 So welcome to the third screencast of chemical kinetics. We're going to do something a bit more data driven and a bit more process driven than before where we were talking about concepts and Specifically we're going to try and figure out how do we get a rate law? So we know rates and we can define that by dy and dx and so on and then that's proportional to k Times some particular concentrations and these are usually raised to some kind of power of some description doesn't matter what these Vessels are numbers actually are How do we get access to them? How can we work that out? So this is something called the initial rates method. The name isn't really important We just need to know that it is how do we get this when we talk about lab practice and so I mean again, we Think about that whole Johnston's triangle thing. This is really a relationship between Microscopic world and the symbolic world. So we're looking at the abstract maths And we're looking at what data can we get from the lab and this is kind of connecting the two It's clever, isn't it? so the order of reaction if we Can define our rate now this rate zero means that we are Interested in the initial rate that is rate one time is equal to zero This is where we are really interested in so that rate is defined as dA by dt or dB by dt or dC by dt With the positive or negatives representing whether they are products or reactants and that is equal to kind of a generic rate law here, so this concentration again time zero and Two exponents that we don't know yet. We don't know what these are like. I keep saying this can only be derived Experimentally, so this is the experiment we would do for it. So here we keep b constant and a is varied So what we will find is that we get three different rates out Because as the concentration of a increases the rate will increase that's what this rate law says, but how does it increase and by how much? So let me just scribble those out of the way This is what we've done. We've done Three experiments you might want more in reality, but there is the red rate here There's our first experiment where the concentration of a was low There's a yellow one where it's kind of in the middle and then the green one which is really fast So you can say these three arrows here Represent the rate at time equals zero We've sort of got a gradient hole We've plotted down a graph try to get a straight line and we've worked out the rate from that So we get three values. These are just three numbers representing the speed of the reaction defined this way So let's look at what we would do if it was just straight a to b so our rate is equal to k a to the power of something Now we want to take a logarithm of it. So if you're not familiar with logarithms This is kind of what's happening. We take lun something. This is the natural logarithm I would just basically means putting ln behind it It's about exponents and you'll realize that this exponent now comes down to here So it becomes a linear equation. We've taken a log of rate and then we've got this it is a linear equation So if it's a linear equation, it becomes a straight line now physical chemists and particularly when we're in thermodynamics and kinetics We like linear equations if we can linearize an equation We can do a lot more with it and we can figure out a lot of things So there is a recurring theme in kinetics of taking logarithms mostly because Things do react in terms of exponential decays and so on. So if we take a log things work out quite nicely So let's look at our a times b idea We take the logarithm of it our exponents come down to the To the front here and we take log of the rate is equal to log a plus the log a plus log b So that's the other law of logarithms when we multiply things they become addition down here Now our experiment we said b was constant we kept b as constant So what happens is this entire part of the equation becomes constant We also find that this part of the equation because it is a part about the rate constant also becomes constant Okay, so the only variable in this equation that we produced is log a this shouldn't theory be a constant as well Because this is the one we're getting from the rate law So let's rearrange that a little bit and show you what's happening here What we get is a straight line graph y equals mx plus c again physical chemists Love y equals mx plus c when we put that we can put an equation into that form It's beautiful. We can get any kind of result We like out of it and it's really easy to find and what it turns out is why is log of our rates? We did three experiments so we can get three values of that This is a constant which don't worry about it and we also got three concentrations here So if we plot the data, we can work out this and to a degree we can also work out that So m here on gradients is equal to the exponent so Try to synthesize the two things we've done here. We have done an experiment with Three rates that we got out three initial rates at the very beginning at time equals zero and Then we did this with the equation here. So this is what we did in the lab and This is our set of equations. So we try to convert our hypothetical Rate law into y equals mx plus c and we've got data we can outplot So our first rate plots to here our second rate plots to here and our third rate plots to here And that gradient is equal is equal to a are Exponent so if rate is equal to Concentration risk to a certain power the gradient is this And so all the consequences of the fact we've linearized this equation by taking logarithms and then plug some data into it So Let's see what that looks like in terms of Hmm actual data when we put some real numbers into this so Here are three sets of numbers three concentrations that we have and we also picked out three rates So these are three things that we've done. This is an actual experiment for now Okay, what it means and what chemicals we use don't matter This is just a kind of an example, but we'll put your numbers into it anyway and Now if you take logs of all of the numbers involved you get this minus five on three five minus four point one minus three point five and fourteen thirteen and eleven and so on and Now if you type this into a calculator, you might not get these values So you need to make sure that you're taken into account the fact that this is in fact Millimoles and times ten to the minus seven now that's not a hundred percent important You'll find out if if you plot the data anywhere without taking that into account You still get the right answer. You just won't get these negative numbers That's it. It's nice thing about logs, you know, we can kind of ignore orders of magnitude. They sometimes tend to cancel it So let's have a look at the actual data here. Here are our X values Here are our Y values and if we actually plot these we get a straight line. So I did this Excel You can do whatever you like Learn how to use Excel. It's a really powerful tool for this sort of thing and the gradient when we actually worked out is 2.008 to Pretty much close to two it rounds off quite nicely. So within experimental error we can get What is Our rate law So here's some more data and we'll cover this in the lecture exactly what's happening We've changed the concentration of a we've changed the concentration of B And we've got two concentrations of B that are the same So we can use that initial rates method to try and work out the exponent a and then we can plug the rest of the data Into get B. So we'll cover this a bit later So Let's go through this process again Right. The initial rates method is used to find a rate law. That's K a B and we're interested in what are these numbers here? So we vary one and we keep the other one constant and so if we plot the natural logs of This equation we can get a y equals mx plus c graph So our multiple bits of data should form a straight line where the gradient is equal to Whatever exponent is up here. So a bit of a convoluted procedure Well, maybe it's quite a simple procedure to you Other way we will meet up in the lecture and do a bit more on it and try and You know plugs more data into this and try and get used to it