Enthalpy

Another important aspect (definition) of internal energy change follows from the first law of thermodynamics, according to which q = ∆U + P∆V If the process is carried out at constant volume, ∆V = 0. The above equation then reduces to the form q = ∆U where subscript v indicates constant volume. Hence, Internal energy change is the heat absorbed or evolved at constant volume.

It may be mentioned further that as ∆U is a state function, therefore, q, is also a state function.

If a process is carried out at constant pressure (as is usually the case, because most of the reactions are studied in vessels open to the atmosphere or if a system consists of a gas confined in a cylinder fitted with a piston, the external pressure acting on the piston is the atmospheric pressure), the work of expansion is given by

w = - P∆V ...(i)

where,

∆V is the increase in volume and P is the constant pressure.

According to first law of thermodynamics, we know that

q = ∆U –w ...(ii)

where q is the heat absorbed by the system, ∆U is the increase in internal energy of the system and w is work done by the system.

Under condition of constant pressure, putting w = - P∆V and representing the heat absorbed by q_p (subscript p indicating constant pressure), we get

q_p = ∆U + P∆V ...(iii)

Suppose when the system absorbs q_p joules of heat, its internal energy increases from U₁ to U₂ and the volume increases from V₁ to V₂. Then, we have

∆U = U₂ – U₁ ...(iv)

∆V = V₂- V₁ ...(v)

Relationship between ∆H and ∆U

∆H = ∆U +P∆V

=∆U +P (V₂-V₁)

= ∆U + (PV₂-PV₁)

=∆U + (n₂RT-n₁RT)

= ∆U + (n₂-n₁) RT

= ∆U + ∆_ng RT.

Thus ∆H = ∆U + ∆_ng RT where ∆_ng RT = n₂-n₁ is the difference in the number of moles of gaseous products and gaseous reactants.