Module 5 Chemical Kinetics Lesson 3. Rate LAw and its components The Rate Law ● The rate constant (k) conveys important information about the kinetics of a chemical reaction. ● If the rate constant is small, the reaction is likely to proceed slowly. ● If the rate constant is large, the reaction is likely to proceed quickly. ● The value of the rate constant, k, depends on the temperature and describes temperature dependence of the reaction rate. Lesson 3. Rate LAw and its components The Rate Law ● The units for the rate constant, k, depends on the overall order of the reaction and must be chosen to balance the units in the rate law. ❖ ❖ ❖ ❖ The The The The rate has units of mol L-1 s-1. concentration has units of mol L-1. unit of k for a first order reaction is s-1. unit of k for a second order reaction is L mol-1 s-1. Lesson 3. Rate LAw and its components The Rate Law Lesson 3. Rate LAw and its components Determination of the Rate Law The rate law can be determined two ways: 1) Measuring the initial rate of the reaction while adjusting the concentrations of the various reactants. 1) Using a series of graphs to compare data to various possible rate laws. Lesson 3. Rate LAw and its components Determination of the Rate Law ● For a reaction with only one reactant, A, the rate of the reaction is rate = k[A]n. ❖ The common possible orders with respect to A are 0, 1, 2. ● If the concentration of A is doubled experimentally, the rate of the reaction will change in a simple and predictable way: ❖ If n = 0, doubling [A] does not change the reaction rate. ❖ If n = 1, doubling [A] doubles the reaction rate. ❖ If n = 2, doubling [A] quadruples the reaction rate. Lesson 3. Rate LAw and its components Determination of the Rate Law: Rate = k[A]n Zero Order: Rate=k[A]0 = k ● The rate does not depend on the concentration. Whatever you do to the concentration, the rate will not change. First Order: Rate=k[A]1=k[A] ● The rate is directly proportional to the concentration. ● If you double the concentration, you double the rate. ● If you triple the concentration, you triple the rate. ● If you halve the concentration, you halve the rate, and so on. Second Order: Rate=k[A]2 ● The rate is proportional to the square of the concentration. ● If you double the concentration, you multiply the rate by four. ● If you triple the concentration, you multiply the rate by nine. ● If you halve the concentration, you divide the rate by four, and so on. Lesson 3. Rate LAw and its components Sample Problem 5: Consider the following data for the kinetics of this reaction: 2 N2O5(g) → 4 NO2(g) + O2(g) Determine the rate law and rate constant for this reaction at the temperature of these experiments. Lesson 3. Rate LAw and its components Sample Problem 5: Lesson 3. Rate LAw and its components Sample Problem 5: Lesson 3. Rate LAw and its components Determination of the Rate Law: ● For a reaction with two reactants, A and B, the rate of the reaction is rate = k[A]m[B]n. ● To separate the influence of one reactant concentration from the other, one reactant concentration is held constant while changing the other to determine its effect on the rate. ● To determine the order with respect to A and B, at least three experiments must be carried out. Lesson 3. Rate LAw and its components Sample Problem 6: The study of the kinetics of real systems can be complicated. For example, there are several ways by which O3 can be converted to O2. One such reaction is NO2 + O3 → NO3 + O2. Three experiments were run, and the following data obtained. Determine the rate law and rate constant for this reaction. Lesson 3. Rate LAw and its components Sample Problem 6: Lesson 3. Rate LAw and its components Sample Problem 6: Lesson 3. Rate LAw and its components Sample Problem 6: Lesson 3. Rate LAw and its components Sample Problem 7: Ozone in the upper atmosphere is depleted when it reacts with nitrogen oxides. The rates of the reactions of nitrogen oxides with ozone are important factors in deciding how significant these reactions are in the formation of the ozone hole over Antarctica. One such reaction is the combination of nitric oxide, NO, with ozone, O3: NO(g)+O3(g)⟶NO2(g)+O2(g) This reaction has been studied in the laboratory, and the following rate data were determined at 25°C. Determine the rate law and the rate constant for the reaction at 25 °C. Lesson 3. Rate LAw and its components Integrated Rate Law ● Because the concentrations of reactants change over time, the rate law does not let us easily predict the concentrations or rate of a reaction at some later time. ● The integrated rate law, derived from the rate law itself, explicitly determines concentration as a function of time. ● The form of the integrated rate law depends on the order of the reaction. Lesson 3. Rate LAw and its components Integrated Rate Law For a zero-order reaction, the rate law is: The zero-order integrated rate law, If a plot of [A] versus time is linear, the overall order is zero order and the slope equals –k. Lesson 3. Rate LAw and its components Integrated Rate Law • A reaction is zero-order if a concentration versus time is linear. • The slope of the plot is –k. plot of reactant Lesson 3. Rate LAw and its components Integrated Rate Law Lesson 3. Rate LAw and its components Integrated Rate Law Lesson 3. Rate LAw and its components Integrated Rate Law Lesson 3. Rate LAw and its components Integrated Rate Law (Zero- Order Reaction) Lesson 3. Rate LAw and its components Integrated Rate Law (First- Order Reaction) Lesson 3. Rate LAw and its components Integrated Rate Law (Second- Order Reaction) Lesson 3. Rate LAw and its components Sample Problem 8: The photodissociation of ozone by ultraviolet light in the upper atmosphere is a first order reaction with a rate constant of 1.0 x 10-5 s-1 at 10 km above the planet’s surface. Consider a laboratory experiment in which a vessel of ozone is27 exposed to UV radiation at an intensity chosen to mimic the conditions at that altitude. If the initial O3 concentration is 5.0 mM, what will the concentration be after 1.0 day? 27 Lesson 3. Rate LAw and its components Sample Problem 8: 28 28 Lesson 3. Rate LAw and its components Sample Problem 9: Lesson 3. Rate LAw and its components Sample Problem 9: Lesson 3. Rate LAw and its components Sample Problem 9: Lesson 3. Rate LAw and its components Sample Problem 9: Lesson 3. Rate LAw and its components Sample Problem 9: Lesson 3. Rate LAw and its components Lesson 4. Half- Life ● The half-life (t1/2) of a reactant is the time it takes for its concentration to fall to one-half its original value. ● When a reaction has proceeded for one half-life, concentration of the reactant must be [A]t = 0.5[A]o. the Lesson 4. Half- Life Sample Problem 10: 36 Lesson 4. Half- Life Sample Problem 11: 37 Lesson 4. Half- Life Sample Problem 12: 38 Lesson 5. Collision theory Collision theory is a model used to explain the factors that affect the rate of chemical reactions. According to collision theory, chemical reactions occur when reactant molecules collide with enough energy and proper orientation. The energy required for a reaction to occur is called activation energy. If reactant molecules do not have enough energy, they will not be able to overcome the activation energy barrier and the reaction will not proceed. Lesson 5. Collision theory ● Increasing the concentration of reactants or decreasing the size of reactant particles can increase the frequency of collisions and therefore increase the rate of reaction. ● Increasing temperature increases the kinetic energy of molecules and the likelihood of successful collisions. ● Collision theory provides a framework for understanding how different factors affect the rate of chemical reactions and can be used to optimize reaction conditions in the laboratory or industrial settings. Lesson 5. Collision theory Factors Affecting Collision Theory 1) Temperature: As the temperature increases, the kinetic energy of the particles also increases, leading to more frequent collisions and higher collision rates. 1) Concentration: Higher concentration of reactants increases the chances of their collision and hence, the collision rate. 1) Surface Area: Smaller particle size or increased surface area increases the number of collisions per unit time. 1) Catalysts: Catalysts increase the collision rate by lowering the activation energy required for a reaction to occur. Lesson 5. Collision theory Applications: Combustion: In combustion reactions, oxygen molecules collide with hydrocarbon molecules and have enough energy to break their bonds and form carbon dioxide and water vapor. Cooking: When food is cooked, the heat increases the kinetic energy of the food particles, leading to more frequent collisions and the formation of new compounds. Corrosion: Corrosion occurs when metal surfaces are exposed to oxygen and moisture, leading to the formation of new compounds. Lesson 6. catalysis Catalysis is the process by which a catalyst speeds up a chemical reaction without being consumed in the process. Catalysis is a fundamental concept in chemistry that is involved in numerous industrial, biological, and environmental processes. Lesson 6. catalysis Types of Catalysis: Homogeneous Catalysis: Catalyst and reactants are in the same phase (e.g. gas or liquid). Heterogeneous Catalysis: Catalyst and reactants are in different phases (e.g. solid catalyst and gas or liquid reactants). Enzymatic Catalysis: Catalysis performed by enzymes, which are biological catalysts. Lesson 6. catalysis Examples on the Types of Catalysis: Homogeneous Catalysis: Include acid-catalyzed reactions, such as the reaction between ethanol and acetic acid to form ethyl acetate. Heterogeneous Catalysis: Include the use of platinum catalysts in catalytic converters to reduce harmful emissions in vehicles, and the use of zeolites in petroleum refining to break down complex hydrocarbons. Enzymatic Catalysis: Include photosynthesis, and respiration. enzymes involved in digestion, Lesson 6. catalysis How Catalysis Work: ● Catalysts lower the activation energy of a chemical reaction, making it easier for the reaction to occur. ● Catalysts provide an alternative reaction pathway with a lower activation energy, allowing the reaction to proceed at a faster rate. ● Catalysts are not consumed in the reaction repeatedly, making them highly efficient. and can be used Lesson 6. catalysis Examples of Catalysts: ● ● ● ● ● ● Enzymatic detergents: Enzymes such as protease and amylase are used as catalysts in laundry detergents to break down protein and starch stains. Baking soda: Baking soda (sodium bicarbonate) can act as a catalyst in the process of baking by releasing carbon dioxide gas, which helps the dough to rise. Hydrogen peroxide: Hydrogen peroxide can act as a catalyst in the reaction with yeast to produce oxygen gas, which is used in the process of baking. Enzymatic cleaners: Enzymes such as protease, amylase, and lipase are used as catalysts in household cleaners to break down stains and odors from carpets, upholstery, and clothing. Oxygen bleach: Oxygen bleach (sodium percarbonate) can act as a catalyst in the process of stain removal by releasing oxygen gas, which helps to break down the stain. Vinegar: Vinegar (acetic acid) can act as a catalyst in the process of cleaning by reacting with baking soda to produce carbon dioxide gas, which helps to lift dirt and grime. Resources ● ● ● Purdue University Department of Chemistry. (n.d.). Integrated rate laws. LibreTexts. (2021, August 24). 2.5: Reaction Rate. In Physical and Theoretical Chemistry Textbook Maps: Supplemental Modules (Physical and Theoretical Chemistry): Kinetics: Reaction Rates. LibreTexts. Commission on Higher Education. (2019). General Chemistry 2 Teaching Guide for Senior High School. CHED.