Chemical Kinetics Chapter 13 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Chemical Kinetics Thermodynamics – Does a reaction take place? (Chapter 18) Kinetics – How fast does a reaction proceed? Concentration Temperature Catalyst GOAL - Understand behavior of reaction and how we can control it Determine Rate Law Determine Reaction Mechanism 13.1

Reaction rate is the change in the concentration of a reactant or a product with time (M/s). A B rate = -  [A] tt rate =  [B] tt  [A] = change in concentration of A over time period  t  [B] = change in concentration of B over time period  t Because [A] decreases with time,  [A] is negative Reaction rate units = concentration time concentration usually in units of Molarity or moles time usually in units of seconds or minutes

A B 13.1 rate = -  [A] tt rate = [B][B] tt time

Br 2 (aq) + HCOOH (aq) 2Br - (aq) + 2H + (aq) + CO 2 (g) time 393 nm light Detector  [Br 2 ]   Absorption 393 nm Br 2 (aq) 13.1

Br 2 (aq) + HCOOH (aq) 2Br - (aq) + 2H + (aq) + CO 2 (g) average rate = -  [Br 2 ] tt = - [Br 2 ] final – [Br 2 ] initial t final - t initial slope of tangent slope of tangent slope of tangent instantaneous rate = rate for specific instance in time 13.1

Since the slope is constantly changing, it may be hard to get an accurate measure of rate from a plot (there is error in measuring the slope of experimental data) => Scientists look for ways to plot data so that a straight line is produced Remember the Clausius-Clapeyron Equation - plot ln P vs 1/T NOT P vs T => Different reactions may need to be plotted in different ways

rate  [Br 2 ] rate = k [Br 2 ] k = rate [Br 2 ] 13.1 = rate constant = 3.50 x s -1 Look for trends in data as [Br 2 ] is cut in half, so is rate => linear relationship

The units of the rate constant (k) depend on the order of the reaction Rate = k [A] x [B] y [C] z or k = Rate. [A] x [B] y [C] z Rate is units of concentration time concentration = M or moles time = seconds or minutes Reaction Order Rate k units 0 k M/s 1 k [A] (M/s)/M = 1/s 2 k [A] 2 (M/s)/M 2 = 1/Ms 3 k [A] 3 (M/s)/M 3 = 1/M 2 s 4 k [A] 4 (M/s)/M 4 = 1/M 3 s Always have time units in denominator

2H 2 O 2 (aq) 2H 2 O (l) + O 2 (g) PV = nRT P = RT = [O 2 ]RT n V [O 2 ] = P RT 1 rate =  [O 2 ] tt RT 1 PP tt = measure  P over time 13.1

2H 2 O 2 (aq) 2H 2 O (l) + O 2 (g) 13.1

Reaction Rates and Stoichiometry A B Two moles of A disappear for each mole of B that is formed. rate =  [B] tt rate = -  [A] tt 1 2 aA + bB cC + dD rate = -  [A] tt 1 a = -  [B] tt 1 b =  [C] tt 1 c =  [D] tt 1 d

Write the rate expression for the following reaction: CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (g) rate = -  [CH 4 ] tt = -  [O 2 ] tt 1 2 =  [H 2 O] tt 1 2 =  [CO 2 ] tt 13.1

The Rate Law 13.2 The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers. aA + bB cC + dD Rate = k [A] x [B] y reaction is xth order in A reaction is yth order in B reaction is (x +y)th order overall

F 2 (g) + 2ClO 2 (g) 2FClO 2 (g) rate = k [F 2 ] x [ClO 2 ] y design experiments to reveal trends in data Double [F 2 ] with [ClO 2 ] constant Rate doubles x = 1 Quadruple [ClO 2 ] with [F 2 ] constant Rate quadruples y = 1 rate = k [F 2 ] [ClO 2 ] units of k = 1/Ms 13.2 Initial rate = rate at time = 0

F 2 (g) + 2ClO 2 (g) 2FClO 2 (g) rate = k [F 2 ][ClO 2 ] Rate Laws Rate laws are always determined experimentally. Reaction order is always defined in terms of reactant (not product) concentrations. The order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical equation

Determine the rate law and calculate the rate constant for the following reaction from the following data: S 2 O 8 2- (aq) + 3I - (aq) 2SO 4 2- (aq) + I 3 - (aq) Experiment [S 2 O 8 2- ][I - ] Initial Rate (M/s) x x x rate = k [S 2 O 8 2- ] x [I - ] y Double [I - ], rate doubles (experiment 1 & 2) y = 1 Double [S 2 O 8 2- ], rate doubles (experiment 2 & 3) x = 1 k = rate [S 2 O 8 2- ][I - ] = 2.2 x M/s (0.08 M)(0.034 M) = 0.081/M s 13.2 rate = k [S 2 O 8 2- ][I - ]

First-Order Reactions 13.3 A product rate = -  [A] tt rate = k [A] k = rate [A] = 1/s or s -1 M/sM/s M =  [A] tt = k [A] - [A] is the concentration of A at any time t [A] 0 is the concentration of A at time t=0 Rearrange:  [A] = -k  t [A] Integrate: t = 0 => t and [A] 0 = [A] t [A] t t d[A] = -k dt [A] [A] 0 0 ln [A] t - ln [A] 0 = - k t or ln [A] t = - k t + ln [A] 0 y = mx + b

First-Order Reactions 13.3 A product [A] = [A] 0 exp(-kt)ln[A] = ln[A] 0 - kt ln [A] t - ln [A] 0 = - k t or ln [A] t = - k t + ln [A] 0 y = mx + b

Decomposition of N 2 O

First-Order Reactions A product ln [A] t - ln [A] 0 = - k t or ln [A] t = - k t [A] 0. at t 1/2 [A] t 1/2 = 1/2 [A] 0 half-life or t 1/2 = ln 2 k half-life is independent of initial concentration

The reaction 2A B is first order in A with a rate constant of 2.8 x s -1 at 80 0 C. How long will it take for A to decrease from 0.88 M to 0.14 M ? ln[A] = ln[A] 0 - kt kt = ln[A] 0 – ln[A] t = ln[A] 0 – ln[A] k = 66 s [A] 0 = 0.88 M [A] = 0.14 M ln [A] 0 [A] k = ln 0.88 M 0.14 M 2.8 x s -1 = 13.3

First-Order Reactions 13.3 The half-life, t ½, is the time required for the concentration of a reactant to decrease to half of its initial concentration. t ½ = t when [A] = [A] 0 /2 ln [A] 0 [A] 0 /2 k = t½t½ ln2 k = k = What is the half-life of N 2 O 5 if it decomposes with a rate constant of 5.7 x s -1 ? t½t½ ln2 k = x s -1 = = 1216 s = 20 minutes How do you know decomposition is first order? units of k (s -1 )

A product First-order reaction # of half-lives [A] = [A] 0 /n

Second-Order Reactions A product rate = -  [A] tt rate = k [A] 2 k = rate [A] 2 = 1/M s M/sM/s M2M2 =  [A] tt = k [A] 2 - [A] is the concentration of A at any time t [A] 0 is the concentration of A at time t=0 1 [A] = 1 [A] 0 - kt t ½ = t when [A] = [A] 0 /2 t ½ = 1 k[A] 0 Rearrange:  [A] = -k  t [A] 2 Integrate: t = 0 => t and [A] 0 => [A] t [A] t t d[A] = - k dt [A] 2 [A] [A] t = 1 [A] 0 kt + -- y = mx + b 1 = 1 [A] 0 kt ½ + [A] 0 /2 Half-life is inversely proportional to initial concentration Note negative signs

Zero-Order Reactions A product rate = -  [A] tt rate = k [A] 0 = k k = rate [A] 0 = M/s  [A] tt = k - [A] is the concentration of A at any time t [A] 0 is the concentration of A at time t=0 t ½ = t when [A] t = [A] 0 /2 t ½ = [A] 0 2k2k [A] t = - kt + [A] 0 Rearrange:  [A] = -k  t Integrate: t = 0 => t and [A] 0 => [A] t [A] t t d[A] = - k dt [A] 0 0 [A] t - [A] 0 = - kt y = mx + b [A] 0 /2 = - kt ½ + [A] 0 Half-life is proportional to initial concentration

Summary of the Kinetics of Zero-Order, First-Order and Second-Order Reactions OrderRate Law Concentration-Time Equation Half-Life rate = k rate = k [A] rate = k [A] 2 ln[A] = ln[A] 0 - kt 1 [A] = 1 [A] 0 + kt [A] = [A] 0 - kt t½t½ ln2 k = t ½ = [A] 0 2k2k t ½ = 1 k[A]