Chapter 12: Chemical Kinetics

1. Reaction Rates2. Rate Laws: Differential and Integrated3. Experimental Determination of the Rate Law4. Integrated Rate Laws & Concentration/Time Problems5. Collision Theory of Reactions

a. Reaction Mechanismsb. Predicting Rate Lawsc. Temperature Dependence of the Rate Constant

(Arrhenius)

6. Catalysis

As the reaction proceeds, it gets slower because the rate depends on [NO2].

Figure 12.1

Example 1:Determine the rate law and the value of the rateconstant for the following reaction.

2 NO + Br2 2 NOBr

Experiment [NO]0 [Br2]0 Initial Rate (mole/L s)

1 1.1 x 10-5 1.2 x 10-5 0.37

2 1.9 x 10-5 1.3 x 10-5 0.69

3 3.5 x 10-5 1.2 x 10-5 1.2

4 4.0 x 10-5 3.0 x 10-5 3.4

Example 2:Determine the rate law and the value of the rate constant for the following reaction in whichOH- is a catalyst.

OCl- + I- OI- + Cl-

Experiment [OCl-]0

(mole/L)

[I-]0

(mole/L)

[OH-]0

(mole/L)

Rate(mole/L s)

1 .0040 .0020 1.00 4.8x10-4

2 .0020 .0040 1.00 5.0x10-4

3 .0020 .0020 1.00 2.4x10-4

4 .0020 .0020 0.50 4.6x10-4

5 .0020 .0020 0.25 9.4x10-4

Example 3:Time (s) [H2O2]

(mole/L)

0.000 1.000

120.000 0.910

300.000 0.780

600.000 0.590

1200.000 0.370

1800.000 0.220

2400.000 0.130

3000.000 0.082

3600.000 0.050

The concentration of H2O2 was monitored over time for the following reaction at 25°C:

H2O2 (aq) H2 (g) + O2 (g)

Find the rate law and the value of the rate constant for this reaction at 25°C.

Example 4:

The decomposition of N2O5 to NO2 and O2 is

first order with a rate constant of

4.80 x 10-4 /s at 45°C.

(a) If the initial concentration of N2O5 is 1.65 x 10-2 mole/L, what is the concentration after 825 s?

(b) How long would it take for the concentration of N2O5 to decrease to 1.00 x 10-2 mole/L if its initial concentration wa that given in (a)?

Example 5:

For the following decomposition reaction,AB A + BRate = k[AB]2

k = 0.20 L/mole sHow long will it take for [AB] to reach one third of its original value of 1.50 M? Whatis [AB] after 10.0 seconds?

Zero Order Half Life

t1/2 = [A]0 / 2k

Note that this half life decreases as the reaction proceeds.

There is less reactant to Consume and it runs at thesame rate so it takes less

time.

Time (s) [A]

0 1.6

2.0 .80

3.0 .40

3.5 .20

3.75 .10

3.88 .050

First Order Half Lifet1/2 = 0.693 / k

The first order half life is

constantfor a particular reaction and temperature.

As the initial concentration decreases, the decrease in rate is compensated for by less reactant to consume.

Time (s) [A]

0 1.6

24.0 .80

48.0 .40

72.0 .20

96.0 .10

120.0 .050

Second Order Half Life

t1/2 = 1 / [A]0 k

Note that this half life increases as the reaction proceeds.

The rate decreases so strongly with concentration that it still takes longer to use up half ofthe reactant even though the amount of reactant consumed isless for each half life.

Time (s) [A]

0 1.6

10.2 .80

30.6 .40

71.4 .20

153 .10

Example 6:

A decomposition reaction has a rate constant of 0.0012 yr-1.

a) What is the half life of this reaction?

b) How long does it take for the concentration of the reactant to reach 12.5% of its original value?

Example 7:

It took 143 s for 50.0% of a particular substance to decompose. If the initial concentration was 0.060 M and the decomposition reaction follows second order kinetics, what is the value for the rate constant?

Example 8:

For the reaction A products, successive half-lives are observed to be 10.0, 20.0, and 40.0 min for an experiment in which [A]0=0.10M.

Calculate the concentration of A at the following times.

a) 80.0 minb) 30.0 min

Example 9:(Book #34)

O + NO2 NO + O2

[NO2] in large excess,

1.0x1013molecules/cm3

a) Find the order of the reaction with respect to [O].

b) Reaction is known to be first order in [NO2]. Find the value of k.

Time (s) [O] atoms/cm3

0 5.0x109

1.0x10-2 1.9x109

2.0x10-2 6.8x108

3.0x10-2 2.5x108

Example 10:

Consider the following mechanism for the decomposition of hydrogen peroxide catalyzed by I-.

Step 1: H2O2 + I- H2O + IO- ; fast equilibrium

Step 2: IO- + H2O2 H2O + O2 + I- ; slow

a) Write a balanced equation for the overall reaction.b) List any catalysts.c) List any intermediates.d) Write the rate law for the overall reaction.

Example 11:

Consider the following mechanism for the production of phosgene:

Step 1: Cl2 (g) 2 Cl (g); fast equilibriumStep 2: Cl (g) + CO (g) COCl (g); fast equilibriumStep 3: COCl (g) + Cl2 (g) COCl2 (g) + Cl (g); slow

Step 4: 2 Cl (g) Cl2 (g); fast

a) Write a balanced equation for the overall reaction.b) List any catalysts.c) List any intermediates.d) Write the rate law for the overall reaction.

Figure 12.11: 2BrNO 2 NO + Br2

Figure 12.13:

Example 12:

The activation energy of the following reaction is 3.5 kJ/mole and the change in enthalpy is -66.6 kJ/mole. Calculate the activation energy of the reverse reaction. OH + HCl H2O + Cl

Hint: Sketch the reaction energy diagram.

Example 13:

For the reaction A2 + B2 2 AB, the activation energy of the forward reaction is 125 kJ/mole and that of the reverse reaction is 85 kJ/mole. Assuming the reaction occurs in one step:a) Draw a reaction energy diagram.b) Calculate H for the reaction.c) Sketch a possible transition state.

Figure 12.2: Collisions with E > Ea at two temperatures.

Example 14:

The isomerization reaction:CH3NC CH3CN

has an activation energy of 161 kJ/mole.If the rate constant at 600K is 0.41/s, what is it at 1000K?

Example 15:

The rate constant of a reaction is 4.50x10-5 L/mol•s at 195°C and 3.20 x 10-3 L/mol•s at 258°C. What is the activation energy of this reaction?

Catalysts Work by Lowering Ea:They provide a different mechanism!

More collisions result in reaction when Ea is lower.

Homogeneous Catalysis

In the stratosphere, NO catalyzes the

destruction of ozone, O3.

NO (g) + O3 (g) NO2 (g) + O2 (g)

O (g) + NO2 (g) NO (g) + O2 (g)

Overall:

O3 (g) + O (g) 2 O2 (g)

C2H4 + H2 C2H6

with a metal catalyst

a. reactantsb. adsorptionc. migration/reactiond. desorption

Heterogeneous Catalysis

Enzymes are Catalysts

Enzymes are Catalysts

Enzymes are Catalysts