 Alright, in this problem-solving video I'm going to show you how to solve an equilibrium constant problem using algebra. Alright, so here's our problem. We've got the equilibrium between hydrogen, iodine and hydrogen iodide. And we know that KEQ for this process at some given temperature is 45.9. And we're told that 2 moles of hydrogen and 3 moles of iodine are put into a sealed glass vessel with a volume of 2 litres and allowed to come to equilibrium. And we're asked to calculate the concentration of hydrogen iodide at equilibrium. Alright, the first two steps in any problem like this are to write the equilibrium expression and to construct an equilibrium table like the ones that you've seen before. So let's do the equilibrium expression first. We'll do it over here. So KEQ is going to do products over reactants. So we've got the concentration of hydrogen iodide squared over the concentration of hydrogen times the concentration of iodine. And next we're going to construct an equilibrium table. So we can do initial moles, or initial concentration rather, and final or equilibrium concentration. And we might as well put the change row in there just for clarity. And we have our equilibrium up the top here. Okay, now we know already that the initial, we're going to do this in concentration. So our volume is 2 litres. And we know that we have 2 moles of hydrogen in 2 litres. So c equals n over v. So the concentration of hydrogen is 2 over 2, which equals 1 mole per litre. So we've got 1 mole per litre of hydrogen. And we've got 3 moles of iodine in 2 litres. So that's going to be 3 divided by 2, which is 1.5 moles per litre. And we have no hydrogen iodide to start off with. Okay, now that's all we know. We're not told how much of any of the species is formed at equilibrium. We have to work that out. And this is where we start using the tables with a variable like x. Okay, so let's say that the, as it comes to equilibrium, let's say that x amount of hydrogen is used up. So I'm going to say that the change in hydrogen is x. So then the equilibrium concentration of hydrogen will be 1 minus x. Now there's a 1 to 1 ratio between hydrogen and iodine. So if x moles per litre of hydrogen is used, then x moles per litre of iodine must also be used. So at equilibrium, we have 1.5 minus x for iodine. And it's a 1 to 2 ratio between hydrogen and hydrogen iodide. So if x moles of hydrogen is used, then 2x moles of hydrogen iodide must have been made. So at equilibrium, we have 2x hydrogen iodide. Okay, so we now have equilibrium concentrations. They're in terms of x, but we do have expressions for them. And we also have an equilibrium expression that relates those concentrations to each other. So what we're going to do now is substitute our concentrations in terms of x into our equilibrium expression. So we're going to get keq. We already know that this equals 45.9. So I'm going to put that equals 45.9, which equals... So we put the concentration of hydrogen iodide on top. That's 2x and it's squared over the concentrations of hydrogen and iodine multiplied together. So we've got 1 minus x and 1.5 minus x. All right, so what we're going to do now is we're going to sort of multiply this out and simplify it, and you'll find that it turns into a quadratic equation of the same type that you've been solving for at least the last two years. Okay, so what I'm going to do is I'm going to bring the bottom of this equation, the 1 minus x times 1.5 minus x, across to the other side, and multiply it all by 45.9. So it's going to look like this. And 2x all squared is 4x squared. So we now have this. And I'm then going to do a couple of steps to simplify it. I'm not going to write them all out here. I'd like you to try and do them for yourself. It's good algebra practice. But what you should end up with is something that looks like... And this looks like a standard quadratic, which you can solve, of course, using the standard formula that you all know and love. x equals minus b, plus or minus the square root of b squared minus 4ac, all over 2a. In this case, your a is 41.9, your b is minus 114.75, and your c is 68.85. So you plug those in and you solve it. And what you will find is that x comes out with two values because of the plus or minus x. The two x's that solve this equation are either x equals 0.89 or x equals 1.85. Now, you go back to your equilibrium table and you look at what your final equilibrium concentrations were going to be. Hydrogen was going to be 1 mole per litre minus x. So it actually makes no physical sense for x to be 1.85 because you can't use more than you started with. We had 1 mole per litre to start with and if x is 1.85, then we've used more than we had. So this is always the case with these equilibrium problems. When you solve them using the x equals minus b, plus or minus the square root of b squared minus 4ac, all over 2a, you will always find that there are two solutions and you'll always find that one of the two solutions is physically impossible. So you find out which one is physically impossible and you cross that out and then x is the one that remains. So in this case, x equals 0.89. So we can now answer the question that we were asked in the first place, which was to calculate the concentration of hydrogen iodide at equilibrium. We know from our table that the concentration of hydrogen iodide was 2x. We know that x is 0.89. Therefore, concentration of hydrogen iodide equals 2 times 0.89, which equals 1.78 molar. All right, now this, it looks like a long and difficult problem and there are quite a few steps that you have to follow, but each step individually is not that difficult and it uses skills that you already had, mostly algebra. This does tend to be a very algebraic course. And if you're feeling worried about this kind of problem, the best thing that you can do is practice a whole heap of them.