Relation between enthalpy 'H' and internal energy 'U'

Relation between enthalpy `H' and internal energy `U'

When the system at constant pressure undergoes changes from an initial state with H1, U 1 and V 1 to a final state with H2, U2 and V2 the change in enthalpy ΔH, can be calculated as follows:

H=U + PV

In the initial state

H1=U1 + PV1−−−−−(7.10)

In the final state

H2 = U2 + PV2−−−−(7.11)

change in enthalpy is (7.11) - (7.10)

(H2−H1) = (U2−U1) + P(V2−V1)

∆H=∆U + P∆V ---------      (7.12)

As per first law of thermodynamics,

∆U = q+w

Equation 7.12 becomes

∆H = q + w + P∆V

w= -P∆V

∆H = qp - P∆V + P∆V

∆H = qp–––––––––– (7.13)

qp− is the heat absorbed at constant pressure and is considered as heat content.

Consider a closed system of gases which are chemically reacting to form gaseous products at constant temperature and pressure with Vi and Vf as the total volumes of the reactant and product gases respectively, and ni and nf as the number of moles of gaseous reactants and products, then,

For reactants (initial state) :

PVi = ni RT ---------(7.14)

For products (final state) :

PVf = nf RT --------- (7.15)

(7.15) - (7.14)

(Vf -Vi) = (nf - ni) RT

P∆V= Dn(g) RT –––––––(7.16)

Substituting in 7.16 in 7.12

∆H = âˆ†U +Dn(g) RT–––––––(7.17)