Derive a formula, similar to equation 5.90, for the shift in the freezing temperature of a dilute solution. Assume that the solid phase is pure solvent, no solute. You should find that the shift is negative: The freezing temperature of a solution is less than that of the pure solvent. Explain in general terms why the shift should be negative.

${\mu }_{A.liq}={\mu }_{A.solid}$

Here, ${\mu }_{A.solid}$is the chemical potential of $A$in the solid phase and ${\mu }_{A.liq}$is the chemical potential of $A$in liquid phase.

Use the equation 5.69, the chemical potential of $A$in liquid phase is as follows.

${\mu }_{A.liq}={\mu }_0(T,P)-\frac{N_{B}kT}{N_A}$

Here, ${\mu }_0$is the chemical potential of the pure solvent and $k$is the Boltzmann's constant.

Substitute${\mu }_0(T,P)-\frac{N_{B}kT}{N_A}={\mu }_{A.solid}(T,P)......\left(1\right)$

$$\begin{gathered} {\mu }_0(T,P)={\mu }_0(T_0,P)+(T-T_0)\frac{\partial {\mu }_0}{\partial T} \\ {\mu }_{A.solid}(T,P)={\mu }_{A.solid}(T_0,P)+(T-T_0)\frac{\partial {\mu }_{A.solid}}{\partial T} \end{gathered}$$

Substitute the above two equations in the equation (1) and simplify.

${\mu }_0(T_0,P)+(T-T_0)\frac{\partial {\mu }_0}{\partial T}-\frac{N_{B}kT}{N_A}={\mu }_{A.solid}\left(T_0,P\right)+(T-T_0)\frac{\partial {\mu }_{A.solid}}{\partial T}.....\left(2\right)$

At the temperature $T_0$, pure liquid phase is equilibrium with the solid phase. Hence,

${\mu }_0(T_0,P)={\mu }_{A.solid}(T_0,P)$

The Gibbs free energy for pure solvent A is,

$G=N_A{\mu }_0$

Here, $N_A$is the number of particle in that phase.

Partial differentiate the equation on both sides with respect to the temperature.

$$\begin{gathered} \frac{\partial G}{\partial T}=N_A\frac{\partial {\mu }_0}{\partial T} \\ -S=N_A\frac{\partial {\mu }_0}{\partial T}\left(Since\frac{\partial G}{\partial T}=-S\right) \\ \frac{\partial {\mu }_0}{\partial T}=-\frac{S}{N_A} \end{gathered}$$

Substitute ${\mu }_0(T_0,P)$for ${\mu }_{A.solid}(T_0,P)$and $-\frac{S}{N_A}$for $\frac{\partial {\mu }_0}{\partial T}$in the equation (2) and simplify.

$$\begin{gathered} {\mu }_0(T_0,P)+(T-T_0)\left(-\frac{S}{N_A}\right)-\frac{N_{B}kT}{N_A}={\mu }_0(T_0,P)+(T-T_0){\left(-\frac{S}{N_A}\right)}_{solid} \\ (T-T_0){\left(-\frac{S}{N_A}\right)}_{liq}-\frac{N_{B}kT}{N_A}=(T-T_0){\left(-\frac{S}{N_A}\right)}_{solid} \\ (T-T_0)\left(S_{liq}-S_{solid}\right)=-N_{B}kT \\ T-T_0=\frac{-N_{B}kT}{S_{liq}-S_{solid}} \end{gathered}$$

$S_{liq}-S_{solid}=\frac{L}{T_0}$

Here, $L$is the latent heat of condensation and $T\approx T_0$.

Substitute $\frac{L}{T}$for $S_{liq}-S_{solid}$in the equation $T-T_0=\frac{-N_{B}kT}{S_{liq}-S_{solid}}$and solve for $T-T_0$

$$\begin{gathered} T-T_0=\frac{-N_{B}kT}{\left({\displaystyle \frac{L}{T}}\right)} \\ =\frac{N_{B}k{T}^2}{L} \end{gathered}$$

Hence, the shift in the freezing temperature of the dilute solution is negative.