Chemical Kinetics Kinetics – how fast does a reaction proceed? Reaction rate Rate Law Change of concentration with time Rate and temperature Reaction mechanism Catalysis

I. Reaction Rate A. Rate is the change in the concentration of a reactant or a product with time (M/s). - Average rate Instantaneous rate Initial rate 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 Because [A] decreases with time, D[A] is negative.

A B time rate = – D[A] Dt rate = D[B] Dt

Ex: Br2 (aq) + HCOOH (aq) -> 2Br- (aq) + 2H+ (aq) + CO2 (g) slope of tangent slope of tangent slope of tangent instantaneous rate = rate for specific instance in time average rate = - D[Br2] Dt = - [Br2]final – [Br2]initial tfinal - tinitial average rate = - (0.00296 M-0.0120 M) (400.0 s - 0.0 s) =2.3 x 10-5 M/s From 0.0 s - 400 s

B. Rate 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

Q: 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 Q: If the CH4(g) is burning at a rate of 0.44 mol/s, what is the rate of consumption of O2(g)? 0.44 mol/s = – D[CH4] Dt = – D[O2] Dt 1 2 – D[O2] Dt = 2 x 0.44 mol/s =0.88 mol/s

C. Rate involving gas measure DP over time 2H2O2 (aq) 2H2O (l) + O2 (g) PV = nRT P = RT = [O2]RT n V [O2] = P RT 1 rate = D[O2] Dt RT 1 DP Dt =

D. Factors affecting reaction rate Physical states of reactants Homogeneous reaction Heterogeneous reaction Concentration Temperature Catalyst

II. Rate law 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 k : rate constant

2 Example: F2 (g) + 2ClO2 (g) 2FClO2 (g) rate = k [F2]x[ClO2]y Rate data for the reaction between F2 and ClO2 rate = k [F2]x[ClO2]y Double [F2] with [ClO2] constant Rate doubles x = 1 Quadruple [ClO2] with [F2] constant Rate quadruples y = 1 k = rate [F2][ClO2] = 1.2 M-1 s-1 rate = k [F2][ClO2] 2 Overall reaction order :

II. Rate Laws Rate laws are always determined experimentally. 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]

S2O82- (aq) + 3I- (aq) 2SO42- (aq) + I3- (aq) Q: 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) Q: 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-] From experiment 1 & 2: double [I-], rate doubles Rate1 Rate2 2.2x10-4M/s 1.1x10-4M/s = = 2 = k(0.08M)x(0.034M)y k(0.08M)x(0.017M)y = 2y 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

III. Change in concentration with time (Integrated rate law) A Product Order Rate Law Integrated rate law Plot rate = k [A] = [A]0 - kt [A] vs. t 1 rate = k [A] ln[A] = ln[A]0 - kt ln[A] vs. t 1 [A] = [A]0 + kt 1 [A] 2 rate = k [A]2

A. First-order reaction [A] = [A]0exp(-kt) A product rate = - D[A] Dt rate = k [A] ln[A] = ln[A]0 - kt rate [A] M/s M = k = = 1/s or s-1 [A] is the concentration of A at any time t [A]0 is the concentration of A at time t=0

Example: Decomposition of N2O5

Q: The reaction 2A –> B is first order in A with a rate constant of 2.2 x 10-2 s-1 at 80oC. (a) How long will it take for A to decrease from 0.72 M to 0.14 M. (b) What is the concentration of A after 20 s if the initial concentration of A is 0.42M? 1st order reaction : ln[A] = ln[A]0 - kt (a) [A]0 = 0.72 M [A] = 0.14 M t = ? kt = ln[A]0 – ln[A] ln [A]0 [A] k = ln 0.72 M 0.14 M 2.2 x 10-2 s-1 = ln[A]0 – ln[A] k t = = 74 s (b) [A]0 = 0.42 M t = 20 s [A] = ? ln[A] = ln[A]0 - kt = ln(0.42) - 2.2 x10-2 s-1 x 20 s =-1.3 [A] = exp(-1.3) = e-1.3 =0.27 M

Half-life, t1/2, for first-order reaction The half-life, t1/2, is the time required for the concentration of a reactant to decrease to half of its initial concentration. t1/2= t when [A] = [A]0/2 ln [A]0 [A]0/2 k = t1/2 ln2 k = 0.693 k = (*indep of concentration) Q: What is the half-life of N2O5 if it decomposes with a rate constant of 5.7 x 10-4 s-1? t1/2 ln2 k = 0.693 5.7 x 10-4 s-1 = = 1200 s = 20 minutes How do you know decomposition is first order? units of k (s-1)

First-order reaction A product # of half-lives [A] = n[A]0 1 1/2 2 1/4 3 1/8 4 1/16

B. Second-order reaction 1 [A] t A product rate = - D[A] Dt rate = k [A]2 1 [A] = [A]0 + kt rate [A]2 M/s M2 = k = = 1/M•s t1/2= t when [A] = [A]0/2 t1/2 = 1 k[A]0 [A] is the concentration of A at any time t [A]0 is the concentration of A at time t=0

C. Zeroth-order reaction A product rate = - D[A] Dt rate = k [A]0= k [A] = [A]0 - kt rate [A]0 k = = M/s t1/2= t when [A] = [A]0/2 t1/2 = [A]0 2k [A] is the concentration of A at any time t [A]0 is the concentration of A at time t=0

Integrated rate law – summary Order Rate Law Concentration-Time Equation Half-Life t/2 = [A]0 2k rate = k [A] = [A]0 - kt t1/2 ln2 k = 1 rate = k [A] ln[A] = ln[A]0 - kt 1 [A] = [A]0 + kt t1/2 = 1 k[A]0 2 rate = k [A]2

Determine graphically the rate law for the reaction of 4 PH3(g) -> P4 (g) + 6 H2(g) Q: Time(s) [PH3] (M/s) ln[PH3] 1/[PH3] 10 30 50 90 150 200 1.000 0.819 0.549 0.368 0.165 0.050 0.018 0.0 -0.2 -0.6 -1.0 -1.8 -3.0 -4.0 1.00 1.22 1.82 2.72 6.06 20.0 55.6 Straight line is obtained when ln[PH3] is plotted against time 1st order reaction

IV. Reaction Rate and Temperature A. Collision theory Based on the kinetic-molecular theory Collisions between reactants leads to reaction Requires proper orientation for reaction to occur Requires minimum energy (activation energy) for reaction to occur

The energy change during the chemical reaction B. Activation energy, Ea Ea: The minimum amount of energy required to initiate a chemical reaction Transition state, activated complex The energy change during the chemical reaction Transition state Transition state Exothermic Reaction Endothermic Reaction

C. Temperature dependence of rate constant Arrhenius equation k = A • exp( -Ea/RT ) Ea is the activation energy (J/mol) R is the gas constant (8.314 J/K•mol) T is the absolute temperature A is the frequency factor lnk = - Ea R 1 T + lnA

lnk = - Ea R 1 T + lnA ln = ( - ) Ea R 1 T2 k1 k2 T1

Q: The rate constant of a certain reaction triples when the temperature increases from 25oC to 35oC. What is the activation energy of this reaction ln = ( - ) Ea R 1 T2 k1 k2 T1 T1 = 25oC = 298 K k2 = 3 k1 Ea=? T2 = 35oC = 308 K ln k1 3k1 = 1 308 K 8.314 J/K.mol Ea ( - ) 298 K 8.314 J/K.mol Ea (-1.09 x 10-4 1/K ) -1.10 = -1.10 x 8.314 J/K.mol (-1.09 x 10-4 1/K) Ea = = 83.9 kJ/mol

(a) Slowest reaction to fastest: (1) < (2) < (3) (fastest) Q: Consider the following potential energy profile. (a) Rank the rate of the reaction. (Assume frequency factor is about the same). (b) Which reaction(s) are endothermic? (1) (2) (3) 34 kJ/mol 23 kJ/mol 17 kJ/mol 30 kJ/mol 56 kJ/mol Potential energy 40 kJ/mol R R P P R P reaction progress reaction progress reaction progress Ea= 40 kJ/mol Ea= 34 kJ/mol Ea= 23 kJ/mol DE= 10 kJ/mol DE=-22 kJ/mol DE= 6 kJ/mol (a) Slowest reaction to fastest: (1) < (2) < (3) (fastest) (b) Endothermic reactions: (1) and (3)

N2O2 is detected during the reaction! V. Reaction Mechanism Detailed molecular process by which reaction occurs Consists of 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) + O2 (g) 2NO2 (g) N2O2 is detected during the reaction! Elementary step: NO + NO N2O2 Overall reaction: 2NO + O2 2NO2 + Elementary step: N2O2 + O2 2NO2

V. Reaction Mechanism Elementary step: each step in a reaction mechanism Elementary step: NO + NO N2O2 N2O2 + O2 2NO2 Overall reaction: 2NO + O2 2NO2 + Intermediates Species that appear in a reaction mechanism but not in the overall balanced equation Always formed in an early elementary step and consumed in a later elementary step Ex. _____ is the intermediate in the above reaction. N2O2 Molecularity Number of molecules participate in an elementary step Unimolecular, bimolecular, termolecular reactions

Rate laws for elementary steps exponent in the rate law = coefficient in the elementary step Unimolecular reaction A products rate = k [A] Bimolecular reaction A + B products rate = k [A][B] Bimolecular reaction A + A products rate = k [A]2 Ex. Write the rate law for the the elementary step NO (g) + O3(g) NO2 (g) + O2 (g) rate = k [NO][O3]

Writing plausible reaction mechanisms aA + bB cC + dD ? When the exponent in rate law ≠ coeff in chemical reaction => multistep reaction 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.

Q: The experimental rate law for the reaction between NO2 and CO to produce NO and CO2 is rate = k[NO2]2. The reaction is believed to occur via two steps: Step 1: NO2 + NO2 NO + NO3 Step 2: NO3 + CO NO2 + CO2 What is the equation for the overall reaction? NO2+ CO NO + CO2 What is the intermediate? NO3 What can you say about the relative rates of steps 1 and 2? rate = k[NO2]2 is the rate law for step 1 so step 1 must be slower than step 2

rate = k[NO2][F2] is the rate law if step 1 is rds Q: The experimental rate law for the reaction between NO2 and F2 to produce NO2F is rate = k[NO2][F2]. What is the reaction mechanism? What is the equation for the overall reaction? 2 NO2+ F2 2 NO2F One possible mechanism? rate = k[NO2][F2] is the rate law if step 1 is rds rds Step 1: NO2 + F2 NO2F + F Step 2: NO2 + F NO2F Overall reaction: 2 NO2+ F2 2 NO2F What is the intermediate? F

Change the reaction rate without being used in the reaction VI. Catalysis Change the reaction rate without being used in the reaction Homogeneous catalysis Heterogeneous catalysis Enzyme Biological catalyst

A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed. k = A • exp( -Ea/RT ) Ea k uncatalyzed catalyzed ratecatalyzed > rateuncatalyzed Ea < Ea ’

In heterogeneous catalysis, the reactants and the catalysts are in different phases. - Haber synthesis of ammonia - Ostwald process for the production of nitric acid - Catalytic converters In homogeneous catalysis, the reactants and the catalysts are dispersed in a single phase, usually liquid or gas. - Acid catalysis - Base catalysis

Haber Process Fe/Al2O3/K2O catalyst N2 (g) + 3H2 (g) 2NH3 (g)

Enzyme Catalysis

Enzyme Catalysis E +S ES ES P + E D[P] rate = Dt rate = k [ES] enzyme uncatalyzed enzyme catalyzed rate = D[P] Dt E +S ES ES P + E rate = k [ES]

Rate law of enzyme catalysis uncatalyzed enzyme catalyzed E +S ES (fast equilibrium) k1 k-1 k1[E][S]= k-1[ES] [ES] = k1[E][S] k-1 k2 ES P + E rate = = k2 [ES] = D[P] Dt k1k2[E][S] k-1