Now that we have the Law of Mass Action and the Equilibrium Constant, it's very common in Chemistry for us to use these to predict the amounts of materials which are present at equilibrium. We start off with certain amounts of material. And we want to know how much winds up in the products. Many cases, this is used in other areas other than gas phase reactions, but this is a very simple way for us to apply the concepts that we have developed. So here again is the Law of Mass Action. Here's a general chemical reaction. Let's imagine that I've given some amount of moles of a and some amount of moles of b, and I'm interested in how much c and d will be present in equilibrium? I should be able to plug in the unknown pressures of all four of these materials and solve this equation for the, the pressures of each of the reactants and the products. Let's illustrate how that's done, and there's a standard procedure that chemists have developed to make this work. And I'll illustrate it with this particular example here. We're going to take H2 gas and I2 gas reacting to form 2HI. We've already measured that at equilibrium, at 700 Kelvin, that equilibrium constant is 1.98 times 10 to the minus 2. Remember, what is the equilibrium constant? For this particular reaction, according to Law of Mass Action, that equilibrium constant will be the product HI pressure squared, divided by the pressure of the first reactant to its stoichiometric coefficient, 1, and the pressure of the iodine. And the product, or the ratio, of those pressures has to be 1.98 times 10 to the minus 2. What if then we start with not some of the H2 or the I2, but rather start with the product HI, in the amount of 0.1 moles, put that into a five liter container, and ask the question at equilibrium how much or what would be the pressure of the H2, and what would be the pressure of the I2. To illustrate this approach chemists very commonly build a table in which the first row of the table is simply the reaction itself as means of labeling the columns of the table. In the first row of the table here, we're going to consider the conditions that we've started off with, which are therefore the initial conditions. Well, let's see what did we start off with? Well, we were given 0.100 moles of HI. We were given no hydrogen and no iodine. So this is the number of moles that we started off with. Now we know we're going to proceed towards equilibrium, and as we proceed to equilibrium, we're going to create some H2 and we're going to create some I2, and we're going to lose a little bit of the HI. That means each one of these materials is going to change. How much is it going to change? Well, we are going to produce some H2. I don't know how much. So, I'm going to call that x because it's an unknown. Now, I also don't know how much I2 I'm going to produce, so that's also an unknown. Well, we can use the chemistry knowledge we have to say that every time two HIs react they produce equal amounts, one H2 and one I2, equal amounts of H2 and I2. So if I have 2x moles of HI reacting, they will give me x moles of H2 and the same number, x moles of I2. It has to be by the chemical equation. Furthermore, the amount of change of the HI has to also be related to x. To produce x moles of I2, I must destroy 2x moles of HI. So the change in the HI is minus 2x. At equilibrium the amount of moles present of each of these is simply the amount I started off with plus the change. So there are x moles of H2 at equilibrium and x moles of I2 at equilibrium. I started off with 0.1 moles of HI and I've used up 2x of them. So in fact I have 0.100 minus 2x moles of the HI, a, at equilibrium. Now just as a little side note here. If l Iabel the first row of this table as the reaction, here at Rice University we like to make note of the fact that the first letters of these words in this table spell out RICE, so we call this a RICE Table. At other universities without that name, they typically leave the R off and call it an ICE Table. But this is a lecture from Rice, so we're going to call this the RICE Table. Now how are we going to solve for x? To solve for x, we need some equation that would relate x to some known. Well, that's going to be this equation up above: the Law of Mass Action. But notice the Law of Mass Action has the pressures of HI, H2, and I2, and here we have the numbers of moles. So we need one more row of our table, which will be the partial pressures of each of these components. And that can be found from the ideal gas law because the number of moles times RT over V is equal to the pressure. So for the H2 the pressure is xrt over V. For the I2, the pressure is xrt over V. And for the HI, the pressure is 0.100 minus 2x, because that's the number of moles of HI, multiplied by RT and divided by v. So those are the pressures in each case. Let's now take each of those pressures and put them into our equilibrium constant expression because now I know all of those pressures. So kp, which is the pressure of hi squared in the numerator. Let's put that in. 0.100 minus 2x times rt over v. All of that squared. Divided by the pressure of the h2, which is x r t over v. And divided by the pressure of the i2, which is x r t over v. And if I take all of those things together that has to be equal to the equilibrium constant which is 1.98 times 10 to the minus 2. Now at this point we're done with the chemistry basically because this is just an algebraic equation with a single unknown in it, which is x. And if you already know how to do this you can proceed. Well, we're going to solve through this problem. And the first step in the pro-, in the solution is to notice that we have xrt over v squared in the denominator. And we have this term squared in the numerator. So let's just take the square root of both sides of this equation. That would make it simpler for us to solve. And we wouldn't require the quadratic equation. So the numerator becomes 0.1 minus 2x over, times rt over v. And the denominator just becomes x times rt over v. The right side of the equation is the square root of 1.98 times 10 to the minus 2, which turns out to be 0.141. Now in solving this equation, ordinarily we might need to plug in the values of R, T, and Z. But here we don't need to do that, because they actually cancel out, and that makes the solution quite simple. Now we're ready to just rearrange this and do the algebra to solve. And if we do that, we'll actually discover that x is equal to 0.047 moles. And we basically solve the problem at this point. We now know how much h2 is present at equilibrium, how much i2 is present at equilibrium, because that's equal to x. And we can figure out the number of moles of HI present at equilibrium by simply taking 0.1 and subtracting 2x. Perhaps though, what we would really like to know here in the end is, what is the pressure of the H2 at equilibrium, and according to our table up above, that pressure is simple x times rt over v. And we are given, back over here, the temperature, and we are given the volume, so it should be relatively straightforward for us to take the gas constant, r, and plug in the numbers here. And when we do that, we will discover that pressure of the H2 at equilibrium is 0.54 atmospheres, which incidentally, is also the pressure of the of the I2. This is an example of how we can use equilibrium constants to calculate the pressures or concentrations or numbers of moles of the reactants in products at equilibrium. Notice to do so, we needed to know some initial conditions. We needed to use the stoichiometric equation, which told us the relationship of the changes. And we needed to use the Law of Mass Action here, to give us an equation to solve for those changes. And in the end, using this Rice table, we're able to make predictions of what the amounts of materials which are present at equilibrium will be. Lots and lots of equilibrium calculations can be done using this approach.