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UI CHEM 1120 - Rate Laws, Reaction Orders, and Half Life
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Chem 1120 1st Edition Lecture 6Outline of Last Lecture I. Reaction RatesII. Concentration and RateIII. Average Rate vs Instantaneous RateA. Definitions, ExamplesB. Reaction Order and Units of Rate Constants Outline of Current Lecture I. Method of Initial RatesA. Deriving the Rate Law II. Concentration and TimeA. Quick Logarithm ReviewB. Reaction Orders III. Half LifeCurrent LectureI. Rate laws must be determined experimentally, and what will usually happen is that your professor will give you a bunch of data from an ex-periment and you will have to use the data given to figure out the rate law, etc. The method of initial rates determines reaction orders from the effect of changing a reactant’s concentration on the initial rate of the reaction. Example: Given the reaction, 2N2O5 ——> 4NO2 + O2 (g) These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.EXPERIMENT # [N2O5]O (mol/L) INITIAL RATE (mol/(L*s))1 0.45 M 2.7 X 10^-42 0.90 M 5.4 X 10^-4====> Rate = k [N2O5]A. The easiest way to explain how to derive the rate law is to show an example. I will debrief it first and then once you read over the example, it willmake more sense. To find the rate law for a reaction, you must use the ex-perimental data and use the reactants and products to put the law into the correct form. Let’s say the equation is A + B ——> C. The rate law would look something like this:r = k [reactant A]n [reactant B]m [product C]pr = ratek = constantn = 1, m = 1, p = 2 = order (we will learn how to find these ex-ponents below)Example: Deduce the rate law for the reaction from the following experimen-tal data:BrO3- (aq) + 5Br- (aq) + 6H+ (aq) ——> 3Br2 (l) + 3H2O (l)EXPERIMENT # [BrO3-]O[Br-]O[H+]OINITIAL RATE (mol/(L*s))1 0.10 M 0.10 M 0.10 M 8 X 10^ -42 0.20 M 0.10 M 0.10 M 1.6 X 10^ -33 0.20 M 0.20 M 0.10 M 3.2 X 10^ -34 0.10 M 0.10 M 0.20 M 3.2 X 10^ -3Solution: Assume a rate law of the form,rate = k [BrO3-]n [Br-]m [H+]p How can n, m, and p be found? ——> use two experiments’ data:EXPERIMENT #[BrO3-]O[Br-]O[H+]OINITIAL RATE (mol/(L*s))1 0.1 M 0.1 M 0.1 M 8 X 10^ -42 0.2 M 0.1 M 0.1 M 1.6 X 10^ -3 II. A lot of times we are interested in the rate or speed of an equation, but sometimes instead we want to know how the concentration of a reactant or product change with time, not the rate. Calculus can help us derive these re-lationships between the reaction rates and concentrations as a function of time.A. In case you do not remember logarithms,B. Equations expressing the relationship between concentration and time can be derived by calculus. The resulting equations depend on the rate law used.Zero Order Reactions: A + bB ——> ProductsFirst Order Reactions: A + bB ——> ProductsSecond Order Reactions: A + bB ——> ProductsExample: At 300 oC, is the reaction, NO2 (g) ——>NO (g) + 1/2 O2 (g), 1st or 2nd Order? III. Half life = t1/2 = the time required for the concentration of a reactant to decrease to one half of its initial valueAfter one half life, [A] = 1/2 [A]o mZero Order Reactions:First Order Reactions: Second Order Reactions:REVIEW OF ZERO, FIRST, AND SECOND ORDERS ON NEXT


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UI CHEM 1120 - Rate Laws, Reaction Orders, and Half Life

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