Chapter 14 Chemical Kinetics

Overview: Reaction Rates Rate Equations Graphical Methods Stoichiometry, Conditions, Concentration Rate Equations Order Initial Rate Concentration vs. Time First Order Rxns. Second Order Rxns. Graphical Methods

Cont’d Molecular Theory Reaction Mechanisms Catalysts Activation Energy Concentration Molecular Orientation Temperature Arrhenius Equation Reaction Mechanisms Elementary Steps, Reaction Order, Intermediates Catalysts

Reaction Rates What Affects Rates of Reactions? Concentration of the Reactants Temperature of Reaction Presence of a Catalyst Surface Area of Solid or Liquid Reactants

Reaction Rates (graphical): Average Rate = D[M] Dt [M] for reaction A  B D[M] time Dt

Rates for A  B - D[A] = D[B] Dt Dt Rate of the disappearance of A is equal in magnitude but opposite in sign to the rate of the appearance of B

Average Rate--D mol (or concentration) over a period of time, Dt Instantaneous Rate-- slope of the tangent at a specific time, t Initial Rate-- instantaneous rate at t = 0 [M] tangent at time, t t time

Average Rate = [A]final time - [A]initial time Dtfinal - Dtinitial for A  B

Instantaneous Ratetime, t = slope of the tangent at time = t

Stoichiometry 4PH3 => P4 + 6H2 - 1D[PH3] = + 1 D[P4] = + 1 D[H2] 4 Dt 1 Dt 6 Dt - D[PH3] = + 4 D[P4] = + 2 D[H2] Dt Dt 3 Dt

General Relationship Rate = - 1 D[A] = - 1 D[B] = + 1 D[C] = + 1 D[D] a D t b D t c D t d D t aA + bB  cC + dD

Conditions which affect rates Concentration concentration  rate Temperature temperature  rate Catalyst substance which increases rate but itself remains unchanged

Rate Equations: aA + bB  xX rate law rate = k[A]m[B]n m, n are orders of the reactants extent to which rate depends on concentration m + n = overall rxn order k is the rate constant for the reaction

Examples: 2N2O5 => 4NO2 + O2 rate = k[N2O5] 1st order 2NO + Cl2 => 2NOCl rate = k[NO]2[Cl2] 3rd order 2NH3 => N2 + 3H2 rate = k[NH3]0 = k 0th order

Determination of Rate Equations: Data for: A + B => C Expt. # [A] [B] initial rate 1 0.10 0.10 4.0 2 0.10 0.20 4.0 3 0.20 0.10 16.0 rate = k[A]2[B]0 = k[A]2 k = 4.0 Ms-1 = 400 M-1s-1 (0.10)2 M2

Exponent Values Relative to DRate Exponent Value [conc] rate 0 double same 1 double double 2 double x 4 3 double x 8 4 double x 16

Problem: Expt. # [NO] [H2] rate Data for: 2NO + H2 => N2O + H2O Expt. # [NO] [H2] rate 1 6.4x10-3 2.2x10-3 2.6x10-5 2 12.8x10-3 2.2x10-3 1.0x10-4 3 6.4x10-3 4.5x10-3 5.0x10-5 rate = k[NO]2[H2]

Units of Rate Constants units of rates M/s units of rate constants will vary depending on order of rxn M = 1 (M)2 (M) for s sM2 rate = k [A]2 [B] rate constants are independent of the concentration

Concentration vs. Time 1st and 2nd order integrated rate equations First Order: rate = - D[A] = k [A] D t ln [A]t = - kt A = reactant [A]0 or ln [A]t - ln [A]0 = - kt

Conversion to base-10 logarithms: ln [A]t = - kt [A]0 to log [A]t = - kt [A]0 2.303

Problem: The rate equation for the reaction of sucrose in water is, rate = k[C12H22O11]. After 2.57 h at 27°C, 5.00 g/L of sucrose has decreased to 4.50 g/L. Find k. C12H22O11(aq) + H2O(l) => 2C6H12O6 ln 4.50g/L = - k (2.57 h) 5.00g/L k = 0.0410 h-1

Concentration vs. Time Second Order: rate = - D[A] = k[A]2 Dt 1 - 1 = kt [A]t [A]0 second order rxn with one reactant: rate = k [A]2

Problem: 1 - 1 = (0.0113) t t = 102 min (0.300) (0.458) Ammonium cyanate, NH4NCO, rearranges in water to give urea, (NH2)2CO. If the original concentration of NH4NCO is 0.458 mol/L and k = 0.0113 L/mol min, how much time elapses before the concentration is reduced to 0.300 mol/L? NH4NCO(aq) => (NH2)2CO(aq) rate = k[NH4NCO] 1 - 1 = (0.0113) t (0.300) (0.458) t = 102 min

Graphical Methods Equation for a Straight Line y = bx + a ln[A]t = - kt + ln[A]0 1st order 1 = kt + 1 2nd order [A]t [A]0 b = slope a = y intercept x = time

First Order: 2H2O2(aq) ® 2H2O(l) + O2(g) time

First Order: 2H2O2(aq) ® 2H2O(l) + O2(g) slope, b = -1.06 x 10-3 min-1 = - k ln [H2O2] time

Second Order: 2NO2 ® 2NO + O2 1/[NO2] slope, b = +k time

Half-Life of a 1st order process: 0.020 M t1/2 = 0.693 k [M] 0.010 M 0.005 M t1/2 t1/2 time

Problem: SO2Cl2(g) => SO2(g) + Cl2(g) The decomposition of SO2Cl2 is first order in SO2Cl2 and has a half-life of 4.1 hr. If you begin with 1.6 x 10-3 mol of SO2Cl2 in a flask, how many hours elapse before the quantity of SO2Cl2 has decreased to 2.00 x 10-4 mol? SO2Cl2(g) => SO2(g) + Cl2(g)

Temperature Effects Rates typically increase with T increase Collisions between molecules increase Energy of collisions increase Even though only a small fraction of collisions lead to reaction Minimum Energy necessary for reaction is the Activation Energy

Molecular Theory (Collision Theory) Activation Energy, Ea DH reaction Ea forward rxn. Ea reverse rxn. Energy Reactant Product Reaction Progress

Activation Energy Activation Energy varies greatly Concentration almost zero to hundreds of kJ size of Ea affects reaction rates Concentration more molecules, more collisions Molecular Orientation collisions must occur “sterically”

The Arrhenius Equation increase temperature, inc. reaction rates rxn rates are a to energy, collisions, temp. & orient k = Ae-Ea/RT k = rxn rate constant A = frequency of collisions -Ea/RT = fraction of molecules with energy necessary for reaction

Graphical Determination of Ea rearrange eqtn to give straight-line eqtn y = bx + a ln k = -Ea 1 + ln A R T slope = -Ea/R ln k 1/T

Problem: Data for the following rxn are listed in the table. Calculate Ea graphically, calculate A and find k at 311 K. Mn(CO)5(CH3CN)+ + NC5H5 => Mn(CO)5(NC5H5)+ + CH3CN ln k k, min-1 T (K) 1/T x 10-3 -3.20 0.0409 298 3.35 -2.50 0.0818 308 3.25 -1.85 0.157 318 3.14

slope = -6373 = -Ea/R Ea = (-6373)(-8.31 x 10-3 kJ/K mol) = 53.0 kJ y intercept = 18.19 = ln A A = 8.0 x 10 7 -3.20 -2.50 -1.85 ln k k = 0.0985 min-1 3.14 3.25 3.35 x 10-3 1/T

Problem: The energy of activation for C4H8(g) => 2C2H4(g) is 260 kJ/mol at 800 K and k = 0.0315 sec Find k at 850 K. ln k2 = - Ea (1/T2 - 1/T1) k1 R k at 850 K = 0.314 sec-1

Reaction Mechanisms Elementary Step Molecularity equation describing a single molecular event Molecularity unimolecular bimolecular termolecular 2O3 => 3O2 (1) O3 => O2 + O unimolecular (2) O3 + O => 2 O2 bimolecular

Rate Equations Molecularity Rate Law unimolecular rate = k[A] bimolecular rate = k[A][B] bimolecular rate = k[A]2 termolecular rate = k[A]2[B] notice that molecularity for an elementary step is the same as the order

2O3 => 3O2 O3 => O2 + O rate = k[O3] O3 + O => 2O2 rate = k’[O3][O] 2O3 + O => 3O2 + O O is an intermediate

Problem: Write the rate equation and give the molecularity of the following elementary steps: NO(g) + NO3(g) => 2NO2(g) rate = k[NO][NO3] bimolecular (CH3)3CBr(aq) => (CH3)3C+(aq) + Br-(aq) rate = k[(CH3)3CBr] unimolecular

Mechanisms and Rate Equations rate determining step is the slow step -- the overall rate is limited by the rate determining step step 1 NO2 + F2 => FNO2 + F rate = k1[NO2][F2] k1 slow step 2 NO2 + F => FNO2 rate = k2[NO2][F] k2 fast overall 2NO2 + F2 => 2FNO2 rate = k1[NO2][F2]

Problem: Given the following reaction and rate law: NO2(g) + CO(g) => CO2(g) + NO(g) rate = k[NO2]2 Does the reaction occur in a single step? Given the two mechanisms, which is most likely: NO2 + NO2 =>NO3 + NO NO2 => NO + O NO3 + CO => NO2 + CO2 CO + O => CO2

Reaction Mechanisms & Equilibria 2O3(g) 3O2(g) overall rxn 1: O3(g) O2(g) + O(g) fast equil. rate1 = k1[O3] rate2 = k2[O2][O] 2: O(g) + O3(g) 2O2(g) slow rate3 = k3[O][O3] rate 3 includes the conc. of an intermediate and the exptl. rate law will include only species that are present in measurable quantities k1 k2 k3

Substitution Method at equilibrium k1[O3] = k2[O2][O] rate3 =k3[O][O3] [O] = k1 [O3] k2 [O2] rate3 = k3k1 [O3]2 or k2 [O2] overall rate = k’ [O3]2 [O2] substitute

Problem: Derive the rate law for the following reaction given the mechanism step below: OCl - (aq) + I -(aq) OI -(aq) + Cl -(aq) OCl - + H2O HOCl + OH - fast I - + HOCl HOI + Cl - slow HOI + OH - H2O + OI - fast k1 k2 k3 k4

Cont’d rate1 = k1 [OCl -][H2O] = rate 2 = k2 [HOCl][OH -] [HOCl] = k1[OCl -][H2O] k2[OH -] rate 3 = k3 [HOCl][I -] rate 3 = k3k1[OCl -][H2O][I -] k2 [OH -] overall rate = k’ [OCl -][I -] [OH -] solvent

Catalyst Facilitates the progress of a reaction by lowering the overall activation energy homogeneous heterogeneous

Ea Ea DHrxn Energy Reaction Progress catalysts are used in an early rxn step but regenerated in a later rxn step

Uncatalyzed Reaction: O3(g) <=> O2(g) + O(g) O(g) + O3(g) => 2O2(g) Catalyzed Reaction: Step 1: Cl(g) + O3(g) + O(g) => ClO(g) + O2(g) + O(g) Step 2: ClO(g) + O2(g) + O(g) => Cl(g) + 2O2(g) Overall rxn: O3(g) + O(g) => 2O2(g)

Ea uncatalyzed rxn Ea catalyzed rxn ClO + O2 + O Cl + O3 + O