Reaction Rate Change in concentration of a reactant or product per unit time. [A] means concentration of A in mol/L; A is the reactant or product being considered. Copyright © Cengage Learning. All rights reserved

The Decomposition of Nitrogen Dioxide Copyright © Cengage Learning. All rights reserved

The Decomposition of Nitrogen Dioxide Copyright © Cengage Learning. All rights reserved

Instantaneous Rate 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. Copyright © Cengage Learning. All rights reserved

Rate Law Shows how the rate depends on the concentrations of reactants. For the decomposition of nitrogen dioxide: 2NO2(g) → 2NO(g) + O2(g) Rate = k[NO2]n: k = rate constant n = order of the reactant Copyright © Cengage Learning. All rights reserved

Rate Law Rate = k[NO2]n 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. Copyright © Cengage Learning. All rights reserved

Rate Law Rate = k[NO2]n The value of the exponent n must be determined by experiment; it cannot be written from the balanced equation. Copyright © Cengage Learning. All rights reserved

Types of Rate Laws Differential Rate Law (rate law) – shows how the rate of a reaction depends on concentrations. Integrated Rate Law – shows how the concentrations of species in the reaction depend on time. Copyright © Cengage Learning. All rights reserved

Rate Laws: A Summary Because we typically consider reactions only under conditions where the reverse reaction is unimportant, our rate laws will involve only concentrations of reactants. Because the differential and integrated rate laws for a given reaction are related in a well–defined way, the experimental determination of either of the rate laws is sufficient. Copyright © Cengage Learning. All rights reserved

Rate Laws: A Summary Experimental convenience usually dictates which type of rate law is determined experimentally. Knowing the rate law for a reaction is important mainly because we can usually infer the individual steps involved in the reaction from the specific form of the rate law. Copyright © Cengage Learning. All rights reserved

Determine experimentally the power to which each reactant concentration must be raised in the rate law. Copyright © Cengage Learning. All rights reserved

Method of Initial Rates The value of the initial rate is determined for each experiment at the same value of t as close to t = 0 as possible. Several experiments are carried out using different initial concentrations of each of the reactants, and the initial rate is determined for each run. The results are then compared to see how the initial rate depends on the initial concentrations of each of the reactants. Copyright © Cengage Learning. All rights reserved

Overall Reaction Order The sum of the exponents in the reaction rate equation. Rate = k[A]n[B]m Overall reaction order = n + m k = rate constant [A] = concentration of reactant A [B] = concentration of reactant B Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! How do exponents (orders) in rate laws compare to coefficients in balanced equations? Why? The exponents do not have any relation to the coefficients (necessarily). The coefficients tell us the mole ratio of the overall reaction. They give us no clue to how the reaction works (its mechanism). Copyright © Cengage Learning. All rights reserved

First-Order Rate = k[A] Integrated: ln[A] = –kt + ln[A]o [A] = concentration of A at time t k = rate constant t = time [A]o = initial concentration of A Copyright © Cengage Learning. All rights reserved

Plot of ln[N2O5] vs Time Copyright © Cengage Learning. All rights reserved

k = rate constant First-Order Time required for a reactant to reach half its original concentration Half–Life: k = rate constant Half–life does not depend on the concentration of reactants. Copyright © Cengage Learning. All rights reserved

A first order reaction is 35% complete at the end of 55 minutes A first order reaction is 35% complete at the end of 55 minutes. What is the value of k? k = 7.8 × 10–3 min–1 ln(0.65) = –k(55) + ln(1) k = 7.8 x 10-3 min-1. If students use [A] = 35 in the integrated rate law (instead of 65), they will get k = 1.9 x 10-2 min-1. Note: Use the red box animation to assist in explaining how to solve the problem. Copyright © Cengage Learning. All rights reserved

Second-Order Rate = k[A]2 Integrated: [A] = concentration of A at time t k = rate constant t = time [A]o = initial concentration of A Copyright © Cengage Learning. All rights reserved

Plot of ln[C4H6] vs Time and Plot of 1/[C4H6] vs Time

Second-Order Half–Life: [A]o = initial concentration of A k = rate constant [A]o = initial concentration of A Half–life gets longer as the reaction progresses and the concentration of reactants decrease. Each successive half–life is double the preceding one. Copyright © Cengage Learning. All rights reserved

EXERCISE! For a reaction aA  Products, [A]0 = 5.0 M, and the first two half-lives are 25 and 50 minutes, respectively. Write the rate law for this reaction. rate = k[A]2 b) Calculate k. k = 8.0 × 10-3 M–1min–1 Calculate [A] at t = 525 minutes. [A] = 0.23 M a) rate = k[A]2 We know this is second order because the second half–life is double the preceding one. b) k = 8.0 x 10-3 M–1min–1 25 min = 1 / k(5.0 M) c) [A] = 0.23 M (1 / [A]) = (8.0 x 10-3 M–1min–1)(525 min) + (1 / 5.0 M) Copyright © Cengage Learning. All rights reserved

Zero-Order Rate = k[A]0 = k Integrated: [A] = –kt + [A]o [A] = concentration of A at time t k = rate constant t = time [A]o = initial concentration of A Copyright © Cengage Learning. All rights reserved

Plot of [A] vs Time Copyright © Cengage Learning. All rights reserved

Zero-Order Half–Life: k = rate constant [A]o = initial concentration of A Half–life gets shorter as the reaction progresses and the concentration of reactants decrease. Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! How can you tell the difference among 0th, 1st, and 2nd order rate laws from their graphs? For the zero-order reaction, the graph of concentration versus time is a straight line with a negative slope. For a first-order graph, the graph is a natural log function. The second-order graph looks similar to the first-order, but with a greater initial slope. Students should be able to write a conceptual explanation of how the half-life is dependent on concentration (or in the case of first-order reactions, not dependent). Copyright © Cengage Learning. All rights reserved

Rate Laws To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved

Summary of the Rate Laws Copyright © Cengage Learning. All rights reserved

Zero order First order Second order 4.7 M 3.7 M 2.0 M EXERCISE! Consider the reaction aA  Products. [A]0 = 5.0 M and k = 1.0 × 10–2 (assume the units are appropriate for each case). Calculate [A] after 30.0 seconds have passed, assuming the reaction is: Zero order First order Second order 4.7 M 3.7 M 2.0 M a) 4.7 M [A] = –(1.0×10–2)(30.0) + 5.0 b) 3.7 M ln[A] = –(1.0×10–2)(30.0) + ln(5.0) c) 2.0 M (1 / [A]) = (1.0×10–2)(30.0) + (1 / 5.0) Copyright © Cengage Learning. All rights reserved

Reaction Mechanism Most chemical reactions occur by a series of elementary steps. An intermediate is formed in one step and used up in a subsequent step and thus is never seen as a product in the overall balanced reaction. Copyright © Cengage Learning. All rights reserved

NO2(g) + CO(g) → NO(g) + CO2(g) A Molecular Representation of the Elementary Steps in the Reaction of NO2 and CO NO2(g) + CO(g) → NO(g) + CO2(g) Copyright © Cengage Learning. All rights reserved

Elementary Steps (Molecularity) Unimolecular – reaction involving one molecule; first order. Bimolecular – reaction involving the collision of two species; second order. Termolecular – reaction involving the collision of three species; third order. Very rare. Copyright © Cengage Learning. All rights reserved

Rate-Determining Step A reaction is only as fast as its slowest step. The rate-determining step (slowest step) determines the rate law and the molecularity of the overall reaction. Copyright © Cengage Learning. All rights reserved

Reaction Mechanism Requirements The sum of the elementary steps must give the overall balanced equation for the reaction. The mechanism must agree with the experimentally determined rate law. Copyright © Cengage Learning. All rights reserved

Decomposition of N2O5 To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved

Decomposition of N2O5 2N2O5(g)  4NO2(g) + O2(g) Step 1: N2O5 NO2 + NO3 (fast) Step 2: NO2 + NO3 → NO + O2 + NO2 (slow) Step 3: NO3 + NO → 2NO2 (fast) 2( ) Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! The reaction A + 2B  C has the following proposed mechanism: A + B D (fast equilibrium) D + B  C (slow) Write the rate law for this mechanism. rate = k[A][B]2 rate = k[A][B]2 The sum of the elementary steps give the overall balanced equation for the reaction. Copyright © Cengage Learning. All rights reserved

Collision Model Molecules must collide to react. Main Factors: Activation energy, Ea Temperature Molecular orientations Copyright © Cengage Learning. All rights reserved

Activation Energy, Ea Energy that must be overcome to produce a chemical reaction. Copyright © Cengage Learning. All rights reserved

Transition States and Activation Energy To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved

Change in Potential Energy Copyright © Cengage Learning. All rights reserved

For Reactants to Form Products Collision must involve enough energy to produce the reaction (must equal or exceed the activation energy). Relative orientation of the reactants must allow formation of any new bonds necessary to produce products. Copyright © Cengage Learning. All rights reserved

The Gas Phase Reaction of NO and Cl2 To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved

Arrhenius Equation A = frequency factor Ea = activation energy R = gas constant (8.3145 J/K·mol) T = temperature (in K) Copyright © Cengage Learning. All rights reserved

Linear Form of Arrhenius Equation Copyright © Cengage Learning. All rights reserved

Linear Form of Arrhenius Equation Copyright © Cengage Learning. All rights reserved

EXERCISE! Chemists commonly use a rule of thumb that an increase of 10 K in temperature doubles the rate of a reaction. What must the activation energy be for this statement to be true for a temperature increase from 25°C to 35°C? Ea = 53 kJ Ea = 53 kJ ln(2) = (Ea / 8.3145 J/K·mol)[(1/298 K) – (1/308 K)] Copyright © Cengage Learning. All rights reserved

Catalyst A substance that speeds up a reaction without being consumed itself. Provides a new pathway for the reaction with a lower activation energy. Copyright © Cengage Learning. All rights reserved

Energy Plots for a Catalyzed and an Uncatalyzed Pathway for a Given Reaction Copyright © Cengage Learning. All rights reserved

Effect of a Catalyst on the Number of Reaction-Producing Collisions Copyright © Cengage Learning. All rights reserved

Heterogeneous Catalyst Most often involves gaseous reactants being adsorbed on the surface of a solid catalyst. Adsorption – collection of one substance on the surface of another substance. Copyright © Cengage Learning. All rights reserved

Heterogeneous Catalysis To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved

Heterogeneous Catalyst Adsorption and activation of the reactants. Migration of the adsorbed reactants on the surface. Reaction of the adsorbed substances. Escape, or desorption, of the products. Copyright © Cengage Learning. All rights reserved

Homogeneous Catalyst Exists in the same phase as the reacting molecules. Enzymes are nature’s catalysts. Copyright © Cengage Learning. All rights reserved

Homogeneous Catalysis To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved