The reaction quotient Q

The reaction quotient $Q$‍ is a measure of the relative amounts of products and reactants present in a reaction at a given time.

For reversible reaction $\text{aA}+\text{bB}\rightleftharpoons \text{cC}+\text{dD}$‍, where $\text{a}$‍, $\text{b}$‍, $\text{c}$‍, and $\text{d}$‍ are the stoichiometric coefficients for the balanced reaction, we can calculate $Q$‍ using the following equation:

This expression might look awfully familiar, because $Q$‍ is a concept that is closely related to the equilibrium constant $K$‍. Unlike $K$‍, which is based on equilibrium concentrations, $Q$‍ can be calculated whether we are at equilibrium or not.

The magnitude of $Q$‍ tells us what we have in our reaction vessel. What does that mean exactly? Let's start by thinking about the extremes. For a reaction that has only starting materials, the product concentrations are $[\text{C}]=[\text{D}]=0$‍. Since our numerator is zero, then $Q=0$‍. For a reaction that has only products, we have $[\text{A}]=[\text{B}]=0$‍ in the denominator of our equation, so $Q$‍ is infinitely large. Most of the time, we will have some mixture of reactants and products, but you can still remember that very small values of $Q$‍ tell you that you have mostly reactants and very large values of $Q$‍ result from having mostly products in the reaction vessel.

Comparing $Q$‍ and $K$‍ for a given reaction tells us which direction the reaction needs to go to reach equilibrium. You can think of this as another way to use Le Châtelier’s principle.

From Le Châtelier’s principle, we know that when a stress is applied that moves a reaction away from equilibrium, the reaction will try to adjust to get back to equilbrium. By comparing $Q$‍ and $K$‍, we can see how our reaction is adjusting—is it trying to make more product, or is it consuming product to make more reactant? Alternatively, are we at equilibrium already?

There are three possible scenarios to consider:

Let’s think back to our expression for $Q$‍ above. We have our product concentrations, or partial pressures, in the numerator and our reactant concentrations, or partial pressures, in the denominator. In the case where $Q>K$‍, this suggests that we have more product present than we would have at equilibrium. Therefore, the reaction will try to use up some of the excess product and favor the reverse reaction to reach equilibrium.

In this case, the ratio of products to reactants is less than that for the system at equilibrium. In other words, the concentration of the reactants is higher than it would be at equilibrium; you can also think of it as the product concentration being too low. In order to reach equilibrium, the reaction will favor the forward reaction and try to use up some of the excess reactant to make more product.

Hooray! The reaction is already at equilibrium! Our concentrations won't change since the rates of the forward and backward reactions are equal.

We know that $Q$‍ can have possible values starting from zero (all reactants) to infinitely large (all products). We also know that our reaction will adjust the concentrations to reach equilibrium if it isn't at equilibrium already. Another way we can think about these ideas is to draw out a number line for all possible values of $Q$‍:

To simplify things a bit, the line can be roughly divided into three regions. For very small values of $Q$‍, ~${10}^{-3}$‍ or less, the reaction has mostly reactants. For intermediate values of $Q$‍, between ~${10}^{-3}$‍ and ${10}^3$‍, we have significant amounts of both products and reactants in our reaction vessel. Finally, when $Q$‍ is large, greater than ~${10}^3$‍, we have mostly products.

If we plot both $Q$‍ and $K$‍ on our number line, the direction we move to get from $Q$‍ to $K$‍ tells us about how the reaction is trying to adjust. If we are moving to the right, we are shifting the concentrations to make more products and favoring the forward reaction. If we are moving to the left toward zero, we are moving in the direction of making more reactants and favoring the reverse reaction.

Given the following concentrations, what is $Q$‍?
And, if $K=1.0$‍, which side of the reaction is favored at that value of $Q$‍?

We can calculate $Q$‍ by writing out the equation using the balanced reaction and then using the given concentrations.

If we compare $Q$‍ to $K$‍, we can see that $Q>K$‍. This tells us that we have excess product compared to equililbrium and therefore the reverse reaction will be favored.

If we draw out the number line with our values of $Q$‍ and $K$‍, we get something like this:

We can see that $Q$‍ falls near the region where we have mostly products, which is to the right of $K$‍. Since the reaction will adjust to move closer to $K$‍, we can draw an arrow for the direction of that shift. This arrow starts at $Q$‍ and points toward $K$‍, and it also points to the mostly reactants region. This tells us that our reaction will be favoring the reverse reaction in order to make more reactants and consume excess products.

As you can see, both methods give the same answer, so you can decide which one works best for you!

We can compare the reaction quotient $Q$‍ to the equilibrium constant $K$‍ to predict what a reaction will do to reach equilibrium. In addition, you might see $Q$‍ pop up in other chemistry topics and equations because we are often interested in what happens to various thermodynamic quantities when we are not at equilibrium. Stay tuned for more!