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? 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. b) Calculate k. c) Calculate [A] at t = 525 minutes. 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 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 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