Chapter 12: Chemical Kinetics Reaction Rates Rate Laws: Differential and Integrated Experimental Determination of the Rate Law Integrated Rate Laws & Concentration/Time Problems Collision Theory of Reactions Reaction Mechanisms Predicting Rate Laws Temperature Dependence of the Rate Constant (Arrhenius) Catalysis

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

Figure 12.1

Initial Rate (mole/L s) Example 1: Determine the rate law and the value of the rate constant 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 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 which OH- is a catalyst. OCl- + I-  OI- + Cl- Experiment [OCl-]0 (mole/L) [I-]0 [OH-]0 Rate (mole/L s) 1 .0040 .0020 1.00 4.8x10-4 2 5.0x10-4 3 2.4x10-4 4 0.50 4.6x10-4 5 0.25 9.4x10-4

Example 3: 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. 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

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. If the initial concentration of N2O5 is 1.65 x 10-2 mole/L, what is the concentration after 825 s? 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 + B Rate = k[AB]2 k = 0.20 L/mole s How long will it take for [AB] to reach one third of its original value of 1.50 M? What is [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 the same rate so it takes less time. Time (s) [A] 1.6 2.0 .80 3.0 .40 3.5 .20 3.75 .10 3.88 .050

First Order Half Life t1/2 = 0.693 / k The first order half life is constant for 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] 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 of the reactant even though the amount of reactant consumed is less for each half life. Time (s) [A] 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. What is the half life of this reaction? 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. 80.0 min 30.0 min

Example 9: (Book #34) O + NO2  NO + O2 [NO2] in large excess, 1.0x1013molecules/cm3 Find the order of the reaction with respect to [O]. Reaction is known to be first order in [NO2]. Find the value of k. Time (s) [O] atoms/cm3 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 equilibrium Step 2: Cl (g) + CO (g) COCl (g); fast equilibrium Step 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: Draw a reaction energy diagram. Calculate DH for the reaction. 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)

Heterogeneous Catalysis C2H4 + H2  C2H6 with a metal catalyst a. reactants b. adsorption c. migration/reaction d. desorption

Enzymes are Catalysts

Enzymes are Catalysts

Enzymes are Catalysts