Let us again consider first law of thermodynamics
U
= Q + W
or Q = U – W
If Q is the heat absorbed by the system, U is the increase
in internal energy and W is the work done by the system.
If pressure is constant,
W = – PU
Q = U – (-PU) = U + PU
If internal energy increases from U1 to U2 and volume increases from V1 to V2 then,
U
= U2 – U1 and U = V2 – V1
Substituting we get
Q = (U2 – U1) + P(V2 – V1)
= (U2 + PV2) – (U1 + PV1)
U1, P and V are functions of state, thus the quantity U + PV must also be a state function.
(U + PV) is called the heat content or enthalpy of the system. It is represented by symbol H.
H = U + PV
Now H2 = U2 + PV2
and H1 = U1 + PV1
Q = H2 + H1
Q = V
Enthalpy change of a system is equal to the heat absorbed or evolved by the system at constant pressure.
Enthalpy of a substance or a system is the energy stored within the substance or the system that is available for conversion into heat.
Hess’s law of constant heat summation: The total amount of heat evolved or absorbed in a reaction is the same whether the reaction takes place in one step or in a number of steps.
In other words, the total amount of heat exchange in a reaction depends only upon the nature of the initial reactants and the nature of the final products and is independent of the path or the manner by which the change is brought about.
Thus, the thermo chemical equations can be treated as algebraic equations which can be added, subtracted, multiplied or divided.
Example: calculate the enthalpy of formation of carbon monoxide (CO) from the following data:
(i) C(s) + O2(g) ? CO2(g) —– (1)
H = – 393.3 kJ/mol
(ii) CO(g) + O2(g) ? CO2(g) —– (2)
H = -282.8 kJ/mol
Solution:
We have to obtain
C(s) + O2(g) ? CO(g)
Subtracting equation (2) from equation (1)
C(s) + O2(g) – CO(g) – O2(g) – CO2(g)
C(s) + O2(g) ? CO(g)
H
= -393.3 – (-282.8) = -110.5 kJ/mol
Thus, heat of formation of CO is
Hf
= 110.5 kJ/mol