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

Aug 14, 2018

The van't Hoff reaction isotherm is given by:

${\left(\frac{\partial \overline{G}}{\partial \xi}\right)}_{T , P} = \Delta {\overline{G}}^{\circ} \left(T\right) + R T \ln Q$,

where:

• ${\left(\frac{\partial \overline{G}}{\partial \xi}\right)}_{T , P}$ 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 $\text{mols}$ (it's represented as $x$ in ICE tables you see in general chemistry).
• $\Delta {\overline{G}}^{\circ}$ is the standard change in molar Gibbs' free energy, the reference point, defined at standard pressure ($\text{1 bar}$), and is a function of only temperature.
• $R T \ln Q$ is the deviation from $\Delta {\overline{G}}^{\circ}$ 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 ${\left(\frac{\partial \overline{G}}{\partial \xi}\right)}_{T , P} = \Delta G = 0$ and $Q = K$.
• Standard conditions is if $Q = 1$, i.e. if all the activities ${a}_{i}$ are $1$, so that ${\left(\frac{\partial \overline{G}}{\partial \xi}\right)}_{T , P} = \Delta {G}^{\circ}$.

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 $\Delta G = 0$ 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, ${\mu}_{i} \equiv {\overline{G}}_{i} = {G}_{i} / {n}_{i}$) of substance $i$ in the reaction

${\overbrace{{\left(\frac{\partial \overline{G}}{\partial \xi}\right)}_{T , P}}}^{\text{Reaction Progress}} = {\sum}_{i} {\nu}_{i} {\mu}_{i}$,$\text{ "" } \boldsymbol{\left(1\right)}$

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:

${\mu}_{i} \left(T , P\right) = {\mu}_{i}^{\circ} \left(T\right) + R T \ln {a}_{i}$,$\text{ "" } \boldsymbol{\left(2\right)}$

where ${\mu}_{i}^{\circ} \left(T\right)$ is the chemical potential defined at standard pressure ($\text{1 bar}$).

By substituting the right-hand side of $\left(2\right)$ into the right-hand side of $\left(1\right)$ for ${\mu}_{i}$ (we write ${\mu}_{i}^{\circ}$ from this point to mean it as a function of only temperature):

${\left(\frac{\partial \overline{G}}{\partial \xi}\right)}_{T , P} = {\sum}_{i} {\nu}_{i} \left({\mu}_{i}^{\circ} + R T \ln {a}_{i}\right)$

$\text{ "" "" "" } \setminus = {\sum}_{i} \left({\nu}_{i} {\mu}_{i}^{\circ} + R T {\nu}_{i} \ln {a}_{i}\right)$

$\text{ "" "" "" } \setminus = {\sum}_{i} {\nu}_{i} {\mu}_{i}^{\circ} + R T {\sum}_{i} {\nu}_{i} \ln {a}_{i}$

Now, at standard pressure and the desired temperature,

$\Delta {\overline{G}}^{\circ} = \underline{{\sum}_{i} {\nu}_{i} {\mu}_{i}^{\circ}}$.

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,

${\sum}_{i} {\nu}_{i} \ln {a}_{i} = {\sum}_{i} \ln \left({a}_{i}^{{\nu}_{i}}\right) = \ln \left({\prod}_{i} {\left({a}_{i}\right)}^{{\nu}_{i}}\right)$.

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

$Q = {\prod}_{i} {\left({a}_{i}\right)}^{{\nu}_{i}} = \left({\prod}_{\text{Products" (a_j)^(nu_j))/(prod_"Reactants}} {\left({a}_{i}\right)}^{{\nu}_{i}}\right)$

So, this really means that $\underline{{\sum}_{i} {\nu}_{i} \ln {a}_{i} = \ln Q}$.

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

$\textcolor{b l u e}{\overline{\underline{|}} \stackrel{\text{ ")(" "((del barG)/(del xi))_(T,P) = DeltabarG^@(T) + RTlnQ" }}{|}}$