(T )  bB(l )(T ) 
 cC (l )(T )  dD( s)(T )
1
2
4
5
H3
aA(s)(298)  bB(s)( 298) D

cC (s)( 298)  dD(s)(298)
DH T  DH1  DH 2  DH 3  DH 4  DH 5
DH1  a 
298
T
C P ( A( s )) dT
Tm ( B )
298

DH 2  b 
C P ( B ( l )) dT  DH m ( B )  a  C P ( B ( s )) dT 
 T

Tm ( B )
o
DH 3  DH 298
Tm ( C )
T

DH 4  c 
C P ( C ( s )) dT  DH m (C )  a  C P ( B (l )) dT 
 298

Tm ( C )
T
DH 5  d  C P ( D ( s )) dT
298
Non-isothermal chemical processes
HT
aA( s )(T1 )  bB(l )(T2 ) D

 cC (l )(T3 )  dD( s )(T3 )
1
2
4
5
H3
aA(s)(298)  bB(s)( 298) D

cC (s)( 298)  dD(s)(298)
DH T  DH1  DH 2  DH 3  DH 4  DH 5
DH1  a 
298
T1
C P ( A( s )) dT  a ( H T1  H 298 ) A( s )
Tm ( B )
298

DH 2  b 
C P ( B (l )) dT  DH m ( B )  a  C P ( B ( s )) dT   b( H T 2  H 298 ) B (l )
 T2

Tm ( B )
o
o
o
DH 3  DH 298
  H 298
( prod.)   H 298( react.)
Tm ( C )
T3

DH 4  c 
C P ( C ( s )) dT  DH m (C )  a  C P ( C (l )) dT   c( H T 3  H 298 ) C (l )
 298

Tm ( C )
T3
DH 5  d  C P ( D ( s )) dT  d ( H T 3  H 298 ) D ( s )
298
Example – A furnace that is designed to melt silver/copper scrap is to be
fired with propane and air. The propane vapor mixes with dry air at 298 K.
Flue gases are expected to exit the furnace at 1505 K under steady state
conditions. How long will a 45.5 kg container of propane maintain the
furnace temperature if heat is conducted through the brickwork at the rate
of 10000 kJ/hour ?
DH
C3 H 8 ( g )  5O2 ( g )  18.8 N 2 ( g ) 
3CO2 ( g )  4 H 2O( g )  18.8 N 2 ( g
298
1505
1
2
3CO2 ( g )  4 H 2O( g )  18.8 N 2 ( g )
298
Substance
C3H8(g)
CO2(g)
H2O(l)
H2O(g)
N2(g)
ΔHv(H2O) =
HT-H298 (J/mole)
-16476+44.25*T+0.0044*T2+8.62*105 T-2
34660+30.01*T+0.00536*T2-0.33*105 T-2
-8502+27.88*T+0.00213*T2
40897 J/mole
Air: 21% O2 + 79% N2
ΔHo298 (kJ/mole)
CP (J/mole K)
-103.55
-393.5
44.14+9.04*10-3 *T-8.54*105*T-2
-285.85
75.47
-241.95
30.01+10.72*10-3 *T+0.33*105*T-2
-27.88+4.27*10-3 *T
Flame temperature
The maximum temperature the gaseous products can reach upon
proceeding of an exothermic reaction is called the flame temperature
The furnace is considered as adiabatic for no heat loss to the surroundings
and maximum flame temperature
DH T
aA( s )(T1 )  bB( g )(T2 ) 
 cC ( g )(T3 )  dD( g )(T3 )
1
2
4
5
DH 3
aA( s )(298)  bB( g )(298) 
 cC (l )(298)  dD( g )(298)
DH1  DH 2  DH 3  DH 4  DH 5
Tm ( C )
Tflame

DH 4  c 
C P (C (l )) dT  DH v (C )  a 
C P (C ( g )) dT   c( H Tflame  H 298 ) C ( g )
 298

Tm ( C )
DH 5  d 
TTflame
298
C P ( D ( g )) dT  d ( H Tflame  H 298 ) D ( g )
An adiabatic vessel contains 1000 grams of liquid aluminum at 700 °C. Calculate the
moles of Cr2O3 which completely reduces chromium and raises the temperature of
the resulting mixture of Al2O3(s) and Cr(s) to 1000 °C when added to the liquid
aluminum. The initial temperature of Cr2O3 is 25 °C.
Chromium reduction reaction: Al(l) + Cr2O3(s) = Al2O3(s) + Cr(s) ,