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

Properties of the equilibrium constant

The equilibrium constant for a reaction depends on how the balanced equation is written. If the equation is reversed, K is inverted. If the equation is multiplied by a factor n, K is raised to the nth power. If multiple equations are added together, K for the overall equation is the product of the K values for the equations that were summed. Created by Jay.

Want to join the conversation?

  • aqualine ultimate style avatar for user Strange Quark
    Is there an intuitive way of understanding the reason behind why we multiply the equilibrium constant of the 2 added reactions in order to get the resultant equilibrium constant of the new, combined reaction (the last problem tackled in the video)? Sorry, I just find it more satisfying to gain an intuitive "sense" as oppose to plainly memorizing formulae.
    (7 votes)
    Default Khan Academy avatar avatar for user
    • leaf red style avatar for user Richard
      This property of equilibrium constants is an extension of Hess's Law. Hess's Law states that the total change of energy of an overall reaction is the sum of the energy changes of the individual reactions. Equilibrium constants take into account thermodynamics and can be related to Gibbs free energies. And Hess's Law also applies to Gibbs free energy. So an overall reaction's equilibrium constant will be a combination of the individual reaction's equilibrium constants similarly how the change in Gibbs free energy of the overall reaction is the combination of the change in Gibbs free energy for the individual reactions.

      Hope that helps.
      (5 votes)
  • hopper cool style avatar for user JPOgle ✝
    Why does Kc change when you multiply each coefficient by two? Isn't it just the same reaction written in a different way?
    (3 votes)
    Default Khan Academy avatar avatar for user
    • leaf red style avatar for user Richard
      An equilibrium expression, and therefore the equilibrium constant, depends on the coefficients of a chemical reaction. So if we change the coefficients, even by the same factor for all chemical species, the constant changes too.

      For example a simple reaction like: A → B, yields an equilibrium expression of:
      K1 = [B]/[A].
      If we change to coefficients by doubling them then the reaction becomes: 2A → 2B. And so the equilibrium expression becomes:
      K2 = [B]^(2)/[A]^(2), since we raise the terms by their coefficients in the chemical equation.
      So [B]^(2)/[A]^(2) can be written as ([B]/[A])^(2), which is K1 squared or K1^(2). So K1^(2) = K2.

      Hope that helps.
      (6 votes)
  • leaf blue style avatar for user Christopher
    The third problem seems to be the same as hess's law, except we are calculating for Kc instead of enthalpy of reaction. Is this the same principle going into play?
    (1 vote)
    Default Khan Academy avatar avatar for user
    • leaf red style avatar for user Richard
      It’s similar, but I would say that they are different. Hess’s Law involves using a sum of different reaction’s quantities, but for equilibrium constants we combine them by multiplication instead. So it is similar in that we are combining different reaction’s quantities into a single quantity, but the operations are different.

      Hope that helps.
      (2 votes)
  • blobby green style avatar for user Dekun
    At , when you reverse the Kc 3.8*10^-6, it becomes 1/3.8*10^-6, then where did 2.6*10^5 come from?
    (1 vote)
    Default Khan Academy avatar avatar for user

Video transcript

- [Instructor] An equilibrium constant has one value for a particular reaction at a certain temperature. For example, for this reaction, we have oxalic acid turning into two H plus ions and the oxalate anion. The equilibrium constant, K C, for this reaction is equal to 3.8 times 10 to the negative six at 25 degrees Celsius. However, the value of the equilibrium constant changes if we change how we write the balanced equation. For example, let's reverse everything. So, instead of having oxalic acid on the reactant side, let's have it on the product side. And instead of having H plus ions in the product side and oxalate on the product side, let's put them on the reactant side. Since we've reversed the reaction, there's a new equilibrium constant. So we're going to write K C for the reverse reaction, is equal to, and to figure out the value for the new equilibrium constant, since we reversed everything, we take the inverse of the original equilibrium constant. So to find the new K C value, we take one over the original equilibrium constant of 3.8 times 10 to the negative six. And this is equal to 2.6 times 10 to the fifth, also at 25 degrees Celsius. Let's look at another example of how changing how we write the equation changes the value for the equilibrium constant. And we start by looking at the ionization of hydrofluoric acid, turning into an H plus ion and a fluoride anion. The K C value for this reaction is equal to 6.8 times 10 to the negative fourth at 25 degrees Celsius. This time we're going to multiply everything through by a factor of two. So it would be two H F turning into two H plus, plus two F minus. Our goal is to find the value for the new equilibrium constant K C. And since we multiplied everything through by a factor of two, we're going to take the old equilibrium constant and we're going to raise it to the second power. So this is K C is equal to 6.8 times 10 to the negative fourth. And we're going to square that. So this is equal to 4.6 times 10 to the negative seventh at 25 degrees Celsius. So since we multiplied through by a factor of two, we raised the equilibrium constant to the second power. If we had multiplied through by a factor of three, we would have raised the old equilibrium constant to the third power. Let's use the two reactions we've just looked at here. So we have oxalic acid turning into H plus and oxalate. And the other one we looked at was hydrofluoric acid ionizing to turn into H plus and F minus. Let's use those two reactions to calculate the equilibrium constant for a third reaction. And this third reaction shows oxalate reacting with hydrofluoric acid to form oxalic acid and the fluoride anion. Our first step is to get these first two reactions to look more like the third reaction. For example, we have oxalic acid on the reactant side, in the first reaction here, and oxalic acid is on the product side for the third reaction. Also notice, oxalate is on the product side for the first and over here for the third, it's on the reactant side. So we need to reverse the first reaction to make it look more like the third one. Here we have that reaction reversed. And since we reversed the reaction, we would have to take the inverse of the equilibrium constant, which we did in the earlier example. So let's go ahead and cross out this first one here, because we're not going to need it anymore. And let's move on to our next reaction and try to make it look more like the third one. So here we have hydrofluoric acid turning into H plus and F minus. And for our third reaction, we have hydrofluoric acid on the reactant side, but notice there is a coefficient of two in the balanced equation, and we have only a one in how we have it written. So we need to multiply everything through by a factor of two to give us two H F and two H plus and two F minus. And since we multiplied everything through by a factor of two, we would need to square the equilibrium constant for the original reaction. And once again, this is from our previous example. So we already know that equilibrium constant value. Next, let's add these two reactions together and make sure they give us the third reaction. So we have all of our reactants on one side and we have all of our products on the other side here. And we look for things that are the same on both the reactants and the products side. Well, there's two H plus on the left and there's two H plus on the right, so that would cancel out. And now, notice that we would have oxalate plus two H F going to oxalic acid plus two F minus. So adding our two reactions together does give us our third reaction. Once we've confirmed that adding our two reactions together gives us our third reaction, we can use the equilibrium constants for the first two reactions to figure out the equilibrium constant for the third. So the equilibrium constant for this first reaction we're gonna call K one, for the second reaction we're gonna call K two, and the equilibrium constant that we're trying to find for the third reaction we'll call K C. To find the equilibrium constant for the third reaction K C, we need to multiply the equilibrium constants for the first two reactions together. So K C is equal to K one times K two. We've already calculated the value for our K one from an earlier example, it was 2.6 times 10 to the fifth, and we also calculated K two, and it was 4.6 times 10 to the negative seven. So to find K C, the equilibrium constant for the third reaction, we simply multiply those two together and we get that K C is equal to point one two at 25 degrees Celsius. So whenever you add reactions together to get a new reaction, to find the equilibrium constant for the new reaction, you simply need to multiply the equilibrium constants of the reactions that you added together.