okay let's make this just a little bit more complicated but still well within your abilities to algebra so if we're solving for equilibrium concentrations and this means from the equilibrium constant and the initial values so the problem is going to say here's the equilibrium constant for this reaction this is how much i'm starting with how much do i have at equilibrium and could be of everything could be of a specific thing depends on the problem so like so many other things i'm going to give you a set of steps this set of six steps are what you should follow in order when you're setting up an equilibrium problem it will also rate as we get to harder things you'll still be able to use these steps it's just the process of going from one step to another could get updated so the steps are fill in the initial conditions in the rice table so i guess step zero is set up a rise table step two calculate q to determine the direction of the reaction three fill in the change row with values of x that match the stoichiometry four complete the equilibrium row five substitute values from the equilibrium row into the equilibrium constant expression and then solve for x and sub the numerical value for x back into the equilibrium row examples these six steps i my recommendation take a screenshot write these on an index card have this next to you when you solve equilibrium problems until you're able to do it without the steps on the card so example one you have been asked to find the concentration of a and b at equilibrium if the initial concentration of a is one molar and the equilibrium constant is 0.33 so rice table my reaction is that a is in equilibrium with 2b and i'll put a 1 in front of the a just to be explicit then if i calculate q to determine the direction well okay my i guess let me say my initial conditions are in here one molar for a and zero for b so calculate q to determine their direction well products over reactants my products are zero so i mean i guess we could set up that zero is less than 0.33 so my reaction has to move to the right but also you could and are allowed to just look at this and say hey there's no product i have to make some so my reaction has to move to the right now the amount by which it moves to the right is the unknown that's why we're putting in x's in the change row now i know what the balance is going to be at the end which is what i use to solve for x but i don't know how much it's going to go down but i do know because my reaction is moving right i'm losing a certain amount of reactant so minus x and i'm making twice as much in product so plus 2x this means that at equilibrium my a concentration is 1 minus x so i just added those together i just wasn't able to combine the terms and then under b 0 plus 2 x so at equilibrium 2x is my b concentration so my equilibrium constant expression is concentration of b squared over concentration of a so what i am doing is taking things written in the equilibrium row and plugging them directly in for concentration of b or concentration of a so for b it's 2x in parentheses squared and then for a it's 1 minus x and 2x quantity squared divided by 1 minus x is equal to 0.33 because that's my equilibrium constant now i have to algebra so the getting things in parentheses super important because i need this to say 4x squared right 2x whole quantity squared is 4x squared divided by 1 minus x is equal to 0.33 which i've put over 1 just to make it easier to cross multiply so 4x squared equals 0.33 times 1 minus x if i distribute that 0.33 and then move everything over to one side i have 4x squared plus 0.33 x minus 0.33 equals zero and i have done it this way so that i can solve it with the quadratic formula now you may not have done quadratic formula in a while i'm not asking you to memorize quadratic formula but you have to be able to use it so right quadratic formula is x is equal to the opposite of b plus or minus the square root of b squared minus 4ac all over 2a where a b and c are the coefficients in this version of the equation that's set equal to zero right a is the 4 b is 0.33 c is minus 0.33 when you do quadratic formula you get two results so my x value is equal to 0.2489 and negative 0.33 when this happens when you get two answers in a case where there can only be one right the reaction can only change by one amount one of the values can be rejected usually it's obvious which one could be rejected right you can't if the change is negative then you did something wrong in the setup because it's telling you that the reaction would move the direction opposite of what you said and you did it right to begin with also right if x is negative 0.33 and the concentration of b is 2x that's a negative concentration for something which which cannot happen it doesn't exist so we reject that value so negative 0.33 gets rejected which means our answer has to be 0.2489 however that's just our answer for x we do not have the final answer to this problem yet the problem asked us to find the equilibrium concentrations so we have to take 2.2489 and plug it back into the equilibrium row where for a it was 1 minus x so 1 minus 0.2489 is 0.7511 molar and for b 2 times 0.2489 is 0.498 so my concentration of b at equilibrium is 0.498 molar with units box around it final answer okay let's do an example that has real chemicals instead of generic chemicals but the same principle we have an equilibrium constant of 0.36 at 100 degrees the initial no2 concentration is one molar and n2o4 is 0.1 molar find the concentration at equilibrium so we have initial values for our reactant and our product i have my rice table set up my reaction is the n2o4 turning into 2no2 my initial values are 0.1 and 204 and 1no2 the sole of the given information all right so figuring out which the way the reaction goes i don't have zero for either reactant or product so i can't just say oh it has to go this way i have to calculate q for this set of conditions to figure out which way it would go so product squared divided by reactant q would be 10 right 1 squared divided by 0.1 is 10. so where i'm at is 10 where i'm going is .36 so 10 is greater than 0.36 which means my reaction is going to move to the left that means in my change row under the reactant my x value will have a plus sign and under the product my x value will have a minus sign i am getting rid of product and making the reactant in order to get the balance to 0.36 so in this looking at my mole ratios i'll make 1 x of n2o4 and lose 2x of no2 so at equilibrium i'm looking at 0.1 x 0.1 plus x and 1 minus 2x so these going into the equilibrium constant expression it's the product concentration squared so 1 minus 2x that whole quantity squared divided by 0.1 plus x and this is equal to 0.36 which i've put over 1 so i can cross multiply i'm also going to expand this thing that is squared so what i'm looking at is 1 minus 2x times 1 minus 2x and that's equal to 0.36 times 0.1 plus x so on the right side i've just distributed that 0.36 on the left side you have to be able to multiply this out so whichever method works for you um a lot of people call it foiling so first outsides insides lasts i can show you how to set up a matrix table that makes this really um like straightforward to layout whatever you need to get it right because right we have to multiply we have to distribute everything through everything so this ends up being 1 minus 2x minus 2x plus 4x squared now you can then combine the minus 2x terms but every single one of those pieces has to come out of this math after rearranging the result is 4x squared minus 4.36 x plus 0.964 equals zero and i've highlighted a b and c here we need to use quadratic formula to solve this when you plug everything in the x values that you get are 0.7817 and 0.3083 so two positive values we still only have one correct answer we need to figure out which one to reject so looking at the equilibrium row with both of them being positive the only way i mean you could plug them both into the 0.1 plus x but then you still just have two positive options it doesn't help you differentiate so let's look at the 1 minus 2x if i use 0.7817 in this calculation i would get a negative value right 1 minus 2 times 0.7817 so 2 times 0.7817 is greater than 1 so i would end up with a negative value now if i did 1 minus 2 times 0.3083 that would give me a positive value so i can reject the 7817 because it would give me negative concentrations so going forward with x equals 0.3083 still not done solving the problem i need to plug this all in and find the equilibrium concentrations that way