Chemical potential and gibbs free energy

Question

I would like to ask why the $\Delta_rG_1 = -RT\ln K + RT\ln Q$ is not equal to $Δ_rG_2$ numerically as shown in the figure where I derived $\Delta_rG_2 = RT\ln Q$ using a cyclic ring since $\Delta_rG$ is a state variable.

I am also aware that $\Delta_rG = \mu_\text{product} - \mu_\text{reactant}$, which should be the same at a particular state, and this is consistent with the constant gradient I have in the graph.

Hence, why is it that numerically, I cannot get $\Delta_rG_1 = \Delta_rG_2$? Thanks

Answer 1

I can't see what you did in your derivation, but, for what it's worth, here's my derivation.

My starting equations are $$G=\sum{n_i\mu_i}$$and, along your contour, at constant temperature and pressure, $$\mathrm dG=\sum{\mu_i\mathrm dn_i}$$The changes in the number of moles of the various species are given by $$n_i=n_{i0}+c_in$$where $c_i$ is the stoichiometric coefficient for species i (a negative integer for a reactant and a positive integer for a product), $n_{i0}$ is the initial number of moles of species i, and $n$ is the molar progress of the reaction (number of times that the balanced reaction equation has been run through).

For an ideal gas mixture, $$\mu_i=\mu_i^{(0)}(T)+RT\ln{p_i}$$where $\mu_i^{(0)}(T)$ is the standard free energy of formation of species i at temperature T and $p_i$ is the partial pressure of species i: $$p_i=P\frac{n_i}{\sum{n_j}}=\frac{n_{i0}+c_in}{\sum{n_{j0}}+(\sum{c_j})n}P$$ If we combine the above equations, we obtain: $$\mathrm dG=\left[\sum{c_i\mu_i^{(0)}}+RT\sum{c_i\ln{p_i}}\right]\mathrm dn$$ But, $$\sum{c_i\mu_i^{(0)}}=\Delta G^{(0)}$$and $$\sum{c_i\ln{p_i}}=\ln{Q}$$Therefore, along the reaction contour, $$\mathrm dG=[\Delta G^{(0)}+RT\ln{Q}]\mathrm dn$$

Answer 2

There are at least two fundamental issues you have to address. First, you have to distinguish between the Gibbs energy $G$ and the Gibbs energy of reaction $\Delta_r G$. In you diagram, one is the value on the y-axis (without defined zero point) and the other is the slope of the line, as labeled in your sketch. The second issue is that in the expressions on the left, the differences in Gibbs free energy of reaction should be distinguished from the values themselves by labeling the former $\Delta \Delta_r G$.

With those two considerations, there are two ways to look at the relationship between the three states. To define these three states in a way that one can turn into the other in a closed system (no transfer of substances into or out of the system, dn = 0), you could start with the standard state ($Q = 1$), choose a certain state while the reaction is running ($Q = Q_t$) and then let it go to equilibrium ($Q = K$).

Gibbs energy of reaction

Here are the expressions for the three states.

For an arbitrary $Q = Q_t$:

$$ \Delta_r G = - R T \ln K + R T \ln Q_t\tag{1}$$

For $Q = 1$ (standard state):

$$ \Delta_r G = - R T \ln K\tag{2}$$

For $Q = K$:

$$ \Delta_r G = 0\tag{3}$$

We can now write the differences between pairs:

$$ \Delta \Delta_r G_{1-2} = R T \ln Q$$

$$ \Delta \Delta_r G_{2-3} = - R T \ln K $$

$$ \Delta \Delta_r G_{1-3} = - R T \ln K + R T \ln Q$$

It all adds up. There is no special meaning to $ \Delta \Delta_r G_{1-2} $ though, and no reason that it should be the same as $ \Delta \Delta_r G_{1-3}$.

Gibbs energy

Specifying the Gibbs energy of the three states is impossible because of the issue with the zero point. It is possible, however, to say how one state is different from another, i.e. find relations for $\Delta G_{1\rightarrow 2}$, $\Delta G_{2\rightarrow 3}$, $\Delta G_{1\rightarrow 3}$. You would have to integrate over the extent of reaction $\xi$ in a way that $Q$ goes from 1 to $Q_t$ to $K$, using the relationships in the previous section. Because $G$ is a state function (and because the integral from state 1 to state 3 can be achieved by first going from state 1 to state 2 and then to state 3), it would add up correctly.