Question

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.

Short answer

The negative sign in the above result shows that adding a solute will lower the temperature of the freezing point of the liquid.

Step-by-step solution

Let us consider a pure solid A. If a solution of solute B in solvent A is equilibrium with solid A, then the chemical potentials of A for the two phases must be equal.

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

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

At a fixed pressure P, the pure liquid is equilibrium with the solid at temperature T0. Then the chemical potential can be written in terms of T0 as follows.

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

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.

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

The difference in the entropy between the liquid and solid is :

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

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.