Enthalpy [Heat content]





Let the change of state of a system be brought about at constant pressure. In such a case, there will be a change of volume. Let the volume increase fromV_A \text{to} V_B at constant pressure ‘P’.

Then the work done (w) by the system will be:

w = P(V_B- V_A)  …………(1)

Substituting the value of w from equation, we get:

dE=q- P(V_B- V_A) V \\[3mm] E_B- E_A = dE = q- P(V_B- V_A) \cdots (2) \\[3mm] \text{or}(E_B + PV_B)- (E_A+PV_A)=q \cdots (3)

 

This quantity E + PV is known as Heat content or Enthalpy of the system and is denoted by H.

Thus,

H = E + PV

Since E, P and V are definite properties, it follows that ‘H’ is also a definite property depending upon the state of the system.

Using the value in equation (3), we get

H_B - H_A = \Delta H = q

So, that the increase of the heat content of a system is equal to the heat absorbed at constant pressure. Substituting the value of AH for q in equation (3), we get

E_B- V_A = \Delta H- P(V_B- V_A) \\[3mm] or \Delta H = (E_B- E_A) + P(V_B- V_A) \\[3mm] or \Delta H = \Delta E + P. \Delta V…..(4)

I.e.

The change in heat content or enthalpy at constant pressure is equal to the sum of the increase in internal energy and mechanical work of the expansion.

The equation (4) may also be written as :

\Delta H = \Delta E + \Delta nRT

Where \Delta n = Difference in no. of moles of gaseous products and reactants

R = gas constant and

T = Temperature in Kelvin.

It is clear that if\Delta n=0, \Delta H=\Delta E; \Delta n>O, \Delta H> \Delta Eand \Delta n 0, \Delta H \Delta E



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