1. Chapter 14
Chemical Kinetics

2. 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

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

4. 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

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

6. 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

7. 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

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

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

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

11. 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

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

13. 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

14. 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

15. 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

16. Exponent Values Relative to DRate
Exponent Value [conc] rate
0 double same double double double x double x double x 16

17. 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]

18. 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

19. 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

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

21. 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

22. 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

23. 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

24. 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

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

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

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

28. 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

29. 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)

30. 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

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

32. 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”

33. 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

34. 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

35. 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

36. 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

37. 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

38. 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

39. 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

40. 2O3 => 3O2 O3 => O2 + O rate = k[O3]
O O => 2O2 rate = k’[O3][O]
2O3 + O => 3O2 + O
O is an intermediate

41. 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

42. 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]

43. 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

44. 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

45. 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

46. 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

47. 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

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

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

50. 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)

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