hey folks welcome to this episode on equilibrium concentration calculations AKA ice tables and we're going to be looking specifically at perfect squares so when we're looking at our equilibrium calculations we can use something called the ice method and this is stands for initial change in equilibrium uh this is when we're given initial concentrations um and we're looking for concentrations at equilibrium so we can use our K equilibrium expression to figure this out so our I stands for initial concentration or could be pressure for we're talking about partial pressures and it's the initial concentration of all species now remember since this is at the beginning of the reaction our reaction has not started yet we'll have products will be equaling to zero uh over here now our change is going to be happening when we change from initial to equilibrium and that's expressed using either a minus sign or a plus sign so usually when we're going to have reactants uh reactants will usually be consumed so they'll be negative and products are going to be uh formed in this in this change period so they're going to be denoted as a plus and the magnitude of this negative or positive is going to be based on the balanced equation so let's say for example we had 2 NH3 being produced that's going to be product we would have plus 2 x that our variable our variable X will give us some magnitude of change and it's going be two times based on the the coefficient of the balanced equation and finally our e is going to be your concentration or pressure at equilibrium and this is going to be the sum between where where it started uh and the change that occurred and that's what over here so if we started off with zero and we had a change of plus 2X then our concentration of equilibrium would be 0 + 2x so we have three different types of of questions or situations that can be solved using ice tables the first one we're going to focus on right now is perfect squares but then we move on to ones that we create a quadratic equation with and then also ones where we can make a simplifying assumption so how do we do this well to step to create an ice table all we do is we first start off with a balanced chemical equations we need that to show uh the coefficients will be our magnitude change um we set up an ice table ice will be the rows the columns uh will include the species we're going to write our equilibrium expression for our balanced chemical equation uh and then we're going to just solve for whatever we need to what we need to solve for okay we're going to use our initial quantities when re U when calculating the reaction quotient which we spoke about before uh and that tell us whether or not the system that we're looking at right now is at equilibrium or if it has to shift to the left or the right to reach equilibrium so let's look at this example and work through this it says the reaction below was carried out at 700 Kelvin if you start with 1.0 mole of Co and 1.0 Mo of H2O uh in a 5 L container how much of each substance is present at equilibrium if K equilibrium is 0.83 okay so here we're given equation we don't know if it's balanced yet so let's just take a look uh we have one carbon one carbon uh two oxygen one two oxygen two hydrogen okay we're balanced so we're good here now what we want to do is we the question is saying how much how much of each substance so that's unless they were saying what this concentration how much basically either means moles or grams in this case it's we're going to assume it's going to mean moles uh we could also do in in terms of grams but let's just keep it at moles as they didn't specify so we need to figure out how to do this we need to figure out what the concentrations are of each of these products okay all these species sorry all the species at equilibrium and right now all we're do all we're given is we're given 1.0 Mo of Co and 1.0 Mo of H2 so we're given the initial concentrations for our reactants nothing for our products we're assuming that there's going to be zero so let's develop our ice table and go from here to see if we can figure out what this change is going to be so like we said ice goes onto the column so I'm just going to make species first so my co My H2O my H2 and my CO2 um so we have Co Gas and this is concentrations our ice tables are all about concentrations H2O and that's gas uh here I have H2 gas and my concentration CO2 gas okay so these are my concentration now the problem that you might be able to see right away is that we're not given concentrations in the question we're actually just given moles so we need to figure out what the concentrations of Co and H2 o o r first uh the way that we do that again is using our equation C isal to n / V our our mole for Co is 1.0 mole and our volume is 5.0 L and so what we get here is our initial concentration is going to be equal to 0.20 molar okay so we're going to work with that in a second let's just keep setting up our ice table so I have i c and e okay so our initial our change and our equilibrium values so once again let's go back uh our concent our initial concentration for Co our initial amount of moles that we had for Co was 1.0 mole divided by our 5 L container we got 0.02 uh 0.20 M excuse me 0.20 mole per liter uh and the same thing goes with the water we we have 1 mole here divid 5 l so this is going to be 0.20 mol now our product since this is the beginning of the reaction they're going to be zero and zero uh we don't have anything produced now now what is going to be our change now our change is some variable uh and it's respective to the mag to the coefficient that they have in the balanced equation now this is easy because we have all of our coefficients are 11 one now the change that's going to be happening is going to be X we're going to call that X and because this this is a reactant this change will be or this will be consumed and we show that using a negative now the same thing for water this has a coefficient of one uh it's being consumed so it will be X our products on the other hand are going to be produced so they are also going to change to some some constant here our X but they're going to be produced so it's going to be plus X and plus X as well now just if we had a coefficient for example two over here we'd also we'd put -2 2 x uh there so our equilibrium is just a summation of our initial and our change so in this case this is going to be 0.20 - x same this 0.20 - x uh for this it's 0 plus X we don't have to write the zero so all we get is plus X and plus X so that's how we set up our ice table let's go see what we do next so this is our ice table here just kind of re Rewritten um what we want to do now is find find out the value for x now we're going to start off to find the value of x by uh creating our K equilibrium uh expression so K equilibrium uh is equal to the concentrations of our products over our reactants and again only in gas or aquous phase now all of them are in gas phase over here so we can all write them down include them in our K equilibrium I'm going to drop the uh the states just to make it a little bit uh easier uh but usually we're going to be including our states so our reactants uh or products over rea reactants we have H2 as our product concentration of CO2 over uh concentration of Co and the concentration of H2O okay so now we're going to sub in our values now remember also if we look at the back of the the question it gives us our our value for K equilibrium 0.83 so we're also going to include that in here all right so 0.83 is going to be equal to now the value for H2 is X the value for CO2 is X so it's going to be x * x uh over now here we have 0.2 - x and 0.2 - x so that's what we're going to plug in there 0.20 - x * 0.20 - x okay now what we want to do is we want to obviously solve for x now one thing that we notice right away is that here we have x * X and we have the same the same uh um sentences or the same phrases or the same operations happening twice so how can we simplify this how can we take the same operation this same operation simplifi well we can square it 0.83 is = to x^2 over 0.20 - x^ 2 okay so kind of we if we were given this this here we could kind of break it apart into this so we're doing the opposite we're condensing it and saying these squares now once we have these squares this makes life so much more more easy because since both the top term and the bottom term are squared we can square root it and get it just to X so I can square root this side and what I do to this side I also have to do to this side over here so if we square root both sides what we end up getting is we get uh 0 write out here 0.911 043 is equal to X over 0.20 - x all right now make it's so much easier we don't have to create a quadratic equation for this because we don't have any squared X's so to isolate for X we're going to bring this up to this side over here multiply by both sides um and then we're going to bring the X all the X's onto one side and solve for x so the way that we get that we're going to get uh 0.911 043 * 0.20 - x and that's going to be equal to x uh in this case that if we foil that out we get and I'm just going to move it to the the other side uh right at the get-go so I'll get uh 1. 191104 3x is going to be equal to 01829 all right so again what I did there is I just foiled this in that multiplied this and multiplied by that then I took the term with the X and brought it over to the same side as the X and I got that I added them up and this is where I get here so now to isolate for X I divide both sides by 1 911 043 and X is going to be equal to 0.953 4 so there's our value for x Now using this we're going to have to figure out what our eventual equilibrium uh concentrations are going to be so now that we know that x x is equal to 0.09 534 um and this is a change this is also going to be in mole per liter uh we can figure out what our our concentrations of our reactants and our products are at equilibrium how do we know this well we have our equations that include x so for our our reactants we have 0.20 - x for both and then for these we have plus X now one thing is remember that it's asking us for how much of the substance So eventually what we're going to have to do remember that we said that these are going to be moles uh we're going to have to go from mole per liter into mole mes so let's first figure out what the concentrations of our reactants are going to be at equilibrium and the way we do that is we have 0.20 - x now X is uh the value that we found before 0.09 534 and we get 01046 mole per liter now to get from mole per liter into moles we have to use our c is equal to n / V and rearrange to find n which is C * V so if we get our 01046 mole per liter and we multiply it by our con our our volume which again from the equation is 5 L our liters cancel out and we get uh 0 oops 0.52 33 mole uh to two significant figures we get 0.52 mole so there's our answer for our reactants now both of them and that's for our concentr or our our value for Co and for H2O because both of them have the same equation okay so we just kind of save some time there both of them have this 0.20 - x and so we just used it for the same thing so now let's see what our equilibrium uh uh values are for H2 and CO2 so the concentration of H2 and CO2 CO2 both equal x and so we already know what x is uh X is going to be 0.095 34 mole per liter um but again it's asking us for how much in terms of moles so we're going to have to multiply this by 5.0 L our lers cancel out and we get 0.476 7 mole to two significant figures it's 0.48 mole and that's the amount for uh H2 and CO2 so there we have it uh we figured out at equilibrium that we have 0.48 Mo of both H2 and CO2 and at equilibrium we have 0.52 mole of uh Co and H2O and that makes sense if we go back to our K equilibrium value remember that for this K equilibrium value for it to be uh a decimal what that means is that our numerator has to be bigger than our uh sorry our denominator has to be bigger than our numerator and remember that our denominator is all about our reactants and our numerator is all about products so at the end what we should have is we should have a greater concentration of reactant then we have products for this number to be a decimal and and we do now this is again in moles if we go back to what our our concentrations are uh 0146 versus uh 095 uh we can see that we have a a LGE much larger concentration of reactants than we do of products and that makes sense because that would give us a value of 0.83 so that's how we attack perfect squares again the main thing here is seeing if we have repeating uh repeating operations we saw x * X and we also saw 0.20 - x uh * 0.20 - x and the way we can simplify that is by squaring it and then Square rooting it so if we're able to do that that's the best way that saves us from doing a quadratic equation next episode we are going to be looking to see how do we do something if we just simply cannot get a square root of that and how do we approach it if we have to form a quadratic equation so thanks very much for watching see you next time