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Chapter 14 Chemical Kinetics
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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
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Cont’d Molecular Theory Reaction Mechanisms Catalysts
Activation Energy Concentration Molecular Orientation Temperature Arrhenius Equation Reaction Mechanisms Elementary Steps, Reaction Order, Intermediates Catalysts
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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
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Reaction Rates (graphical):
Average Rate = D[M] Dt [M] for reaction A B D[M] time Dt
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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
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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
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Average Rate = [A]final time [A]initial time Dtfinal Dtinitial for A B
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Instantaneous Ratetime, t = slope of the tangent at time = t
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Stoichiometry 4PH3 => P4 + 6H2
- 1D[PH3] = D[P4] = D[H2] Dt Dt Dt - D[PH3] = D[P4] = + 2 D[H2] Dt Dt Dt
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General Relationship Rate = 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
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Conditions which affect rates
Concentration concentration rate Temperature temperature rate Catalyst substance which increases rate but itself remains unchanged
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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
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Examples: 2N2O5 => 4NO2 + O2 rate = k[N2O5] 1st order
2NO Cl2 => 2NOCl rate = k[NO]2[Cl2] 3rd order 2NH3 => N H2 rate = k[NH3]0 = k 0th order
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Determination of Rate Equations:
Data for: A B => C Expt. # [A] [B] initial rate rate = k[A]2[B]0 = k[A]2 k = 4.0 Ms = M-1s (0.10)2 M2
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Exponent Values Relative to DRate
Exponent Value [conc] rate 0 double same double double double x double x double x 16
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Problem: Expt. # [NO] [H2] rate
Data for: 2NO H2 => N2O H2O Expt. # [NO] [H2] rate x x x x x x x x x rate = k[NO]2[H2]
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Units of Rate Constants
units of rates M/s units of rate constants will vary depending on order of rxn M = (M)2 (M) for s sM rate = k [A]2 [B] rate constants are independent of the concentration
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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
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Conversion to base-10 logarithms:
ln [A]t = - kt [A]0 to log [A]t = - kt [A]
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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) g/L k = h-1
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Concentration vs. Time Second Order: rate = - D[A] = k[A]2 Dt
= kt [A]t [A]0 second order rxn with one reactant: rate = k [A]2
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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 mol/L and k = L/mol min, how much time elapses before the concentration is reduced to mol/L? NH4NCO(aq) => (NH2)2CO(aq) rate = k[NH4NCO] = (0.0113) t (0.300) (0.458) t = min
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Graphical Methods Equation for a Straight Line y = bx + a
ln[A]t = - kt ln[A]0 1st order = kt nd order [A]t [A]0 b = slope a = y intercept x = time
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First Order: 2H2O2(aq) ® 2H2O(l) + O2(g)
time
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First Order: 2H2O2(aq) ® 2H2O(l) + O2(g)
slope, b = x min-1 = - k ln [H2O2] time
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Second Order: 2NO ® NO O2 1/[NO2] slope, b = +k time
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Half-Life of a 1st order process:
0.020 M t1/2 = k [M] 0.010 M 0.005 M t1/2 t1/2 time
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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)
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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
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Molecular Theory (Collision Theory)
Activation Energy, Ea DH reaction Ea forward rxn. Ea reverse rxn. Energy Reactant Product Reaction Progress
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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”
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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
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Graphical Determination of Ea
rearrange eqtn to give straight-line eqtn y = bx a ln k = -Ea ln A R T slope = -Ea/R ln k 1/T
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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
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slope = -6373 = -Ea/R Ea = (-6373)(-8.31 x 10-3 kJ/K mol) = 53.0 kJ
y intercept = = ln A A = 8.0 x 10 7 ln k k = min-1 x 10-3 1/T
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Problem: The energy of activation for C4H8(g) => 2C2H4(g)
is 260 kJ/mol at 800 K and k = sec Find k at 850 K. ln k2 = - Ea (1/T /T1) k R k at 850 K = sec-1
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Reaction Mechanisms Elementary Step Molecularity
equation describing a single molecular event Molecularity unimolecular bimolecular termolecular 2O => O2 (1) O => O O unimolecular (2) O O => 2 O bimolecular
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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
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2O3 => 3O2 O3 => O2 + O rate = k[O3]
O O => 2O2 rate = k’[O3][O] 2O3 + O => 3O2 + O O is an intermediate
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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
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Mechanisms and Rate Equations
rate determining step is the slow step -- the overall rate is limited by the rate determining step step NO2 + F2 => FNO2 + F rate = k1[NO2][F2] k slow step NO F => FNO2 rate = k2[NO2][F] k fast overall 2NO2 + F2 => 2FNO2 rate = k1[NO2][F2]
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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 + CO CO + O => CO2
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Reaction Mechanisms & Equilibria
2O3(g) O2(g) overall rxn 1: O3(g) O2(g) O(g) fast equil. rate1 = k1[O3] rate2 = k2[O2][O] 2: O(g) O3(g) O2(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
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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] [O2] substitute
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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
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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
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Catalyst Facilitates the progress of a reaction by lowering the overall activation energy homogeneous heterogeneous
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Ea Ea DHrxn Energy Reaction Progress catalysts are used in an early rxn step but regenerated in a later rxn step
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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) O2(g) Overall rxn: O3(g) O(g) => 2O2(g)
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Ea uncatalyzed rxn Ea catalyzed rxn ClO + O2 + O Cl + O3 + O
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