In this unit we will learn to calculate the value of the reaction quotient and determine what it mean for a reaction at equilibrium. Our goal is to determine the direction of the reaction by using the concentrations we are given initially and how the value of the reaction quotient 'q' compares to 'k'. In the previous examples we have worked we have seen the reaction can only precede in one direction. We either had the present of reactants or the presents of products and therefore the reaction had to go in a certain directionb because we could not lose something we did not have. However, we can also see that we have reactions where have both initial concentrations of reactants and products. We are going to have to use something called the reaction quotient which is abbreviated 'q' to determine if a system is at equilibrium and if it is not to determine if a system is at equilibrium and if it is not how we know which direction it is going to go to get to equilibrium. So when we calculate the value of q we actually calculate it the exact same way we calculate k. The reason for the different label is q we actually calculate it the exact same way we calculate k. The reason for the different label is because we do not know if we are at equilibrium and so we cannot call it an equilibrium constant so we call it q which stand for the reaction quotient. Notice that for my reaction A going to B we can write the reaction quotient, the concentration of B over the concentration of A, or products over reactants. I can then calculate the value of q and compare it to the value of k I can then calculate the value of q and compare it to the value of k to determine which way the reaction will proceed. and compare it to the value of k to determine which way the reaction will proceed. The first example we will look at is when q is less then k. So here we have a value of q of 1.3 and k is 3.2. What that tells us is that the value of q needs to increase in order to reach equilibrium. And for this value to increase we need the concentration of b to increase. And we need the value of A to decrease. As a result that mean the reaction will proceed in the forward direction. We are consuming A and producing more B. On the other side of the spectrum when we look at a q value of 4.8 which is great then the k value of 3.2 for t a particular reaction when we look at a q value of 4.8 which is great then the k value of 3.2 for t a particular reaction we see that we need to do just the opposite. Our numerator, is too high our denominator is too low so what we need to happen now is for the value of B, or the amount of B we have present to decrease and the amount of A to increase. And when that happens we will see that ratio of concentration of B over the concentration of A will also reduce the value of q will reduce and it will continue until we get to the value of k. If we calculate the reaction quotient, and we find that q is equal to k. Then we know we are at equilibrium. Now, remember we are dynamic equilibrium. Reactants are still going to products products are still going to reactants. But the rate of those two reactions are exactly the same. That is what we mean by that dynamic chemical equilibrium. So lets look at an example the reaction has an equilibrium constant of 0.078 at 100 degrees Celsius. If the initial concentration of each of the three substances is 0.100 molar which way will the reaction proceed to reach equilibrium. Hopeful you saw, that reaction for each would proceed in the reverse direction. We have to look at our reaction which is a balanced equation that was given to us we have to look at the value of k that was given a 0.078 and we need to compare that to the value of q. So what we are going to look at one the next slide and we need to compare that to the value of q. So what we are going to look at one the next slide is to see how we calculated that value of q. So here I have summarized the information given in the problem. The concentration of each of the three species is 0.100 molar. The concentration of each of the three species is 0.100 molar. The k value is given as 0.078 at 100 degree Celsius. Now what we need to calculate is q, so I look at the concentration of the product so the concentration of SO_2 times the concentration of Cl_2 and I need to divide that by the concentration of SO_2-Cl_2 So here is my reaction quotient expression the concentration of SO_2-Cl_2 So here is my reaction quotient expression I can now plug in the values that I know which are going to be the same numbers for each concentration. Because that is what was give in the problem, but it do not always have to be that way. which are going to be the same numbers for each concentration. Because that is what was give in the problem, but it do not always have to be that way. When I do this I see that my value of q is going to be equal to 0.100. So now I have to look at the comparison between q and k. And when I look at them I see that the value of q, 0.100, is greater then 0.078. So I know that q is greater then k. In order to reach equilibrium, I need the value of q to decrease so I need q to decrease. The way that the q decreases is that the amount of my products decrease and the amount of my reactant needs to increase. So now, if this happens the products decrease, the reactant increases So now, if this happens the products decrease, the reactant increases I am going to see that the value of q decreases the products decrease, the reactant increases I am going to see that the value of q decreases and what I see is that this will continue until q equals k, at which point we will be at chemical equilibrium. Now that we have figured out which way the reaction will proceed we have to also figure out what equilibrium concentrations are Now that we have figured out which way the reaction will proceed we have to also figure out what equilibrium concentrations are we are going to go have to using an ICE table but the problems are going to be a little bit more complicated then what we saw in the previous section.
