Question

Give relation between ∆H and ∆E.

Answer 1

In only pressure-volume work is done at constant pressure, then according to first law of thermodynamics,
∆E = q_p – P∆V
q_p = ∆E + P∆V
q_p = (E₂ – E₁) – (PV₂ + PV₁)
For constant pressure, q is represented by q_p.
Here, E, P and V, all are state functions, E + PV is also a state function. This is known as enthalpy or heat content.
Enthalpy of a system is equal to sum of its internal energy and pressure – volume energy. The enthalpy change for initial and final state are:
H₂ – H₁ = (E₂ -E₁ + PV₂ – PV₁)
∆H = ∆E + P∆V = q_p
Thus, heat absorbed by the system at constant pressure is equal to its enthalpy change. H is a state function. Hence, q_p is also a state function.
If V₂ = V₁ then, ∆H = ∆E
In gaseous state, there is an appreciable distance between H and E, then according to ideal gas equation,
PV_A = n_ART
PV_B = n_BRT
PV_B – PV_A – n_BRT – n_ART = (n_B – n_A )RT
P(V_B – V_A) = (n_B – n_A)RT
P∆V = n_gRT
Here, ∆n_g refers to the number of moles of gaseous products minus the number of moles of gaseous reactants. Substituting the value of P∆V from equation, we get
∆H = ∆E + ∆n_gRT