Question

What expression represents the `DeltaH` for a chemical reaction in terms of the potential energy, `E`, of its products and reactants?

Answer 1

There isn't one I can think of off-hand, so we'll have to derive it. I get:

`barul(|" "stackrel(" ")(DeltaH_(rxn) = DeltaE_(rxn) + PDeltaV_(rxn))" "|)`

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Consider a general potential energy diagram for an endothermic reaction

`A + B -> C + D`,

Let the reactants `A` and `B` at potential energies `E_A` and `E_B` form `C` and `D` at potential energies `E_C` and `E_D`. The change in internal energy for the reaction is given by

`DeltaE_(rxn) = (E_C + E_D) - (E_A + E_B)`

So, if we know the ground-state energies of the reactants and products, we can calculate `DeltaE_(rxn)`.

From the Maxwell Relation for the enthalpy `H` of a reversible process in a thermodynamically-closed system (conservation of mass),

`dH = TdS + VdP`

Now, consider adding and subtracting reversible pressure-volume work `w_(rev)` (done from the perspective of the system), `-PdV`, so that

`dH = TdS - PdV + PdV + VdP`

The first two terms are given in the first law of thermodynamics, i.e. conservation of energy for a reversible process in a thermodynamically-closed system:

`dE = q_(rev) + w_(rev) = TdS - PdV`,

where `q_(rev)` is the reversible heat flow and `S` is the entropy.

Thus, we can rewrite this as:

`dH = dE + PdV + VdP`

Now, in ordinary reactions, we are at a constant atmospheric pressure, so `dP = 0`, and `P` as shown is in a sense the initial pressure:

`dH = dE + PdV`

By integrating this from an initial state to a final state, we obtain:

`DeltaH = DeltaE + PDeltaV`,

which should be familiar from general chemistry. For a reaction, we can write this as:

`color(blue)(barul(|" "stackrel(" ")(DeltaH_(rxn) = DeltaE_(rxn) + PDeltaV_(rxn))" "|))`

Thus, if we also know the pressure and the change in volume (through knowing the masses and the densities of each substance), we can then calculate `DeltaH_(rxn)`.