Chemistry
Chemical thermodynamics-II
Session Objectives
Session objectives
1. Enthalpy
2. Various types of enthalpy of reactions
3. Heat capacities of gases
4. Adiabatic process
5. Hess’s law
6. Bond energy
7. Lattice energy
8. Limitation of first law of thermodynamics
Enthalpy
Enthalpy is the total heat contents
of the system at constant pressure.
Enthalpy is shown by ‘H’.
H = E + PV
Enthalpy change at constant pressure
H = E + PV
H2 – H1 = E2 – E1 + P(V2 – V1)
Where H1, E1 and V1 are the enthalpy, internal energy
and volume respectively in initial state while H2, E2 and V2
are the enthalpy, internal energy and volume respectively
in final state.
Enthalpy
Enthalpy is a state function
PV1 = n1RT
At constant pressure
(for initial state)
PV2 = n2RT
(for final state)
P(V2 – V1) = RT(n2 – n1)
PV = n g RT
Where ng = np–nr (gaseous moles only)
 H = E + ngRT
Enthalpy of formation
It is the change in enthalpy when
one mole of a compound is formed
from its elements in their naturally
occuring physical states.
2C(s)  2H2(g) 
 C2H4 Hf  52kJ
Enthalpy of combustion
It is the change in enthalpy when
one mole of the substance undergoes
complete combustion.
CH4  g  2O2  g  CO2  g  2H2O(g)
; H   890.3 kJ
2C2H6  g  7O2  g 
 4CO2  g  6H2O l
H   745.6 K calmole1
 Heat of combustion
745.6

 – 372.8 Kcal mol1
2
Application of heat of
combustion
Calorific value
Amount of heat produced per gram of
a substance (food or fuel) is completely burnt.
CH4  g  2O2  CO2  g  2H2O l
H = -890.3 kJ/mol
– 890
 – 55.6 kJ / g
Calorific value of CH4 (g) =
16
Hydrogen has the highest calorific value (150 kJ/g)
Enthalpy of neutralization
Strong acid and strong base
HCl (aq) + NaOH (aq)  NaCl (aq) + H2O
H = -13.7 Kcal
HCl  aq
 strong acid

NH4OH
 weak base
 aq

 NH4Cl  aq  H2O l
H  – 12.3Kcal
Difference in energy is used to ionize weak base.
What is the Ionization energy of NH4OH?
Ionization energy of NH4OH = 13.7  12.3  1.4KCal.
Enthalpy of solution
Amount of heat evolved or absorbed
per mole of the substance in excess
of water,
KCl  s   H2O 
 KCl  aq
H  – 4.4 Kcal
KOH  s   H2O 
 KOH  aq H  – 13.3 Kcal
Enthalpy of fusion
One mole of solid substance changes
to its liquid state at its melting point.
Melting
H2O  s   H2O l
Freezing
H  1.44Kcal
H2O l 
 H2O  s  H  – 1.44 Kcal
Enthalpy of vaporization
One mole of the substance changes
from liquid state to gaseous state at
its boiling point.
H2 O  l   H2 O  g
Boiling
H  10.5 Kcal
H2 O  g  H2 O l  H  – 10.5 Kcal
Cooling
Enthalpy of sublimation
Enthalpy change per mole of a solid converts directly to
its vapours
sublimation

NH4Cl (s) 
 NH4Cl  g
H  14.9 Kcal
Heat capacity
Quantity of heat required to raise the
temperature of the system by one degree.
dq
Heat capacity =
dT
 H 
C

Heat capacity at constant pressure
p
 T 

p
(Since at constant pressure dq = dH)
Heat capacity
Heat capacity at constant volume
 E 
CV   
 T  V
(Since at constant volume dq = dE)
The difference between Cp and Cv is equal to the work
done by 1 mole of gas in expansion when heated through
1° C.
 Cp-Cv = R
Heat capacity
Specific heat capacity is the heat
required to raise the temperature of
unit mass by one degree.
q
= c × m × T
m = Mass of the substance
q
= Heat required
T = Temperature difference
c = Specific heat capacity. Specific heat capacity of
water is 4.18 J/g K.
Adiabatic work
For adiabatic process, q = 0 it means
no heat is exchanged with the
surrounding.
E  q  w
i.e. E  w
q0
For a finite change of an ideal gas,
E  Cv .T
 w  E  Cv T
Reversible Adiabatic expansion
Relations for reversible adiabatic
expansion of an ideal gas
PV   cons tan t
...(i)
TV  1  cons tan t
...(ii)
T  V  1  cons tan t ...(iii)
Now, considering this relation
Work done, W  Cv T

R
(T2  T1 )
(  1)
[
(For 1 mole of gas)
R
   1]
Cv
Irreversible Adiabatic expansion
(i)
For free expansion,
w = 0. Since Pext = 0
(ii)
For intermediate expansion,
w   Pext (V2  V1 )
 T2 T1 
  PextR 


P
P
 2
1 
i.e. w  Cv (T2  T1 )   R Pext
 T2 T1 



P
P
 2
1 
Hess’s Law
According to Hess’s law
q = q1 + q 2
q1
C
q2
q
A
B
Applications of Hess’s law
Determination of heat changes of
transformation of rhombic sulphur
into monoclinic sulphur.
Given
S (rhombic) + O2 (g)  SO2 (g); H = - 297.5 kJ -----(i)
S (monoclinic) + O2 (g)  SO2 (g); H = - 300.0 kJ-----(ii)
Subtracting (ii) from (i), we get
S (rhombic) – S (monoclinic)
H = +2.5 kJ
Resonance energy
Resonance energy =
Enthalpy of formation calculated from bond energy –
Experimental value of enthalpy
Question
Illustrative example
Calculate resonance energy of benzene
from the following data:
(i)
(ii)
(iii)
H0 benzene = –358.5 kJ mol–1
f
Heat of atomization of carbon is 716.8 kJ mol-1
Bond energies of C–H, C–C, C = C and H–H bonds
are 490, 340, 620 and 436.9 kJ mol–1
respectively.
Solution
The required equation is
6C(s) + 3H2(g)  C6H6 , H = -358.5 kJ
H =
f
Bond energies of reactants – Bond energies of products
= {6 [  HC(s) c(g)] + 3 (  HH–H) – 3{[  HC–C] + 3 (  HC =C) +
6 (  HC–H )
=6716.8 + (3 436.9) – (3 340) – (3620) – (6490) kJ mol–1
Resonance energy =  Hf (obs.) –  Hf (cal)
= – 358.5 – ( – 208.5)
= – 150.0 kJ mol–1
Determination of lattice energy
 H
MX
   Hsub M 
1
( H
2
)
diss X2
 (IP)M  (  EA)X  (  U)MX
Figure
Determination of bond energies or
bond enthalpies
Energy required to break the bond
or energy released during the bond
formation is called bond energy.
H2S 
 H  g  SH  g
H  100 kJ / mole
SH  g 
 S  g  H  g
H  200 kJ / mole
The average of these two bond dissociation energies
gives the value of bond energy of S — H.
Bond energy of S — H bond 
100  200
 150 kJ / mole
2
Limitations of first law
1. The first law of thermodynamics
states that one form of energy
disappears, an equivalent amount
of another form of energy is produced.
But it is salient about the extent to which such
conversion can take place.
2. It does not tell about the direction of flow of heat.
3. It does not tell about spontaneity of reaction.
Class exercise
Class exercise 1
For the reaction
1
C6H6(l)  7 O2(g) 
 6CO2(g)  3H2O(l)
2
0)
,

H(27
= -780.9 Kcal, E
for the reaction will be
(a) – 781.80 Kcal (b) – 780.009 Kcal
(c) – 780.75 Kcal
(d) 780.05 Kcal
Solution
H = E + nRT
15
3
   1.5
2
2
E = H - nRT
n  6 
= –780.9 – (–1.5 RT)
 1.5  1.987  300 
 780.9  

1000


= – 780.9 + 0.891
= – 780.009 Kcal
Hence, the answer is (b).
Class exercise 2
When 1 gram of methane (CH4) burns
in O2 the heat evolved (measured under
standard conditions) is 13.3 kcal. What is
the heat of combustion?
(a) –13.3 k cals
(b) +213 k cals
(c) – 213 k cals
(d) – 416 k cals
Solution
13.3 Kcal/gm evolved
Ho
  13.3  16  Kcal / mol
comb
= – 213 Kcal/mol
Hence, the answer is (c).
Class exercise 3
When 4.184 J of heat is transferred
to 1 g of water at 20° C, its temperature
rises to 21° C. The molar heat capacity
at this temperature is
18
-1
–1
(a) 18 JK
(b)
JK
4 .184
(c) 75.4 JK–1
(d) 4.184 JK–1
Solution:
n CT = 4.184
C = (4.184) × 18 = 75.4 J/K
Hence, the answer is (c).
Class exercise 4
When 0.532 g of benzene (B.P 80° C)
is burnt in a constant volume system
with an excess of oxygen, 22.3 kJ of
heat is given out. H for the
combustion process is given by
(a) – 21 kJ
(b) – 1234.98 kJ
(c) – 221 kJ
(d) – 3273.26 kJ
Solution
H   22.3 
Mol. wt C6H6 
 22.3  78

0.532
0.532
= – 3269.5 kJ
Hence, the answer is (d).
Class exercise 5
Consider the reaction
1
SO2(g)  O2(g)  SO3(g) H
2
= – 98.3 kJ. If the enthalpy of formation
of SO3(g) is – 395.4 kJ, then the enthalpy of
formation of SO2(g) is
(a) 297.1 kJ
(b) 493.7 kJ
(c) – 493.7 kJ
(d) – 297.1 kJ
Solution:
Ho
 Ho
  98.3 kJ
f  SO3 
f SO2 
 Ho
 98.3  395.4 kJ  297.1 kJ
f SO2 
Hence, the answer is (d).
Class exercise 6
Calculate the heat change for
the following reaction:
CH4(g) + 2O2(g) 
 CO2(g) + 2H2O(l)
H0 for CH4 , H2O and CO2 are –17.89, –68.3
f
and–94.05 kcal/mole.
Solution
CH4  g  2O2 g 
 CO2 g  2H2O l
o
o
o
o
Hreaction
 2Ho


H


H

2

H
f H2O
f CO2
f CH4
f O2
= – 2 × 68.3 – 94.05 + 17.89 – 0
= – 212.76 kcal/mol
Class exercise 7
Calculate the heat of combustion of
benzene from the following data:
6C (s) + 3H2(g) 
 C6H6(l)
H = 11720 cal
1
H2(g) + O2(g) 
 H2O(l) H = -68320 cal
2
C(s) + O2(g) 
 CO2(g) H =-93050 cal
Solution
The required reaction is
C6H6 
15
O2 
 3H2O  6CO2    (1)
2
6C + 3H2  C6H6    (2)
C + O2  CO2    (3)
1
H2  O2 
 H2O
2
3 × (4) + 6 × (3) – (2)
o
o
o
 H1
 3Ho

6

H


H
4
3
2
= 3(– 68320) + 6(– 93050) – (1720 Cal)
= – 774.980 kcal/mol
Class exercise 8
Calculate H for the following
reaction at 27o C.
C2H4 (g) + 3O2(g)  2CO2 (g) + 2H2O(l)
Given H= – 337 kcal R= 1.987 cal deg–1 mole–1
Solution
H = E + PV
= E + nRT
{n = -2}
 E =H - nRT
=– 337 – (–2)(1.987)(300) × 103
= – 335.8078 kcal/mol
Class exercise 9
Calculate the heat of combustion of
acetic acid at 25o C if the heat of
formation of CH3COOH(l),CO2(g) and
H2O(l) are –116.4, –94.0
and –68.3 kcal mole–1 respectively.
Solution:
CH3COOH l  3O2 g 
 2CO2 g  2H2O l
Hreactions = 2 . Hf (CO2 ) + 2 . Hf (H2O)
-Hf (CH3COOH) - 3Hf (O2 )
= 2(– 94.0) + 2(– 68.3) – (–116.4)
= – 208.2 Kcal/mol
Thank you