How to derive the chemical potential of an ideal solution of a non-volatile solute?

Question

$ΔG=-nFe$ (example) and $μ = μ° + RT\ln a$ are used very frequently in chemical thermodynamics.

However, if the Gibbs energy and pressure $P^\circ,G^\circ$ in the standard state of an ideal gas are known, when an isothermal reaction $(P^\circ,G^\circ) \to (P,G)$ occurs, if the pressure P at the end of this reaction is known, the Gibbs energy G at the end state can be expressed by the following formula.

$$G\ =G^\circ+\ nRT\ln\left(\frac{P}{P^\circ}\right)$$

Using the extensivity of G and partially differentiating with respect to n, the "Gibbs energy per mole", that is, the chemical potential μ of this system, is

$$\mu=\mu^\circ+\ RT\ln\left(\frac{P}{P^\circ}\right)$$

So, If we further assume that in an ideal solution where solute A is volatile, and the mole fraction of substance A is $\chi_A$ and the vapor pressure of the pure substance A is $P_A^\bullet$, then the vapor pressure $P_A$ of A satisfies the following equation.

$$P_A=\ \chi_A\ P_A^\bullet$$

Therefore, if substance A is volatile and the vapor is an ideal gas, the chemical potential of the vapor of substance A is:

$$\mu_A=\mu_A^\circ+RT\ln\left(\frac{\chi_A\ P_A^\bullet}{P^\circ}\right)\ \\ =\mu_A^\circ+RT\ln\left(\frac{\ P_A^\bullet}{P^\circ}\right)\ +RT\ln\left(\chi_A\right)$$

When we define$\mu_A^\bullet$ as follows

$$\mu_A^\bullet := \mu_A^\circ+RT\ln\left(\frac{\ P_A^\bullet}{P^\circ}\right)\ $$

we get $$\mu_A = \mu_A^\bullet +RT\ln\left(\chi_A\right)$$

However, if A is non-volatile (for example NaCl),

$$P_A^\bullet =0$$

and so

$$\ln\left(\frac{\ P_A^\bullet}{P^\circ}\right) =\ln (0) = -\infty$$

Of course, for volatile non-ideal solvents an empirical correction using activity values ​​could be made. The chemical potential μ is expressed as a function of the standard chemical potential μ° and the activity a as follows:

$μ = μ° + RT \ln a$

However, in the case of non-volatile solvents such as NaCl, there is the major problem of ln(0)=-∞ mentioned above, so derivation using vapor pressure seems inherently difficult, and even if activity is used, it seems necessary to use a logic that at least avoids "ln(0)=-∞".

Despite this, many textbooks use the same formula for the chemical potential of an ideal solution of a non-volatile solute as for a volatile solute.

My question: Why does the following equation hold even when the A is non-volatile? How does the following equation derive when A is non-volatile?

$μ = μ° + RT \ln a$

Answer 1

I think I was able to solve the problem myself using the following method:

●The chemical potential of an ideal gas is: $$\mu=\mu_0+RT\ln\left(\frac{P}{P_0}\right) \tag{0-1}$$

Where,

• $\mu$: Chemical potential
• $\mu_0$: Standard state chemical potential
• $R$: Gas constant
• $T$: Temperature
• $P$: System pressure
• $P_0$: Standard state pressure

●In the case of volatile components (volatile solutes and solvents)

When the mole fraction of a volatile component is $x$, applying Henry's law under the condition that "the volatile components of the dilute solution are in equilibrium with the gas phase," the saturated vapor pressure P of the volatile component is given as follows:

$$P=K_{H}x \tag{1-1}$$

Here, $K_\mathrm{H}$ is Henry's constant.

Substituting Henry's law into the formula (0-1) for the chemical potential of an ideal gas,

$$\mu = \mu_0 + RT \ln \left( \frac{K_H x}{P^0} \right)\tag{1-2}$$

Rearranging this,

$$\mu = \mu_* + RT \ln x\tag{1-3}$$

Here, $$\mu_* = \mu_0 + RT \ln \left( \frac{K_H}{P^0} \right)\tag{1-4}$$

Expression using activity: To take into account the difference between ideal and actual behavior, we introduce activity a instead of concentration $x$,

$$\mu = \mu_* + RT \ln a\tag{1-5}$$

●In the case of non-volatile solutes
For non-volatile solutes, the vapor pressure is $P=0$, and direct substitution into the basic formula is mathematically inappropriate. For simplicity, consider the case where one non-volatile solute is dissolved in a volatile solvent.

Therefore, consider the application of the Gibbs-Duhem equation. When the total amount of substance is constant and the temperature and pressure are constant, the Gibbs-Duhem equation can be written as follows;

$$n_1d\mu_1 + n_2 d\mu_2 = 0\tag{2-1}$$

Here, $n_1$ and $n_2$ are the amounts of solute and solvent, respectively, and $\mu_1$ and $\mu_2$ are the chemical potentials of the solute and solvent, respectively. If this is rewritten using the mole fractions $x_1$ and $x_2$ of the solute and solvent, it becomes as follows.

$$x_1 d\mu_1 + x_2 d\mu_2 =0\tag{2-2}$$

here,the mole fraction $x_1$ of solvent 1 and the mole fraction $x_2$ of solute 2 are defined as follows:

$$x_1 = \frac{n_1}{n_1 + n_2}\tag{2-3}$$ $$x_2 = \frac{n_2}{n_1 + n_2}\tag{2-3a}$$

From (2-2), we get $$d\mu_2 = - \frac{x_1}{x_2} d\mu_1 \tag{2-4}$$

Assuming that Raoult's law holds true, the chemical potential of the solvent can be written as follows using the same discussion as in the previous section. $$\mu_1 = \mu_1^* + RT \ln x_1\tag{2-5}$$

Differentiating both sides gives us the following:

$$d\mu_1 = RT \frac{dx_1}{x_1}\tag{2-6}$$

From (2-4)＆(2-6), we get: $$d\mu_2 = - \frac{x_1}{x_2} d\mu_1 = - \frac{x_1}{x_2} RT \frac{dx_1}{x_1} = - RT \frac{dx_1}{x_2}\tag{2-7}$$

On the other hand, (2-3) shows the fact that the ratio of solvent to solute is always 100% (or 1) overall. $$x_1 + x_2 =1\tag{2-8}$$

Differentiating both sides of (2-8) gives us the following:

$$d{x}_1=−d{x}_2\tag{2-9}$$

From the above (2-6),(2-9), the following equation holds for the chemical potential of the solute: $$d\mu_2 = RT \frac{dx_2}{x_2}\tag{2-10}$$

Integrating both side of (2-19) gives us: $$\mu_2 = \mu_2^* + RT \ln x_2\tag{2-11}$$

This method allows us to properly describe the chemical potential even for non-volatile solutes.

Answer 2

This answer does not address the derivation of the solute chemical potential, but points out the important concept of definitions in science.

The IUPAC explicit definition of relative thermodynamic activity is

$$a = \exp{\dfrac{\mu_{T} - \mu^{\circ}_{T} }{RT}},\tag{1}$$

or in its logarithmic implicit form

$$\mu_{T} = \mu^{\circ}_{T} + RT \ln{a} \tag{2}$$

Being a definition, it cannot be derived nor confirmed nor refuted nor formally proved, unless using it explicitly or implicitly in an invalid circular respective action. It is nothing specific to activity. The same applies e.g. on the enthalpy definition.

The activity definition says nothing about how to obtain particular values of chemical potentials, it only defines the relation of activity to other three state variables.

Due $\mu^{\circ}$ choice convention, the (relative) activity of the pure substance is $a=1$. That is also the reason why pure condensed substances are not explicitly involved in expressions of thermodynamic equilibrium constants.

For ideal solutions $x=a$, so:

$$\mu_{T} = \mu^{\circ}_{T} + RT \ln{x},\tag{3}$$

For non-ideal solution, or if the standard chemical potential is defined in other way (like for ionic compound solutions), $x \ne a$.

Introducing activity coefficient for non-ideal solutions:

$$\mu_{T} = \mu^{\circ}_{T} + RT \ln{(\gamma \cdot x)} \tag{4}$$

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For ionic solution context, where activity is aligned amount concentration:

The approximate conversion of amount fraction to amount concentration for diluted water solutions, using water amount density:

$$\mu_{T} \overset{diluted}{\approx} \mu^{\circ}_{T} + RT \ln{\left(\frac{\gamma \cdot c[\pu{mol L-1}]}{\pu{55.5 mol L-1}}\right)} \tag{5}$$ $$\mu_{T} = \mu^{\circ}_{T} - RT \ln{\left(\frac{\pu{55.5 mol L-1}}{\pu{1 mol L-1}}\right)} + RT \ln{\left(\frac{\gamma \cdot c[\pu{mol L-1}]}{\pu{1 mol L-1}}\right)}\tag{6}$$

Redefining the reference state for diluted water ionic solutions so $$\lim_{(c \to 0)}{\gamma}=1\tag{7}$$ $$\lim_{(c \to 0)}{a}=c\tag{8}$$ $$\mu^{\circ \text{, ion}}_{T}= \mu^{\circ}_{T} - RT \ln{\left(\frac{\pu{55.5 mol L-1}}{\pu{1 mol L-1}}\right)} \tag{9}$$ $$\mu_{T} = \mu^{\circ \text{, ion}}_{T} + RT \ln{a} \tag{10}$$ $$\mu_{T} = \mu^{\circ \text{, ion}}_{T} + RT \ln{\left(\frac{ \gamma \cdot c[\pu{mol L-1}]}{\pu{1 mol L-1}}\right)}\tag{11}$$