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## Chemical Kinetics

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**Chemical Kinetics**AP Chem Unit 12**Chemical Kinetics**• Reaction Rates • Rate Laws: An Introduction • Determining the Form of the Rate Law • The Integrated Rate Law • Reaction Mechanisms • A Model for Chemical Kinetics • Catalysis**Characteristics of a Reaction**• Identities of products and reactants • Stoichiometric quantities • Spontaneity • Refers to the inherent tendency for the process to occur. Does not imply anything about speed. Spontaneous does not mean fast. A reaction can be considered spontaneous but take years to occur.**Characteristics of a Reaction**• The area of chemistry that concerns rates is called chemical kinetics. • One of the main goals of chemical kinetics is to understand the steps by which a reaction takes place. • This series of steps is called the reaction mechanism. • Understanding the mechanism allows us to find ways to change or improve the rate of a reaction.**Example**We start with a flask of NO2 gas at 300°C. But NO2 dioxide decomposes to nitric oxide (a source of air pollution) and O2. • 2NO2(g) 2NO(g) + O2(g) • If we were to measure the concentrations of the three gases over time we would see a change in the amount of reactants and products over time.**Example**• The reactant NO2 decreases with time and the concentrations of the products (NO and O2) increase with time. • 2NO2(g) 2NO(g) + O2(g)**Reaction Rate**The speed, or rate, of a process is defined as the change in a given quantity (concentration in Molarity) over a specific period of time. • Reaction rate = [concentration of A at time(t2) – concentration of A at time (t1)]/ (t2 – t1) • (Final – initial)**Reaction Rate**• The square brackets indicate concentration in mol/l • In Kinetics, rate is always defined as positive. Since the concentration of reactants decreases over time, a negative sign is added to the equation.**Example Problem1**• Looking at the NO2 table, calculate the average rate at which NO2 changes over the first 50 seconds of the reaction. • 4.2 x 10-5 mol/ls**Reaction Rates**Reaction rates are often not constant through the course of a reaction. For example, average rates for NO2 are not constant but decreases with time.**Instantaneous Rate**The value of the rate at a particular time can be obtained by computing the slope of a line tangent to the curve at that point in time.**Reaction Rates**• When considering rates of a reaction you must also take into account the coefficients in the balanced equation for the reaction. • The balanced reaction determines the relative rates of consumption of reactants and generation of products.**Reaction Rates**2NO2(g) 2NO(g) + O2(g) • In this example, both the reactant NO2 and the product NO have a coefficient of 2, so NO is produced at the same rate NO2 is consumed.**Reaction Rates**2NO2(g) 2NO(g) + O2(g) • The product O2 has a coefficient of 1, which means it is produced half as fast as NO.**Nature of Reactions**Chemical reactions are reversible. Often times as products are formed, they accumulate and react to form what was the reactant(s). • The previous example: • 2NO2(g)<-> 2NO(g) + O2(g) • As NO and O2 accumulate, they can react to re-form NO2.**Nature of Reactions**Chemical reactions are reversible. Often times as products are formed, they accumulate and react to form what was the reactant(s). • Now the Δ[NO2] depends on the difference in the rates of the forward and reverse reactions. • In this unit we will not take into account the reverse reactions.**Rate Law**If the reverse reaction can be neglected, the reaction rate will depend only on the concentrations of the reactants. • A rate law shows how concentrations relate to the rate of a reaction. • A is a reactant. k is a proportionality constant and n is called the order of the reactant. Both are usually determined by experiment.**Rate Law**Most simple reactions, the rate orders are often positive integers, but they can be 0 or a fraction. • The concentrations of the products do not appear in the rate law because the reaction rate is being studied under conditions where the reverse reaction does not contribute to the overall rate.**Rate Law**Most simple reactions, the orders are often positive integers, but they can be 0 or a fraction. • The value of the exponent n must be determined by experiment; it cannot be written from the balanced equation. • The rate law constant is dependent upon species in a reaction.**Types of Rate Laws**There are two types of rate laws: • The differential rate law (often called simply the rate law) shows how the rate of a reaction depends on concentration. • The integrated rate law shows how the concentrations of species in the reaction depend on time.**Types of Rate Laws**• The differential and integrated rate laws for a given reaction are related and knowing the rate law for a reaction is important because we can usually infer the individual steps involved in a reaction from the specific form of the rate law.**Determining Rate Law Form**The first step in understanding how a given chemical reaction occurs is to determine the form of the rate law. • Reaction rate form is described as orders. • Example: first order, second order, zero order • A first order reaction: concentration of the reactants are reduced by half, the overall rate of the reaction will also be half. • First Order: A direct relationship exists between concentration and rate.**Rate Form and Initial Rates**• One common method for experimentally determining the form of the rate law for a reaction is the method of initial rates. • The initial rate of a reaction is the instantaneous rate determined just after the reaction begins. Before the initial concentrations of reactants have changed significantly.**Method of Initial Rates**NH4+(aq)+ NO2-(aq) → N2(g)+ 2H2O(l) • Rate 1 = 1.35 x10-7mol/ls = k(.100M)n(.0050M)m**Method of Initial Rates**NH4+(aq)+ NO2-(aq) → N2(g)+ 2H2O(l) • n and m can be determined by dividing known rates.**Method of Initial RatesExample Problem 2**NH4+(aq)+ NO2-(aq) → N2(g)+ 2H2O(l) • Example: Find the form of the rate law for each reactant and the overall reaction order: • n and m are both 1 (unrelated), overall reaction order is 2 (n+m=2)**Method of Initial Rates**NH4+(aq)+ NO2-(aq) → N2(g)+ 2H2O(l) • Example: Find the rate constant for this reaction • k = 2.7 x 10-4 L/mols**Practice Problem 1**The reaction between bromate ions and bromide ions in acidic aqueous solution is given by the equation: • BrO3-(aq) + 5Br-(aq) + 6H+(aq) 3Br2(l) + 3H2O(l) • Using the experimental data, determine the orders for all three reactants, the overall reaction order, and the value of the rate constant.**Practice Problem 1**• BrO3-(aq) + 5Br-(aq) + 6H+(aq) 3Br2(l) + 3H2O(l) • n=1, m=1, p=2, overall = 4, k=8.0 L3/mol3s**Integrated Rate Law**The rate laws we have considered so far express the rate as a function of the reactant concentrations. The integrated rate law expresses the reactant concentrations as a function of time.**First Order Reactions**First order rate law: • The above rate law can be put into a different form using calculus (integration) • Integrated rate law for first order: • this equation is of the form y = mx + b • first order slope = -k, ln[A] vs t is a straight line.**First Order Reactions**First order rate law: • Integrated rate law for first order can also be written:**Practice Problem 2**The decomposition of N2O5 in the gas phase was studied at constant temperature: 2N2O5(g) 4NO2(g) + O2(g) The following results were collected: Verify that this is first order for N2O5. Calculate k.**Practice Problem 2**2N2O5(g) 4NO2(g) + O2(g) • k= -slope= 6.93 x 10-3s-1**Practice Problem 3**2N2O5(g) 4NO2(g) + O2(g) Using the data from practice problem 2, calculate [N2O5] at 150 s after the start of the reaction. k= 6.93 x 10-3s-1 • .0353 mol/L**Half-Life of a First Order Rxn**The time required for a reactant to reach half its original concentration is called the half life of a reactant. • t1/2**Half-Life of a First Order Rxn**Using the data from the previous example we can observe half-life.**Half-Life of a First Order Rxn**The general equation for the half-life of a first-order reaction is • The half-life does not depend on concentration**Practice Problem 4**A certain first-order reaction has a half-life of 20.0 minutes. Calculate the rate constant for this reaction and determine how much time is required for this reaction to be 75% complete? • k=3.47 x 10-2min-1or 5.78 x 10-4s-1 • t= 40 min.**Second-Order Rate Laws**• second order rate law: • integrated second order rate law: • A plot of 1/[A] vs t is a straight line, slope =k • half-life of a second order reaction:**Practice Problem 5**Butadiene reacts to form its dimer according to the equation: 2C4H6(g)C8H12(g) The following data was collected: Is this reaction first or second order? What is the rate constant? What is the half-life?**Practice Problem 5**2C4H6(g)C8H12(g)**Practice Problem 5**2C4H6(g)C8H12(g) • second order, k =6.14x10-2L/mols • half life= 1630s**Half life**It is important to recognize the difference between the half-life for a first-order reaction and the half-life for a second-order reaction. • First order half life depends only on k • half life remains constant throughout the rxn. • Second order half life depends on k and the initial concentration • each successive half-life doubles the preceding one.**Half life**• First order • Second order