CHAPTER Four(13) Chemical Kinatics

Chapter 4 / Chemical Kinetics Chapter Four Contains: 4.1 The Rate of a Reaction 4.2 The Rate Law 4.3 The Relation Between Reactant Concentration and Time 4.4 Activation Energy and Temperature Dependence of Rate Constants 4.5 Reaction Mechanisms 4.6 Catalysis Chapter Four outline 2

4.1 The Rate of a Reaction Chapter 4 / Chemical Kinetics 3 Thermodynamics – does a reaction take place? Kinetics – how fast does a reaction proceed? Chemical kinetics is the area of chemistry concerned with the speeds, or rates, at which a chemical reaction occurs. 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.

4.1 The Rate of a Reaction Chapter 4 / Chemical Kinetics 4 A B rate = -  [A] tt rate = [B][B] tt time

4.1 The Rate of a Reaction Chapter 4 / Chemical Kinetics 5 Br 2 (aq) + HCOOH (aq) 2Br - (aq) + 2H + (aq) + CO 2 (g) time 393 nm light Detector  [Br 2 ]   Absorption

4.1 The Rate of a Reaction Chapter 4 / Chemical Kinetics 6 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

4.1 The Rate of a Reaction Chapter 4 / Chemical Kinetics 7 rate  [Br 2 ] rate = k [Br 2 ] k = rate [Br 2 ] = rate constant = 3.50 x s -1

4.1 The Rate of a Reaction Chapter 4 / Chemical Kinetics 8 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

4.1 The Rate of a Reaction Chapter 4 / Chemical Kinetics 9

4.1 The Rate of a Reaction Chapter 4 / Chemical Kinetics 10 Reaction Rates and Stoichiometry 2A 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

The Rate of a Reaction Chapter 4 / Chemical Kinetics Example: 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

Example: Consider the reaction 4NO 2 ( g ) + O 2 ( g ) 2N 2 O 5 ( g ) Suppose that, at a particular moment during the reaction, molecular oxygen is reacting at the rate of M/s. (a) At what rate is N 2 O 5 being formed? (b) At what rate is NO 2 reacting? The Rate of a Reaction Chapter 4 / Chemical Kinetics rate = -  [NO 2 ] tt = -  [O 2 ] tt =  [N 2 O 5 ] tt =  [N 2 O 5 ] tt 1 2 = x 2 = M/s  [N 2 O 5 ] tt  [NO 2 ] tt 1 4 =  [NO 2 ] tt = x 4 = M/s -

The Rate Law Chapter 4 / Chemical Kinetics 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

The Rate Law Chapter 4 / Chemical Kinetics F 2 (g) + 2ClO 2 (g) 2FClO 2 (g) rate = k [F 2 ] x [ClO 2 ] y 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 ]

The Rate Law Chapter 4 / Chemical Kinetics 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. 1

The Rate Law Chapter 4 / Chemical Kinetics Example: Drive rate Law and k for CH 3 CHO (g)  CH 4 (g) + CO (g) For experimental data for rate of disappearance of CH 3 CHO Exp[CH3CHO]R(mol / L.s ) rate = k [CH 3 CHO] x Rate 2 Rate 1 = = 4 = k (0.2) x k (0.1) x 4= ( 2 ) x x = 2 rate = k [CH 3 CHO] 2 R 1 = 0.02 = k ( 0.1) = 0.01 kk = 0.02 / 0.01 = 2. 0 L / mol.s

The Rate Law Chapter 4 / Chemical Kinetics 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) Experimen t [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) = 0.08/M s rate = k [S 2 O 8 2- ][I - ]

The Rate Law Chapter 4 / Chemical Kinetics Rate 1 Rate 2 = 2.2 x x = 2 = k (0.08) x (0.034) y k (0.08) x (0.017) y = 2 y = 2 (0.034) y (0.017) y Then y =1 Rate 3 Rate 2 = 2.2 x x = 2 = k (0.16) x (0.017) y k (0.08) x (0.017) y = 2 x = 2 (0.16) x (0.08) x Then x=1 To determine x and y we follow the following method

The Relation Between Reactant Concentration and Time Chapter 4 / Chemical Kinetics First-Order Reactions A product rate = -  [A] tt rate = k [A] k = rate [A] = 1/s or s -1 M /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 [A] = [A] 0 exp(-kt) ln[A] = ln[A] 0 - kt and

4.3 The Relation Between Reactant Concentration and Time Chapter 4 / Chemical Kinetics 2N 2 O 5 4NO 2 (g) + O 2 (g) 20

4.3 The Relation Between Reactant Concentration and Time Chapter 4 / Chemical Kinetics 21 For gas-phase reactions we can replace the concentration terms with the pressures of the gaseous reactant ln[A] = ln[A] 0 - kt ln P = ln P 0 - kt

The Relation Between Reactant Concentration and Time Chapter 4 / Chemical Kinetics Example :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 =

The Relation Between Reactant Concentration and Time Chapter 4 / Chemical Kinetics Example: The conversion of cyclopropane to propene in the gas phase is a first-order reaction with a rate constant of 6.7x10 -4 s -1 at 500°C. (a)If the initial concentration of cyclopropane was 0.25 M, what is the concentration after 8.8 min? (b)How long (in minutes) will it take for the concentration of cyclopropane to decrease from 0.25 M to 0.15 M? (c)How long (in minutes) will it take to convert 74 percent of the starting material? a) [A] 0 = 0.25 M, t= 8.8 min = 8.8x60 = 528 s, k= 6.7x10 -4 s -1, [A] = ? ln[A] = ln[A] 0 – kt = ln (0.25 ) – ((6.7x10 -4 s -1 ) ( 528 )) = [A] = e = 0.18M

The Relation Between Reactant Concentration and Time Chapter 4 / Chemical Kinetics Example: The conversion of cyclopropane to propene in the gas phase is a first-order reaction with a rate constant of 6.7x10 -4 s -1 at 500°C. (a)If the initial concentration of cyclopropane was 0.25 M, what is the concentration after 8.8 min? (b)How long (in minutes) will it take for the concentration of cyclopropane to decrease from 0.25 M to 0.15 M? (c)How long (in minutes) will it take to convert 74 percent of the starting material? b) [A] 0 = 0.25 M, [A] = 15M, k= 6.7x10 -4 s -1, t= ? ln[A] = ln[A] 0 – kt ln[A] – ln[A] 0 = – kt t = ln[A] – ln[A] 0 -k ln [A] 0 [A] k = = ln[A] 0 – ln[A] k = 7.6 x10 2 s ln x10 -4 = = (7.6 x10 2 ) / 60 = 13 min

The Relation Between Reactant Concentration and Time Chapter 4 / Chemical Kinetics Example: The conversion of cyclopropane to propene in the gas phase is a first-order reaction with a rate constant of 6.7x10 -4 s -1 at 500°C. (a)If the initial concentration of cyclopropane was 0.25 M, what is the concentration after 8.8 min? (b)How long (in minutes) will it take for the concentration of cyclopropane to decrease from 0.25 M to 0.15 M? (c)How long (in minutes) will it take to convert 74 percent of the starting material? C) The converted concentration will be 0.25 x (74/100) = M The remaining concentration will be = = M [A] 0 = 0.25 M, [A] = 0.065M, k= 6.7x10 -4 s -1, t= ? ln[A] = ln[A] 0 – kt ln[A] – ln[A] 0 = – kt t = ln[A] – ln[A] 0 -k ln [A] 0 [A] k = = ln[A] 0 – ln[A] k = 2.0 x10 3 s ln x10 -4 = = (2.0 x10 2 ) / 60 = 33 min

The Relation Between Reactant Concentration and Time Chapter 4 / Chemical Kinetics First-Order Reactions 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 = = 1200 s = 20 minutes How do you know decomposition is first order? units of k (s -1 )

The Relation Between Reactant Concentration and Time Chapter 4 / Chemical Kinetics A product First-order reaction # of half-lives [A] = [A] 0 /n

The Relation Between Reactant Concentration and Time Chapter 4 / Chemical Kinetics Second-Order Reactions A product rate = -  [A] tt rate = k [A] 2 k = rate [A] 2 = 1/M s M /s M 2 =  [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 and

The Relation Between Reactant Concentration and Time Chapter 4 / Chemical Kinetics Example :Iodine atoms combine to form molecular iodine in the gas phase I( g ) + I( g ) I 2 ( g ) This reaction follows second-order kinetics and has the high rate constant 7.0 x 10 9 /M.s at 23°C. (a) If the initial concentration of I was M, calculate the concentration after 2.0 min. (b) Calculate the half-life of the reaction if the initial concentration of I is 0.60 M and if it is 0.42 M. [A] 0 = M, k= 7.0x10 9 /M.s, t= 2 min = 2 x 60= 120 s,[A] = ? 1 [A] = 1 [A] 0 + kt 1 [A] = (7.0x10 9 x 120) [A] = 1.2x M a) t ½ = 1 k[A] 0 t ½ = 1 7.0x10 9 x0.60 = 2.4x s t ½ = 1 k[A] 0 t ½ = 1 7.0x10 9 x0.42 = 3.4x s b)

The Relation Between Reactant Concentration and Time Chapter 4 / Chemical Kinetics 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] = [A] 0 /2 t ½ = [A] 0 2k2k [A] = [A] 0 - kt and

The Relation Between Reactant Concentration and Time Chapter 4 / Chemical Kinetics 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] 0

Activation Energy and Temperature Dependence of Rate Constants Chapter 4 / Chemical Kinetics Exothermic ReactionEndothermic Reaction The activation energy (E a ) is the minimum amount of energy required to initiate a chemical reaction. A + B AB C + D + +

Activation Energy and Temperature Dependence of Rate Constants Chapter 4 / Chemical Kinetics Collision Theory Molecules must collide to react. More collisions per unit time should lead to faster reaction Colliding molecules must have a minimum amount of kinetic energy for a collision result in products ( otherwise they just bounce off each other ). Minimum collision energy needed is called the activation energy, E a Raising the temperature of a reaction raises the kinetic energy of the reactants and increases the number of collisions per unit time. Raising the temperature should lead to faster reactions. At some point, when two molecules collide and react, there is a highest energy state called the transition state This barrier is the activation for the reaction to occur.

Activation Energy and Temperature Dependence of Rate Constants Chapter 4 / Chemical Kinetics

Activation Energy and Temperature Dependence of Rate Constants Chapter 4 / Chemical Kinetics Temperature Dependence of the Rate Constant k = A exp( -E a / RT ) E a is the activation energy (J/mol) R is the gas constant (8.314 J/Kmol) T is the absolute temperature A is the frequency factor lnk = - EaEa R 1 T + lnA (Arrhenius equation)

Activation Energy and Temperature Dependence of Rate Constants Chapter 4 / Chemical Kinetics lnk = - EaEa R 1 T + lnA

Activation Energy and Temperature Dependence of Rate Constants Chapter 4 / Chemical Kinetics lnk 1 = - EaEa R 1 T1T1 + lnA lnk 2 = - EaEa R 1 T2T2 + lnA ln = EaEa R T 1 – T 2 T 1 T 2 k1k1 k2k2 () Example : The rate constant of a first-order reaction is 3.46 x10 -2 s -1 at 298 K. What is the rate constant at 350 K if the activation energy for the reaction is 50.2 kJ/mol? ln = EaEa R T 1 – T 2 T 1 T 2 k1k1 k2k2 () ln = 50.2 x – x x10 -2 k2k2 () ln = x10 -2 k2k2 = e = x10 -2 k2k2 k 2 = s -1

38 Chapter 4 / Chemical Kinetics Example: The activation energy of a certain reaction is 76.7 KJ/mol How many times faster will the reaction occur at 50  C than at 0  C ? ln = EaEa R T 1 – T 2 T 1 T 2 k1k1 k2k2 () ln = 76.7 x – x 323 () k1k1 k2k2 ln = k1k1 k2k2 k1k1 k2k2 = k 1 = k 2 k 2 = 187 k 1 / Activation Energy and Temperature Dependence of Rate Constants

Reaction Mechanisms Chapter 4 / Chemical Kinetics The overall progress of a chemical reaction can be represented at the molecular level by a series of simple elementary steps or elementary reactions. The sequence of elementary steps that leads to product formation is the reaction mechanism. 2NO (g) + O 2 (g) 2NO 2 (g) N 2 O 2 is detected during the reaction! Elementary step:NO + NO N 2 O 2 Elementary step: N 2 O 2 + O 2 2NO 2 Overall reaction: 2NO + O 2 2NO 2 +

Reaction Mechanisms Chapter 4 / Chemical Kinetics 2NO (g) + O 2 (g) 2NO 2 (g)

Reaction Mechanisms Chapter 4 / Chemical Kinetics Elementary step:NO + NO N 2 O 2 Elementary step: N 2 O 2 + O 2 2NO 2 Overall reaction: 2NO + O 2 2NO 2 + Intermediates are species that appear in a reaction mechanism but not in the overall balanced equation. An intermediate is always formed in an early elementary step and consumed in a later elementary step. The molecularity of a reaction is the number of molecules reacting in an elementary step. Unimolecular reaction – elementary step with 1 molecule Bimolecular reaction – elementary step with 2 molecules Termolecular reaction – elementary step with 3 molecules

Reaction Mechanisms Chapter 4 / Chemical Kinetics Unimolecular reactionA productsrate = k [A] Bimolecular reactionA + B productsrate = k [A][B] Bimolecular reactionA + A productsrate = k [A] 2 Rate Laws and Elementary Steps 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.

Reaction Mechanisms Chapter 4 / Chemical Kinetics Initial rate determining stepFinal rate determining step Position of rate determining step in the reaction mechanism

Reaction Mechanisms Chapter 4 / Chemical Kinetics Sequence of Steps in Studying a Reaction Mechanism

Reaction Mechanisms Chapter 4 / Chemical Kinetics The experimental rate law for the reaction between NO 2 and CO to produce NO and CO 2 is rate = k[NO 2 ] 2. The reaction is believed to occur via two steps: Step 1:NO 2 + NO 2 NO + NO 3 Step 2:NO 3 + CO NO 2 + CO 2 What is the equation for the overall reaction? NO 2 + CO NO + CO 2 What is the intermediate?NO 3 What can you say about the relative rates of steps 1 and 2? rate = k[NO 2 ] 2 is the rate law for step 1 so step 1 must be slower than step 2

Catalysis Chapter 4 / Chemical Kinetics A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed. k = A exp( -E a / RT ) EaEa k rate uncatalyzed < rate catalyzed E a > E a ‘ UncatalyzedCatalyzed

Catalysis Chapter 4 / Chemical Kinetics In heterogeneous catalysis, the reactants and the catalysts are in different phases. In homogeneous catalysis, the reactants and the catalysts are dispersed in a single phase, usually liquid. Haber synthesis of ammonia Ostwald process for the production of nitric acid Catalytic converters Acid catalysis Base catalysis

Catalysis Chapter 4 / Chemical Kinetics 2N 2 O(s)  2N 2 (g) + O 2 (g), Ea =245 KJ, No catalyst

Catalysis Chapter 4 / Chemical Kinetics N 2 (g) + 3H 2 (g) 2NH 3 (g) Fe/Al 2 O 3 /K 2 O catalyst Haber Process

Catalysis Chapter 4 / Chemical Kinetics Ostwald Process Hot Pt wire over NH 3 solution Pt-Rh catalysts used in Ostwald process 4NH 3 (g) + 5O 2 (g) 4NO (g) + 6H 2 O (g) Pt catalyst 2NO (g) + O 2 (g) 2NO 2 (g) 2NO 2 (g) + H 2 O (l) HNO 2 (aq) + HNO 3 (aq)

Catalysis Chapter 4 / Chemical Kinetics Catalytic Converters CO + Unburned Hydrocarbons + O 2 CO 2 + H 2 O catalytic converter 2NO + 2NO 2 2N 2 + 3O 2 catalytic converter

Catalysis Chapter 4 / Chemical Kinetics Enzyme Catalysis

Catalysis Chapter 4 / Chemical Kinetics uncatalyzed enzyme catalyzed rate =  [P] tt rate = k [ES]

54 Chapter 4 / Chemical Kinetics