What is the equilibrium expression for the following reaction at equilibrium? 3A aqueous plus 2B liquid in equilibrium with 2C aqueous plus D solid.
An equilibrium expression is a means to calculate the equilibrium constant. The equilibrium constant expresses the balance of products and reactants for a specific equilibrium. A high equilibrium constant indicates we’re getting more products. While a lower equilibrium constant indicates we’re retaining more of the reactants. We can easily write the equilibrium expression for an equilibrium where all components are in the aqueous phase.
This is what the equilibrium expression looks like for this reaction. We take the concentration of each of our products and raise it to the power of their stochiometric coefficient and multiply them together. And then divide by the same for the reactants. The reasons for this have to do with the rates of the forward and reverse reactions. But it’s a bit more complicated than that. So based on this alone, it looks like we’d take the concentration of C from our equilibrium and raise it to the power of two and multiply it by the concentration of D. And then divide by the concentration of A to the power of three multiplied by the concentration of B to the power of two.
This would correspond with answer B. But there’s one more rule we need to observe to get the right answer. When figuring out the equilibrium expression, we ignore anything that’s in the solid or liquid state. The reasons for this are again quite complicated. But we can boil it down to the fact that changing the concentration of a solid or liquid has a much smaller effect on the rate of a reaction than changing the concentration of a dissolved ion or a gas. So actually, we ignore the presence of solid D and liquid B, leaving us with answer C. So our equilibrium expression for the equilibrium is the concentration of C to the power of two divided by the concentration of A to the power of three.