If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Using the reaction quotient

AP.Chem:
TRA‑7 (EU)
,
TRA‑7.F (LO)
,
TRA‑7.F.1 (EK)
,
TRA‑8 (EU)
,
TRA‑8.B (LO)
By comparing the reaction quotient to the equilibrium constant, we can predict the direction a reaction will proceed to reach equilibrium. If Q < K, the reaction will proceed towards the products. If Q > K, the reaction will proceed towards the reactants. If Q = K, the reaction is already at equilibrium and will not change. Created by Jay.

Want to join the conversation?

  • female robot grace style avatar for user Forever Learner
    I don't see how you could get a different Qc value than your Kc value if you calculate them the exact same way. I guess sometimes the actual experimental Kc value is different than the Kc calculated? I just want to know why it would be different.....A big bucket of thanks to anybody who helps me with this! ;)
    (1 vote)
    Default Khan Academy avatar avatar for user
    • leaf red style avatar for user Richard
      So the reaction quotient, Qc, the equilibrium constant, Kc, are calculated using the same equation for any particular reaction, but the key is that we input different concentrations at different time periods for the two values.

      The equilibrium constant is a single value (at a particular temperature) in which the forward and reverse reaction rates are equal. Realistically this is observed at constant concentrations for all chemical species involved in the reaction. So it's a single set of concentrations which yields a single unchanging value.

      The reaction quotient meanwhile is taken at any instant of time and can have infinitely many possible values. The purpose of the reaction quotient is just to compare to the equilibrium constant to judge which reaction rate will be favored; the forward or the reverse.

      Hope that helps.
      (2 votes)
  • duskpin ultimate style avatar for user THE WATCHER
    How do you know what direction a reaction will travel to reach equilibrium.?
    Also, I heard that if Q < K, the reaction will proceed towards the products and that If Q > K, the reaction will do the same. Is this correct or did they miss something?

    Cheers.

    - THE WATCHER
    (0 votes)
    Default Khan Academy avatar avatar for user

Video transcript

- [Instructor] 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 A gas 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 QC is equal to, and this has the same form as the equilibrium constant expression. So we would put 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 one 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 one liter to get a concentration of 0.3 moles. So we can go ahead and plug in 0.3 mole for the concentration of B. Next, we divide that by the concentration of A, and since there are five red particles, and each particle represents 0.1 moles, five times 0.1 is 0.5 moles of A divided by a volume of one liter, it gives a concentration of 0.5 mole 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 QC is equal to three fists or 0.6. And KC remember was equal to three. So Q is not equal to K. In this case, QC is less than KC. Since Q is not equal to K, at this moment and time, the reaction is not at equilibrium. In order for this reaction to reach equilibrium, Q needs to be equal to K. And 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. And so the net reaction, if 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. And the first particulate 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 QC at this moment and time for our second particular diagram to see if we've reached equilibrium yet. Well, we have five blue particles and only three reds. So QC would be equal to five over three or five thirds. Remember that the equilibrium constant KC, is equal to three. Therefore, QC is still not equal to KC. And therefore the reaction is not at equilibrium. And 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 particulate diagram. And the second particulate diagram, we had five blues and three reds. And then the third one here, we have one, two, three, four, five, six blues, and two reds. So we've increased in the amount of blue and we've decreased in the amount of red. Let's calculate QC for the moment of time represented by our third particulate diagram. Well, there are six blues and only two reds. So QC is equal to six divided by two, which is equal to three. So QC is equal to three and remember K is also equal to three. So QC is equal to KC and therefore this reaction is now at equilibrium. And at equilibrium, the reactions turn into the products at the same rate the products turn back into the reactants. And therefore the concentration of both reactants and products remained 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. KC for this reaction is equal to 2.2 times 10 to the negative 10th at 100 degrees Celsius. And 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. And if the reaction's not at equilibrium, we need to predict which direction the net reaction will go to reach equilibrium. Our approach is gonna be to calculate QC at that moment in time, and then compare QC to KC. So first we need to write our QC expression, and this is equal to the concentration of carbon monoxide raised to the first power times the concentration of chlorine raised to the first power and that's divided by the concentration of phosgene. So the concentration of COCL2. So at this moment of time, the concentration of carbon monoxide is 3.4 times 10 to the negative six mole. The concentration of chlorine is 6.0 times, times 10 to the negative six mole, and the concentration of phosgene is equal to 2.0 times 10 to the negative third mole. When we plug all those into our Q expression and solve, we get that QC is equal to 1.0 times 10 to the negative eighth. So in this case, QC is greater than KC because QC is equal to 1.0 times 10 to the negative eight and KC is equal to 2.2 times 10 to the negative 10th. And when QC is greater than KC, 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.