 So let's do an example problem. The example I've given you so far has been dissolution equilibrium, but this works just as well for chemical reactions as opposed to just physical equilibrium. So let's take this one we've used it in several of our examples before, the equilibrium that hydrogen gas and iodine gas form with hydrogen iodide. We're told that the KEQ for this reaction at 80 degrees doesn't really matter what the temperature is as long as it's not changing. Remember that the only thing that can change the equilibrium constant for a given reaction is temperature. So the KEQ at this temperature is 45.9 and we're told that in a three litre flask we have 3.2 moles of hydrogen 0.6 moles of iodine and 2.1 moles of hydrogen iodide gas they're put into that flask and we have to predict which way the reaction is going to have to shift to reach equilibrium. So the first thing I would do would be to write the equilibrium expression that's always a good place to start. Okay, KEQ products over reactants, our product is and it has a stoichiometric coefficient of 2 so I'm raising it to the power of 2. Our reactants are hydrogen and iodine. Okay so there's our equilibrium expression. The second thing I would do is to work out the concentrations of each of the species. I've given you moles and the volume of the flask but not the actual concentrations. So we've got hydrogen, concentration of hydrogen, we've got moles over volume, so 3.2 moles over a volume of 3 litres which gives you a concentration of about 1.1 moles per litre. Our iodine is 0.6 over 3 so that's 0.2 moles per litre and our hydrogen iodide is 2.1 over 3 which is 0.7 moles per litre. Okay to work out which way the reaction is going to shift we have to evaluate Q so that we can compare it to KEQ. So we evaluate Q by simply sticking these values into the equilibrium expression and seeing how it turns out. So our Q is going to the concentration of hydrogen iodide squared so that's 0.7 squared over the concentration of hydrogen times the concentration of iodine so that's 0.1 times 0.2. And if you evaluate that you get a value of Q of 2.2. Okay so clearly in this case our Q being 2.2 is less than our KEQ which was 45.9. Now let's have a look at the equilibrium expression. If Q is less than KEQ okay the values that I put in are less than KEQ. It means that the bottom of this fraction either the top of this fraction is too small or the bottom of this fraction is too big. Either way it kind of means the same thing. If the top of the fraction is too small it means there's not enough products and if the bottom of this fraction is too big it means there's too many reactants. Either way what the reaction has to do to regain equilibrium is turn some of the reactants some of the bottom of the fraction into some products which is the top of the fraction that will make the overall value of this fraction bigger and it will eventually get to KEQ. So Q is less than KEQ this means either way the way that this is solved is by the equilibrium shifting to the right to use up some reactants and produce more products. If we had just for argument's sake if we had found the opposite situation that Q was bigger than KEQ that would mean that the fraction was top heavy that there were too many products or not enough reactants and in that case Q being bigger than KEQ the way that that would be solved would be to use up some product and turn it back into reactants so the reverse reaction would be favoured and that would be called the equilibrium shifting to the left.