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1.13.6: Equilibrium - Depression of Freezing Point of a Solvent by a Solute

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    A given homogeneous liquid system (at pressure \(\mathrm{p}\)) comprises solvent \(\mathrm{i}\) and solute \(\mathrm{j}\) at temperature \(\mathrm{T}\) and pressure \(\mathrm{p}\). In the absence of solute \(\mathrm{j}\), the freezing point of the solvent is \(\mathrm{T}_{1}^{0}\). But in the presence of solute \(\mathrm{j}\) the freezing point is temperature \(\mathrm{T}\) where \(\mathrm{T} < \mathrm{T}_{1}^{0}\). The depression of freezing point \(\theta\left[=\mathrm{T}_{1}^{0}-\mathrm{T}\right]\) is recorded for a solution where the mole fraction of solvent is \(\mathrm{x}_{1}(\mathrm{sln})\). If the solution is dilute, we can assume that the thermodynamic properties of the solution are ideal. From the Schroeder-van Laar equation,

    \[-\ln \left[x_{1}(s \ln )\right]=\frac{\left[\Delta_{f} H_{1}^{0}(T)\right]}{R} \,\left[\frac{1}{T}-\frac{1}{T_{1}^{0}}\right]\]

    \[-\ln \left[\mathrm{x}_{1}(\mathrm{~s} \ln )\right]=\frac{\Delta_{\mathrm{f}} \mathrm{H}_{1}^{0}}{\mathrm{R}} \, \frac{\theta}{\left(\mathrm{T}_{1}^{0}-\theta\right) \, \mathrm{T}_{1}^{0}}\]

    If

    \[\mathrm{T}_{1}^{0}-\theta \cong \mathrm{T}_{1}^{0},-\ln \left[\mathrm{x}_{1}(\mathrm{~s} \ln )\right]=\frac{\Delta_{\mathrm{f}} \mathrm{H}_{1}^{0}}{\mathrm{R}} \, \frac{\theta}{\left(\mathrm{T}_{1}^{0}\right)^{2}}\]

    Or,

    \[\ln \left[\frac{1}{x_{1}(s \ln )}\right]=\frac{\Delta_{\mathrm{f}} H_{1}^{0}}{R} \, \frac{\theta}{\left(T_{1}^{0}\right)^{2}}\]

    Hence [2]

    \[\theta=\left[\frac{\mathrm{R} \,\left(\mathrm{T}_{1}^{0}\right)^{2} \, \mathrm{M}_{1}}{\Delta_{\mathrm{f}} \mathrm{H}_{1}^{0}}\right] \, \mathrm{m}_{\mathrm{j}}\]

    The quantity enclosed in the […] brackets is characteristic of the solvent.

    Footnotes

    [1] \(\theta=\mathrm{T}_{1}^{0}-\mathrm{T}\); \(\frac{1}{\mathrm{~T}}-\frac{1}{\mathrm{~T}_{1}^{0}}=\frac{\mathrm{T}_{1}^{0}-\mathrm{T}}{\mathrm{T} \, \mathrm{T}_{1}^{0}}=\frac{\mathrm{T}_{1}^{0}-\mathrm{T}}{\left(\mathrm{T}_{1}^{0}-\theta\right) \, \mathrm{T}_{1}^{0}}=\frac{\theta}{\left(\mathrm{T}_{1}^{0}-\theta\right) \, \mathrm{T}_{1}^{0}}\)

    [2] \(\frac{1}{x_{1}}=\frac{1}{1-x_{j}}=\frac{1}{1-\left[n_{j} /\left(n_{1}+n_{j}\right)\right]}=\frac{n_{1}+n_{j}}{n_{1}+n_{j}-n_{j}}\) For a solution where the molality of solute \(j=m_{j}\) \(\mathrm{m}_{\mathrm{j}}=\frac{\mathrm{n}_{\mathrm{j}}}{\mathrm{n}_{1} \, \mathrm{M}_{1}}\)
    Then, \(\frac{1}{\mathrm{x}_{1}}=\frac{\mathrm{n}_{1}+\mathrm{n}_{1} \, \mathrm{M}_{1} \, \mathrm{m}_{\mathrm{j}}}{\mathrm{n}_{1}}\)
    \(-\ln \left[\mathrm{x}_{1}(\mathrm{~s} \ln )\right]=-\ln \left[1+\mathrm{M}_{1} \, \mathrm{m}_{\mathrm{j}}\right]\);
    \(-\ln \left[x_{1}(\operatorname{sln})\right]=-\ln \left[1-x_{j}(s \ln )\right] \approx x_{j}\)
    \(x_{j}=\frac{m_{j}}{\left(1 / M_{1}\right)+m_{j}} \approx m_{j} \, M_{1}\)

    [3] see I Prigogine and R Defay, Chemical Thermodynamics, trans. D. H. Everett, Longmans Green, London, 1953.

    This page titled 1.13.6: Equilibrium - Depression of Freezing Point of a Solvent by a Solute is shared under a Public Domain license and was authored, remixed, and/or curated by Michael J Blandamer & Joao Carlos R Reis.