# Wright CHM 1220 - CHM 1220 SI_Chemical Kinetics II_Rev 2_ANSWERS (2 pages)

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**View the full content.**## CHM 1220 SI_Chemical Kinetics II_Rev 2_ANSWERS

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## CHM 1220 SI_Chemical Kinetics II_Rev 2_ANSWERS

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- Chm 1220 - General Chemistry II

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Questions Comments Suggestions Email me at stevens 161 wright edu ANSWERS Chapter 13 SI Activity Chemical Kinetics II CHM 1220 Problems 1 Beta blockers are drugs that are used to manage hypertension It is important for doctors to know how rapidly a beta blocker is eliminated from the body A certain beta blocker is eliminated in a first order process with a rate constant of 7 6 10 3 min 1 at normal body temperature 37 C A patient is given 20 mg of the drug What mass of the drug remains in the body 5 0 h after administration Need first order integrated rate law ln X kt ln X 0 Use the drug mass m in place of the concentration ln mt kt ln m0 Solving the expression for mt ln mt ln m0 kt m ln t kt m0 mt e kt m0 mt m0 e kt Plugging in given values kt mt m 0e 20 mg e 60 min 7 6 10 3 min 1 5 0 hr 1 hr 2 0 mg 2 The following mechanism has been proposed for the gas phase reaction between HBr and NO2 Step 1 HBr NO2 HOBr NO slow Step 2 HBr HOBr H2O Br2 fast Write the overall net reaction Write the rate law for each elementary step and indicate the molecularity Indicate which step is the rate determining step Identify the reaction intermediate s if any Adding the two steps to find net reaction HBr HBr 2 HBr NO 2 HOBr NO 2 HOBr H 2O H 2O NO Br2 Br2 NO Since these are elementary steps rate laws are pulled from balanced chemical eqn Step 1 Rate1 k1 HBr NO 2 Molecularity Bimolecular Step 2 Rate 2 k2 HBr HOBr Molecularity Bimolecular Rate determining step is the slowest Step 1 Intermediates are consumed as soon as they are created HOBr 3 A reaction rate increases 1000 fold in the presence of a catalyst at 298 K The activation energy of the original pathway is 98 kJ mol What is the activation energy of the new pathway all other factors being equal Arrhenius eqn Will need a system of Arrhenius eqns Defining variables k1 k T 298 E ln k a ln A k2 1000 k Ea 1 98 kJ mol RT Plugging each pathway into its own Arrhenius eqn The two temperature Ea 1 form of the Arrhenius Original Pathway ln k ln A equation was not used RT because the change in k Ea 2 is NOT due to a change ln A New Pathway ln 1000k RT in temperature Solving this system by subtracting new old 1000 k Ea 1 Ea 2 E E Ea 2 Ea 1 RT ln 1000 ln 1000k ln k a 2 ln A a 1 ln A ln RT k RT RT Plugging in variables Ea 2 90 kJ mol 0 00831447 kJ mol K 298 K ln 1000 80 9 kJ mol Potential Energy J 4 Refer to the reaction coordinate diagram below How many steps does this reaction have Which is the rate determining step in this reaction Which step is the fastest How many intermediates must form in the reaction A catalyst is added that accelerates the third step only What effect if any will the catalyst have on the rate of the overall reaction Is this reaction endothermic or exothermic Activated Complexes Ea H Progress of Reaction Number of Steps Three RDS Step 1 Fastest Step Step 3 Intermediates Two Catalytic Effect None Enthalpy Exothermic Number of steps number of activated complexes Rate Determining Step RDS slowest step highest activation energy barrier Fastest step lowest activation energy barrier Intermediates must exist in multi step reactions represented by dips between activated complexes marked with Since the catalyst does not affect the rate determining step there is not effect for the overall reaction Enthalpy is negative exothermic because the final energy state is lesser than the initial Therefore heat was lost 5 For the reaction A B a plot of 1 A vs time in seconds is perfectly linear with a slope of 1 75 10 6 If the reaction begins with A 0 125 M how many years will it be until B 0 123 M Since 1 A is linear this is a second order rxn Therefore A remaining k slope 1 75 10 6 M 1s 1 If B 0 123 M then 0 123 M of A has reacted So A 0 A reacted A 0 B 0 125 M 0 123 M 0 002 M Solving the second order integrated rate law for time and plugging in variables 1 1 1 1 A A 0 0 002 M 0 125 M 2 81 108 s 8 91 yr 1 1 kt t t k 1 75 10 6 M 1s 1 A A 0 Rev 2 Fall 2017 J Stevens

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