Créer une présentation
Télécharger la présentation

Télécharger la présentation
## Kinetics Part III: Integrated Rate Laws

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -

**Kinetics Part III:Integrated Rate Laws**Jespersen Chap. 14 Sec 4 Dr. C. Yau Spring 2013 1 1**Integrated Rate Laws**How are “integrated rate laws” different from “rate laws?” A rate law gives the speed of reaction at a given point in time, and how it is affected by the concentration of the reactants. An integrated rate law gives the change in concentration from time zero up to a given point in time. 2 2**Integrated Rate Law for First-Order Reactions**We already know the rate law for a 1st order reaction is Rate = k[A] Using calculus, the integrated rate law is Where [A]o = molarity of A at time zero [A]t = molarity of A at time t and k = rate constant for the reaction 3**Integrated Rate Law of First-Order Reactions**It is often more useful to rewrite the equation by finding the anti natural log of both sides of the equation: 4**Quick Reminder of Treatment of Sig. Fig. in Log**Log 3.4 x 10-1 = -0.468521083 = -0.47 2 sig.fig. 2 decimal places antilog -3.213 = 10-3.213= 6.123503x10-4 = 6.12x10-4 3 decimal places 3 sig. fig.**Treatment of Sig. Fig. in Ln**Ln 3.4 x 10-1 = -1.078809 = - 1.08 antiln -4.273 = e-4.273 = 1.393990x10-2 = 1.39x10-2 2 sig. fig. 3 decimal places**Use of the Integrated Rate of 1st Order Rxns**Example 14.7 p. 655Dinitrogen pentoxide is not very stable. In the gas phase or dissolved in a nonaqueous solvent, like CCl4, it decomposes by a 1st order rxn into dinitrogen tetroxide and molecular oxygen. 2 N2O5 2 N2O4 + O2 The rate law is Rate = k[N2O5] At 45oC, the rate constant for the rxn is 6.22 x 10-4s-1. If the initial concentration of N2O5 at 45oC is 0.500 M, what will its concentration be after exactly one hour? Ans. 0.053 M Do Practice Exercises 15 & 16 p.656 7**Determination of the Rate Constantfor the Integrated Rate**Law The rate constant can be determined graphically. Again we manipulate the rate law to allow us to obtain a linear graph: Variables are [A]t and t How does this allow us to determine the rate constant, k, graphically? First determine which are variable & which are constants. 8**Determination of the Rate Constantfor the Integrated Rate**Law Slope is calculated to be - 6.0x10-4 s-1. So, k = 6.0x10-4 s-1 Rate = k[N2O5] Units are correct! What is the slope equal to? slope = - k So, k = - slope Note: (-) does not mean k is negative but it has the opposite sign of the slope.**Half-life (t½ ) of a Reactant**"Half-life" of a reactant is the time it takes for ½ of the reactant to disappear. It is NOT half the time it takes for all of the reactant to disappear. For example, t½ of the radioisotope I-131 is 8.0 days. Starting with 20 g of I-131, after 8.0 days, there would be 10 g left. After a total of 16.0 days (two half-lives), there would be…… 5 g left**Half-life is a measure of the speed of reaction: The shorter**the half-life, the faster is the reaction. • By definition, after one half-life, [A]t = ½ [A]o For a 1st order rxn, Note that half-life isnot affected byconcentration. How do you know that? There is no concentration term in the equation.**Fig.14.7 p.658**Application of Half-life I-131 has a half-life of 8 days. After 4 half-lives (32 days), it is down to 1/16 of its original concentration.**If we start with 20.0 g of I-131, after 40 days, how much of**I-131 is still there? (Half-life of I-131 = 8 days.) What percent of I-131 remains after 24 days? What fraction of I-131 remains after 48 days?**Example 14.8 A patient is given a certain amt of I-131 as**part of a diagnostic procedure for a thyroid disorder. What fraction of the initial I-131 would be present in a patient after 25 days if none of it were eliminated through natural body processes? t1/2 = 8.02 days What to do if # days is not an exact multiple of the half-life? We solve this problem using the integrated rate law for 1st order rxn?**Half-life = 8.02 days, fraction after 25 days?**After 25 days, 1/9 of the original initial I-131 would be present. Do Practice Exercises 15, 16, 17 p. 659**The half-life of I-132 is 2.295h. What percentage remains**after 24 hours? (Previous problem was for I-131.) % is based on 100 Ans. 0.071 % I-132 remains.**C-14 dating: Determination of the age of organic**substances. • When object is still living, it is ingesting C with a constant ratio of C-14/C-12 = 1.2x10-12 (given) = ro • Once the object dies, the amount of C becomes constant and no more C-14 is incorporated. As a result the C-14/C-12 ratio begins to decrease due to the decay of C-14 (to give rt = ratio at time t). • By measuring the C-14/C-12 ratio (rt) of the object, we can estimate how long it has been dead.**REMEMBER THIS!**C-14 has a half-life of 5730 years and radioactive decay is a first-order process. This means… and = 1.21x10-4 yr -1 What is the rate constant, knowing that the half-life is 5730 yrs?**Example 14.9 p.660**A sample of an ancient wooden object was found to have ratio of 14C to 12C equal to 3.3x10-13 as compared to a contemporary biological sample, which has a ratio of 1.2x10-12. What is the age of the object? Calculate k from t1/2 for C-14 = 1.21x10-4 yr-1 Find t. Ans. 1.07x104 years Do Pract Exer 20, 21 p.661**Integrated Rate Law of 2nd Order Reactions**• They are of several types: Rate=k[A]2, Rate=k[A]1[B]1 and Rate=k[A]2[B]0, etc… • 2nd order means the powers must add up to 2. • The integrated equation is of the form**Example 14.10 p.661**Nitrosyl chloride, NOCl, decomposes slowly to NO and Cl2. 2NOCl 2NO + Cl2 The rate law shows that the rate is second order in NOCl. Rate = k[NOCl]2 The rate constant k equals 0.020 L mol-1 s-1 at a certain temp. If the initial conc of NOCl in a closed reaction vessel is 0.050 M, what will the concentration be after 35 minutes? Do Pract Exer 22 & 23 p.662 Ans. 0.016 M**Integrated Rate Law of 2nd Order Reactions**How can we determine rate constant graphically? Which are the variables? Rearrange y = mx + b What do we plot on the x-axis? y-axis? What does the slope represent?**Derived from**Half-life of 2nd Order Reactions Note that unlike 1st order reactions, half-life of 2nd order rxns depend on the initial conc of the reactant. Ex 14.11 p.664 2HI(g) H2(g) + I2 (g) has the rate law, Rate = k[HI]2 with k = 0.079 L mol-1s-1 at 508oC. What is the half-life of this rxn at this temp when the initial HI concentration is 0.10 M? Ans. 1.3x102s Do Prac Ex 24 & 25 p.664**Summary of Integrated Rate Laws**1st Order Rxn 2nd Order Rxn