Chemical potential and gibbs free energy

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

• I suspect you are attaching the wrong meaning to the different equations. The change in G at equilibrium is zero. The change in G when products and reactants are in standard states is $\Delta G^\circ$ Sep 11 '20 at 18:23
• Hi, thanks for the reply, I understood this, but my issue was that why ΔrG1 from a particular state to standard state not equals to ΔrG2 when that particular state moves to equilibrium. This sis because ΔrG is the gradient of the graph as shown in the main diagram. So, at any point, if you want to have a different end state (regardless of whether if it is standard state or equilibrium state), the gradient of the graph should not change, but numerically in my derivation, somehow it did. So, I am unsure where I went wrong. Sep 12 '20 at 1:41

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, $$\mathrm dG=[\Delta G^{(0)}+RT\ln{Q}]\mathrm dn$$

• Hi, thanks for your explanation, I learnt a lot from it. However when ΔG= dG/dn, the reaction from particular state to equilibrium, ΔG1=-RTlnK + RTlnQ and from standard state to equilibrium, ΔG(0)=-RTlnK. Since ΔG is a state variable, I derived ΔG2 from particular state to standard state as ΔG2 = ΔG1 - ΔG(0) = RTlnQ. From the graph of summation G against extent of reaction, the gradient = ΔG. Hence, at particular state at a certain moment of extent of reaction, the gradient is the same, so ΔG1 = ΔG2 should be the case. But he numerical calculation I have above, shows ΔG1 different from ΔG2. Sep 12 '20 at 1:52
• You can't get from the standard state to any arbitrary state (and vice versa) unless the arbitrary state contains stoichiometric quantities of reactants and products to begin with and the total pressure in the arbitrary state is 1 bar. In other words, the standard state does not lie along the reaction trajectory of an arbitrary starting state. Sep 12 '20 at 2:03
• @ChetMiller In my answer, I started with the standard state, defined an arbitrary state along the reaction trajectory as Qt, and let it go to equilibrium. This way, you can ensure those three states are accessible in a single reaction mixture. Sep 12 '20 at 11:54
• @KarstenTheis That's, of course, a valid starting state, but it is only at 1 bar. Still, maybe that's closer to the path the OP envisioned. I guess the more general path I treated includes this as a special case. Sep 12 '20 at 12:30

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.