Learn about Advanced Placement Chemistry, Chemical kinetics 1, in this comprehensive video by bannanaiscool.
Read the full transcript »
Male Speaker: Who wants to state rates of reactions? Well you do of course and look all the scientists all over the world need to know rates of reactions from the industrial guys who are saying, well alright, how much Tylenol is coming down to conveyor belt at any one time being produced, manufactured in the plant. Well, there is a certain rate of reaction that needs to be studied so they can maximize how much they can bottle every day. Rates are important in chemistry to be able to determine and that's part of well how molecules react together their kinetics. So, chemical kinetics deals with the rate of reactions. Now let's take for instance certain equation and let's elaborate on what we are talking about in terms of rates. The decomposition of carbon-dioxide and the carbon monoxide and oxygen goes according to this balanced equation, 2, 2, 1. Okay, let's say we start half of the quantity of CO2 and we have to turn it into CO and O2, and how that's going to look graphically representing. Well, if a graph concentration versus time, the carbon-dioxide is going to decrease in concentration over given time and the carbon-dioxide is going to increase and so was the O2 where the O2 balance in a 1-2 ratio isn't going to increase as much as the CO, correct that makes sense. Okay, now what's the rate of this being lost and these two chemicals being gained? Well, the rate can be established by taking the change in the concentration of the quantity of CO2 over time, how many moles per liter lets say in concentration especially how many moles per liter is it losing per unit time, moles per liter per second that would be your rate. So, just dividing concentration by time gives us the rate. Now here is the -- if this was a straight line, straight down like this, it will be easy to be able to calculate the rate because for the straight line we just take two points on the y and two points from that x axis and then take the difference between the two rise over one is going to be able to give you the slope of the line and that will be unit moles per liter per second or how much CO2 we will be losing but if not a straight line, it's a curve so how do we figure out how many moles per liter per second that this is losing. Well, we can't do that necessarily right away straight off for a curve here, we can do for any particular one point in time, that's possible, you know, if you say -- be able to determine what the instantaneous rate is there. Well, when you draw a line that is tangent to the curve, you remember doing that and then for that line there get the slope of that line, two points on the y, two points on the x, subtract the ys from each other, the xs from each other and you have got then the slope of that line you will get the instantaneous rate at that point. That's how you figure data. Here is what we are interested in chemical kinetics and what I am going to show you and what most university textbooks are showing too. We are going to be interested in the rate of decomposition of the reactant only, we are not going to be worried of the products because actually you know the products can actually reform the reactants again and messes all up to, we don't even want to consider that at the moment. So, it's going to be, the rate is going to be the change in the concentration here of the CO2 over the change in time that gives us our rate but we are going to take this and we are going to turn it into a little bit of a calculus type of thing and be able to write something called the rate law which will explain to us a lot better, look at the better in this here on how the rates of decomposition are going to be actually shown. Now more accurately that last equation we can call the rate of it equal to the change in concentration of CO2 over the change in time but we know that the CO2 because it's a reactant is losing in amounts so its going to actually reduce its concentration. We put a negative in front of this negative quantity because it's b
Copyright © 2005 - 2014 Healthline Networks, Inc. All rights reserved for Healthline.