This problem seems fairly lengthy and it has lots of parts, but we're going to learn a lot about chemical kinetics by working through this problem. First, let's look at the reaction that's given. It says it's a decomposition reaction so I'm going to write the reaction up here in this space. It's SO2Cl2 decomposing, that means it's breaking apart. And it gives me the products since breaking apart in two. Both gases, and it's balanced as written. We're wanting to figure out the order of this reaction, and many times students will look at a overall reaction and think that they can get the order from that, but you cannot determine an order by looking at a reaction. You have to experimentally determine the reaction, and that is what is taking place here. They monitored this reaction at a certain temperature. Now we're not told what the temperature is. But it's some temperature they're monitoring SO2CL2 as a function of time, meaning as time goes by, they're taking readings of how much SO2CL2 is present and they found that when they plotted the natural log of that concentration versus time, they got a straight line. So along the y-axis is a natural log of SO2 Cl2, and along the x-axis is time in seconds. And they tell me it is a linear plot so it's a straight line. This is a slope intercept form of a line so it's y equals mx plus b. We are seeing right here the slope. It's a negative slope so I want to draw my line in this direction, and the slope of this line is a -3.10. Times 10 to the minus 4 and I'm going to give it units. It's changing Y over change in X to get a slope. Y has no units but X is second so the units of that slope is 1/seconds. So all the information in the first half of this problem is obtained and given in that little graph. Now we can start answering the questions. The first question is what is the order of the reaction? Any time the natural log of concentration versus time gives a linear plot, this tells me it is first order. If we know it's first order, we can write the rate law. Ray is equal to K. Reactant S02CL2 to the 1st power. Question number two says what is the rate constant. Well that is obtained from the slope. The slope of the line is equal to -K. So K will equal 3.10 times 10 to the minus 4. Rate constants are always positive, and that is the value of that rate constant. So that's question two. Now let's move on to question number three. It starts off by saying at this temperature. The reason they say that is because k changes with temperature. So if we want to know something about this reaction and we want to use this k that we have here we better make sure we're running it the same temperature. We're given time information and concentration information and that tells me that I would like to use the integrated law for first order reaction. There would be the natural law that the concentration of SO2 Cl2 at same time divided by concentration of SO2 Cl2 Initially is equal to minus k times time. Now let's plug in what we know. We know, we don't know this that's what we're trying to determine. But we know we're starting with 0.2 molar. The value of k was 3.10 times 10 to the minus 4 1 over seconds. And the time is five minutes, but I'm going to need to convert that five minutes to seconds because the rate constant was in seconds. So sixty seconds is the same as one minute. The right hand side of this equation is equal to a negative zero point 0 930. And that will have no units. And that's equal to the natural log of X over 0.200 molar. Trying to solve for X, so I'm going to take e to both sides. That will cancel that part and have X over 0.200 m is equal to 0.911. If I multiply both sides by 0.200, I will have X equal to 0.180 molar so this is a concentration of SO2 CO2 after five minutes has gone by. Part four, part four we want to know the concentration of the products at five minutes. Now a lot of times students will look at the balanced equation and say well they all have the same coefficients so their concentrations all have to be the same at the end. Well that is not true we know that however much of SO2 reacts that's how much of SO2 and CO2 are produced and 0.180 is not how much reacted it's how much is still there. So let's write the reaction with some room to work under it to explain how you will figure out the amount of the s02 and the sl2 that are produced. Let's think about this reaction at the start. At the start of the reaction we were told that we had 0.200 molar of this. And of course there's not going to be any product. After 5 minutes has gone by, we learned from the previous problem that we have 0.180 molar still there. So right here let's think about how much reactant or formed, depending on whether you're talking about reactants or products. Now, if I start with 0.2 and I finish with 0.18, did I not have to use up .020 molar? I'm going to use up that much, think about that. If I started with 0.2 and finish with 0.18, 0.02 had to of reacted. Now this is how much actually reacted; so we can go up to this balanced equation now and say every time one of these reacts I produced one of these. But one of these didn't react, 0.02 did. So I have to produce 0.02 of this. Same thing for the Cl2, the amount of Cl2 matches the amount of SO2 so it is also 0.020 and those are all in molar. So, if I start with none and I added 0.02 then, after five minutes these are the amounts of the SO2 and the Cl2 that get produced at the end of the reaction. So now we know how much of this remains or is produced once the reaction takes place.
