Molar form of Enthalpy equation – how to derive?

The Enthalpy equation is represented as ΔH = ΔU + pΔV …(1)

where ΔU is the change in internal energy and ΔV is the change in volume of the system.

The difference between ΔH and ΔU is not usually significant for solids and liquids. However, when the systems involve gases, the difference becomes significant.

Here, in this post, we will derive the molar form of the Enthalpy equation.

We will use the general form of the enthalpy equation given above (equation 1) to derive the molar form.

Deriving the molar form of the Enthalpy equation

Consider a chemical reaction taking place at a constant temperature (T) and pressure (p).

Let VA is the total volume of the gaseous reactants and VB is the total volume of the gaseous products, nA is the number of moles of gaseous reactants and nB is the number of moles of gaseous products.

According to ideal gas equation, pV = nRT

For reactants,

pVA = nART (at constant T and p) …(2)

For products,

pVB = nBRT (at constant T and P) …(3)

Subtracting Eq. (3) from Eq. (2), we get

p(VB − VA) = nB RT − nA RT = (nBnA) RT

or pΔV = Δng RT …(4)

where Δng is the change in the number of gaseous moles of products and gaseous moles of reactants.

Thus, Eq. (1) becomes

ΔH = ΔU + Δng RT

Energy change at constant p = Energy change at constant volume + Change in the number of moles x RT

This relation is very useful for converting ΔH into ΔU or ΔU into ΔH.

summary

So here are the 2 equations for ΔH, and the last one is the molar form.

ΔH = ΔU + pΔV

ΔH = ΔU + Δng RT

Molar form of Enthalpy equation – how to derive?
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