 All right, so let's talk a little more about the left side of this equilibrium condition for chemical reactions. We know that when a reaction is at equilibrium, this left-hand side, the amount of each species raised to the stoichiometric coefficients is equal to the right-hand side. Partition functions raised to the stoichiometric coefficients. The right-hand side, we've gotten used to calling K the equilibrium constant. And as the name implies, that's a constant at a particular temperature for a particular reaction. Left-hand side also has a name. We also assign a variable to that left-hand side. We call that one Q, and that's called the reaction quotient. So whether we call it Q or whether we write it as this ratio or product of amounts of each species, it's called the reaction quotient. In equilibrium, we say the reaction quotient is equal to the equilibrium constant. And the important thing to remember about the reaction quotient is it's not a constant. In fact, it depends on the extent of reaction as we saw when we worked out an example of what the extent of reaction was that allowed our reaction to reach equilibrium. So as the reaction proceeds forwards, proceeds backwards, the value of the reaction quotient changes. There's one particular value of the extent of reaction that makes this reaction quotient equal to the equilibrium constant. So what that means is the reaction quotient is useful not just for helping us understand exactly where the position of an equilibrium is, but also in helping us determine at a quick glance whether a reaction is going to proceed forwards or proceed backwards. Sometimes we don't want to know exactly how far a reaction proceeds. We just want to know, is it going to go forwards or is it going to go backwards? So if we have a chemical reaction in general, some chemical reaction like reactants turning into some products, what this reaction quotient looks like, that's the, this particular ratio of amounts raised to their stoichiometric coefficients, products in the numerator. So moles of C, moles of D, or molecules of C, molecules of D in the numerator, in the denominator, molecules of A, molecules of B, and so on. These would be raised to some power in my made up chemical reaction. I've got one molecule of A, one molecule of B turning into one molecule of C and one molecule of D. So these powers are all one in my example. The way we connect the amounts of each of these molecules to the extent of reaction, remember, moles of product or molecules of product is some initial amount plus the extent of reaction, again, with some stoichiometric coefficient that might or might not be a one. Molecules of D is its initial amount plus extent of reaction because it's on the product side and so on in the numerator. In the denominator it looks very similar, but because the reactants are in the denominator, I'm subtracting the extent of reaction. So I've got plus squiggle in the numerator, minus squiggle in each one of these terms in the denominator. What that means is if the reaction proceeds in the forward direction, meaning the value of the extent of the reaction, the value of the reaction coordinate is increasing when I move the reaction forwards. What happens to Q, this ratio, is because squiggle is increasing. This number has gotten larger, this number has gotten larger. All of the terms in the denominator have gotten larger. With the minus sign here, this term has gotten smaller. This term has also gotten smaller, all the terms in the denominator have gotten smaller. So I've made the numerator larger, I've made the denominator smaller. So that means the value of the reaction quotient goes up. When the reaction proceeds forwards, the reaction quotient goes up. On the other hand, the exact opposite is true when the reaction proceeds backwards. Squiggle is becoming smaller or more negative. I'm decreasing the terms in the numerator, I'm increasing the terms in the denominator, and the reaction quotient decreases. So we can predict what direction Q is going to change when the reaction shifts forwards or backwards. So if we happen to have Q equal to K, we know what that means. That means the reaction is at equilibrium. It's not going to move. It's already at equilibrium. It's not going to shift forward or backward. If we have Q bigger than or Q less than K, on the other hand, then the reaction will shift. If at any current set of conditions, Q happens to be less than K, in the process of reaching equilibrium, Q needs to get larger in order to become equal to K. So Q needs to increase. The only way to make Q increase would be to have the reaction go forwards. So if we ever find that the reaction quotient is less than K, then the reaction proceeds in the forward direction. On the other hand, if at a current set of conditions, the reaction quotient happens to be larger than the equilibrium constant, the only way to make Q go down and become equal to the equilibrium constant is to have the reaction proceed backwards. So that's what I meant when I said Q is useful not just in solving for the exact position of an equilibrium, but being able to predict based on whether it's currently larger or smaller than the equilibrium constant, whether the reaction is going to proceed backwards or forwards. So just to make sure that's clear, let's work a quick example with our favorite reaction, hydrogen gas and bromine gas forming HBr gas. Let's suppose, again, we're not at equilibrium. We are not interested in exactly where the equilibrium lies, but let's say we just pick some amounts of these molecules to throw into a container, throw a tiny amount of H2, a slightly larger amount of Br2, and a full mole of HBr into a container. Those are our initial amounts. Question is, we know the equilibrium lies pretty far towards product. We know for this reaction, at least at some temperature, the temperature we've considered previously when we solve this equilibrium, the equilibrium lies pretty far toward product. We're going to end up with more HBr than H2 and Br2, but is this already enough HBr or not enough? Do we need to convert more H2 and Br2 into HBr in order to reach equilibrium? Or in fact, if we started out too far towards product and we're going to have to convert some HBr back to H2 and Br2, the way to find that out is to do a quick calculation of the reaction quotient. If I just calculate at these conditions products divided by reactants, moles of HBr squared over moles of H2 and moles of Br2, if I calculate one mole squared divided by 0.01 moles of H2, also divide by 0.05 moles of Br2. That math is not too difficult. 1 over 0.01 is 100. If I divide by 0.05, that's like multiplying by 20, so that's going to be, again, units are going to cancel conveniently. Mole squared cancels a mole and a mole. One times 100 times 20, that works out to a value of 2,000, a unit list value of 2,000. We know the equilibrium constant is K. That value of Q was close to K, but it's not exactly equal to K. It's a little larger than K. Since the reaction quotient is already larger than K, the way to reach equilibrium is for that reaction to proceed backwards. We know if these were our initial amounts of H2 and Br2 and HBr, what's going to happen next is a small amount of HBr is going to be consumed, converted to H2 and Br2 in order for that system to reach equilibrium.