 We're now going to have a look at a variation on equilibrium tables where instead of knowing all our quantities explicitly we have some of them in terms of a variable called x. So here we've got this. Ignore that straight line there that was the cursor when I took a screenshot. You fill them in exactly the same way as the ones we were doing just before. They are just as easy. The only thing to take into account is that you're dealing with a variable called x, but the way of thinking, the logic, is exactly the same. So we'll take this first one. We've again got nitrogen reacting with hydrogen to give ammonia, and we're told that initially we have, in this case, two moles per litre. We're doing it in concentration, but the mass is identical. When you do these reactions, you're generally doing them in a constant volume, and that means that the way that the moles changes is exactly the same as the way that the concentration changes. So these tables can be written in terms of moles, as we did on the last slide, or they can be written in terms of concentrations, and you just treat them in exactly the same way. Okay, so we have two moles per litre of nitrogen, and we have two moles per litre of hydrogen, and no ammonia to start off with. And we're told that when it reaches equilibrium, we have 2x moles per litre of ammonia. So let's put in our change row. Okay, we started with zero ammonia, and then end up with 2x, so our change is positive 2x. Whatever x is, we have produced 2x worth of ammonia. Let's do nitrogen. Nitrogen has a 1 to 2 ratio with ammonia. So if 2x moles per litre of ammonia was produced, then half as much nitrogen must have been produced. So half of 2x is x, and of course nitrogen is a reactant, so it's being used, so that will actually be minus x. So we started with 2, we used up x, so at equilibrium we have 2 minus x. Now let's have a look at hydrogen, and rather than comparing hydrogen with ammonia, that's a 2 to 3 ratio, which is mathematically a little bit harder to deal with. It's simpler to just look at nitrogen and hydrogen, which is a 1 to 3 ratio. That's a bit simpler. So we know that x amount of nitrogen was used, and we know that 3 times as much hydrogen must have been used. So 3 times x is 3x, so we have used 3x worth of hydrogen. Notice this is because nitrogen to hydrogen in the reaction is a 1 to 3 ratio. So if you use 1 mole of nitrogen, you must use 3 moles of hydrogen. Down here we used x moles of nitrogen, so we must have used 3x moles, or moles per litre rather, of hydrogen. So we started with 2 moles of hydrogen initially, we used 3x, so our equilibrium concentration is 2 minus 3x. Now it looks a bit weird having all your equilibrium concentrations in terms of x, but you will find as we go on and do more complicated equilibrium problems, that it actually means you can solve quite interesting problems without having to know exactly what the concentrations were at equilibrium straight off. Okay, let's do the other one. Okay, that's all for now.