We are ready to look at our second learning objective in our chemical kinetics unit. We are going to see that not all substances appear and disappear at the same rate. In our previous learning objective we didn't have any coefficients other then 1. But coefficients will start we didn't have any coefficients other then 1. But coefficients will start playing a role, and we will see how they play a role. We are going to learn how to write, what is called a rate expression. We are going to use it to compare the rate of appearance or disappearance of one substance. in a reaction to the rate of appearance or disappearance of another substance. To demonstrate how coefficients play a role, we are going to do a demonstration. We are going to end the power point and you will see me making little booklets and then we will resume the power point and see how the numbers play out and how the coefficients come into play. We are going to simulate a chemical reaction. The reaction we are going to simulate is 2 A going to B. This is my A, this is my reactants. I have 18 sheets of paper here. I am going to take this, and take two of them and produce a booklet. It will be a little 4 page booklet, and that will be B. We are going to see that we can either monitor the reactants, to determine the rate of the reaction. Or we can monitor the product and we are going to see how the coefficients come into play as we do that. So what we are going to look at now is how we can monitor the reactant, which was our sheets of paper and our product which is the booklet, and determine the rate of reaction. What we need to look at is how much we have now. We had started with 18 sheets What we need to look at is how much we have now. We had started with 18 sheets then as we count through these I now have 10 remaining. If I look at the booklets I started with no booklets, and I now have four that are there. So we are going to see how you can either monitor the reactants or monitor products and determine the rate of the reaction. So you saw me do the demonstration in which I took two pieces of paper and made one booklet. So this reaction might and made one booklet. So this reaction might be used to represent that process. Two pieces of paper [we will call that A] turning into a product of B [the booklet]. Now we recorded this process for 30 seconds. Initially, I had 18 piece of paper. I counted when it was all said and done and there was still 10 piece of paper left over for the booklets, I didn't have any booklets to begin with but I produced 4 booklets. So if we take those numbers and we ponder the change So if we take those numbers and we ponder the change so we could talk about the change in A and we could talk about the change in B we can put a little delta in here for the change in B as time goes by. If you think about those numbers and I will let you do that before we switch slides you will be able to make a comparison of the rate of the disappearance of A, verse the rate of the appearance of B. Now for the demonstration did B appear at twice the rate that A is disappearing? Or did A disappear at twice the rate that B is appearing? Or did they disappear and appear at the same rate. So take those numbers and I will go back to that slide and think about the value of the chance in concentration. And the rate in which is appears and disappears. Did you say that A disappears at twice the rate that B is appearing? If you did, then that would be correct. Now we can use this right here to compare A to B. The minus sign is there because the change in A with respect to time was negative. It was the change in A with respect to time was negative. It was 10 - 18 here. That would be a negative 8 [-8]. We put a minus sign in there to make it a positive 8. That would be a negative 8 [-8]. We put a minus sign in there to make it a positive 8. We had over here 4 minus 0, it is already positive so that is 4. So we do not need a minus sign. Since this was 8 and this was four we are going to have to divide by 2 and that was right here in order to make these two portions here equal to each other. That is how we define the rate of the reaction. We will include that coefficient of 2 down here in the denominator to make define the rate of the reaction. So when I use of 2 down here in the denominator to make define the rate of the reaction. So when I use the word rate here, I am talking about about the rate of the reaction. We can monitor in terms of A we can monitor in terms of B. For this reaction as we look at it, lets put the full numbers in here. We have a negative 1/2 change in concentration in A, that was. We have a negative 1/2 change in concentration in A, that was. The value of 10-18 or negative 8. and that was divided by the 30 seconds. Over here we had 4 - 0, again that is divided by 30 seconds and whether you monitor it in terms of A or monitor it in terms of B you are going to get the same number. That is the rate of the reaction. Now I have a numerator here, that I have written down as one because it is not molarity, it is the number of pieces that I have, but the values are going to be exactly the same because I incorporated the coefficient of 2 into the expression. Now we can look as any generic reaction. No matter what the coefficients are and we can define the rate of the reaction with what we call a rate expression. This is the rate expression. So it a change in concentration of the reactants. You put the minus sign in for reactants. We include the coefficients You put the minus sign in for reactants. We include the coefficients all along the way. So we could define the rate of the reaction all along the way. So we could define the rate of the reaction in term of the reactants. or in terms of the products, it does not matter. It will always give us the same rate for the reaction, and we call this a rate expression. It will always give us the same rate for the reaction, and we call this a rate expression. If a problem says, write the rate expression for a reaction, this is what you are going to be writing. Now we need to look and see how we use this. Here I have a reaction in which there is all sorts of different coefficients. Ones and Fives and sixes and threes. in which there is all sorts of different coefficients. Ones and Fives and sixes and threes. Somebody goes in, and they monitor the rate at which BrO3 minus is disappearing and that is what this is saying, the change in concentration which respect to time of BrO3 how fast is it disappearing? It is disappearing at a rate of 1.5 x 10^-2 molarity per second. And this is at some instance in time, because remember the rate change is not the same through out the course of the reaction. And this problem, wants us to say it is disappearing at this rate, at what rate is the Br- disappearing. Well we can come up with a rate expression the Br- disappearing. Well we can come up with a rate expression and work this problem. Now the rate, in terms of the BrO3 is the change of concentration of BrO3- with respect to time put a minus sign in there, and that would define the rate of the reaction. Or we could do it in terms of the Br-. It would be the change in concentration of Br- with respect to time. Put a minus sign in there so it is positive and also put a 5 in there so they can be equal. Now they gave me this value, I am going to put a box around it. It is 1.5 x 10 ^ -2 molarity per second. There asking me for this portion. And I am going to put a circle around it. They are asking me for that portion. So how do I solve for that portion? I will have to take, and I am going to write the whole thing again. and multiply both sides by 5. If I multiply both side by 5. The five cancel and I will have solved for portion I have put a circle around. I will have solved for a negative Br- over delta t and that would be 7.5 x 10 ^ 2 Lets use some language. Instead of saying the value of and they give you this. They might use the words the rate of disappearance of BrO3- is 1.5 x 10 ^ -2. So let me say that in words and write it down. The rate of disappearance of BrO3- is that is a way of writing in expressions with symbols, but the words would be the rate of disappearance. If they said instead of these symbols here they might say, 'What is the rate of disappearance of Br-? That would be how you use words for that expression. OK we are going to work this problem. It is based on the same reaction as the previous problem. Wording is a little different, we are solving for a component of the reaction. Lets look at how it is written here. There are actually using the words now Br- disappears at a certain rate. There are actually using the words now Br- disappears at a certain rate. At what rate is Br2 appearing? Let turn those words into symbols. The rate of disappearance Let turn those words into symbols. The rate of disappearance of Br- would be the change on concentration of Br- with respect to time. Now that would be a negative value, because it is disappearing. It is telling me, by using the word disappearing, wouldn't give me a positive number, they have include that minus sign. it is 7.5 times 10 ^ -2. They want to know as what rate Br2 is appearing. So symbols for that would be change in concentration So symbols for that would be change in concentration of BR2 with respect to time it is a product, it is appearing it is already positive. I do not need that negative sign to turn it positive. That is what I am trying to find. I will being with a rate expression. Rate is equal to. I am only going to do it for the substances I am interested in, there is no point for everything in there. But to compare those substances to the rate of the reaction and to each to other we have to include the coefficients. So, the negative change in the concentration of Br- divided 5 delta t would equal the change of concentration of Br2 would equal the change of concentration of Br2 and lets include the 3 delta t. If we look carefully at the symbols that are represented right here. Where is that found here in this expression? It is right here. Everything except for the five. So I am going to put that value in 7.5 x 10 ^ -2 and I will include that 5 so that the rate of the reaction is equal to that. I am going to go ahead and put these symbols back in with the 3 and think about where is this portion found, it is right here. I want to solve for that, that i just put a box around. So I am going to multiply both sides by 3. 3 times 7.5 x 10 ^ -2 divided by 5 would give me what I am looking for. The value for that, is 4.5 x 10 ^ -2 molarity per second. That is equal to the change in concentration of Bromine with respect to time so this is how fast it is appearing. So that is how you use a rate expression to compare if you know the rate of change for one substance and compare it to the rate of change for the other substance you can also obtain the rate of the reaction. The rate of the reaction would simply be plugging this portion right here The rate of the reaction would simply be plugging this portion right here divide by 5, and that would give me the rate of the reaction. So that is the end of our learning objective 2. We are using a rate expression which we have learned how to write to compare the rate of the appearance or disappearance of any one substance to the rate of appearance or disappearance of another substance.
