Chapter 14 Chemical Kinetics Lecture Presentation Chapter 14 Chemical Kinetics James F. Kirby Quinnipiac University Hamden, CT

14.1 Factors Affecting Reaction Rates Chemical kinetics is the study of rate or speed of chemical reactions. Chemical kinetics also gives us information about the reaction mechanism, that is, what really goes on during a chemical reaction.

14.2 Reaction Rates Reaction rate is the change in the concentration of a reactant or a product with time (M/s). D[A] = change in concentration of A over time period Dt A B rate = − D[A] Dt Because [A] decreases with time, D[A] is negative. rate = D[B] Dt D[B] = change in concentration of B over time period Dt Average rate is the rate over a period of time Dt. Instantaneous rate is the rate at a particular time t. Initial rate is the rate at time t ≈ 0.

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

Consider the reaction 4NO2 (g) + O2 (g) 2N2O5 (g). EXERCISE #1 Consider the reaction 4NO2 (g) + O2 (g) 2N2O5 (g). a) Write the rate expression for the reaction. − 1 4 ∆[NO 2 ] ∆t = − ∆[O 2 ] ∆t = 1 2 ∆[N 2 O 5 ] ∆t rate = Suppose that, at a particular moment during the reaction, molecular oxygen is reacting at the rate of 0.024 M/s. b) At what rate is N2O5 being formed? − ∆[𝑂 2 ] ∆𝑡 = 0.024 M/s − ∆[O 2 ] ∆t = 1 2 ∆[N 2 O 5 ] ∆t = 0.024 M/s ∆[N 2 O 5 ] ∆t = 2 (− ∆[O 2 ] ∆t ) = 2(0.024 M/s) = 0.048 M/s c) At what rate is NO2 reacting? − 1 4 ∆[NO 2 ] ∆t = − ∆[O 2 ] ∆t = 0.024 M/s − ∆[NO 2 ] ∆t = 4 (− ∆[𝑂 2 ] ∆𝑡 ) = 4(0.024 M/s) = 0.096 M/s

14.3 Rate Law 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 is xth order in A reaction is yth order in B reaction is (x +y)th order overall

14.3 Determining the Form of the Rate Law Method of Initial Rates F2 (g) + 2ClO2 (g) 2FClO2 (g) Rate law: rate = k [F2]x[ClO2]y Double [F2] with [ClO2] constant Quadruple [ClO2] with [F2] constant Rate doubles Rate quadruples x = 1 y = 1 rate = k [F2][ClO2]

14.3 Determining the Form of the Rate Law Conclusion about Rate Law 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) rate = k [F2][ClO2] 1

EXERCISE #2 Consider the reaction of nitric oxide with hydrogen at 1280oC: The following data were collected at this temperature: a) Determine the rate law. rate = k[NO]x[H2]y = 10.0 x 10 −3 M 5.0 x 10 −3 M k [H2] 3 k [H2] 2 = 4.0 x 10 −3 M 2.0 x 10 −3 M k [NO] 2 k [NO] 1 = 2 = 2 = 5.0 x 10 −5 M/s 1.3 x 10 −5 M/s rate 2 rate 1 rate 3 rate 2 = 10.0 x 10 −5 M/s 5.0 x 10 −5 M/s = 3.85 = 2 2x = 3.85 x = 1.94 ≈ 2 2y = 2 x = 1 rate = k[NO]2[H2]

EXERCISE #2 b) Determine the rate constant. rate = k[NO]2[H2] rate NO 2[H2] = 5.0 x 10 −5 M/s 10.0 x 10 −3 M 2(2.0 x 10 −3 M) k = = 2.5 x 102 M−2 s−1 c) Determine the rate of the reaction when [NO] = 12.0 x 10−3 M and [H2] = 6.0 x 10 −3 M. rate = k[NO]2[H2] = (2.5 x 102 M−2 s−1) (12.0 x 10−3 M)2 (6.0 x 10−3 M) = 2.2 x 10−4 M/s d) Determine the order of the reaction in regards to NO and to H2, respectively, and the overall order of the reaction. 2nd order in NO 1st order in H2 3rd order overall

14.4 First-Order Rate Law rate = − D[A] Dt A product [A]t = concentration of A at time t [A]o = initial concentration of A k = rate constant t = time Rate law: rate = k[A] Integrated rate law: ln[A]t = ln[A]o – kt Half-life t1/2 Time required for the concentration of a reactant to decrease to half of its initial concentration. t ½ = ln(2) k Half–life does not depend on the concentration of reactants.

14.4 Second-Order Rate Law rate = − D[A] Dt A product [A]t = concentration of A at time t [A]o = initial concentration of A k = rate constant t = time Rate law: rate = k[A]2 Integrated rate law: 1 A 𝑡 = 1 [A] o + kt Half-life: t ½ = 1 k [A] o Half–life gets longer as the reaction progresses and the concentration of reactants decrease.

14.4 Zero-Order Rate Law rate = − D[A] Dt A product [A]t = concentration of A at time t [A]o = initial concentration of A k = rate constant t = time Rate law: rate = k[A]0 = k Integrated rate law: [A]t = [A]o − kt Half-life: t ½ = [A] o 2k Half–life gets shorter as the reaction progresses and the concentration of reactants decrease.

EXERCISE #3 The reaction 2A B is first order in A with a rate constant of 2.8 x 10-2 s-1 at 800C. How long will it take for [A] to decrease from 0.88 M to 0.14 M ? ln[A]t = ln[A]0 – kt ln A 0 − ln A t k = ln 0.88 − ln⁡(0.14) 2.8 x 10 −2 𝑠 −1 kt = ln[A]0 – ln[A]t t = = 66 s EXERCISE #4 What is the half-life of N2O5 if it decomposes as first order reaction with a rate constant of 5.7 x 10−4 s−1? t1/2 = ln(2) k = 0.693 5.7 𝑥 10 −4 𝑠 −1 = 1200 s = 20 minutes

Concentration-Time Equation 14.4 The Rate Laws Summary of the Kinetics of Zero-Order, First-Order and Second-Order Reactions Order Rate Law Concentration-Time Equation Half-Life t½ = [A]0 2k rate = k [A]t = [A]0 − kt t½ ln(2) k = 1 rate = k [A] ln[A]t = ln[A]0 − kt 1 [A]t = [A]0 + kt t½ = 1 k[A]0 2 rate = k [A]2

14.4 Determining the Form of the Rate Law 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) rate = k [F2][ClO2] 1 Why?

NO2(g) + CO(g) NO(g) + CO2(g) 14.6 Reaction Mechanisms Reaction Mechanisms NO2(g) + CO(g) NO(g) + CO2(g) NO2 NO3 NO Elementary step: + NO3 CO NO2 CO2 Elementary step: Overall reaction: NO2 + CO NO + CO2 The sequence of elementary steps that leads to product formation is the reaction mechanism. An intermediate is not reflected in the rate law or in the overall reaction equation. It is always formed in an early elementary step and consumed in a later elementary step. ---- NO3

14.6 Reaction Mechanisms Writing plausible reaction mechanisms: The sum of the elementary steps must give the overall balanced equation for the reaction. The rate-determining step should predict the same rate law that is determined experimentally. The rate-determining step is the slowest step in the sequence of steps leading to product formation. EXERCISE #5 The proposed mechanism for the reaction NO2 + CO NO + CO2 is NO2 + NO2 NO3 + NO (slow) NO3 + CO NO2 + CO2 (fast) Write a plausible rate law for the reaction. rate-determining step rate = k[NO2]2

EXERCISE #6 The experimental rate law for the reaction between NO2 and O3 to produce N2O5 and O2 is rate = k[NO2][O3]. The proposed reaction mechanism is as follows: Step 1: NO2 + O3 NO3 + O2 Step 2: NO3 + NO2 N2O5 a) Write the equation for the overall reaction. 2NO2 + O3 N2O5 + O2 b) Identify the intermediate/s. NO3 c) Identify the rate-determining step. The rate law reflects step 1; therefore step 1 is the rate-determining step.

14.5 A Model for Chemical Kinetics We have seen how concentrations of the reactants affect the rate of the reaction. Let us see how the rate constant can affect the rate of the reaction. rate ∝ k Collision Model Collision --- Molecules must collide to react. Activation energy (Ea) --- Collision must involve enough energy to produce a reaction. Orientation --- Relative orientation of the reactant molecules must allow formation of any new bonds necessary to produce products. Temperature --- The rate constant is temperature dependent.

14.5 Activation Energy and Transition State Activation Energy (Ea) The minimum amount of energy required to initiate a chemical reaction. A + B AB C + D + Exothermic Reaction Endothermic Reaction

14.5 Temperature Dependence of the Rate Constant Linear form of Arrhenius equation Arrhenius equation ln 𝑘=− E a R 1 T + ln 𝐴 Taking into account the collision model, the rate constant k can be represented as follows: k=A e − E a / RT Arrhenius equation Ea = activation energy (J/mol) R = gas constant (8.314 J/K∙mol) T = absolute temperature (in K) A = frequency factor

14.5 Temperature Dependence of the Rate Constant Arrhenius equation at two different temperatures k1 = rate constant at temperature T1 ln k 1 k 2 = E a R T 1 − T 2 T 1 T 2 k2 = rate constant at temperature T2 Ea = activation energy (J/mol) R = gas constant (8.314 J/K∙mol) EXERCISE #7 T1, T2 = absolute temperatures (in K) When the temperature of a certain reaction was increased from 25oC to 35oC, the rate of the reaction doubled. Calculate the activation energy of this reaction. rate ∝ k k1 = 1 k2 = 2 T1 = 25oC = 298 K T2 = 35oC = 308 K ln k1 k2 R T 1 − T 2 T 1 T 2 = 298 K− 308 K (298 K)(308 K) Ea = ln 1 2 (8.314 J/K∙mol) = 52893 J = 53 kJ

14.7 Catalysis Catalyst A substance that speeds up a reaction without being consumed itself. Provides a new pathway for the reaction with a lower activation energy. ln 𝑘=− E a R 1 T + ln 𝐴 Ea k Ea (catalyzed) < Ea (uncatalyzed) ratecatalyzed > rateuncatalyzed

END OF CHAPTER 14