Calculating Equilibrium Constants (Part II)

hello and welcome to the chemistry solution this tutorial is called calculating equilibrium constants part two so in part one of calculating equilibrium constants we used our equilibrium constant expression and the known concentrations of our reactants and products at equilibrium to calculate our equilibrium constant we're going to take it one step further in this tutorial today and we are going to look at how to calculate equilibrium constants if we're not given the concentrations of all of our reactants and products at equilibrium and how we can use stoichiometry to do that so for more basic problems you'll want to watch calculating equilibrium constants part one and we'll cover problems where you don't know all of the concentrations in this video here so if we don't know the equilibrium concentrations of everything involved what we're going to start by doing is making a table with known concentrations and typically these are known initial and maybe a known equilibrium concentration if we know both the initial and the equilibrium concentration for at least one species in our chemical reaction we can use this information to calculate the change as our reaction proceeds to equilibrium and then we can use the stoichiometry of our balanced chemical equation to calculate the changes for all of the other species in our chemical reaction and then from that information we can calculate all of our equilibrium concentrations and use these to calculate the equilibrium constant so let's try a problem here we have a closed system that initially contains 5 times 10 to the negative 4th molar h2 and 1 third molar i2 so these are going to be our initial concentrations of our reactants we then allow the system to reach equilibrium and at equilibrium determine the concentration of hi to be 9. 35 times 10 to the negative fourth molar let's calculate our equilibrium constant for this reaction so i know that i'm going to need to write my equilibrium constant expression and i know that to calculate the equilibrium constant i need concentrations of my reactants and products at equilibrium and what we're seeing in this problem is that we have only one concentration at equilibrium that's provided to us that's the concentration of hi so i'm going to start by writing this chemical equation and i'm going to construct what i'm going to call an ice chart so i stands for initial change and equilibrium concentrations and then i'm going to fill in what i know that initially i have five times ten to the negative fourth molar h2 and i have one times 10 to the negative third molar i2 these are the only two species present initially so i know that my initial concentration of hi is zero now i don't know how these concentrations change yet but i do know that at equilibrium my concentration of h i is 9. 35 times 10 to the negative fourth molar and now what i'm going to do is i'm going to look at the stoichiometry of my equation and use that to help me calculate the change in these concentrations so i know that h2 is going to decrease by some amount and i'm gonna call that amount x now based on the coefficients for my balanced chemical equation i know that for every one h2 i'm going to need one i2 so i'm going to also say that however much the concentration of h2 decreases by the concentration of i2 is gonna decrease by that same amount but now looking over here on the product side for every one h2 that reacts with one i2 i get 2 hi so if the concentration of h decreases by some amount x i know that the concentration of h i is going to increase by twice that amount so i'm just looking at the coefficients in my balanced chemical equation for every 1 h2 i get 2i2 so i'm using stoichiometry here

well now look at what this tells us i know that the concentration of h i increased from zero to nine point three five times ten to the negative fourth molar and i know that change is equal to two x so what i can do now is i know that 2x is equal to this change in concentration 9 0. 35 times 10 to the negative fourth molar and so what we can do is we can solve for x and x is going to be equal then to four point six seven five times ten to the negative fourth molar and i'm not going to do any rounding yet so now that i know what x is i know how much my hydrogen concentration decreased by and iodine concentration decreased by so i'm going to plug x in here and so my concentration of h2 at equilibrium is going to be equal to 5 times 10 to the negative 4th molar minus 4. 675 leaving me with 3. 25 times 10 to the negative 5th molar h2 at equilibrium i know that my concentration of iodine decreased by the same amount because these are going to react in a one to one ratio based on my balanced chemical equation and so i can take 1 times 10 to the negative third molar minus 4. 675 times 10 to the negative 4th molar and my concentration of i2 at equilibrium is five point three two five we'll save our rounding for the end times ten to the negative fourth molar well now i have all of my concentrations at equilibrium and so if i know my equilibrium concentrations i can plug those into my equilibrium constant expression and solve for my equilibrium constant so i know that the value of kc is going to be equal to the concentration of h i squared and if you're unsure of how to write equilibrium constant expressions there's another tutorial for that called writing equilibrium constant expressions but it's equal to the concentration of our products raised to their respective coefficients from our balanced chemical equation divided by the concentration of our reactants raised to the power of their respective coefficients and so our equilibrium constant is going to be equal to 9. 35 times 10 to the negative fourth molar squared divided by the concentration of h2 3. 25 times 10 to the negative fifth molar raised to the first power and the concentration of i2 about 5. 325 times 10 to the negative fourth molar again we report our equilibrium constant values without units and so we get an equilibrium constant for this reaction that's equal to about 50. 5 okay i thought it might be helpful to practice one more of these problems so this would be a great time to pause the video and try this on your own and then come check your work with me so i have two moles of n2o4 added to a one liter flask this is going to be my initial concentration so i'm just going to rewrite this chemical equation here and construct an ice chart initial change and equilibrium concentration so i have two moles in a one liter flask that's an initial concentration of n2o4 that's equal to two molar i don't have any no2 to start with now i know that as this reaction proceeds to equilibrium that my concentration of my reactant is going to decrease by some amount and i am going to call that amount x using the stoichiometry of my chemical reaction i know that for every 1 and 204

i get 2 no2 so if n2o4 decreases by some amount x i know that the amount of no2 is going to increase by twice that amount so it's 2x now you can define x with respect to any reactant or product that you'd like to i think it's always helpful to define x with respect to a species that has a coefficient of 1 in my chemical equation and then relate everything else in terms of multiples of x so i like to say that n2o4 is going to decrease by some amount x and then based on the coefficient 2 here by my product no2 i know that for every 1 n2o4 i'm going to get 2no2 so i know that my concentration of no2 is going to increase by double the concentration that n2o4 decreased by okay the other very important piece of information here is that at equilibrium we have 0. 734 moles of no2 so i'm going to put that in as my equilibrium concentration 0. 734 moles in 1 liter gives us a concentration of 0. 734 molar so i know the concentration of no2 increased from 0 to 0. 734 molar and in my ice chart i've defined that as increasing by 2x so let's solve for x then so 2x is equal to 0. 734 molar we'll solve for x that makes x equal to about 0. 367 and so i know now that i started with 2 molar n2o4 and it decreased by this amount x by 0. 367 molar so at equilibrium i have a concentration of n2o4 that's equal to 1. 633 molar and we'll save our rounding for the very end so now that i know these equilibrium concentrations i can use them in my equilibrium constant expression to solve for my equilibrium constant so my equilibrium constant is going to be equal to the concentration of my products no2 raised to the power of their respective coefficients so no2 squared divided by the concentration of n2o4 again these are concentrations at equilibrium so the concentration of no2 at equilibrium is 0. 734 molar squared divided by the concentration of n2o4 at equilibrium 1. 633 molar again we report our equilibrium constants without units and so this gives us an equilibrium constant that's equal to about three okay i hope that was helpful check back for more videos on equilibrium and we'll consider le chatelier's principle the reaction quotient and how we can also use ice charts to solve for concentrations of species at equilibrium if we're given the equilibrium constant