slide 1 of 61 chemistry ninth edition general principles and modern applications petrucci harwood...
TRANSCRIPT
Slide 1 of 61
CHEMISTRYNinth
Edition GENERAL
Principles and Modern Applications
Petrucci • Harwood • Herring • Madura
Chapter 14: Chemical Kinetics
Slide 2 of 61
Contents
14-1 The Rate of a Chemical Reaction
14-3 Effect of Concentration on Reaction Rates: The Rate Law
14-4 Zero-Order Reactions
14-5 First-Order Reactions
14-6 Second-Order Reactions
14-7 Reaction Kinetics: A Summary
Slide 3 of 61
Contents
14-8 Theoretical Models for Chemical Kinetics
14-9 The Effect of Temperature on Reaction Rates
14-10 Reaction Mechanisms
14-11 Catalysis
Slide 4 of 61
14-1 The Rate of a Chemical Reaction
Rate of change of concentration with time.
2 Fe3+(aq) + Sn2+ → 2 Fe2+(aq) + Sn4+(aq)
t = 38.5 s [Fe2+] = 0.0010 M
Δt = 38.5 s Δ[Fe2+] = (0.0010 – 0) M
Rate of formation of Fe2+= = = 2.610-5 M s-1Δ[Fe2+]
Δt
0.0010 M
38.5 s
Slide 5 of 61
Rates of Chemical Reaction
Δ[Sn4+]Δt
2 Fe3+(aq) + Sn2+ → 2 Fe2+(aq) + Sn4+(aq)
Δ[Fe2+]
Δt=
1
2
Δ[Fe3+]
Δt = -
1
2
Slide 6 of 61
General Rate of Reaction
a A + b B → c C + d D
Rate of reaction = rate of disappearance of reactants
=Δ[C]
Δt1c
=Δ[D]
Δt1d
Δ[A]
Δt1a
= -Δ[B]
Δt1b
= -
= rate of appearance of products
Slide 7 of 61
14-3 Effect of Concentration on Reaction Rates: The Rate Law
a A + b B …. → g G + h H ….
Rate of reaction = k[A]m[B]n ….
Rate constant = k
Overall order of reaction = m + n + ….
Slide 8 of 61
Establishing the Order of a reaction by the Method of Initial Rates. Use the data provided establish the order of the reaction with respect to HgCl2 and C2O2
2- and also the overall order of the reaction.
EXAMPLE 14-3
Slide 9 of 61
Notice that concentration changes between reactions are by a factor of 2.
Write and take ratios of rate laws taking this into account.
EXAMPLE 14-3
Slide 10 of 61
R2 = k[HgCl2]2m[C2O4
2-]2n
R3 = k[HgCl2]3m[C2O4
2-]3n
R2
R3
k(2[HgCl2]3)m[C2O42-]3
n
k[HgCl2]3m[C2O4
2-]3n
=
2m = 2.0 therefore m = 1.0
R2
R3
k2m[HgCl2]3m[C2O4
2-]3n
k[HgCl2]3m[C2O4
2-]3n
= = 2.0=2mR3
R3
= k(2[HgCl2]3)m[C2O42-]3
n
EXAMPLE 14-3
Slide 11 of 61
R2 = k[HgCl2]21[C2O4
2-]2n = k(0.105)(0.30)n
R1 = k[HgCl2]11[C2O4
2-]1n = k(0.105)(0.15)n
R2
R1
k(0.105)(0.30)n
k(0.105)(0.15)n =
7.110-5
1.810-5= 3.94
R2
R1
(0.30)n
(0.15)n = = 2n =
2n = 3.94 therefore n = 2.0
EXAMPLE 14-3
Slide 12 of 61
+ = Third Order
R2 = k[HgCl2]2 [C2O4
2-]2
First order
1
Second order
2
EXAMPLE 14-3
Slide 13 of 61
Worked Examples Follow:
Slide 14 of 61
Slide 15 of 61
CRS Questions Follow:
Slide 16 of 61
Consider the following reaction,
2 2 22NO Br(g) 2NO (g) + Br (g)
whose rate can be expressed as
2d[Br ]r =
dt
Equivalent expressions are…
2 2d[NO ] d[NO Br]1. r = =
dt dt
2 21 d[NO ] 1 d[NO Br]2. r = =
2 dt 2 dt
2 21 d[NO ] 1 d[NO Br]3. r = =
2 dt 2 dt
2 21 d[NO ] 1 d[NO Br]4. r = =
2 dt 2 dt
Slide 17 of 61
Consider the following reaction,
2 2 22NO Br(g) 2NO (g) + Br (g)
whose rate can be expressed as
2d[Br ]r =
dt
Equivalent expressions are…
2 2d[NO ] d[NO Br]1. r = =
dt dt
2 21 d[NO ] 1 d[NO Br]2. r = =
2 dt 2 dt
2 21 d[NO ] 1 d[NO Br]3. r = =
2 dt 2 dt
2 21 d[NO ] 1 d[NO Br]4. r = =
2 dt 2 dt