Chapter 13 Lecture 1 Chemical KineticsI. Kinetics = The area of chemistry that is concerned with reaction rates.

A. What do Kinetics tell us?

1) Stoichiometry tells us what and how many reactants/products there are

2) Thermodynamics tells us if the reaction is energetically favorable (-H)

3) Kinetics describes how fast or slow a reaction occurs

2 H2 + O2 2 H2O H = -242 kJ/mol

This spontaneous reaction is so slow, it essentially doesn’t happen!

4) We need Kinetics to fully describe chemical reactions

5) Kinetics help us determine how a reaction occurs = Mechanism

B. Reaction Rates = Change in concentration (conc) of a reactant or product per unit time.

1) For generic reaction: aA + bB cC + dD

t

][H

t-t

][H-][HRate 2

12

t2t2 12

t

[D]1

t

[C]1

t

[B]1

t

[A]1Rate

dcba

If you find the rate for onereactant or product, you’vefound them all!

This is an average rateover this time period.

1) Example: 2 NO2 2 NO + O2

Time (s) [NO2] [NO] [O2]

0 .0100 0 0

50 .0079 .0021 .0011

100 .0065 .0035 .0018

150 .0055 .0045 .0023

200 .0048 .0052 .0026

250 .0043 .0057 .0029

300 .0038 .0062 .0031

350 .0034 .0066 .0033

400 .0031 .0069 .0035

2) [NO2] = mol/L of NO2. This is decreasing with time.

3) [NO] and [O2] are increasing with time

4)

a) Rates are always positive for product ([A]2 > [A]1)

b) Rates are always negative for reactants ([A]2 < [A]1)

c) To let us work with only + numbers, we always put a (-) before the rate if it is describing a reactant

5) The rate changes over the course of the reaction

a) Instantaneous Rate at a point = slope of a tangent line at the point

b) At t = 100 s, what is the instantaneous rate?

12

12

t-t

]A[[A]

t

[A]Rate

smol/L 102.4s 0s 50

]0100[.]0079.0[

t-t

]A[[A]

t

[A]][NO Rate 5

12

122

smol/L 104.2s110

]0026.0[

50s-160s

0.0076][]005.[0

t

]NO[

x

ySlope 52

Rate Time (s)

4.2x10-5 0-50

2.8x10-5 50-100

2.0x10-5 100-150

1.4x10-5 150-200

1.0x10-5 200-250

6) Rates of Product Formation 2 NO2 2 NO + O2

a) 2 NO produced for every 2 NO2 consumed

i. Rate same at any point in the reaction

ii. Curve is same, only inverted

iii. Rate [NO] at 250 s = 8.6 x 10-6 mol/L.s

b) 1 O2 produced for every NO2 consumed

i. Rate = ½ the rate of NO2 consumption at any point

ii. Rate = ½ the rate of NO production at any point

iii. Rate [O2] at 250 s = 4.3 x 10-6 mol/L.s

iv. O2 curve has a different shape

c) We can write an equation of the rates from the balanced equation

7) We must be specific when we talk about a reaction rate.

a) Rate depends on which reactant or product

b) Rate depends on how long the reaction has been going

t

]O[

1

1

t

[NO]

2

1

t

][NO

2

1Rate 22

II. Rate Laws

A. Background1) Chemical reaction are reversible. Products recombine to give reactant.

2 NO2 2 NO + O2

2) When you start with only reactants, the reaction only goes one way

1) [NO2] at this point depends on only the forward reaction2) This makes equations simple

3) After some products are formed, the reaction goes both directions

a) [NO2] at this point depends on both forward and reverse reactionsb) This makes equations complicated

4) We will consider reactions only at times and conditions that have only the forward reaction of any significance

B. Rate Law Basics1) A Rate Law describes how concentrations are changing in a reaction2) Rate = k[A]n

a) k = rate constant = different for each reactionb) n = the order of the reactant A

c) Products do not appear in the rate law if we use conditions giving the forward reaction only

d) The order n must be determined experimentally; it can’t be written directly from the balanced equation

e) We must specify which species we are using [A] in each rate law

3) Differential Rate Law tells us how rate changes with concentration

a) This is what we will call the rate law

b)

4) Integrated Rate Law tells us how concentrations depend on time

a) This can be derived from the differential rate law (and vice versa)

b) We will look at integrated rate laws later

5) What do we learn from rate laws?

a) They help us determine if a reaction is fast enough to be useful

b) They help us figure out the exact steps (mechanism) of a reaction

i. Slowest step determines the overall rate

ii. Chemists speed up reactions by changing that step

t

]A[k[A]rate n

C. Determining the Form of a Rate Law

1) We must do experiments to determine the order of each reactant in the rate law

2) Example: 2 N2O5 4 NO2 + O2 (g)

a) O2 gas escapes, so there is no reverse reaction

b) We can determine rate at various times from the data[N2O5] Time (s)

1.00 0

0.88 200

0.78 400

0.69 600

0.61 800

0.54 1000

0.48 1200

0.43 1400

0.38 1600

0.34 1800

0.30 2000

c) Examine how the rate changes as concentration changes

[N2O5] = 0.90 M rate = 5.4 x 10-4 mol/L.s

[N2O5] = 0.45 M rate = 2.7 x 10-4 mol/L.s

d) How does the [A] ratio compare to the rate ratio?

i. If the ratios are the same, the order of the reactant is 1

ii.

iii. When the order is 1, we call this a First Order Reaction

e) The general expression for a first order reaction is:

1

2

45.0

90.0

1

2

107.2

104.54

4

152

52 ]Ok[Nt

]O[N- rate

1k[A]t

[A]- rate

3) The Method of Initial Rates

a) Initial rate = rate just after a reaction starts (t = 0). This is used because we don’t have to worry about reverse reaction.

b) We start the reaction several time with different initial concentrations of reactants.

c) If the rate changes in the same ratio as the concentration, n = 1

d) If the rate changes as the square of the concentration ratio, n = 2

e) If the rate changes as the cube of the concentration ratio, n = 3, etc…

f) If the rate doesn’t change at all, n = 0

4) Example with 2 reactants: NH4+ + NO2

- N2 + 2 H2O

5) First order in both reactants, and overall second order (n + m = 2)

Exp. # [NH4+]0 [NO2

-]0Rate (mol/L.s)

1 0.100 0.0050 1.35 x 10-7

2 0.100 0.0100 2.70 x 10-7

3 0.200 0.0100 5.40 x 10-7

Rate = k[NH4+]n[NO2

-]m

Exp1Exp2 [NH4+] same, [NO2

-] doublesrate doubles so m = 1

Exp2Exp3 [NH4+] doubles, [NO2

-] samerate doubles so n = 1

Rate = k[NH4+]1[NO2

-]1 = k[NH4+][NO2

-]

6) Now we can actually find what k is from any of the experiments since we know that: Rate = k[NH4

+][NO2-]

7) Example: Find rate law, k, and order of the reaction

BrO3- + 5 Br- + 6 H+ 3 Br2 + 3 H2O

sL/mol 107.2]mol/L 005.0][mol/L 100.0[

smol/L 1035.1

]][NO[NH

ratek 4

7

-24

Exp. # [BrO3-] [Br-] [H+] rate

1 0.10 0.10 0.10 8.0 x 10-4

2 0.20 0.10 0.10 1.6 x 10-3

3 0.20 0.20 0.10 3.2 x 10-3

4 0.10 0.10 0.20 3.2 x 10-3