Hi there! I’m Jeremy Krug, and welcome to my review of AP Chemistry Unit 5 – which covers Chemical Kinetics. If you learn something from this video, please hit the like button, and share this video with the rest of your class! It really does help! And if you’re looking for the full review of Kinetics and ALL NINE units, then check out my AP Chemistry Ultimate Review Packet – for LOTS of practice free response, AP-level multiple-choice questions, full-length unit study guides – head over to Ultimate Review Packet dot com. Now, let’s get started with our 10-minute review of Kinetics…. Kinetics is all about how fast or slow a chemical reaction proceeds. Four main factors affect the rate of a reaction: the concentration of the reactants, the particle size of the reactants, the temperature, and the presence of a catalyst. In a chemical reaction, we can determine the relative rates of the reactants and products by looking at the coefficients of the balanced equation. For example, in this reaction, 2 NO + O2 → 2 NO2 , since the coefficients of NO and NO2 are equal, we can say that the rate of disappearance of NO will be equal to the rate of appearance of NO2. On the other hand, since NO has a coefficient twice as large as that of O2, we can say the rate of disappearance of O2 will be only half that of NO. When we calculate reaction rate, we normally use units of concentration, per unit time, so moles per liter per second would be a fairly common unit to use. A rate law is an equation that relates the concentrations of the reactants to the initial rate of reaction. For this reaction, we’d write the rate law in the format Rate equals k, which is the rate constant, times the concentration of ClO2 raised to a power, we’ll call it x for right now, times the concentration of hydroxide, raised to a power, which we’ll call y for right now. This is the basic format for every rate law. For determine the powers, or orders, of the reaction for each reactant, we need experimental data, like we have here. So for ClO2, we find two experiments where ClO2 is the only reactant changing, so let’s use experiments 1 and 2. Notice that as the concentration of ClO2 triples, the rate increases by a factor of nine. Since 3 to the second power is 9, the order for ClO2 is 2. For hydroxide, let’s use experiments 1 and 3. As hydroxide triples, initial rate also triples. Since 3 to the 1st power is 3, the order of hydroxide is 1. So the rate law looks like this. The overall order is just the sum of the individual orders of the reactants, so it’s overall 3rd order in this case. We can calculate the rate constant very easily by taking the data for any one of the three experiments and plugging it into the rate law and solving for k, using correct units. We can also determine the order with respect to a reactant graphically. We do that by monitoring the concentration of the reactant as it decreases over time. Then make three graphs. One graph will be plotting time on the x axis and the concentration of the reactant on the y axis. Another graph will plot time on the x axis, and the natural logarithm of concentration on the y axis. The last graph will plot time on the x, and the reciprocal of concentration on the y axis. If the concentration versus time graph is a straight line, it will be zero order. If the natural log versus time graph is a straight line, it’s first order. And if the reciprocal of concentration versus time graph is straight, then it’s second order. For any straight line graph, the absolute value of the slope is equal to the rate constant. We use integrated rate laws to relate the rate constant, the initial concentration of a reactant, the time elapsed, and the concentration left after that amount of time. These are the integrated rate laws for 0, 1st, and 2nd order processes. If you know any three of the values, you can solve for the fourth. We can derive the half-life equation for any of these, but in AP Chem, you’ll only be asked about half-life for first-order processes. For first order processes, half-life equals 0.693 divided by the rate constant. Many chemical reactions take place in multiple steps. Let’s take this chemical reaction that takes place in two steps. Each of those steps has its own little rate law. The rate law for the first step, just as an example, is Rate equals k sub one times NO times Cl2. You could do the same thing for the second step. Notice that each of these steps has two molecules colliding. It’s fairly common for one or two molecules to react with each other in an elementary step. In fact, it would be very unusual for any single step in a reaction to have more than two molecules colliding at once. In order for molecules to react with each other, two things have to happen. First, they have to collide with sufficient energy. We call that the activation energy. Secondly, they have to collide in the correct orientation, lined up in the right direction so that old bonds can break and new bonds can form. The Boltzmann distribution here shows that at higher temperatures, a greater fraction of the molecules will have at least the minimum amount of energy to react. We can draw a graph that represents the energy of a reaction as it progresses from reactants to the transition state, then on to the products. Notice some key points on the graph: the distance from the beginning up to the peak is the activation energy of the forward reaction. The distance between energy of reactants and products is the change in enthalpy, which tells us if the reaction is exothermic or endothermic. The Arrhenius equation gives us a way to relate the rate constants for a reaction at different temperatures to the activation energy. If you plot the reciprocal of temperature on the x axis and the natural log of the rate constant on the y axis, you’ll find that the slope is equal to negative activation energy divided by the gas constant. Just multiply the slope by –8.314 Joules per mole Kelvin to determine the activation energy. Back to multistep reactions, notice that the individual steps add up to the overall balanced equation. In this mechanism, we have a temporary molecule created in the early step that gets used up in the later step. That’s called a reaction intermediate. And sometimes, we see molecules present at the beginning of the process and also at the end. It participates in the reaction, but never gets consumed. We call that a catalyst. If we want to build evidence that this is the actual mechanism for the reaction, then we should look for evidence of this IO negative ion popping up somewhere when we do our chemical analysis. Let’s go back to this reaction mechanism. The rate for any reaction is determined by the slowest step in the mechanism. The rate law for that slow step is the rate law for the whole reaction. So in this case, it would be Rate equals k times N2O. When we have a mechanism where the slow step isn’t the first step, this causes a little problem in writing the rate law. The slow step still determines the rate, but notice that if we write that as the rate law, we end up with a reaction intermediate in the rate law, and you can’t have a reaction intermediate in the rate law, because it doesn’t show up in the overall balanced equation. So we have to determine the rate law for the formation of that intermediate, and plug it in where that intermediate showed up in our proposed rate law. So by doing this, we end up with the actual rate law for the reaction. A multistep reaction will have an energy profile with multiple humps, like you see here. The hump that rises the highest is the slowest step, so that’s the first step in this example. The lower peak is the faster step. Since there are two steps, we will have two transition states, and it will be almost impossible to isolate either one. The reaction intermediate will exist in this valley. And once the intermediate is formed, it will be easier for it to form the products than the reactants, since the activation energy for the forward second step is less than the activation energy for going backwards toward the reactants. A catalyst speeds up a reaction without being used up. It does this by lowering the activation energy of the reaction, and often it also increases the number of collisions that lead to a reaction. Most commonly, a catalyst will be present at the beginning of a reaction, and it will participate in an early step, likely get bound into a reaction intermediate, but then that intermediate reacts again and spits out the catalyst molecule again, just like we see in this mechanism. Other catalysts work differently, they’re called surface catalysts, and they allow a reacting molecule to form a bond with the surface of the catalyst. So that’s my quick review of Unit 5 over Kinetics! Don’t forget that the full 30-minute review video for Unit 5 is over at Ultimate Review Packet dot com, along with lots of resources designed to help take you to the head of YOUR class and get that FIVE on the AP Chem exam. I’m Jeremy Krug, Thanks for watching, and don’t forget to join me for my 10 minute review of Unit 6 – Thermochemistry! See you soon!