CHEM 102 1nd Edition Lecture 16 Outline of Last Lecture I. Transition State Theory II. Arrehenius EquationIII. Reaction MechanismOutline of Current Lecture I. Rate Determining StepII. Guidelines for Constructing MechanismsIII. Mechanisms with Fast First StepsIV. CatalystsCurrent LectureI. Rate Determining Stepa. The rate determining step is the slowest step in the mechanismb. If the first step is the slow step, the reaction rate of the entire reaction is the reaction rate of the first step.c. Examplei. Reaction: NO2 + CO NO + CO2ii. Experimental Rate = k[NO2]21. Proposed two step mechanism:2. NO2 + NO2 NO3 + NO slow step3. NO3 + CO NO2 + CO2 fast step4. Combined reactions = NO2 + CO NO + CO25. The slowest step is the first step; therefore, the reaction rate of the entire reaction is the same as the rate of the first6. So, the reaction rate would be: k[NO2]2 which is the same as the reaction rate observed in the experimentII. Guidelines for Constructing Mechanismsa. If given the choice, always choose a bimolecular equation over a trimolecular equation as the probability of a trimolecular equation actually reacting is small.b. If on a test, the question is which mechanism will match the experimental reaction rate, compare which mechanisms will have the right reaction ratesc. ExampleThese 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.i. 2NO2 + F2 2NO2Fii. Experimental Reaction Rate: k[NO2][F2]iii. One Step Mechanism: NO2 + NO2 + F2 2NO2Fiv. However, the reaction rate of the mechanism is k[NO2]2[F2] which does not equal the experimental reaction rate.d. Example:i. If the following reaction were a one step reaction what would the overall order be?ii. 2 ADP + 2 PI + C6H12O6 2 ATP + H2O + 2 C3H3O2iii. Answer: 5iv. Rate = k[ADP]2[PI]2[Glucose]v. 2 + 2+ 1 = 5III. Mechanisms with Fast First Stepsa. You are given the reaction: 2NO + Br2 2NOBrb. The experimental reaction rate is k[NO]2[Br]i. Mechanism:ii. NO + Br2 NOBr2 (eq) fast stepiii. NOBr2 + NO 2NOBr slow stepiv. Write the rate of the second step: k2[NOBr2][NO]v. The overall reaction rate cannot have an intermediate term (a compound that is made then used) in it, substitute for NOBr2vi. A step marked with a double ended arrow or (eq) means that the reactionrate forward will equal the reaction rate backwards; therefore, [NO][Br2]=[NOBr2].vii. So, the reaction rate overall = [NO][Br2][NO] or [NO]2[Br2] which equals the experimental reaction rate.viii. The same process would be used if faced with a three step mechanism, identify the slowest step, write the reaction rate, and substitute any intermediate compoundsIV. Catalystsa. Substance which increases the reaction rate without being consumedb. The catalyst will show on the reactant side and the product side of the equation.c. Speeds the forward and reverse reactiond. Can be part of the rate
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