If we consider the following reaction:

$H_2 + Cl_2 \Longleftrightarrow 2HCl$

We might be tempted to decide that the reaction will proceed until we have minimized the energy of the system. In order to examine that, we consider the bond strengths of each of the bonds:

$D_{H-H} = 4.5 eV$, $D_{Cl-Cl} = 2.5 eV$, $D_{H-Cl} = 4.5 eV$

so our three species have, respectively

$H_2 = 4.5 eV$, $Cl_2 = 2.5 eV$, $2xHCl = 2x4.5 eV = 9.0eV$

In other words, we need to put in 7$eV$ to break the bonds of the "reactants", but regain 9$eV$ when they reform as 2 molecules of $HCl$. This would suggest that the reaction would (at equilibrium) proceed completely to the right.

However, having two moles of *differing* material (or
better yet, all three!) gives us higher entropy than if we make a
system of pure $HCl$.

This entropic argument becomes particularly powerful when we either gain (higher entropy) or lose (lower entropy) moles as the reaction proceeds (i.e., $\sum_i\nu_i \ne 0$).

In reality, we need to *balance* our change in energy
with the change in entropy (trying to minimize energy, while
maximizing entropy), in much the same way that we did with phase
equilibrium.

Mathematically, we can see this by considering the Gibbs Free Energy of the system (which is a mixture!):

$G = \sum_i n_i \bar G_i = \sum_i n_i \mu_i$

we recall that we need a reference state for $\bar G_i$ or $\mu_i$.

We will use the **standard state** as our reference
point, which is chosen to be the T of interest in our problem and a
pressure of 1 bar.

this gives us a $\mu_i$ expression of

$\mu_i = g_i^o + RT \ln \left [\frac{\hat f_i}{f_i^o} \right ]$

For simplicity, let's use an ideal gas for this example, so we reduce our expression to:

$\mu_i = g_i^o + RT \ln \left [\frac{p_i}{1 bar} \right ]= g_i^o + RT \ln \left [\frac{y_iP_{tot}}{1 bar} \right ]$

Plugging this into our total $G$ expression, we can expand each of the ln terms and make use of the ln product rule to get

$G = \sum_i n_i g_i^o + RT \sum_i n_i\ln \left [P_{tot} \right ] + RT \sum_i n_i\ln \left [y_i \right ]$

where we can note that:

- all terms depend on $\xi$ through the $n_i$ variable
- the first term otherwise depends only on the bond energies
- the second term otherwise is constant
- the third term looks like the $\Delta G_{mix}$

If we plot this for the reaction in question, we get

The minimum Gibbs Free Energy balances the entropy gained
through diversifying the material present with the energy gained by
changing the bond energies. We note that the
**minimum** free energy as a function of $\xi$ is
found where $\displaystyle{\frac{dG}{d\xi}=0}$.

Explain the relationship between energy and entropy in reacting systems