Suggested languages for you:

Americas

Europe

Q.5.81

Expert-verifiedFound in: Page 208

Book edition
1st

Author(s)
Daniel V. Schroeder

Pages
356 pages

ISBN
9780201380279

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.

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

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

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 ${\mu}_{0}(T,P)-\frac{{N}_{B}kT}{{N}_{A}}={\mu}_{A.solid}(T,P)......\left(1\right)$

${\mu}_{0}(T,P)={\mu}_{0}({T}_{0},P)+(T-{T}_{0})\frac{\partial {\mu}_{0}}{\partial T}\phantom{\rule{0ex}{0ex}}{\mu}_{A.solid}(T,P)={\mu}_{A.solid}({T}_{0},P)+(T-{T}_{0})\frac{\partial {\mu}_{A.solid}}{\partial T}\phantom{\rule{0ex}{0ex}}$

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.

$\frac{\partial G}{\partial T}={N}_{A}\frac{\partial {\mu}_{0}}{\partial T}\phantom{\rule{0ex}{0ex}}-S={N}_{A}\frac{\partial {\mu}_{0}}{\partial T}\left(Since\frac{\partial G}{\partial T}=-S\right)\phantom{\rule{0ex}{0ex}}\frac{\partial {\mu}_{0}}{\partial T}=-\frac{S}{{N}_{A}}$

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.

${\mu}_{0}({T}_{0},P)+(T-{T}_{0})\left(-\frac{S}{{N}_{A}}\right)-\frac{{N}_{B}kT}{{N}_{A}}={\mu}_{0}({T}_{0},P)+(T-{T}_{0}){\left(-\frac{S}{{N}_{A}}\right)}_{solid}\phantom{\rule{0ex}{0ex}}(T-{T}_{0}){\left(-\frac{S}{{N}_{A}}\right)}_{liq}-\frac{{N}_{B}kT}{{N}_{A}}=(T-{T}_{0}){\left(-\frac{S}{{N}_{A}}\right)}_{solid}\phantom{\rule{0ex}{0ex}}(T-{T}_{0})\left({S}_{liq}-{S}_{solid}\right)=-{N}_{B}kT\phantom{\rule{0ex}{0ex}}T-{T}_{0}=\frac{-{N}_{B}kT}{{S}_{liq}-{S}_{solid}}$

${S}_{liq}-{S}_{solid}=\frac{L}{{T}_{0}}$

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 $T-{T}_{0}$

$T-{T}_{0}=\frac{-{N}_{B}kT}{\left(\frac{L}{T}\right)}\phantom{\rule{0ex}{0ex}}=\frac{{N}_{B}k{T}^{2}}{L}$

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

94% of StudySmarter users get better grades.

Sign up for free