# Conceptualizing Multiple Equilibria

Take the reaction: $$(1) \; \ce{A + B <=>X + Y}, \; K_1=\frac{[X][Y]}{[A][B]}$$ Now assume that X isomerizes to Z according to: $$(2) \; \ce{X <=> Z}, \; K_2=\frac{[Z]}{[X]}$$ We can then add the equations to obtain: $$(3) \; \ce{A + B <=> Z + Y}, \; K_3=K_1K_2=\frac{[Z][Y]}{[A][B]}$$ It's pretty easy to show that $K_3=K_1K_2$ when considering the ratio of products and reactants, but this makes less sense when considering the actual numeric value of $K_3$. The fact that $K_3=K_1K_2$ can be explained by the statement of multiple equilibria: all equilibria must be satisfied simultaneously. This is what I'm having trouble conceptualizing. Does the equilibrium constant in $(1)$ already take into account $(2)$, and vice versa? Equation $(3)$ will reach a state of maximum entropy in which $K_3$ applies. This must also occur when both $K_2$ and $K_3$ are satisfied? I can do the math all day, but I'm not quite getting the link between the three equilibrium constants.

When there is a situation of multiple equilibria, all of them must be satisfied. If this is the case, the whole system will be at equilibrium.

For your example and as you report, the maths let you write that

$K_{3}=K_{1}\times K_{2} \quad (1)$

This final equation, beyond being the mere mathematical combination of two initial equations, is obtained by considering that Gibbs free energy is a function of state: iIt describes "quantitatively an equilibrium state of a thermodynamic system, irrespective of how the system arrived in that state" (Wikipedia). One may get to equilibrium via an unique straight step, or through several steps. But the final destination will be the same.

This definition allows us to write:

$\Delta G_{3}^{o}=\Delta G_{1}^{o} + \Delta G_{2}^{o} \quad (2)$

Equation (2) can be converted into equation (1), by remembering that $\Delta G^{o}=-RT\text{ln} K$.

We can first of all say that $K_{1}$ and $K_{2}$ relate to the two "pieces" of the reaction path one has to cover to reach the overall equilibrium. Knowing them is essential, if one wants to obtain the quantities for all the chemical species, at equilibrium.

$K_{3}$, on the other hand, relates to the "shortcut" (the unique "ideal" step connecting the initial and final states of a system) and is linked to $K_{1}$ and $K_{2}$, through equation (1).

Using the examples you provide and assuming initially 1 mole for $A$ and 1 mole for $B$, we write:

$K_{1}=\frac{(y-z)y}{(1-y)^2}$

$K_{2}=\frac{z}{y-z}$

We could more generally write:

$K_{1}=f(y,z)$

$K_{2}=g(y,z)$

The two constants are different functions: but they both depend on the same parameters, $y$ and $z$, which are temperature and pressure (for gases) dependent. If temperature changes, both constants will accordingly change.

To determine the equilibrium quantities, at given temperature and pressure, we'll need the values for $K_{1}$ and $K_{2}$, since knowing only $K_{3}$ won't be enough.

Beyond all this, kinetics will affect the time needed to reach system equilibrium: but that's another story.

• Ok, I can accept that $(1)$ is true due to $(2)$. Is there any way to explain why $K_1$ and $K_2$ are also both satisfied? That is, why are ALL equilibria satisfied? You could prove that both $K_1$ and $K_2$ have to be true in order to reach $K_3$ due to $(2)$ being true. I feel like this is more along the lines of working backwards from the solution, however. Is there any way to explain this in terms of energy/entropy? Apr 10, 2014 at 4:06
• @David Wouldn't be correct to say that since all equilibria have to satisfied, each $\Delta G$ (related to a single equilibrium) has to be $0$ ? (short reminder: $\Delta G \neq \Delta G^{0}$). $\Delta G=0$ is the condition that always applies once a chemical equilibrium is established: $G=H-TS$, so in the definition of $G$ you find enthalpy and entropy.... Apr 10, 2014 at 7:04
• The only thing I can think of is the following: $∆G_3 = ∆G_1 + ∆G_2$. Since $∆G_3 = 0$, $∆G_1 + ∆G_2 = 0$ and $∆G_1 = -∆G_2$. This is clearly true for $∆G_1 = ∆G_2 = 0$, but I'm not sure how to prove this is always the case. If $∆G_i = 0$, then $K_i$ must apply. Apr 10, 2014 at 12:22
• @David one further point: if $\Delta G_{3}=0$, then $\Delta G_{1}=-\Delta G_{2}$. One option is that both are equal to $0$ and that means that the two reactions are at equilibrium. Another option is that $\Delta G_{1}=-\Delta G_{2} \neq 0$: so if $\Delta G_{1} < 0$, then $\Delta G_{2} > 0$. That can't work: reaction $1$ would be spontaneous, while reaction $2$ would not. In these conditions, the overall reaction $3$ would never reach equilibrium, since reaction $2$, essential in order to complete reaction $3$, would never occur. Apr 10, 2014 at 14:33
• You say reaction $(2)$ would never occur for $∆G > 0$, but I feel like it would since it is coupled to reaction $(1)$. This is essential to the idea of spontaneous processes forcing non-spontaneous ones to occur. i.e. The coupling of ATP hydrolysis to non-spontaneous enzymatic reactions makes the overall process spontaneous. Am I wrong, or is there another explanation? Does the coupling I've explained react differently when the coupled reaction is at equilibrium, as opposed to being simply positive or negative? Apr 10, 2014 at 15:34