How is dG=0 employed to describe equilibrium in constant volume systems?

Question

GIBBS ENERGY

Here's the derivation of how dG=0 describes equilibrium state:

$$\Delta S_{\text{system}} + \Delta S_{\text{surrounding}} > 0$$

$\implies\Delta S_{\text{system}} + \frac{-Q}{T_{\text{surr}}} > 0$

$\implies T_{\text{surr}} dS_{\text{system}} - dQ > 0$

$\implies dQ - TdS < 0$, assuming Tsurr = Tsystem, surrounding temp is constant, so system temp is constant

$\implies dH - TdS < 0$, as dQ can be written as dH when P_ext=P_i=P_f, which is P_ext = P_int at an instant.

$\therefore d(H - TS) < 0$ or $dG < 0$, this is the condition for a reaction to move forward

Since we assumed dQ=dH, the validity of the final expression should hold only under the conditions that justify this assumption, right? So how can this expression be applied, even at an instant, in a constant volume reaction?

CHEMICAL POTENTIAL

Another problem I'm having is writing dG at an instant for constant volume reaction.

$$chemical potential \mu = \frac{\partial G}{\partial n}_{\text{ n',P,T}}$$

The chemical potential of a substance is the partial derivative of total Gibbs energy of the system with respect to moles of the substance being observed, keeping moles of other substances, pressure, and temperature constant.

$G = n\mu + n'\mu'$, total Gibbs energy of a binary mixture

Eqn II. $dG = dn\mu + dn'\mu' + nd\mu + n'd\mu'$, instantaneous change in Gibbs energy of a binary mixture

At constant temperature and total pressure, $nd\mu + n'd\mu' = 0$ (Gibbs-Duhem Equation)

At constant n', $u'dn' = 0$

This further validates, that at constant temperature, total pressure and n', $\mu dn = dG$ (from Eqn. II)

using chemical potential in a reaction A-->2B

This reaction can be viewed at an instant as:

1. the removal of dn moles of A (process 1)
2. the addition of 2dn moles of B (process 2)

dn is positive

Let's consider the reaction to occur at constant temperature and total pressure:

1. The change in Gibbs energy due to process 1 is simply $dG_{\text{1}} = -\mu dn$, where $\mu$ is the chemical potential of A.

2. The change in Gibbs energy due to process 2 is simply $dG_{\text{2}} = 2\mu' dn$, where $\mu'$ is the chemical potential of B.

Change in Gibbs energy through the entire reaction instantaneously $dG = dG_{\text{2}} + dG_{\text{1}} = dn(2\mu' -\mu)$

For this reaction to move forward at an instant, $$2\mu' -\mu < 0$$

Therefore, it is said that at equilibrium, dG=0, the sum of the chemical potential of the products multiplied by their stoichiometric coefficients is equal to the sum of the chemical potentials of the reactants multiplied by their stoichiometric coefficients.

Even though I have shown this for isobaric and isothermal, the same is used for describing a constant volume and temperature equilibrium. This is my second doubt.

The problem in using this same result for a isochoric system (assuming that dG=0 can be used to describe constant volume's equilibrium in the first place) is shown ahead:

Using the same two processes, but now in constant volume

1. The change in Gibbs energy due to process 1 is no longer simply $dG_{\text{1}} = -\mu dn$, where $\mu$ is the chemical potential of A. It is now, $dG_{\text{1}} = -\mu dn + nd\mu_{\text{process 1}} + n'd\mu'_{\text{process 1}}$, because $nd\mu + n'd\mu' ≠ 0$, for an isochoric process.

2. The change in Gibbs energy due to process 2 is no longer simply $dG_{\text{1}} = 2\mu' dn$, where $\mu'$ is the chemical potential of B. It is now, $dG_{\text{2}} = 2\mu' dn + nd\mu_{\text{process 2}} + n'd\mu'_{\text{process 2}}$, because $nd\mu + n'd\mu' ≠ 0$, for an isochoric process.

$\implies$ Change in Gibbs energy through the entire reaction instantaneously:

$dG = + 2\mu' dn + nd\mu_{\text{process 2}} + n'd\mu'_{\text{process 2}}$

$-\mu dn + nd\mu_{\text{process 1}} + n'd\mu'_{\text{process 1}}$

and as long as the above dG<0, the reaction moves forward

Above expression for dG, is clearly not what we obtained for isobaric isothermal. However, $2\mu' -\mu = 0$ is still used to describe isochoric equilibrium. Why?

CONCLUSION

To conclude, my doubts are why in the first place is dG=0 valid for constant volume reactions (explained in the first section) and if dG=0 is valid for constant volume reactions, how can dG be written for an isochoric reaction in the same way as for an isobaric reaction?

I'm a Standard XII student preparing for an entrance exam.

Answer 1

The chemical potential appears in the definitions of all of the thermodynamic potentials:

$$U=-pV+TS+\sum_i \mu_i n_i$$ $$H= TS+\sum_i \mu_i n_i$$ $$A=-pV+\sum_i \mu_i n_i$$ $$G=\sum_i \mu_i n_i$$

The differential form of the potentials is

$$dU=-pdV+TdS+\sum_i \mu_i d n_i$$ $$dH= Vdp + TdS+\sum_i \mu_i d n_i$$ $$dA=-pdV-SdT +\sum_i \mu_i d n_i$$ $$dG=Vdp -SdT +\sum_i \mu_i d n_i$$

From these the chemical potential can be evaluated by applying appropriate constraints when taking the partial derivative of the potential:

$$\mu_i = \left( \frac{\partial U}{\partial n_i} \right)_{V,S,j\neq i} $$ $$\mu_i = \left( \frac{\partial H}{\partial n_i} \right)_{p,S,j\neq i} $$ $$\mu_i = \left( \frac{\partial A}{\partial n_i} \right)_{V,T,j\neq i} $$ $$\mu_i = \left( \frac{\partial G}{\partial n_i} \right)_{p,T,j\neq i} $$

When you consider a process at constant volume and temperature the appropriate potential to employ is the Helmholtz free energy $A$. Note that under isochoric conditions, and absent other forms of work but pressure-volume, $dq=dU$. The corresponding condition for spontaneity is then $dA<0$.