Question

Give the thermodynamic derivation of van't Hoff reaction isotherm, and explain its significance?

Answer 1

The van't Hoff reaction isotherm is given by:

`((del barG)/(del xi))_(T,P) = DeltabarG^@(T) + RTlnQ`,

where:

• `((delbarG)/(del xi))_(T,P)` describes the infinitesimal change in the molar Gibbs' free energy as the reaction proceeds at constant temperature and pressure.
• `xi` is the extent of reaction in terms of `"mols"` (it's represented as `x` in ICE tables you see in general chemistry).
• `DeltabarG^@` is the standard change in molar Gibbs' free energy, the reference point, defined at standard pressure (`"1 bar"`), and is a function of only temperature.
• `RTlnQ` is the deviation from `DeltabarG^@` at the same temperature.

It allows you to find the deviation of the Gibbs' free energy away from equilibrium, or away from standard conditions, at the same temperature.

• Equilibrium is if `((del barG)/(del xi))_(T,P) = DeltaG = 0` and `Q = K`.
• Standard conditions is if `Q = 1`, i.e. if all the activities `a_i` are `1`, so that `((del barG)/(del xi))_(T,P) = DeltaG^@`.

An important distinction is that standard conditions has `Q = 1`, but equilibrium does not require `Q = 1`, since `K` does not have to be `1`. Equilibrium just needs `DeltaG = 0` and `Q = K`. Thus, the two ways to use this equation described above are indeed different.

---

To derive this, we begin from the definition of reaction progress based on the chemical potential `mu_i` (the molar Gibbs' free energy, `mu_i -= barG_i = G_i//n_i`) of substance `i` in the reaction

`overbrace(((delbarG)/(del xi))_(T,P))^"Reaction Progress" = sum_i nu_i mu_i`,`" "" "bb((1))`

where `nu_i` is the unitless stoichiometric coefficient of substance `i`, and is negative for reactants and positive for products.

The deviation of the chemical potential due to changes in activities `a_i` (nonideal concentrations) is given by:

`mu_i(T,P) = mu_i^@(T) + RTln a_i`,`" "" "bb((2))`

where `mu_i^@(T)` is the chemical potential defined at standard pressure (`"1 bar"`).

By substituting the right-hand side of `(2)` into the right-hand side of `(1)` for `mu_i` (we write `mu_i^@` from this point to mean it as a function of only temperature):

`((del barG)/(del xi))_(T,P) = sum_i nu_i (mu_i^@ + RTln a_i)`

`" "" "" "" " \ = sum_i (nu_imu_i^@ + RT nu_iln a_i)`

`" "" "" "" " \ = sum_i nu_imu_i^@ + RT sum_i nu_iln a_i`

Now, at standard pressure and the desired temperature,

`DeltabarG^@ = ul(sum_i nu_i mu_i^@)`.

You may have seen this in general chemistry as:

`DeltaG_(rxn)^@ = overbrace(sum_"products" nu_P DeltaG_(f,P)^@ - sum_"reactants" nu_R DeltaG_(f,R)^@)^"Gibbs' free energies of formation"`

Using the properties of logarithms,

`sum_i nu_i ln a_i = sum_i ln (a_i^(nu_i)) = ln (prod_i (a_i)^(nu_i))`.

The definition of the reaction quotient `Q` in terms of activities is

`Q = prod_i (a_i)^(nu_i) = (prod_"Products" (a_j)^(nu_j))/(prod_"Reactants" (a_i)^(nu_i))`

So, this really means that `ul(sum_i nu_i ln a_i = lnQ)`.

Therefore, we obtain the van't Hoff reaction isotherm:

`color(blue)(barul|stackrel(" ")(" "((del barG)/(del xi))_(T,P) = DeltabarG^@(T) + RTlnQ" ")|)`