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

Previewing page 1 of 2 page document View the full content.
View Full Document

CHM 1220 SI_Chemical Kinetics II_Rev 2_ANSWERS



Previewing page 1 of actual document.

View the full content.
View Full Document
View Full Document

CHM 1220 SI_Chemical Kinetics II_Rev 2_ANSWERS

32 views


Pages:
2
School:
Wright State University
Course:
Chm 1220 - General Chemistry II

Unformatted text preview:

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



View Full Document

Access the best Study Guides, Lecture Notes and Practice Exams

Loading Unlocking...
Login

Join to view CHM 1220 SI_Chemical Kinetics II_Rev 2_ANSWERS and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view CHM 1220 SI_Chemical Kinetics II_Rev 2_ANSWERS and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?