hey guys this is michael from concord chemistry in today's video we'll be going over some equilibrium ice table practice problems this will be part one in the series where i'll show you all the different ways that ice table problems can be asked so you can be prepared for any problems that appear on your exam let's start with the first example in this question you're given the balanced chemical reaction you're also given the kc value of 7.2 you're given the temperature although that's not too relevant and then you're given the initial concentration of both the reactants are 0.2 and that there are no products and then you're asked to determine the equilibrium concentration of the product so since you're given an initial condition and equilibrium condition that's indication that we're going to have to use an ice table so i'm just going to start by rewriting the chemical reaction so it's a little bit bigger then we're going to set up a nice table if you're unfamiliar with what ice table is i stands for initial c stands for change and e stands for equilibrium this is a good way to allow you to keep track of what you started with and what you end with at equilibrium the initial concentration of both the reactants were 0.200 molar and there was no product to start with so we'll put this as zero molar since the reaction started with just reactants and no products that means this reaction is going to proceed towards the product side so that means the reactants will be consumed and the products will be produced so for c line this would be minus x and minus x we're using x's here because we don't know exactly how much will be consumed since this reaction goes to equilibrium and not completion and then on the product side this will be plus 2x plus because it's a it's it's getting larger and then 2x because there's a 2 in the front and then the e line is just the sum of the i and c line you can think of this it's just these two lines add together and that'll give you the e line so this will be 0.200 minus x this will be 0.200 minus x and then this would just be 2x now that ice table is complete let's set up the kc expression so kc would just be concentration products over reactants so it'll be concentration of brcl squared because there's a coefficient of 2 divided by the concentration of reactants br 2 multiplied by the concentration of cl2 then we're going to substitute the kc value in as well as the equilibrium values equilibrium variables into the expression so kc is 7.20 the concentration of brcl was 2x and then we got a square because it's squared here divided by the concentration the product of these two and both those are exactly the same so we'll just write this as 0.200 minus x squared since it's squared on the top and on the bottom we can make our calculation a little easier by taking the square root of both sides and then when we do that we get 2.68 equals 2x divided by 0.200 minus x then we can cross multiply and we'll get 0.537 minus 2.6 x equals 2x add 2.68 x to both sides then divide both sides by 4.68 to isolate x and that'll give us x equals 0.115 molar so now we have x we can substitute it back into the e line here we're just asked to solve for the equilibrium concentration of brcl you can see that the equilibrium concentration brco is 2x so then the equilibrium concentration of brcl is just going to be 2 multiplied by the value of x that we got and that will give us 0.230 approximately if you kept other numbers without rounding then you might get 0.229 molar if instead this question was asking us for the equilibrium concentration everything you can just take the x value that you got and plug it in to point plug into here to get the concentration of br2 and plug into this line to get the concentration of cl2 let's take a look at the next problem so in this problem you're again given the balanced chemical reaction you're given the amount of so3 and the volume and then you're also and then you're given the equilibrium values of the so2 and then this time you're asked to solve for k this is a little different in the previous question we were given the initial values and then given the k values and have to solve for the equilibrium values this time we're given the initial values and the equilibrium values and we have to solve for k so let's uh start by again writing rewriting the reaction to so3 gas in equilib proceeds to so2 gas plus o2 gas again because we're given an initial condition and an equilibrium condition we can set up an ice table here since all these are gases we can solve for either kp or kc but because we're given moles and liters it's going to be easier to solve for kc given the fact that concentration which is molarity will be equal to moles per liter and we already have the moles in the liters so we can write the kc expression here kc is going to equal concentration so2 squared because it's two multiplied by the concentration of o2 divided by the concentration of so3 squared so what we just have to figure out the equilibrium concentration of these and we can plug it in and we'll get the kc value since this is moles you can't plug that in we have to convert that to molarity and molarity is moles of liters so the initial the initial concentration of so3 is going to equal 12.0 moles divided by the 3.0 liters and they'll give us 4.0 molar the question doesn't mention anything about so2 or o2 initially so we can assume that we didn't have any products start so this would be zero molar and that'll be zero molar once again because we're starting with all reactants and no products this reaction will proceed to the right so then this will be minus two x because of the coefficient of two this will be plus two x and then this will be plus x then the e line is just the sum of the i and c line 4.0 minus 2x 2x and x it tells us that at equilibrium there are 3.0 moles of so2 present because our ice table was in molarity we should also convert this to molarity so we can say that the co equilibrium concentration so2 equals the 3.0 moles divided by the 3.0 liters which will give us 1 molar so that means that this is going to equal to 1 which means that x is going to equal 0.5 so then once we have our x we can just plug the x back into the rest the rest of the e line this would be 4.0 minus 2 times 0.5 which is just 4.0 minus 1. so it'll give us 3 molar this right here was what's one molar because it because it's two point two x so two times point five and then this right here would be point five molar then now that we have all the cons the concentration of all the reactants products we can plug into kc so kc will be the concentration so2 squared 1 squared times the concentration o2 which is 0.5 divided by the concentration sl3 squared which is 3 squared then we can enter that to the calculator so it'll be 1 1 times 0.5 divided by 3 squared and that'll give us 0.02 five how many sig figs should we have three sig fig two sig fig two sig fix our final answer should have two sig fig it'll be point zero five six and k c and k c k p doesn't have any units so that would just be our final answer so that's it for the first part in the next part we'll be covering more scenarios about how ice tables could be could be formatted if you want to learn how to ace chemistry if you want to learn what's the best way to study for this class if you want to learn some neat tricks and tips to take into your exam and do better on them then you should head over to my website and get this free guide 12 secrets to ace and chemistry you can head over to www.conquerchemistry.comchemsecrets i'm gonna include a link in the description below check it out i think it's really gonna help you and you're gonna you're gonna like it until next time keep working hard and continue the good work