1.4.3: Chemical Equilibria- Solutions- Simple Solutes
A given chemical equilibrium involves association of two solutes \(\mathrm{X}(\mathrm{aq})\) and \(\mathrm{Y}(\mathrm{aq})\) to form solute \(\mathrm{Z}(\mathrm{aq})\).
\[2 X(a q)+Y(a q) \rightleftharpoons 4 Z(a q) \nonumber \]
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.
\[\text { At equilibrium, } 2 \, \mu_{\mathrm{X}}^{\mathrm{eq}}(\mathrm{aq})+\mu_{\mathrm{Y}}^{\mathrm{eq}}(\mathrm{aq})=4 \, \mu_{\mathrm{Z}}^{\mathrm{eq}}(\mathrm{aq}) \nonumber \]
At fixed \(\mathrm{T}\) and \(p\), assuming ambient pressure is close to the standard pressure \(p^{0}\),
\[\begin{aligned}
&2 \,\left[\mu_{\mathrm{X}}^{0}(\mathrm{aq})+\mathrm{R} \, \mathrm{T} \, \ln \left(\mathrm{m}_{\mathrm{X}} \, \gamma_{\mathrm{X}} / \mathrm{m}^{0}\right)^{\mathrm{eq}}\right] \\
&\quad+\left[\mu_{\mathrm{Y}}^{0}(\mathrm{aq})+\mathrm{R} \, \mathrm{T} \, \ln \left(\mathrm{m}_{\mathrm{Y}} \, \gamma_{\mathrm{Y}} / \mathrm{m}^{0}\right)^{\mathrm{eq}}\right] \\
&\quad=4 \,\left[\mu_{\mathrm{Z}}^{0}(\mathrm{aq})+\mathrm{R} \, \mathrm{T} \, \ln \left(\mathrm{m}_{\mathrm{Z}} \, \gamma_{\mathrm{Z}} / \mathrm{m}^{0}\right)^{\mathrm{eq}}\right]
\end{aligned} \nonumber \]
\[\text { where } \Delta_{\mathrm{r}} \mathrm{G}^{0}=-\mathrm{R} \, \mathrm{T} \, \ln \left(\mathrm{K}^{0}\right)=4 \, \mu_{\mathrm{Z}}^{0}(\mathrm{aq})-2 \, \mu_{\mathrm{X}}^{0}(\mathrm{aq})-\mu_{\mathrm{Y}}^{0}(\mathrm{aq}) \nonumber \]
For this equilibrium at temperature \(\mathrm{T}\) and pressure \(p\),
\[\mathrm{K}^{0}=\frac{\left(\mathrm{m}_{\mathrm{Z}}^{\mathrm{eq}} \, \gamma_{\mathrm{Z}}^{\mathrm{eq}} / \mathrm{m}^{0}\right)^{4}}{\left(\mathrm{~m}_{\mathrm{X}}^{\mathrm{eq}} \, \gamma_{\mathrm{X}}^{\mathrm{eq}} / \mathrm{m}^{0}\right)^{2} \,\left(\mathrm{m}_{\mathrm{Y}}^{\mathrm{eq}} \, \gamma_{\mathrm{Y}}^{\mathrm{eq}} / \mathrm{m}^{0}\right)} \nonumber \]
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
\[\mathrm{K}^{0}=\left(\mathrm{m}_{\mathrm{eq}}^{\mathrm{eq}} / \mathrm{m}^{0}\right)^{4} /\left(\mathrm{m}_{\mathrm{X}}^{\mathrm{eq}} / \mathrm{m}^{0}\right)^{2} \,\left(\mathrm{m}_{\mathrm{Y}}^{\mathrm{eq}} / \mathrm{m}^{0}\right) \nonumber \]
\(\mathrm{K}^{0}\) 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.
\[\text { Thus, } \mathrm{K}_{\mathrm{m}}^{0}=\left(\mathrm{m}_{\mathrm{Z}}^{\mathrm{eq}}\right)^{4} /\left(\mathrm{m}_{\mathrm{X}}^{\mathrm{eq}}\right)^{2} \,\left(\mathrm{m}_{\mathrm{Y}}^{\mathrm{eq}}\right) \nonumber \]
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.