In this unit we will learn how to calculate equilibrium concentraions. Based on having some concentrations values and the value of k we can use this information to calculate the equilibrium concentrations of our substances. In order to find the equilibrium concentrations there are two things we can do and the choice of those depends on what information we have. In order to find the equilibrium concentrations there are two things we can do and the choice of those depends on what information we have. If we know k and all but one of our equilibrium concentrations or pressures we can plug in our known values and solve for that one unknown. Many of the problems we will look at we will know the k value and we will know the initial concentrations because we can measure those when we do experiments but we are going to have to calculate the equilibrium concentrations. We will be using an ICE table that we introduced previously. and by the time you are done with this unit you should probably be dreaming on ICE tables. and by the time you are done with this unit you should probably be dreaming on ICE tables. If not, you probably have not done enough practice problems. Here is an example of the first scenario. We have a balanced chemical equation of ammonia decomposing to nitrogen and hydrogen. We have the value of k notice that is has K_eq, that just stands for equilibrium. We see a variety of subscripts when we look at K values. In this case K_eq is the same as K_c. We don't need to worry about the initial and change rows in our ICE table because we know some of the equilibrium concentrations and we are just solving for the one unknown. So when we read through the problem if the equilibrium concentrations of N_2 and H_2 So when we read through the problem if the equilibrium concentrations of N_2 and H_2 are 0.500 molar if the equilibrium concentrations of N_2 and H_2 are 0.500 molar and 0.100 molar respectively what is the equilibrium concentration of NH_3? Now, given that information and knowing we can set up the k expression we can actually solve for that unknown concentration. Our k expression, the concentration of N_2 times the concentrations of H_2 cubed divided by the concentrations of NH_3 squared comes from our balanced chemical equation. Remember it is products over reactants and the coefficients become powers in the expression. Now we can plug in the values that we know and rearrange to solve for the NH_3. We have NH_3 square equals 0.500 and rearrange to solve for the NH_3. We have NH_3 square equals 0.500 times 0.100 cubed divided by our k value of 3.81. We can solve that to find NH_3 squared is equal to 1.31 times 10 ^ -4. Then the value of NH_3 as 0.0115 Molar. If we were plug this value back into our k expression with our concentrations of N_2 and H_2 we should find the value of k is 3.81. Now lets look at a little bit more complicated situation. Here we are given a balanced chemical equation N_2 plus O_2 yield 2 NO. We are given the k value 4.10 x 10 ^ -4 at 2000 Kelvin. It does not really matter what the temperature is what is important is that the temperature is constant through out the experiment. I look at the information given in the problem. 0.500 moles of N_2 gas is mixed with 0.860 moles of O_2 gas in a 2.00 liter tank at 2000 Kelvin. What are the equilibrium concentrations of each substance? So before I can do this problem before I can even worry about the equilibrium concentrations I need to know how to deal with the initial concentrations. Note, that I am not given the concentration if N_2 and O_2 initially. I am given the moles of N_2 and O_2 and I am given the volume. So I can find the concentration in unit of molarity. So for N_2 I take the 0.500 divided by 2 and get 0.250. For Oxygen take 0.860 moles divided by 2 and get 0.430 molar. I see there is no mention of the NO initially so I can assume that value is 0. Now, I have to look at the change row. Remember that when we deal with the change row we have Now, I have to look at the change row. Remember that when we deal with the change row we have to go back to our balanced chemical equation and look at the stoichiometry. If I have look at the change in nitrogen and treat it as x what I will see is that the change in nitrogen will be -x. If I have look at the change in nitrogen and treat it as x what I will see is that the change in nitrogen will be -x. The change in oxygen will be -x. And the change of NO will be plus +2x. Again, I am having to worry about whether I am gaining or losing because I can only gain on the NO because I have nothing to lose there. Now I have my change row now I can set up my equilibrium row. and it is simply the addition of the initial row and the change row so I get 0.250 - x for nitrogen 0.430 - x for oxygen and 2x for my NO concentration. 0.430 - x for oxygen and 2x for my NO concentration. Now, I can take the k expression and my equilibrium row of values and solve for the value of x and therefore find the concentrations of N_2, O_2, and NO at equilibrium. So that is the next step, we have our data table now I need to set up our k expression. So k is going to equal to NO squared because I have NO as a product. now I need to set up our k expression. So k is going to equal to NO squared because I have NO as a product. 2 is the coefficient of it, so that becomes the power. N_2 is an reactant 2 is the coefficient of it, so that becomes the power. N_2 is an reactant and O_2 is a reactant so they both go into the denominator. They both have coefficients of 1 so I do not have any powers associated with them. Now, I have k equals NO squared over N_2 times O_2 I can plug in the values I know over N_2 times O_2 I can plug in the values I know and I find that I get 4.10 x 10 ^ -4 equals 2x squared because I had 2x in my equilibrium row I still have to remember to square that because 2x represents the concentration. The square is part of the law of mass action. On the bottom, I have 0.250 - x The square is part of the law of mass action. On the bottom, I have 0.250 - x and 0.430 - x. Now if I were to solve this and rearrange to solve time for x what I am going to find, is that I have a quadratic equation. This makes the calculation a little more complicated. what I am going to find, is that I have a quadratic equation. This makes the calculation a little more complicated. However, in some cases we can make a simplifying assumption. That simplifying assumption is if k is small That simplifying assumption is if k is small then we can assume x is much less then the original concentration. So for example, if my concentration is 0.39 and I know the value of x is going to very small then I can basically treat that number 0.39. The value of x is not zero but it is so small that it does not reflect the value once I take significant figures into account. So for example if the value of x was 0.0023 then what I would find is 0.39 - x is equal to 0.3877 and I would still end up with same initial value of 0.39. Note that k must have a small value. After we do the calculation we will also look at how we can test to make sure we made a valid assumption. So why do we simplify, and when can we simplify? Well, it is a small value of x. The reaction favors the reactants. That tells us, that if we start with reactants and we have a small k value very little of the product will actually be formed. and we have a small k value very little of the product will actually be formed. If very little of the product is formed that means the value lost from the reactants side that change row, the values with respect to x. Will be small, and therefore the concentration at equilibrium Will be small, and therefore the concentration at equilibrium will not be that different then the concentration initially. Our check at the end is to check to see if x is less Our check at the end is to check to see if x is less than 5% of the value from which it was subtracted. If we look at the example we looked at on the previous slide we had the value of 0.0023 and we were comparing that to 0.39. We see that this number is much much less then 0.39 we also see that is less then 5% of the value of .39 and so we have an acceptable approximation, we can use the simplifying assumption. If the value is greater then 5% we will have to redo our calculation If the value is greater then 5% we will have to redo our calculation using the quadratic equation. So now we are going to show our law of mass action here. We have our values substituted in that we have seen before. Now we are going show how we use this simplifying assumption. And the first thing I am going to do is say that lets assume that x is much much less the 0.250. Now by default if x much much less then 0.250 it must also be much much less then 0.430. Making this simplifying assumption makes our problem much easier to solve. We can avoid the quadratic equation which greatly simplifies the mathematical calculations needed. Because we have assumed that it is much much smaller we actually can cancel out the minus x. [-x]. We assume that the value 0.250... We assume that the value 0.250 - x is approximately equal to 0.250 and likewise for the 0.430. Now we can rewrite our expression. We still have the same value of k. We still have the value of x on the top because even though it is a small value it will still have a mathematically We still have the value of x on the top because even though it is a small value it will still have a mathematically significance when it is numerator multiplied by 2. On the bottom, we have now simplified to having 0.250 and 0.430. Now, we multiply 4.10 x 10 ^ -4 by both 0.250 and 0.430 and we get 4.41 x 10 ^ -5 by both 0.250 and 0.430 and we get 4.41 x 10 ^ -5 equals 2x notice this is in brackets squared. So everything inside those brackets needs to be squared. notice this is in brackets squared. So everything inside those brackets needs to be squared. I end up with x equals 3.32 x 10 ^ -3. squared. So everything inside those brackets needs to be squared. I end up with x equals 3.32 x 10 ^ -3. So I ended up with this being equal to 4 x squared. I end up with x equals 3.32 x 10 ^ -3. So I ended up with this being equal to 4 x squared. I divided both side by 4 So I ended up with this being equal to 4 x squared. I divided both side by 4 and then I took the square root of the number and got my x value to be 3.32 x 10 ^ -3. We also have to say, is our assumption valid? Was it reasonable to make the approximation that x was much much less then 0.250? We look at the value of x and divide by value from which it was subtracted. So 3.32 x 10 ^ -3 over .250 times 100 and we find that it is less then 5%. It is only 1.33%. So this lets us know that our assumption was valid. We don't need to check the assumption for .430 because if it was less then 5% of .250 this number getting larger, our denominator getting larger. is only going to be a smaller percentage. Now we go back to our ice table we see the values of our equilibrium concentrations 0.250 - x 0.430 - x, and 2x. And we plug in our value of x that we solved for in our previous slide into these expressions and find the equilibrium concentrations of nitrogen, oxygen and NO. So for nitrogen, 0.250 - x is equal to 0.250 - 3.32 x 10 ^ -3 = .247. For Oxygen we have .430 - x and we end up with .427 molar. For Oxygen we have .430 - x and we end up with .427 molar. And for NO we we end up with 2x being equal to 6.54 x 10 ^ -3 molar. Now, we have the equilibrium concentration for each of our substances. We still have subtract x from our concentration of N_2 and O_2 even thought we made the simplifying assumption that they were much We still have subtract x from our concentration of N_2 and O_2 even thought we made the simplifying assumption that they were much smaller then the value, in this case they still changed the value slightly. not a lot, but it does change it, and we need to include it. smaller then the value, in this case they still changed the value slightly. not a lot, but it does change it, and we need to include it. But because it is such a small amount the simplifying approximation made in the previous calculations is valid. There are other examples of equilibrium problems as worked examples that you can review to see some different example to see how these problems are worked. We will not , however, do any problems involving the quadratic equation. On all of our problems, we will be able to use the simplifying assumption. We will not , however, do any problems involving the quadratic equation. On all of our problems, we will be able to use the simplifying assumption. In the next unit we will look at LaSalle principle. This idea helps us determine how a system as equilibrium responds when the equilibrium is disturbed.
