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1.4.3: Chemical Equilibria- Solutions- Simple Solutes

  • Page ID
    352525
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    A given chemical equilibrium involves association of two solutes \(\mathrm{X}(\mathrm{aq})\) and \(\mathrm{Y}(\mathrm{aq})\) to form solute \(\mathrm{Z}(\mathrm{aq})\).

    \[2 X(a q)+Y(a q) \rightleftharpoons 4 Z(a q)\]

    Phase Rule. The aqueous solution is prepared using two chemical substances: substance \(\mathrm{Z}\) and solvent water. Hence \(\mathrm{C} = 2\). There are 2 phases: vapour and solution so \(\mathrm{P} = 2\). Then \(\mathrm{F} = 2\). Hence at fixed temperature and in a system prepared using mole fraction \(x_{\mathrm{Z}}\) of substance \(\mathrm{Z}\) (an intensive composition variable), the equilibrium vapour pressure and the equilibrium amounts of \(\mathrm{X}(\mathrm{aq})\), \(\mathrm{Y}(\mathrm{aq})\) and \(\mathrm{Z}(\mathrm{aq})\) are unique.

    \[\text { At equilibrium, } 2 \, \mu_{\mathrm{X}}^{\mathrm{eq}}(\mathrm{aq})+\mu_{\mathrm{Y}}^{\mathrm{eq}}(\mathrm{aq})=4 \, \mu_{\mathrm{Z}}^{\mathrm{eq}}(\mathrm{aq})\]

    At fixed \(\mathrm{T}\) and \(p\), assuming ambient pressure is close to the standard pressure \(p^{0}\),

    \[\begin{aligned}
    &2 \,\left[\mu_{\mathrm{X}}^{0}(\mathrm{aq})+\mathrm{R} \, \mathrm{T} \, \ln \left(\mathrm{m}_{\mathrm{X}} \, \gamma_{\mathrm{X}} / \mathrm{m}^{0}\right)^{\mathrm{eq}}\right] \\
    &\quad+\left[\mu_{\mathrm{Y}}^{0}(\mathrm{aq})+\mathrm{R} \, \mathrm{T} \, \ln \left(\mathrm{m}_{\mathrm{Y}} \, \gamma_{\mathrm{Y}} / \mathrm{m}^{0}\right)^{\mathrm{eq}}\right] \\
    &\quad=4 \,\left[\mu_{\mathrm{Z}}^{0}(\mathrm{aq})+\mathrm{R} \, \mathrm{T} \, \ln \left(\mathrm{m}_{\mathrm{Z}} \, \gamma_{\mathrm{Z}} / \mathrm{m}^{0}\right)^{\mathrm{eq}}\right]
    \end{aligned}\]

    \[\text { where } \Delta_{\mathrm{r}} \mathrm{G}^{0}=-\mathrm{R} \, \mathrm{T} \, \ln \left(\mathrm{K}^{0}\right)=4 \, \mu_{\mathrm{Z}}^{0}(\mathrm{aq})-2 \, \mu_{\mathrm{X}}^{0}(\mathrm{aq})-\mu_{\mathrm{Y}}^{0}(\mathrm{aq})\]

    For this equilibrium at temperature \(\mathrm{T}\) and pressure \(p\),

    \[\mathrm{K}^{0}=\frac{\left(\mathrm{m}_{\mathrm{Z}}^{\mathrm{eq}} \, \gamma_{\mathrm{Z}}^{\mathrm{eq}} / \mathrm{m}^{0}\right)^{4}}{\left(\mathrm{~m}_{\mathrm{X}}^{\mathrm{eq}} \, \gamma_{\mathrm{X}}^{\mathrm{eq}} / \mathrm{m}^{0}\right)^{2} \,\left(\mathrm{m}_{\mathrm{Y}}^{\mathrm{eq}} \, \gamma_{\mathrm{Y}}^{\mathrm{eq}} / \mathrm{m}^{0}\right)}\]

    If the solution is quite dilute, \(\gamma_{\mathrm{X}}^{\mathrm{eq}}\), \(\gamma_{\mathrm{Y}}^{\mathrm{eq}}\) and \gamma_{\mathrm{Z}}^{\mathrm{eq}} are effectively unity in the real solution at equilibrium. Then

    \[\mathrm{K}^{0}=\left(\mathrm{m}_{\mathrm{eq}}^{\mathrm{eq}} / \mathrm{m}^{0}\right)^{4} /\left(\mathrm{m}_{\mathrm{X}}^{\mathrm{eq}} / \mathrm{m}^{0}\right)^{2} \,\left(\mathrm{m}_{\mathrm{Y}}^{\mathrm{eq}} / \mathrm{m}^{0}\right)\]

    \(\mathrm{K}^{0}\) is dimensionless. But the latter statement signals a common problem in this subject because chemists find it more convenient and informative to define a quantity \({\mathrm{K}_{\mathrm{m}}}^{0}\) in which the m0 terms in equation (f) [or its equivalent] have been removed.

    \[\text { Thus, } \mathrm{K}_{\mathrm{m}}^{0}=\left(\mathrm{m}_{\mathrm{Z}}^{\mathrm{eq}}\right)^{4} /\left(\mathrm{m}_{\mathrm{X}}^{\mathrm{eq}}\right)^{2} \,\left(\mathrm{m}_{\mathrm{Y}}^{\mathrm{eq}}\right)\]

    Hence the units for \({\mathrm{K}_{\mathrm{m}}}^{0}\) signal the stoichiometry of the equilibrium whereas the dimensionless \(\mathrm{K}^{0}\) does not [1,2].

    Footnotes

    [1] As a consequence of the removal of the \(\mathrm{m}^{0}\) terms, \(\mathrm{K}^{0}\) quantities have units unless the equation for the chemical equilibrium is stoichiometrically balanced: e.g. n-moles of reactants form n-moles of products.

    But from equation (g), \(\mathrm{K}_{\mathrm{m}}^{0}=\left[\mathrm{mol} \mathrm{kg}^{-1}\right]^{4} \,\left[\mathrm{mol} \mathrm{kg}{ }^{-1}\right]^{-2} \,\left[\mathrm{mol} \mathrm{kg}^{-1}\right]^{-1}\) or \(\mathrm{K}_{\mathrm{m}}^{0}=\left[\mathrm{mol} \mathrm{} \mathrm{kg}^{-1}\right]\)

    If we write, \(\Delta_{\mathrm{r}} \mathrm{G}^{0}=-\mathrm{R} \, \mathrm{T} \, \ln \left(\mathrm{K}_{\mathrm{m}}^{0}\right)\)

    Then \(\Delta_{\mathrm{r}} \mathrm{G}^{0}=\left[\mathrm{J} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\right] \,[\mathrm{K}] \, \ln \left[\mathrm{mol} \mathrm{kg}{ }^{-1}\right]\)

    ln There is clearly a slight problem in handling a logarithm of a composition unit. There are two approaches to this problem. The first approach ignores the problem, which is unsatisfactory practice. The second approach is to ask - what happened to the composition unit and trace the problem back through the equations.

    [2] The impact of composition units in quantitative analysis of data was addressed by E. A. Guggenheim, Trans. Faraday Soc., 1937,33,607.


    This page titled 1.4.3: Chemical Equilibria- Solutions- Simple Solutes is shared under a Public Domain license and was authored, remixed, and/or curated by Michael J Blandamer & Joao Carlos R Reis via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.