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Equilibrium Constants - K


When a chemical reaction reaches equilibrium, all the concentrations of all the species stop changing. We say that all concentrations are constant. Yes, they are constant and non-changing. However, the forward reaction and the reverse reaction are still going. It's just that at equilibrium they are going at the same exact rates which means no net change on either side of the equation.

Below is our truly favorite generic reaction of two reactants making two products with generic number of moles of all species (the coefficients that you balance the reaction with).

\[a{\rm A} + b{\rm B} \rightleftharpoons c{\rm C} + d{\rm D}\]

Now let's assume that all those species are in aqueous solution (aq) which means they will have concentrations in molarity units of mol/L. We can now write the equation for the equilibrium expression for \(K_{\rm c}\) - the subscript "c" means we are using concentration terms in the expression.

\[K_{\rm c} = {[\rm C]^c\;[\rm D]^d\over [\rm A]^a\;[\rm B]^b}\]

It is always the product of the product concentrations raised to their coefficients over the product of the reactant concentrations raised to their coefficients. The name of that part of the equation (the part to the right of the equal sign) is the mass action expression or equation. It is also important to point out that if there are any pure solids in the reaction (s) we don't really have a concentration term for them so they drop out of the expression. The same goes for pure liquids - like water in all aqueous chemistry. Even though water might react, the whole reaction is in water and the concentration is effectively constant itself - so all aqueous chemistry is normalized to it and the term drops out. Take home lesson: Do NOT ever include pure solids or pure liquids in your concentration based mass action expressions. Only include the dissolved species in the expression.

Oh, and if you happen to be doing all gas phase reactions, then you follow the same procedure but with pressures since pressure is our standard way of showing gas concentration. Note we show \(K_{\rm p}\) now because we are using partial pressures "p" instead of concentrations.

\[K_{\rm p} = {P_{\rm C}^c \; P_{\rm D}^d\over P_{\rm A}^a \; P_{\rm B}^b}\]

Same rule applies about NOT including pure solids and liquids in the expression - only show the partial pressures of the gases. And yes, the part to the right of the equal sign is the mass action expression.

K's are Temperature Dependent

The equilibrium constant is constant - but only at a given temperature. Anytime you change the temperature, the equilibrium constant is going to change with it. It can go higher or lower with temperature. At this stage of the game - just know it changes with temperature. We'll tackle HOW it changes elsewhere.


Reaction Quotient, Q

Hey, while you're here, lets go ahead and point out that we often will use the mass-action expression for concentrations and pressures that aren't even at equilibrium. It is actually a handy number that sort of tells us what the current progress of a reaction is. When ANY concetration or pressure can be put into the mass action expression, the value is called the reaction quotient and is shown as a capital Q. So here is the equation for Q in our generic reaction above:

\[Q = {[\rm C]^c\;[\rm D]^d\over [\rm A]^a\;[\rm B]^b}\]

Looks just like the Kc equation up top, right? Well not really. The Kc version HAS to have all the concentrations be equilibrium concentrations, whereas the Q version can use ANY concentrations we dream up. When Q is really really small, we know that the reaction mixture is mostly all reactants. When Q is really really big (say 1000 or more) we know that most of the mixture is products. And when Q is around 1 the mixture is about half reactants and half products. This is an important concept in chemistry. You need to know how far along a reaction is sometimes. Q is our best single reference number for us to use that indicates how far along a reaction is. You'll see more of Q in our electrochemistry chapter.

A really cool thing that Q and K can do for us is predicting the direction of a reaction that is not at equilibrium for some reason. Realize that Q = K for any reaction that is at equilibrium. If Q does not equal K, then the reaction is not at equilibrium and it must either go more forward or more reverse to finally reach equilibrium. This is easier to just show as a set of three possibilities for Q vs Q. Each of them implies a specific outcome as well:

Q < K the rxn will proceed forward
Q = K the rxn is at equilibrium
Q > K the rxn will proceed backward

So a simple comparison will allow you to predict the direction a reaction will go. Also know that all reactions will try to reach equilibrium the best way avaialable to them. So think this way: Q wants to equal K.





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