Chemical Kinetics Chapter 15 Copyright © The McGraw-Hill Companies, Inc.  Permission required for reproduction or display.

Chapter 15; Kinetics Factors Controlling Rate Two Mathematical Rate Descriptions Change Concentration/ Change in Time Rate Law Experimental Determination Reaction Order

Kinetics – how fast does a reaction proceed? Chemical Kinetics Kinetics – how fast does a reaction proceed? Chemical Reaction; Reactant Bonds Broken, Product Bonds Made NO2 (g) + CO (g) -> NO (g) + CO2 (g) N O C + 13.1

How to increase the rate of reaction? http://www.youtube.com/watch?v=OttRV5ykP7A

Molecular Collision Theory Incorrect Orientation Reactants must collide with correct orientation enough energy, Ea O C N O O Correct Orientation O C N

13.4 A + B C + D Exothermic Reaction Endothermic Reaction The activation energy (Ea) is the minimum amount of energy required to initiate a chemical reaction. 13.4

Activation Energy.. ..is energy needed so there is enough energy to break reactant bonds. ..is the energy needed to get molecules in the correct orientation. ..is the energy needed to reach the transition state or activated complex

Factors for Reaction Rate Temperature Concentration

Concentration Reaction rate is the change in the concentration of a reactant or a product with time (M/s). A B rate = - D[A] Dt D[A] = change in concentration of A over time period Dt rate = D[B] Dt D[B] = change in concentration of B over time period Dt 13.1

Reaction Rates and Stoichiometry 2A B Two moles of A disappear for each mole of B that is formed. Rate = - D[A] Dt 1 2 rate = D[B] Dt aA + bB cC + dD rate = - D[A] Dt 1 a = - D[B] Dt 1 b = D[C] Dt 1 c = D[D] Dt 1 d 13.1

Write the rate expression for the following reaction: CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g) rate = - D[CH4] Dt = - D[O2] Dt 1 2 = D[CO2] Dt = D[H2O] Dt 1 2 13.1

A B time rate = - D[A] Dt rate = D[B] Dt 13.1

Algebra Review of Slopes of Straight Lines y = mx +b; m = slope Dy/Dx (+) positive slope (-) negative slope y y x x

Algebra Review of Slopes of Straight Lines y = mx +b; m = slope Dy/Dx More vertical; higher slope More horizontal; lower slope y y x x Note; both are positive slopes

Curved Rather than Straight Line Tell Us That..... Rate is not Constant Throughout Reaction Reaction Rate is Higher with Higher Concentration (-) slope means concentration is decreasing.

Two Mathematical Expressions to Describe Reaction Rate A B Rate = -A/ t =; + B/ t Determined from stoichiometry Uses both reactants & products Rate Law; rate =k[A]x Determined by experimental data- Stoichiometry of equation is irrelevant Only reactants in rate law

The Rate Law 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 k is the Rate Law Constant x and y are determined experimentally, and do not depend on stoichiometric coefficients from balanced equation 13.2

The Rate Law 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 Order reaction is xth order in A reaction is yth order in B reaction is (x +y)th order overall Reaction order tells how quickly rate will increase when concentration increases 13.2

Reaction Order Tells How Changing Reactant Concentration Affects Rate Oth Order; rate =k[A]0; double [A], rate same 1st Order; rate = k[A]1; double [A], double rate 2nd Order; rate =k[A]2; double [A]; quadruple rate 3rd Order; rate =k[A]3; double [A]; rate 8X

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. F2 (g) + 2ClO2 (g) 2FClO2 (g) 1 rate = k [F2][ClO2] 13.2

Determining the Form of the Rate Law

How Data is created

Method of Initial Rates Used to find the form of the rate law Choose one reactant to start with Find two experiments where the concentration of that reactant changes but all other reactants stay the same Write the rate laws for both experiments Divide the two rate laws Solve for the order Follow the same technique for other reactants

Example Choose one reactant to start with NH4+ Find two experiments where the concentration of that reactant changes but all other reactants stay the same Exp 2 & 3

Example Write the rate laws for both experiments Exp 2: Rate = 2.70x10-7 = k(0.100)x(0.010)y Exp 3: Rate = 5.40x10-7 = k(0.200)x(0.010)y Divide the two rate laws 0.50 = 0.50x Use log rules to solve for the order x = 1 so the order for NH4+ is one

Example Follow the same technique for other reactants NO2-: Exp 1 & 2 Exp 1: Rate = 1.35x10-7 = k(0.100)1(0.0050)y Exp 2: Rate = 2.70x10-7 = k(0.100)1(0.010)y 0.5 = 0.5y y = 1 So Rate = k[NH4+]1[NO2-]1 Overall Reaction Order – sum of orders of reactants

Finding k We can find k using values from any of the experiments given Units will be different for k depending on order of reactants

Example BrO3- : Exp 1 & 2 Exp 1: Rate = 8.0x10-4 = k(0.10)x(0.10)y(0.10)z Exp 2: Rate = 1.6x10-3 = k(0.20)x(0.10)y(0.10)z 0.50 = 0.50x x = 1

Example Br- : Exp 2 & 3 Exp 2: Rate = 1.6x10-3 = k(0.20)1(0.10)y(0.10)z Exp 3: Rate = 3.2x10-3 = k(0.20)1(0.20)y(0.10)z 0.50 = 0.50y y = 1

Example H+ : Exp 1 & 4 Exp 1: Rate = 8.0x10-4 = k(0.10)1(0.10)1(0.10)z 0.25 = 0.50z OR ¼ = (½)z z = 2

Example So Rate = k[BrO3-]1[Br-]1[H+]2 Solve for rate constant, k Overall order of reaction = 4 Solve for rate constant, k

Double [F2] with [ClO2] constant F2 (g) + 2ClO2 (g) 2FClO2 (g) rate = k [F2]x[ClO2]y Double [F2] with [ClO2] constant Rate doubles x = 1 rate = k [F2][ClO2] Quadruple [ClO2] with [F2] constant Rate quadruples y = 1 13.2

S2O82- (aq) + 3I- (aq) 2SO42- (aq) + I3- (aq) Determine the rate law and calculate the rate constant for the following reaction from the following data: S2O82- (aq) + 3I- (aq) 2SO42- (aq) + I3- (aq) Experiment [S2O82-] [I-] Initial Rate (M/s) 1 0.08 0.034 2.2 x 10-4 2 0.017 1.1 x 10-4 3 0.16 rate = k [S2O82-]x[I-]y y = 1 x = 1 rate = k [S2O82-][I-] Double [I-], rate doubles (experiment 1 & 2) Double [S2O82-], rate doubles (experiment 2 & 3) k = rate [S2O82-][I-] = 2.2 x 10-4 M/s (0.08 M)(0.034 M) = 0.08/M•s 13.2

Types of Rate Laws Differential Rate law - describes how rate depends on concentration. Integrated Rate Law - Describes how concentration depends on time. For each type of differential rate law there is an integrated rate law and vice versa. Rate laws can help us better understand reaction mechanisms.

Integrated Rate Laws Consider a simple 1st order rxn: A  B Differential form: How much A is left after time t? Integrate:

Integrated Rate Laws The integrated form of first order rate law: Can be rearranged to give: [A]0 is the initial concentration of A (t=0). [A]t is the concentration of A at some time, t, during the course of the reaction.

Integrated Rate Laws Manipulating this equation produces… y = mx + b …which is in the form y = mx + b

First-Order Processes If a reaction is first-order, a plot of ln [A]t vs. t will yield a straight line with a slope of -k. So, use graphs to determine rxn order.

First-Order Processes Consider the process in which methyl isonitrile is converted to acetonitrile. CH3NC CH3CN How do we know this is a first order rxn?

First-Order Processes CH3NC CH3CN This data was collected for this reaction at 198.9°C. Does rate=k[CH3NC] for all time intervals?

First-Order Processes When ln P is plotted as a function of time, a straight line results. The process is first-order. k is the negative slope: 5.1  10-5 s-1.

Second-Order Processes Similarly, integrating the rate law for a process that is second-order in reactant A: Rearrange, integrate: y = mx + b also in the form

Second-Order Processes So if a process is second-order in A, a plot of 1/[A] vs. t will yield a straight line with a slope of k. First order: If a reaction is first-order, a plot of ln [A]t vs. t will yield a straight line with a slope of -k.

Determining rxn order The decomposition of NO2 at 300°C is described by the equation NO2 (g) NO (g) + 1/2 O2 (g) and yields these data: Time (s) [NO2], M 0.0 0.01000 50.0 0.00787 100.0 0.00649 200.0 0.00481 300.0 0.00380

Determining rxn order Graphing ln [NO2] vs. t yields: The plot is not a straight line, so the process is not first-order in [A]. Time (s) [NO2], M ln [NO2] 0.0 0.01000 -4.610 50.0 0.00787 -4.845 100.0 0.00649 -5.038 200.0 0.00481 -5.337 300.0 0.00380 -5.573 Does not fit:

Second-Order Processes A graph of 1/[NO2] vs. t gives this plot. This is a straight line. Therefore, the process is second-order in [NO2]. Time (s) [NO2], M 1/[NO2] 0.0 0.01000 100 50.0 0.00787 127 100.0 0.00649 154 200.0 0.00481 208 300.0 0.00380 263

Half-Life Half-life is defined as the time required for one-half of a reactant to react. Because [A] at t1/2 is one-half of the original [A], [A]t = 0.5 [A]0.

Half-Life For a first-order process, set [A]t=0.5 [A]0 in integrated rate equation: NOTE: For a first-order process, the half-life does not depend on [A]0.

Half-Life- 2nd order For a second-order process, set [A]t=0.5 [A]0 in 2nd order equation.

Outline: Kinetics First order Second order complicated Rate Laws Integrated Rate Laws complicated Half-life