F.Y. B. Sc., Semester -I Physical Chemistry Savitribai Phule Pune University, Pune

1. Arts, Commerce & Science College, Tryambakeshwar (Nashik)
Email ID : Manohar@mvptryambakcollege.ac.in
Chapter – 1 : Chemical Energetics

2. 1.5 Variation of Heat of Reaction with
Temperature (Kirchoff's Equation] :
• Heat evolved or absorbed in a reaction depends upon
the temperature at which it is carried out. The relationship
showing the variation of the heat of reaction with
temperature was given by Kirchoff in 1856 and is stated
as,
• “ The change in the heat of reaction at
constant pressure for every degree of
temperature is equal to the change in the
heat capacity at constant pressure,
accompanying the reaction”.
It can be derived as under. Consider the
general reaction. Chapter – 1 : Chemical Energetics

3. • Case 1 When the reaction is carried out at constant
pressure :
• Suppose the heat of reaction (heat evolved) at
temperature T1 = ∆H1, and the heat of reaction (heat
evolved) at temperature T2 = ∆H2 (Fig 1.5)
• Starting from A at temperature T1, we can reach B at temperature
T2 by two different routes
• Route I : We perform the reaction at temperature T, to obtain B at
the same temperature and then heathis up to the temperature T2
• Then Heat evolved in the first stage = ∆H1
• Heat absorbed in the II stage = (Cp)B (T2 – T1)
• Where (Cp)B is is the average molar heat capacity of the products.
• .: Net heat evolved = ∆H1 - (CP)B (T2 – T1)
Chapter – 1 : Chemical Energetics

4. Chapter – 1 : Chemical Energetics
Fig. Kirchoff’s Equation for reaction

5. • Route II :
• We first heat up the reactants to the temperature T2 and
then perform the reaction to give the products at the same
temperature. Then if (Cp)A is the average molar heat capacity of
the reactant A,
• Heat absorbed in the first stage = ((Cp)A (T2-T1)
• Heat evolved in the second stage = ∆H2
• .: Net heat evolved = ∆H2 -(CP)A (T2-T1)
• Initial and the final states are the same in both the cases. Therefore, by law of
conservation of energy, we must have.
Net heat evolved by route I = Net heat evolved by route II
i.e. ∆H2 -(CP)A (T2-T1) = ∆H1 - (CP)B (T2 – T1)
Or ∆H2−−
∆H1
T2 – T1
= ∆ CP
Chapter – 1 : Chemical Energetics

6. • When reaction is carried out at constant volume : We have the We
have the thermodynamic relation.
∆ H = ∆ U + P ∆ V
Under condition of constant volume. ∆V = 0
i.e. ∆ H = ∆ U
Hence equation derived above takes the form
∆U2−−
∆U1
T2 – T1
= ∆ CV
Where CV is molar heat capacity at constant Volume
• Thus Kirchoff's equation may also be expressed as follows:
“The change in the heat of reaction at constant volume for every
degree change of temperature is equal to the change in the heat
capacity at constant volume, accompanying the reaction”.
Chapter – 1 : Chemical Energetics

7. Problem 1.12
• 1.12 : Calculate the heat of formation of H2O (l) at 383 K if ∆H for the
reaction
H2 (g)+ O2 (g) H2O (l)
at 298 K is - 298.06 kJ. The average values of heat capacities between
the for H2 (g), O2 (g) and H2O (l) are 27.61, 29.50 and 75.31 JK-1mol-1
respectively.
Solution : H2 (g)+ O2 (g) H2O (l)
Formula: ∆CP = CP (H2O) – (H2) -
1
2
(O2)
= 75.31 -27.61-
1
2
29.50
Chapter – 1 : Chemical Energetics

8. = 32.95 JK-1mol-1
= 32.95 x 10-3 kJK-1mol-1
Applying Kirchoff’s equation,
∆H2−−
∆H1
T2 – T1
= ∆ CP
substituting values, we have
∆H2−(−298.06)
383 –
= 32.95 x 10-3
∆H2 + 298.06
𝟖𝟓
= 32.95 x 10-3
∆H2 + 298.06 = 85 x 32.95 x 10-3
∆H2 = 2.80-298.06
∆H2 = -295.26 kJ
Chapter – 1 : Chemical Energetics

9. Problem 1.13
1.13 : Calculate the heat of formation of HCl at 348 K from
the following data :
𝟏
𝟐
H2 (g) +
𝟏
𝟐
Cl2 (g) HCl (g) ∆H0
298 =-92300 J
• The mean heat capacities over this temperature range are
H2 (g); CP = 28.53 JK-1mol-1
Cl2 (g); CP = 32.26 JK-1mol-1
HCI (g); CP = 28.49 JK-1mol-1
• Solution : The reaction under consideration is
𝟏
𝟐
H2 (g) +
𝟏
𝟐
Cl2 (g) HCl (g)
Chapter – 1 : Chemical Energetics

10. • For this reaction,
∆CP of reaction = CP (HCl) –
1
2
CP(H2) -
1
2
CP(Cl2)
= 29.49 -
1
2
(28.53) -
1
2
(32.26)
= -1.91 JK-1 mol -1
It is given that, ∆H1 = -92300 J, T1 = 298 K , ∆H2
∆H2−−
∆H1
T2 – T1
= ∆ CP
∆H2 = ∆H1 + ∆ CP (T2 – T1)
= -92300 + (-1.91) x ( 348 – 298 )
= -92300 - 95.5
∆H2 = -92395.5 J
Chapter – 1 : Chemical Energetics