CHEM 1415 1st EditionLecture 14Unit: Integrated Rate Laws and ApplicationsOutline of Current Lecture I. Forms of Rate EquationII. Half LifeIII. First Order ProcessIV. Example ProblemV. Reaction Between two gaseous moleculesVI. Arrhenius EquationCurrent Lecture Forms of Rate EquationDifferential form of Rate Equation Integrated FormZero Orderrate=k[A]0[ A ]t=− kt+[ A ]0First Orderrate=k[A]1[ A ]t=[ A ]0e−ktSecond Orderrate=k[A]2rate=k[A][B]1[ A ]t−1[ A]0=kt Half life- of a reactant is the time it takes for its concentration to get to one half of its original valueo[ A ]t1/2=12[ A ]0 First Order Processo[A]t=[A]0e−kto[A]t12=[A]0e−k t12o12[A]0=[A]0e−k t12o12=e−k t12These 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.oln(12)=− k t1 /2 ot1 /2= .693k Example problemo You take two Aspirin that are 325 mg each. If the half life for Aspirin is 3 hours how much Aspirin is left in the bloodstream after 4 hours?o (2 aspirin)(325 mg) = 650 mgo4 hrs3hrshalflife=1.33 half liveso(650 mg) (.5)1.33=258 mg For a reaction to occur between two gaseous moleculeso 1) must cloto 2) have enough energy Electron repulsions Break bonds in the reactantso 3) right geometric orientation Arrhenius Equationok =A e−EaRToEa = energy of activationo R = gas constanto T = temperatureo A = frequency factorso K = rate constanto Increase T →increase k → increased rateo Increase Ea→ decresed k → decreased
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