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Chapter 12: Chemical Kinetics 1.Reaction Rates 2.Rate Laws: Differential and Integrated 3.Experimental Determination of the Rate Law 4.Integrated Rate.

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Presentation on theme: "Chapter 12: Chemical Kinetics 1.Reaction Rates 2.Rate Laws: Differential and Integrated 3.Experimental Determination of the Rate Law 4.Integrated Rate."— Presentation transcript:

1 Chapter 12: Chemical Kinetics 1.Reaction Rates 2.Rate Laws: Differential and Integrated 3.Experimental Determination of the Rate Law 4.Integrated Rate Laws & Concentration/Time Problems 5.Collision Theory of Reactions a.Reaction Mechanisms b.Predicting Rate Laws c.Temperature Dependence of the Rate Constant (Arrhenius) 6.Catalysis

2 As the reaction proceeds, it gets slower because the rate depends on [NO 2 ].

3 Figure 12.1

4 Example 1: Determine the rate law and the value of the rate constant for the following reaction. 2 NO + Br 2  2 NOBr Experiment [NO] 0 [Br 2 ] 0 Initial Rate (mole/L s) 11.1 x x x x x x x x

5 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 (mole/L) [ OH - ] 0 (mole/L) Rate (mole/L s) x x x x x10 -4

6 Example 3: Time (s)[H 2 O 2 ] (mole/L) The concentration of H 2 O 2 was monitored over time for the following reaction at 25°C: H 2 O 2 (aq)  H 2 (g) + O 2 (g) Find the rate law and the value of the rate constant for this reaction at 25°C.

7 Example 4: The decomposition of N 2 O 5 to NO 2 and O 2 is first order with a rate constant of 4.80 x /s at 45°C. (a)If the initial concentration of N 2 O 5 is 1.65 x mole/L, what is the concentration after 825 s? (b)How long would it take for the concentration of N 2 O 5 to decrease to 1.00 x mole/L if its initial concentration wa that given in (a)?

8 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?

9 Zero Order Half Life t 1/2 = [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]

10 First Order Half Life t 1/2 = / 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]

11 Second Order Half Life t 1/2 = 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]

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13 Example 6: A decomposition reaction has a rate constant of 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?

14 Example 7: It took 143 s for 50.0% of a particular substance to decompose. If the initial concentration was M and the decomposition reaction follows second order kinetics, what is the value for the rate constant?

15 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 min b)30.0 min

16 Example 9: (Book #34) O + NO 2  NO + O 2 [NO 2 ] in large excess, 1.0x10 13 molecules/cm 3 a)Find the order of the reaction with respect to [O]. b)Reaction is known to be first order in [NO 2 ]. Find the value of k. Time (s)[O] atoms/cm x x x x x x x10 8

17 Example 10: Consider the following mechanism for the decomposition of hydrogen peroxide catalyzed by I -. Step 1: H 2 O 2 + I - H 2 O + IO - ; fast equilibrium Step 2: IO - + H 2 O 2  H 2 O + O 2 + 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.

18 Example 11: Consider the following mechanism for the production of phosgene: Step 1: Cl 2 (g) 2 Cl (g); fast equilibrium Step 2: Cl (g) + CO (g) COCl (g); fast equilibrium Step 3: COCl (g) + Cl 2 (g)  COCl 2 (g) + Cl (g); slow Step 4: 2 Cl (g)  Cl 2 (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.

19 Figure 12.11: 2BrNO  2 NO + Br 2

20 Figure 12.13:

21 Example 12: The activation energy of the following reaction is 3.5 kJ/mole and the change in enthalpy is kJ/mole. Calculate the activation energy of the reverse reaction. OH + HCl  H 2 O + Cl Hint: Sketch the reaction energy diagram.

22 Example 13: For the reaction A 2 + B 2  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.

23 Figure 12.2: Collisions with E > E a at two temperatures.

24 Example 14: The isomerization reaction: CH 3 NC  CH 3 CN has an activation energy of 161 kJ/mole. If the rate constant at 600K is 0.41/s, what is it at 1000K?

25 Example 15: The rate constant of a reaction is 4.50x10 -5 L/mols at 195°C and 3.20 x L/mols at 258°C. What is the activation energy of this reaction?

26 Catalysts Work by Lowering E a : They provide a different mechanism!

27 More collisions result in reaction when E a is lower.

28 Homogeneous Catalysis In the stratosphere, NO catalyzes the destruction of ozone, O 3. NO (g) + O 3 (g)  NO 2 (g) + O 2 (g) O (g) + NO 2 (g)  NO (g) + O 2 (g) Overall: O 3 (g) + O (g)  2 O 2 (g)

29 C 2 H 4 + H 2  C 2 H 6 with a metal catalyst a. reactants b. adsorption c. migration/reaction d. desorption Heterogeneous Catalysis

30 Enzymes are Catalysts

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