Using the reaction quotient | Equilibrium | AP Chemistry | Khan Academy

The reaction quotient is symbolized by the capital letter Q, and it tells us whether a reaction is at equilibrium or not. If the reaction is not at equilibrium, it also allows us to predict which direction the net reaction will go to reach equilibrium.

For example, let's look at the hypothetical reaction where gas A turns into gas B. Gas A will be represented by red circles or red spheres, and gas B will be represented by blue spheres. The equilibrium constant for this hypothetical reaction is equal to 3 at 25 degrees Celsius.

Let's start by writing out the expression for the reaction quotient. So we would write out here:

$$Q_c = \frac{[B]^1}{[A]^1}$$

This has the same form as the equilibrium constant expression, so we would put the concentration of B to the first power divided by the concentration of A, also to the first power.

Let's look at our first particulate diagram here and let's think about each particle representing 0.1 moles of a substance, and the volume of the container is 1 liter. So let's first find the concentration of B, so we can plug it into our expression for Q.

B are the blue spheres, so we count up one, two, three blue spheres. Each sphere or each particle represents 0.1 moles, so three times 0.1 is 0.3, and then we divide that by the volume of 1 liter to get a concentration of 0.3 molar. So we can go ahead and plug in 0.3 molar for the concentration of B.

Next, we divide that by the concentration of A. Since there are five red particles and each particle represents 0.1 moles, 5 times 0.1 is 0.5 moles of A, divided by a volume of 1 liter, gives a concentration of 0.5 molar for A.

Notice that we could have just counted the number of particles: three blues and five reds, and just done three divided by five and get the same value for the reaction quotient Q.

So,

$$Q_c = \frac{3}{5} = 0.6$$

And K_c, remember, was equal to 3. So Q is not equal to K. In this case,

$$Q_c < K_c$$

Since Q is not equal to K at this moment in time, the reaction is not at equilibrium.

In order for this reaction to reach equilibrium, Q needs to be equal to K. Since Q is a lot smaller than K, if you think about it, we need to increase the numerator and decrease the denominator. So that means we have too many reactants and not enough products.

The net reaction, if we go back up here to the equation here, the net reaction is going to move to the right to make more products. So the net reaction moves to the right to make more blue particles, and therefore the number of red particles would decrease.

We can see that comparing these first two particulate diagrams here. So let's compare the first particular diagram to the second particular diagram. In the first particular diagram, we had three blues and five reds, and the second particulate diagram we have five blues and only three reds. So that shows the reaction has moved to the right to increase the amount of products and to decrease the amount of reactants.

Let's calculate Q_c at this moment in time for our second particulate diagram to see if we've reached equilibrium yet. Well, we have five blue particles and only three reds, so

$$Q_c = \frac{5}{3} = \frac{5}{3}$$

Remember that the equilibrium constant K_c is equal to 3; therefore, Q_c is still not equal to K_c, and therefore the reaction is not at equilibrium.

Q is actually still less than K; therefore, the net reaction is going to move to the right again to increase the amount of products and to decrease the amount of reactants.

Let's compare our second particulate diagram to our third particular diagram. In the second particulate diagram, we had five blues and three reds, and then in the third one here, we have one, two, three, four, five, six blues and two reds. So we've increased the amount of blue and we've decreased the amount of red.

Let's calculate Q_c for the moment of time represented by our third particulate diagram. Well, there are six blues and only two reds, so

$$Q_c = \frac{6}{2} = 3$$

So Q_c is equal to 3, and remember K is also equal to 3. So Q_c is equal to K_c, and therefore this reaction is now at equilibrium.

At equilibrium, the reactants turn into the products at the same rate the products turn back into the reactants, and therefore the concentration of both reactants and products remain constant at equilibrium.

So when Q is less than K, the reaction is not at equilibrium; there are too many reactants and not enough products. Therefore, the net reaction goes to the right to increase the amount of products. The net reaction continues to go to the right until Q is equal to K, and the reaction has reached equilibrium.

At that point, the concentrations of reactants and products stop changing, and they remain constant. It's also possible for Q to be greater than K, and if that's true, the reaction is not at equilibrium, but in this case, you have too many products and not enough reactants.

Therefore, the net reaction goes to the left to increase the amount of reactants and to decrease the amount of products. The net reaction will continue to go to the left until Q is equal to K, and the reaction reaches equilibrium.

Let's look at another reaction, which is the decomposition of phosgene to form carbon monoxide and chlorine gas. K_c for this reaction is equal to 2.2 times 10 to the negative 10th at 100 degrees Celsius.

Let's say we're given concentrations of phosgene, carbon monoxide, and chlorine at a moment in time and asked if the reaction is at equilibrium or not. If the reaction is not at equilibrium, we need to predict which direction the net reaction will go to reach equilibrium.

Our approach is going to be to calculate Q_c at that moment in time and then compare Q_c to K_c. So first, we need to write our Q_c expression, and this is equal to

$$Q_c = \frac{[CO][Cl]}{[Phosgene]}$$

So at this moment of time, the concentration of carbon monoxide is 3.4 times 10 to the negative 6 molar, the concentration of chlorine is 6.0 times 10 to the negative 6 molar, and the concentration of phosgene is equal to 2.0 times 10 to the negative 3 molar.

When we plug all those into our Q expression and solve, we get that

$$Q_c = 1.0 \times 10^{-8}$$

So in this case, Q_c is greater than K_c because

$$Q_c = 1.0 \times 10^{-8}$$

and K_c is equal to

$$2.2 \times 10^{-10}$$

When Q_c is greater than K_c, we have too many products and not enough reactants; therefore, the net reaction is going to go to the left, and there's going to be an increase in the amount of phosgene.

The reaction will continue to go to the left until Q is equal to K, and the reaction reaches equilibrium.