1.14.58: Reversible Chemical Reactions
Two important themes in thermodynamics concern the description of chemical equilibria and the kinetics of chemical reactions in closed systems at fixed temperature and pressure. These two themes are often linked in descriptions of chemical reactions. We comment on this link.
A given aqueous solution (at temperature \(\mathrm{T}\) and pressure \(\mathrm{p}\), which is close to the standard pressure, \(\mathrm{p}^{0}\)) is prepared using chemical substance \(\mathrm{X}\). Spontaneous chemical reaction forms chemical substance \(\mathrm{Z}\) in the following chemical reaction.
\[\mathrm{X}(\mathrm{aq}) \rightarrow \mathrm{Z}(\mathrm{aq}) \nonumber \]
Experiment confirms that the extent of chemical reaction is given by equation (b).
\[\frac{1}{\mathrm{~V}} \, \frac{\mathrm{d} \xi}{\mathrm{dt}}=-\mathrm{k}_{1} \,[\mathrm{X}] \nonumber \]
In this system, \(\mathrm{c}_{\mathrm{x}}\{=[\mathrm{X}]\}=\mathrm{n}_{\mathrm{x}} / \mathrm{V}\). The common assumption is that for dilute solutions both \(\mathrm{k}_{1}\) and volume \(\mathrm{V}\) are independent of time.
\[\mathrm{dc}_{\mathrm{x}} / \mathrm{dt}=-\mathrm{k}_{1} \, \mathrm{c}_{\mathrm{x}} \nonumber \]
Rate constant \(\mathrm{k}_{1}\) is expressed using the unit, \(\mathrm{s}^{-1}\). We consider a system prepared using chemical substance \(\mathrm{Z}\) which undergoes spontaneous chemical reaction to form chemical substance \(\mathrm{X}\). The analogue of equation (c) takes the following form.;
\[\mathrm{dc}_{\mathrm{Z}} / \mathrm{dt}=-\mathrm{k}_{2} \, \mathrm{c}_{\mathrm{Z}} \nonumber \]
We assert that the chemical reaction described by equations (c) and (d) proceed until the properties of the system (at fixed \(\mathrm{T}\) and \(\mathrm{p}\)) are independent of time. In other words the system is in thermodynamic equilibrium with the surroundings with \(\mathbf{c}_{X}=\mathbf{c}_{X}^{e q}\) and \(\mathbf{c}_{Z}=\mathbf{c}_{Z}^{e q}\) where, macroscopically, \(\mathrm{dc}_{\mathrm{X}} / \mathrm{dt}\) and \(\mathrm{dc}_{\mathrm{Z}} / \mathrm{dt}\) are zero. Also \(\operatorname{limit}\left(\mathrm{m}_{\mathrm{z}} \rightarrow 0 ; \mathrm{m}_{\mathrm{x}} \rightarrow 0\right) \gamma_{\mathrm{z}}=1\) and \(\operatorname{limit}\left(\mathrm{m}_{\mathrm{x}} \rightarrow 0\right.\); and \(\left.\mathrm{m}_{\mathrm{z}}=0\right) \gamma_{\mathrm{x}}=1\) Thus
\[\mathrm{k}_{1} \, \mathrm{c}_{\mathrm{x}}^{\mathrm{eq}}=\mathrm{k}_{2} \, \mathrm{c}_{\mathrm{Z}}^{\mathrm{eq}} \nonumber \]
From a thermodynamic point of view, at equilibrium (at fixed \(\mathrm{T}\) and \(\mathrm{p}\)) the affinity for spontaneous change \(\mathrm{A}\) is zero, and the system is at a minimum in Gibbs energy \(\mathrm{G}\). If the molalities of substances \(\mathrm{X}\) and \(\mathrm{Z}\) are \(\mathrm{m}_{\mathrm{X}}^{\mathrm{eq}}\) and \(\mathrm{m}_{\mathrm{Z}}^{\mathrm{eq}}\) respectively, the standard increase in Gibbs energy \(\Delta_{\mathrm{r}} \mathrm{G}^{0}(\mathrm{~T}, \mathrm{p}, \mathrm{aq})\) is related to a (dimensionless) thermodynamic equilibrium constant \(\mathrm{K}(\mathrm{aq} ; \mathrm{T} ; \mathrm{p})\) using equation (f).
\[\Delta_{\mathrm{r}} \mathrm{G}^{0}=-\mathrm{R} \, \mathrm{T} \, \ln (\mathrm{K})=\mu_{\mathrm{Z}}^{0}(\mathrm{aq})-\mu_{\mathrm{x}}^{0}(\mathrm{aq}) \nonumber \]
where,
\[\mathrm{K}=\left(\mathrm{m}_{\mathrm{z}} \, \gamma_{\mathrm{z}} / \mathrm{m}^{0}\right)^{\mathrm{eq}} /\left(\mathrm{m}_{\mathrm{x}} \, \gamma_{\mathrm{x}} / \mathrm{m}^{0}\right)^{\mathrm{eq}} \nonumber \]
Here \(\gamma_{\mathrm{X}}\) and \(\gamma_{\mathrm{Z}}\) are the activity coefficients for solutes \(\mathrm{X}\) and \(\mathrm{Z}\). If the aqueous solution is quite dilute, we can assume that the thermodynamic properties of the solution are ideal. Moreover the ratio \(\left(m_{Z} / m_{X}\right)^{e q}\) is, based on the same approximation, equal to \(\left(c_{Z} / c_{X}\right)^{e q}\). In other words equations (e) and (g) can be written in the following forms. Law of Mass Action
\[\left(c_{Z} / c_{X}\right)^{e q}=\left(k_{1} / k_{2}\right)^{e q} \nonumber \]
Thermodynamics
\[\left(\mathrm{c}_{\mathrm{Z}} / \mathrm{c}_{\mathrm{X}}\right)^{\mathrm{eq}}=\mathrm{K} \nonumber \]
Within the context of the assumptions outlined above , we obtain by comparing equations (h) and (i) the following classic equation.
\[\mathrm{K}=\mathrm{k}_{1} / \mathrm{k}_{2} \nonumber \]
Equation (j) is fascinating because the two sides of the equation have different origins, Law of Mass Action and the Laws of Thermodynamics. Indeed equation (j) is often used in an introduction to the concept of chemical equilibrium, the latter emerging as a ‘balance of rates of reaction’. In a wider context equation (j) is used in treatments of fast chemical reactions where a given closed system is only marginally displaced from equilibrium by transient changes in electric field, magnetic field, pressure or temperature [1-3].
Footnotes
[1] E. Caldin, Fast Reactions in Solution, Blackwell Scientific Publications, Oxford, 1964.
[2] M. J. Blandamer, Introduction to Chemical Ultrasonics, Academic Press, London, 1973.
[3] The analysis takes a similar form in cases where the reaction stoichiometry is more complicated. Consider the case of an association reaction in aqueous solution.
\[\mathrm{X}(\mathrm{aq})+\mathrm{Y}(\mathrm{aq}) \rightarrow \mathrm{Z}(\mathrm{aq}) \nonumber \]
Law of Mass Action For the forward reaction \(\mathrm{dc}_{\mathrm{X}} / \mathrm{dt}=-\mathrm{k}_{1} \, \mathrm{c}_{\mathrm{X}} \, \mathrm{c}_{\mathrm{Y}}\) For the reverse reaction
\[\mathrm{Z}(\mathrm{aq}) \rightarrow \mathrm{X}(\mathrm{aq})+\mathrm{Y}(\mathrm{aq}) \nonumber \]
\[\mathrm{dc}_{\mathrm{Z}} / \mathrm{dt}=-\mathrm{k}_{2} \, \mathrm{c}_{\mathrm{z}} \nonumber \]
For a system where, macroscopically, \(\mathrm{dc}_{\mathrm{X}} / \mathrm{dt}=\mathrm{dc}_{\mathrm{Y}} / \mathrm{dt}=\mathrm{dc}_{\mathrm{Z}} / \mathrm{dt}=0\),
\[\mathrm{k}_{1} \,\left(\mathrm{c}_{\mathrm{X}} \, \mathrm{c}_{\mathrm{Y}}\right)^{\mathrm{cq}}=\mathrm{k}_{2} \,\left(\mathrm{c}_{\mathrm{Z}}\right)^{\mathrm{eq}} \nonumber \]
Or,
\[\mathrm{k}_{1} / \mathrm{k}_{2}=\left(\mathrm{c}_{\mathrm{Z}}\right)^{\mathrm{eq}} /\left(\mathrm{c}_{\mathrm{X}} \, \mathrm{c}_{\mathrm{Y}}\right)^{\mathrm{eq}} \nonumber \]
From a thermodynamic viewpoint, at equilibrium (at fixed \(\mathrm{T}\) and \(\mathrm{p}\)),
\[\Delta_{\mathrm{r}} \mathrm{G}^{0}=-\mathrm{R} \, \mathrm{T} \, \ln (\mathrm{K})=\mu_{\mathrm{Z}}^{0}(\mathrm{aq})-\mu_{\mathrm{X}}^{0}(\mathrm{aq})-\mu_{\mathrm{Y}}^{0}(\mathrm{aq}) \nonumber \]
where
\[\mathrm{K}=\left(\mathrm{m}_{\mathrm{Z}} \, \gamma_{\mathrm{Z}} / \mathrm{m}^{0}\right)^{\mathrm{eq}} /\left(\mathrm{m}_{\mathrm{X}} \, \gamma_{\mathrm{X}} / \mathrm{m}^{0}\right)^{\mathrm{eq}} \,\left(\mathrm{m}_{\mathrm{Y}} \, \gamma_{\mathrm{Y}} / \mathrm{m}^{0}\right)^{\mathrm{eq}} \nonumber \]
In the limit that the solution is dilute, \(\left(\gamma_{\mathrm{Z}}\right)^{e q}=\left(\gamma_{\mathrm{x}}\right)^{\mathrm{eq}}=\left(\gamma_{\mathrm{Y}}\right)^{\mathrm{eq}}=1\). Then
\[\mathrm{K}=\left(\mathrm{c}_{\mathrm{Z}} / \mathrm{c}_{\mathrm{r}}\right)^{\mathrm{eq}} /\left(\mathrm{c}_{\mathrm{X}} / \mathrm{c}_{\mathrm{r}}\right)^{\mathrm{eq}} \,\left(\mathrm{c}_{\mathrm{Y}} / \mathrm{c}_{\mathrm{r}}\right)^{\mathrm{eq}} \nonumber \]
Comparison of equations (e) and (h) allows identification of the ratio \(\mathrm{k}_{1} / \mathrm{k}_{2}\) with the equilibrium constant \(\mathrm{K}\).
Comment
If at equilibrium \(\mathrm{dc}_{\mathrm{X}} / \mathrm{dt}=0\) and \(\mathbf{c}_{\mathrm{x}}^{\mathrm{cq}} \neq 0\) then according to equation (c), \(\mathrm{k}_{1}\) must be zero. The same problem arises from equation (d). To circumvent this objection, the Principle of Microscopic Reversibility states that at Equilibrium, the amount of chemical substance \(\mathrm{X}\) consumed by equation (a) and described by equation (c) equals the amount of chemical substance \(\mathrm{X}\) produced by the reverse reaction described by equation (d).