Chapter 14 Chemical Kinetics

Reaction Rates Combustion of propane (C 3 H 8 ) Rusting of iron (Fe 2 O 3 ) Rate at which reactants disappear / products form KINETICS

Factors governing reaction rates: Chemical nature Contact ability Concentration Temperature Catalyst

Rates: Quantitative aspects X  Y + Z  [X] /  t Units of rates (concentration / time): C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O(g) Know rate of change of [O 2 ]; rates of other species in equation? Use coefficients from balanced equation.

Reaction Rates vs. time Not constant during a reaction (straight-line relationship between conc. & time) Plot conc. vs time, usually obtain a curve 2NO 2 (g)  2NO(g) + O 2 (g) Can plot conc. vs time for NO 2, NO and O 2 Obtain rate wrt NO 2, NO and O 2 Shape of plot for formation of a product vs. shape of plot for consumption of a reactant

Concentration and Rate, Rate Laws Looked at one component of a reaction Rate expression that includes all reactants A + B  Products Rate  [A] m [B] n (exponents m and n determined experimentally- look at very soon) Rate Law: Shows how initial reaction rate changes wrt concentration of species involved; introduces rate constant, k Rate = k [A] m [B] n Calculate rate with k, m and n (any concentration of A and B)

Rate Law calculation H 2 SeO 3 + 6I - + 4H +  Se + 2I H 2 O Rate Law: Rate = k [H 2 SeO 3 ] x [I - ] y [H + ] z Let x=1; y=3; z=2 (found experimentally) for initial rate of reaction k = 5.0 x 10 5 L 5 mol -5 s -1 (at 0 °C – k dependent on T) Note: Units of k (look to be strange – can actually be deduced; see later) Need concentrations of H 2 SeO 3, I - and H +

Exponents in Rate Law - No No’s Exponents in Rate Law  Coefficients in equation 2A + 3B  C + D Rate = k[A] 2 [B] 3 ; Cannot make this assumption (unless experimentally determined) Coincidences do occur: 2HI(g)  I 2 (g) + H 2 (g) Rate =

Exponents and Orders of reaction Exponent = Order of reaction wrt reactant 2N 2 O 5  4NO 2 + O 2 Rate = k[N 2 O 5 ]; 1 st order wrt N 2 O 5 ; 1 st order overall 2HI(g)  I 2 (g) + H 2 (g) Rate = k[HI] 2 ; 2 nd order wrt HI; 2 nd order overall H 2 SeO 3 + 6I - + 4H +  Se + 2I H 2 O Rate = k [H 2 SeO 3 ] [I - ] 3 [H + ] 2 Orders of reaction? Overall Order of reaction = Sum of orders of reaction of each reactant. Overall order for above reaction?

Experimental Determn. - Exponents Study effects of variation of concentration on initial reaction rate: A + B  products; Rate = k[A] n [B] m ; values of n and m? Vary concs. of A and B separately, and observe effect on rate. Expt. #Initial Concs (M)Initial Rate (mol L -1 s -1 ) [A][B] Rate Law: Rate = k If desired, k can now be calculated for this reaction

Integrated Rate Law – 1 st order Rate Law: Shows initial rate varies with concentration. Integrated Rate Law: Shows how conc. varies with time 1 st order reaction; Rate = k[A] ln [A] o / [A] t = kt(ln = natural logarithm) 2N 2 O 5  2N 2 O 4 + O 2 ; Rate = k[N 2 O 5 ] At 45 °C, k = 6.22 x s -1. If initial conc. Of N 2 O 5 is 0.10M, how many minutes for [N 2 O 5 ] to drop to 0.01 M?

Integrated Rate Law – 2 nd order B  A + C’ Rate = k[B] 2 1/ [B] t – 1/[B] o = kt 2NOCl  2NO + Cl 2 ; Rate = k[ At 50 °C, initial conc. of NOCl is M, conc. after 30 min? k = L mol -1 s -1

Zero order reactions Rate is constant (does not change with concentration) A  B + C Rate law: Rate = k[A] 0 = k Integrated Rate Law: [A] t = -kt + [A] o Commonly observed with catalysts, enzymes 2N 2 O(g)  2N 2 (g) + O 2 (g)

Half-life (t 1/2 ) Half-life of a particular chemical t for ½ reactant to disappear 3g A  1 st and 2 nd order reactions (half-life is constant)

Half-life (t 1/2 ) – 1 st order reactions t 1/2 = / k 131 I; t 1/2 = 8 days. 10g sample of 131 I, after 8 days….decayed down to…. Radioactive emission (  -,  -,  -, X-rays) to stable isotope Fraction of 131 I present after 24 days? Amount of 131 I?

Half-life (t 1/2 ) – 2 nd order reactions t 1/2 = 1/ k[X] o 2HI(g)  I 2 (g) + H 2 (g); Rate = k[HI] 2 ; k = Lmol -1 s -1 Initial conc. of HI = 0.050M; t 1/2 = ?

Predicting Rate Laws / Mechanisms Reaction - elementary processes  mechanism of reaction Elementary process : Rate law can be written from chemical equation Overall Reaction: 2NOCl  2NO + Cl 2 Rate Laws can be written for elementary processes. Helps to predict reaction mechanisms. 2NO 2 Cl  2NO 2 + Cl 2 (1) Rate = k[NO 2 Cl] (Experimental determination) Occurs in one step? NO 2 Cl  NO 2 + Cl (a) NO 2 Cl + Cl  NO 2 + Cl 2 (b) (a) + (b) = (1) Rate Determining Step

Predicting Rate Laws / Mechanisms NO 2 (g) + CO(g)  NO(g) + CO 2 (g) Rate = k[NO 2 ] 2 ; Single step? Possible Mechanism: Mechanism : A series of elementary steps that must satisfy 2 requirements: (1) Sum of elementary steps must give overall balanced equation for reaction (2) Mechanism must agree with experimentally derived Rate Law

Elementary Steps

Another reaction 2NO + 2H 2  N 2 + 2H 2 O; Rate = k[NO] 2 [H 2 ]; Single step? Possible Mechanism: Avoid elementary processes (whenever possible) that involve more than 2 molecules. Actual Mechanism:

Mechanism problem 2NO 2 (g) + F 2 (g)  2NO 2 F(g); Rate = k[NO 2 ][F 2 ] Suggested Mechanism: NO 2 + F 2  NO 2 F + F (slow) F + NO 2  NO 2 F (fast) Acceptable mechanism? Does it satisfy the 2 requirements?

How do reactions occur? Model for Chemical Kinetics Collision Model (Collision Theory) Molecules must successfully collide to react. 2 reasons for ineffective collisions 1. Insufficient K.E to overcome E a barrier 2. Orientation issues

Activation Energy (E a ) Changing T changes rate; also changes k Extent of k variation w /T depends on E a. Arrhenius Equation: k=Ae –Ea/RT k & T exponentially related Calculation of E a : ln (k 2 /k 1 ) = -E a /RT (1/T 2 – 1/T 1 ) Decomposition of HI has rate constants: k = L mol -1 s -1 at 508 °C k = 0.24 L mol -1 s -1 at 540 °C; E a of reaction (kJmol -1 )?

Catalysts Change rates of reactions without being used up Lowers activation energy of reaction (changing mechanism) Homogeneous catalysts Heterogeneous catalysts Catalytic converters