 A chemical system is in dynamic equilibrium when the rate of the forward process equals the rate of the reverse process. From a macroscopic viewpoint, a system in equilibrium will look unchanging, but at the molecular level, the forward and reverse processes, whether they are physical or chemical changes, are churning away continuously. In this video, we'll focus on chemical reactions. So let's take a hypothetical reversible reaction, A goes to B. If this reaction were at equilibrium, it would mean that the rate of A turning into B was equal to the rate of B turning into A. What does that mean about the concentration of A at equilibrium? Would they be the same? Well, think back to kinetics. Let's say that we know the rate laws of the forward and reverse reactions. So the forward rate, the rate of the forward reaction, that's how fast B is being produced. That's equal to a rate constant, I'll call it K1, times the concentration of the reactant, which is A. And the reverse rate, the rate of the reverse reaction, that's how fast A is being produced, that's equal to a different rate constant, I'll call it K2, multiplied by the concentration of B, because B is the reactant here. So that's K2B. Okay, note that the rates each depend on two things. The concentration of the reactant and the rate constant. And recall also from kinetics that a high rate constant implies a reaction with a low activation energy, and therefore it must be a reaction that has a high probability of successful collisions. Now let's say we know the values of our rate constants. Let's say that K1 equals 6 and K2 equals 2. As long as we keep the temperature constant, those rate constants won't change. So if this reaction is at equilibrium, then the forward rate equals the reverse rate. And that means that the rate at which B is being produced equals the rate at which A is being produced. So we could write that K1A, or K1 times the concentration of A equals K2 times the concentration of B. And then if we substitute in those values for the two Ks, we get 6 times A equals 2 times B. Now if the concentration of A and B are equal, that would mean that the forward reaction is faster than the reverse reaction, and that means it's not at equilibrium. 6 times 1 concentration doesn't equal 2 times the same concentration. So in order to get the rates to be equal, the concentration of B has to be higher than the concentration of A. And we can work out just how much higher by rearranging that equation. You get the concentration of B over the concentration of A, so the ratio of the product to the reactant concentrations equals K1 over K2, so 6 over 2, which equals 3. So what that means is in order for the forward and reverse reactions to be equal in this situation, you would need the concentration of B to be 3 times as high as that of A. This is because the forward reaction in this case, which relies on the concentration of A, happens more easily. It has a higher K, so in kinetics terms we would say that the probability of a successful collision is higher. For the reverse reaction, K is lower, meaning there's a lower probability of a successful collision. So a higher concentration of that reactant of B is needed to get the same rate of reaction.