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Average=84.2 Standard Deviation=13.4

Average=84.2 Standard Deviation=13.4. Average=161.6 Standard Deviation=24.9. Reaction Rates. Chemical kinetics is the study of reaction rates, how reaction rates change under varying conditions, and what molecular events occur during the overall reaction.

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Average=84.2 Standard Deviation=13.4

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  1. Average=84.2 Standard Deviation=13.4

  2. Average=161.6 Standard Deviation=24.9

  3. Reaction Rates • Chemical kinetics is the study of reaction rates, how reaction rates change under varying conditions, and what molecular events occur during the overall reaction. • What variables affect reaction rate? Surface area of a solid reactant or catalyst.

  4. The experimentally determined rate law is Dependence of Rate on Concentration • Reaction Order • Consider the reaction of nitric oxide with hydrogen according to the following equation. • Thus, the reaction is second order in NO, first order in H2, and third order overall.

  5. Dependence of Rate on Concentration • Reaction Order • Zero and negative orders are also possible. • The concentration of a reactant with a zero-order dependence has no effect on the rate of the reaction. • Although reaction orders frequently have whole number values (particularly 1 and 2), they can be fractional.

  6. Dependence of Rate on Concentration • Determining the Rate Law. • One method for determining the order of a reaction with respect to each reactant is the method of initial rates. • It involves running the experiment multiple times, each time varying the concentration of only one reactant and measuring its initial rate. • The resulting change in rate indicates the order with respect to that reactant.

  7. Dependence of Rate on Concentration • Determining the Rate Law. • If doubling the concentration of a reactant has a doubling effect on the rate, then one would deduce it was a first-order dependence. • If doubling the concentration had a quadrupling effect on the rate, one would deduce it was a second-order dependence. • A doubling of concentration that results in an eight-fold increase in the rate would be a third-order dependence.

  8. A Problem to Consider • Iodide ion is oxidized in acidic solution to triiodide ion, I3- , by hydrogen peroxide. • A series of four experiments was run at different concentrations, and the initial rates of I3- formation were determined. • From the following data, obtain the reaction orders with respect to H2O2, I-, and H+. • Calculate the numerical value of the rate constant.

  9. Initial Concentrations (mol/L) H2O2 I- H+ Initial Rate [mol/(L.s)] Exp. 1 0.010 0.010 0.00050 1.15 x 10-6 Exp. 2 0.020 0.010 0.00050 2.30 x 10-6 Exp. 3 0.010 0.020 0.00050 2.30 x 10-6 Exp. 4 0.010 0.010 0.00100 1.15 x 10-6 A Problem to Consider • Comparing Experiment 1 and Experiment 2, you see that when the H2O2 concentration doubles (with other concentrations constant), the rate doubles. • This implies a first-order dependence with respect to H2O2.

  10. Initial Concentrations (mol/L) H2O2 I- H+ Initial Rate [mol/(L.s)] Exp. 1 0.010 0.010 0.00050 1.15 x 10-6 Exp. 2 0.020 0.010 0.00050 2.30 x 10-6 Exp. 3 0.010 0.020 0.00050 2.30 x 10-6 Exp. 4 0.010 0.010 0.00100 1.15 x 10-6 A Problem to Consider • Comparing Experiment 1 and Experiment 3, you see that when the I- concentration doubles (with other concentrations constant), the rate doubles. • This implies a first-order dependence with respect to I-.

  11. Initial Concentrations (mol/L) H2O2 I- H+ Initial Rate [mol/(L.s)] Exp. 1 0.010 0.010 0.00050 1.15 x 10-6 Exp. 2 0.020 0.010 0.00050 2.30 x 10-6 Exp. 3 0.010 0.020 0.00050 2.30 x 10-6 Exp. 4 0.010 0.010 0.00100 1.15 x 10-6 A Problem to Consider • Comparing Experiment 1 and Experiment 4, you see that when the H+ concentration doubles (with other concentrations constant), the rate is unchanged. • This implies a zero-order dependence with respect to H+.

  12. Initial Concentrations (mol/L) H2O2 I- H+ Initial Rate [mol/(L.s)] Exp. 1 0.010 0.010 0.00050 1.15 x 10-6 Exp. 2 0.020 0.010 0.00050 2.30 x 10-6 Exp. 3 0.010 0.020 0.00050 2.30 x 10-6 Exp. 4 0.010 0.010 0.00100 1.15 x 10-6 • Because [H+]0 = 1, the rate law is: A Problem to Consider • The reaction orders with respect to H2O2, I-, and H+, are 1, 1, and 0, respectively.

  13. Initial Concentrations (mol/L) H2O2 I- H+ Initial Rate [mol/(L.s)] Exp. 1 0.010 0.010 0.00050 1.15 x 10-6 Exp. 2 0.020 0.010 0.00050 2.30 x 10-6 Exp. 3 0.010 0.020 0.00050 2.30 x 10-6 Exp. 4 0.010 0.010 0.00100 1.15 x 10-6 A Problem to Consider • You can now calculate the rate constant by substituting values from any of the experiments. Using Experiment 1 you obtain:

  14. Initial Concentrations (mol/L) H2O2 I- H+ Initial Rate [mol/(L.s)] Exp. 1 0.010 0.010 0.00050 1.15 x 10-6 Exp. 2 0.020 0.010 0.00050 2.30 x 10-6 Exp. 3 0.010 0.020 0.00050 2.30 x 10-6 Exp. 4 0.010 0.010 0.00100 1.15 x 10-6 A Problem to Consider • You can now calculate the rate constant by substituting values from any of the experiments. Using Experiment 1 you obtain:

  15. Change of Concentration with Time • A rate law simply tells you how the rate of reaction changes as reactant concentrations change. • A more useful mathematical relationship would show how a reactant concentration changes over a period of time.

  16. Change of Concentration with Time • A rate law simply tells you how the rate of reaction changes as reactant concentrations change. • Using calculus we can transform a rate law into a mathematical relationship between concentration and time. • This provides a graphical method for determining rate laws.

  17. You could write the rate law in the form Concentration-Time Equations • First-Order Rate Law

  18. Concentration-Time Equations • First-Order Rate Law • Using calculus, you get the following equation. • Here [A]t is the concentration of reactant A at time t, and [A]o is the initial concentration. • The ratio [A]t/[A]o is the fraction of A remaining at time t.

  19. The first-order time-concentration equation for this reaction would be: A Problem to Consider • The decomposition of N2O5 to NO2 and O2 is first order with a rate constant of 4.8 x 10-4 s-1. If the initial concentration of N2O5 is 1.65 x 10-2 mol/L, what is the concentration of N2O5 after 825 seconds?

  20. A Problem to Consider • The decomposition of N2O5 to NO2 and O2 is first order with a rate constant of 4.8 x 10-4 s-1. If the initial concentration of N2O5 is 1.65 x 10-2 mol/L, what is the concentration of N2O5 after 825 seconds? • Substituting the given information we obtain:

  21. A Problem to Consider • The decomposition of N2O5 to NO2 and O2 is first order with a rate constant of 4.8 x 10-4 s-1. If the initial concentration of N2O5 is 1.65 x 10-2 mol/L, what is the concentration of N2O5 after 825 seconds? • Substituting the given information we obtain:

  22. A Problem to Consider • The decomposition of N2O5 to NO2 and O2 is first order with a rate constant of 4.8 x 10-4s-1. If the initial concentration of N2O5 is 1.65 x 10-2 mol/L, what is the concentration of N2O5 after 825 seconds? • Taking the inverse natural log of both sides we obtain:

  23. A Problem to Consider • The decomposition of N2O5 to NO2 and O2 is first order with a rate constant of 4.8 x 10-4 s-1. If the initial concentration of N2O5 is 1.65 x 10-2 mol/L, what is the concentration of N2O5 after 825 seconds? • Solving for [N2O5] at 825 s we obtain:

  24. You could write the rate law in the form Concentration-Time Equations • Second-Order Rate Law

  25. Concentration-Time Equations • Second-Order Rate Law • Using calculus, you get the following equation. • Here [A]t is the concentration of reactant A at time t, and [A]o is the initial concentration.

  26. Graphing Kinetic Data • In addition to the method of initial rates, rate laws can be deduced by graphical methods. • If we rewrite the first-order concentration-time equation in a slightly different form, it can be identified as the equation of a straight line. • This means if you plot ln[A] versus time, you will get a straight line for a first-order reaction. (see Figure 14.9)

  27. Figure 14.9: A plot of log [R] versus time.

  28. Graphing Kinetic Data • In addition to the method of initial rates, rate laws can be deduced by graphical methods. • If we rewrite the second-order concentration-time equation in a slightly different form, it can be identified as the equation of a straight line. y = mx + b

  29. y = mx + b Graphing Kinetic Data • In addition to the method of initial rates, rate laws can be deduced by graphical methods. • If we rewrite the first-order concentration-time equation in a slightly different form, it can be identified as the equation of a straight line.

  30. In one half-life the amount of reactant decreases by one-half. Substituting into the first-order concentration-time equation, we get: Half-life • The half-life of a reaction is the time required for the reactant concentration to decrease to one-half of its initial value. • For a first-order reaction, the half-life is independent of the initial concentration of reactant.

  31. Half-life • The half-life of a reaction is the time required for the reactant concentration to decrease to one-half of its initial value. • Solving for t1/2we obtain: • Figure 14.8 illustrates the half-life of a first-order reaction.

  32. Figure 14.8: A graph illustrating that the half-life of a first-order reaction is independent of initial concentration. Half life, t1/2, is the time it takes for the [R] to decrease by 1/2. This is exactly like radioactive decay.

  33. Substitute the value of k into the relationship between k and t1/2. Half-life • Sulfuryl chloride, SO2Cl2, decomposes in a first-order reaction to SO2 and Cl2. • At 320 oC, the rate constant is 2.2 x 10-5 s-1. What is the half-life of SO2Cl2 vapor at this temperature?

  34. A model of SO2CI2(g)

  35. Half-life • Sulfuryl chloride, SO2Cl2, decomposes in a first-order reaction to SO2 and Cl2. • At 320 oC, the rate constant is 2.20 x 10-5 s-1. What is the half-life of SO2Cl2 vapor at this temperature? • Substitute the value of k into the relationship between k and t1/2.

  36. Half-life • Sulfuryl chloride, SO2Cl2, decomposes in a first-order reaction to SO2 and Cl2. • At 320 oC, the rate constant is 2.20 x 10-5 s-1. What is the half-life of SO2Cl2 vapor at this temperature? • Substitute the value of k into the relationship between k and t1/2.

  37. Again, assuming that [A]t = ½[A]o after one half-life, it can be shown that: Half-life • For a second-order reaction, half-life depends on the initial concentration and becomes larger as time goes on. • Each succeeding half-life is twice the length of its predecessor.

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