# How do you calculate Gibbs free energy of mixing?

##### 1 Answer
Jun 21, 2016

The Gibbs' free energy of mixing for an ideal binary solution is calculated using this equation:

$\setminus m a t h b f \left({\Delta}_{\text{mix"G^"id}} = R T \left[{n}_{i} \ln {\chi}_{i} + {n}_{j} \ln {\chi}_{j}\right]\right)$

It gets more complicated with nonideal solutions though, as you have to incorporate the activity coefficient ${\gamma}_{j} = {a}_{j} / {\chi}_{j} = \frac{{P}_{j}}{{\chi}_{j} {P}_{j}^{\text{*}}}$ (from Raoult's law).

Eventually, we would have gotten:

$\left({\Delta}_{\text{mix"barG^"real")/(w) = (Delta_"mix"G^"real")/(nw) = (RT)/w[stackrel("ideal solutions")overbrace(chi_ilnchi_i + chi_jlnchi_j)] + stackrel("deviation from ideality}}\right) \overbrace{{\chi}_{i} {\chi}_{j}}$

where $w$ is a constant that my textbook states we "don't need to know".

Below, I derive how to get the one for ideal binary solutions, which is easier to understand.

We can start from this equation:

${\mu}_{j}^{\text{soln" = mu_j^"*}} \left(l\right) + R T \ln {\chi}_{j}$ (Eq. 1)

where:

• ${\mu}_{j}$ is the chemical potential of solvent $j$ in the solution, i.e. when solute is present. Chemical potential is analogous to potential energy.
• ${\mu}_{j}^{\text{*}} \left(l\right)$ is the chemical potential of solvent $j$ as a liquid by itself (i.e. pure solvent).
• $R$ and $T$ are typical variables that you know from the ideal gas law.
• ${\chi}_{j} = \frac{{n}_{j}}{{n}_{i} + {n}_{j}}$ is the $\setminus m a t h b f \left(\text{mol}\right)$ fraction of solvent $j$ in the ideal binary solution at any given moment (with solute if solute is present).

That equation would tell you that the chemical potential decreases upon adding more solute.

The Gibbs' free energy of mixing is defined as:

\mathbf(Delta_"mix"G) = \mathbf(G^"soln"(T,P,n_i,n_j) - [G_i^"*"(T,P,n_i) + G_j^"*"(T,P,n_j)]) (Eq. 2)

where:

• ${G}^{\text{soln}}$ is the final Gibbs' free energy state for the solution as a function of temperature, pressure, and $\text{mol}$s of solute $i$ and solvent $j$. This is after mixing!
• ${G}_{i}^{\text{*}}$ is the pure initial Gibbs' free energy state for the solute as a function of temperature, pressure, and the $\text{mols}$ of solute. This is before mixing!
• ${G}_{j}^{\text{*}}$ is the pure initial Gibbs' free energy state for the solvent as a function of temperature, pressure, and the $\text{mols}$ of solvent. This is before mixing!

In essence, what this says is that...

The change in Gibbs' free energy due to mixing the solute into the solvent, ${\Delta}_{\text{mix}} G$, is the difference between the Gibbs' free energy of the solution before mixing (${G}_{i}^{\text{*" + G_j^"*}}$), and after the solution has been made (${G}^{\text{soln}}$).

Next, we note that $\overline{G} = \frac{G}{n} = \mu$, so ${G}_{i} = {n}_{i} {\mu}_{i}$. For an ideal binary solution, we modify Eq. 2 to get:

Delta_"mix"G^"id" = stackrel(G^"soln")overbrace(n_imu_i^"soln" + n_jmu_j^"soln") - stackrel(G^"*")overbrace([n_imu_i^"*" + n_jmu_j^"*"]) (Eq. 3)

Then, note that we use Eq. 1 to modify Eq. 3 to get:

$\textcolor{b l u e}{{\Delta}_{\text{mix"G^"id") = n_i(mu_i^"soln" - mu_i^"*") + n_j(mu_j^"soln" - mu_j^"*}}}$

$= {n}_{i} R T \ln {\chi}_{i} + {n}_{j} R T \ln {\chi}_{j}$

$= \textcolor{b l u e}{R T \left[{n}_{i} \ln {\chi}_{i} + {n}_{j} \ln {\chi}_{j}\right]}$

So, to calculate the Gibbs' free energy of mixing for an ideal binary solution of solute $i$ and solvent $j$, you need to use this equation and the following variables:

• ${n}_{i}$, the $\setminus m a t h b f \left(\text{mol}\right)$s of solute $\setminus m a t h b f \left(i\right)$.
• ${\chi}_{i} = \frac{{n}_{i}}{{n}_{i} + {n}_{j}}$, the $\setminus m a t h b f \left(\text{mol}\right)$ fraction of solute $\setminus m a t h b f \left(i\right)$.
• ${n}_{j}$, the $\setminus m a t h b f \left(\text{mol}\right)$s of solvent $\setminus m a t h b f \left(j\right)$.
• ${\chi}_{j} = \frac{{n}_{j}}{{n}_{i} + {n}_{j}}$, the $\setminus m a t h b f \left(\text{mol}\right)$ fraction of solvent $\setminus m a t h b f \left(j\right)$.
• $R$, the universal gas constant. To get the right units for $G$, use $\text{8.314472 J/mol"cdot"K}$.
• $T$, the temperature of the solution.