Learn about Advanced Placement Chemistry, Chemical kinetics 3, in this comprehensive video by bannanaiscool.
Read the full transcript »
Male Speaker: So you are given a question you have to determine the rate law for this reaction written down below here. So, it's again carbon-dioxide decomposing the carbon monoxide and oxygen. So, oh, I know how to write the rate law, you write rate equals k times the concentration of the CO2 and all we have to do is find that power of order right, the power to which the carbon-dioxide is taking. Yeah, alright so let's look at the data so if you can figure it out. Well, wait a minute, hang out a second. This data isn't the same the before, it doesn't have an initial rate in concentrations, its got time given for a reaction and the various concentrations that CO2 has, this is graph for CO2, I charted for CO2. The concentration CO2 has as it decreases over a given amount of time oh men, now actually that is data that looks a little bit more doable in our labs, it's very difficult to get initial rates but it's easy to time a reaction and measure the concentrations -- for more scientists. Well, how you are going to approach this type of question then, how do you do it because it doesn't fall within the same guidelines as the initial rate method. This requires integrated calculus, that's called the integrated rate law method, now I am going to show you how to determine the order for that carbon-dioxide in this reaction with this data, here is the first step coming up. First step is to take the concentration of the CO2 which is given in the chart that I gave you before that's versus time and find for these concentrations two other types of data, here is what you are going to look for, you have to take the concentration of your reactant and then take the natural log of that concentration, natural log is a button ln on your calculator and what you do is just type that ln and then 0.1 to get negative 2.30. The natural log is just when you know how the number 100 is ten square. So, if you take the log of ten square, you will actually the number 2. Logs are done to the base 10, but natural logs are done to the base 2.1783. So, it sounds kind of 2.7183 and do not sounds kind of crazy but the thing is if you take this number 0.1 and then you want to find the exponent for when you would express this number in the base 2.7183 you are going to get yourself negative 2.30, okay. So, you find the natural log of this number but you also take this concentration and go one over that concentration and record that data and now you have got a new chart, the concentration and time as well and you got a natural log of the concentration one over the concentration this is what you are going to do with this data. If you take the time and put it on the x-axis of a graph and you plug the concentration of the CO2 versus time, the natural log of the concentration of the CO2 versus time or one over the concentration of the CO2 versus time, we get three different graphs. The one that gives you a straight line or base the straight line equation y equals m x plus b if you take our data that we collected and you turn that all of that data into the form y equals m x plus b we get three equations here, y equals m x plus b. Now if you take the natural log of concentration of the CO2 and you graph it versus time and you get a straight line. That means that the data we collected obeys this equation and that will tell you you have got first order in terms of the power to which that reactant CO2 is taking in the rate law, so you determined what that order is. If you don't get a straight line for this, if you want to get a straight line for this one then that means then that when you graph one over the concentration of the CO2 versus time and a straight line is give you have a second order but if you just graph the concentration versus time to get a straight line you have got yourself zero order. So, we take all that data from the chart and we plug it in to our calculators that are real quick if you have a Texas instrument type of calculator the first thing you do is yo
Copyright © 2005 - 2015 Healthline Networks, Inc. All rights reserved for Healthline.