# Equilibrium constant for a solid-solid equilibrium

Suppose there are two equations:

\begin{align} \ce{A(s) &<=> B(g) + C(s)} &\quad &K_1 \tag{R1}\\ \ce{B(g) &<=> D(s) + E(s)} &\quad &K_2 \tag{R2} \end{align}

These reactions have equilibrium constants $$K_1$$ and $$K_2$$, respectively. Both of these reactions are occurring simultaneously inside a container.

When I add these equations, I get

$$\ce{A(s) <=> C(s) + D(s) + E(s)} \tag{R3}$$

Since I added the reactions, I must multiply the equilibrium constants to get the equilibrium constant for above reaction which essentially is the net of the two reactions.

I have two questions:

1. How do I even write equilibrium constant for solid solid equilibrium?

2. Since $$\ce{B}$$ doesn't appear in equilibrium constant, is it safe to say that the concentration of $$\ce{B}$$ doesn't affect the equilibrium concentrations of any other species?

• Where do you get "using the adding of reactions argument K=K1K2 , which is a different value."? Write out the multiplication and you should find that it's the same value of 1, as the B terms cancel. – Andrew Jun 9 '19 at 11:40
• yes indeed , I realised that after I posted the question , but still , the other two questions were my main concern – Aditya Prakash Jun 9 '19 at 11:45
• I excluded that part – Aditya Prakash Jun 9 '19 at 11:47

How do I even write equilibrium constant for solid solid equilibrium?

First, let's write the expression for the reaction quotient:

$$Q = 1$$

Q is a constant, always one, as long as all reactants and products are present. If you run out of one, it is zero or infinity.

Now, there are two cases. If the reaction is at equilibrium, K is also 1. If not, you can't measure the equilibrium concentrations (the reaction will go to completion, that is not an equilibrium). If you want to obtain the equilibrium constant, you would have to determine the Gibbs energy at standard state, and calculate K from that.

This is certainly quite confusing when we contrast it with other reactions (like R1 and R2), which will reach equilibrium. It gets less confusing if we rewrite R3 as:

$$\ce{A(s) -> C(s) +D(s) +E(s)}$$

An example you might be familiar with is ice turning into water. The reaction quotient is always one. Depending on the temperature, the equilibrium constant is smaller than one (all water freezes), is larger than one (all ice melts) or is exactly one (water and ice coexist in an equilibrium). You might think it is unlikely that we adjust the temperature exactly right to hit the melting point, but in fact it self-adjusts because melting ice is endothermic.

Since B doesn't appear in equilibrium constant, is it safe to say that the concentration of B doesn't affect the equilibrium concentrations of any other species?

It is best to un-ask the question, for two reasons.

1. There is no equilibrium
2. The concentration of the other species are constant, so they are not affected by the reaction at all (unless the species is used up)

For reactions where at least one species is a solute or a gas, i.e. a species whose concentration changes continuously and where there is some entropy of mixing, the Gibbs energy will have a minimum somewhere along the way between all reactants and all products. In this special case, everything is a bit different.