This preview shows pages 1–5. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Quiz Solutions: S10 (Section 011, CHEM444) 27th February 2010 1 Quiz 1 1.1 Problem 1 What is the ratio of the following values for a particle of Argon and Krypton at 298K? • ν Ar ave ν Kr ave ν Ar ave ν Kr ave = r m Kr m Ar = r 83 . 798 39 . 948 = 1 . 45 • ν Ar most probable ν Kr most probable ν Ar most probable ν Kr most probable = r m Kr m Ar • ν Ar rms ν Kr rms ν Ar rms ν Kr rms = r m Kr m Ar 1 1.2 Problem 2 What is the ratio of average kinetic energy of the two systems of Problem 1? < KE > Ar < KE > Kr = 1 We will show this the long way. < KE > Ar = Z ∞ 4 π m Ar 2 π k B T 3 / 2 1 2 m Ar ν 2 e ( m Ar ν 2 / 2 k B T ) ν 2 dν = 1 2 m Ar ν 2 rms = 1 2 m Ar 3 k B T m Ar = 3 2 k B T < KE > Kr = Z ∞ 4 π m Kr 2 π k B T 3 / 2 1 2 m Kr ν 2 e ( m Kr ν 2 / 2 k B T ) ν 2 dν = 1 2 m Kr ν 2 rms = 1 2 m Kr 3 k B T m Kr = 3 2 k B T Thus the ratio is 1 . For ideal gases, energy is a function of temperature. Note that the expression for the average kinetic energy we derived in this exercise is identically the result one would obtain from rigorous statistical mechanical treatment. The kinetic energy is related to the rms speed, not the average or the most probable. The following information be of use: • 3D Maxwell Speed Distribution: F ( ν ) = Ω( ν ) = 4 π m 2 π k B T 3 / 2 e m ν 2 / 2 k B T ν 2 dν • ν ave = 8 k B T π m 1 / 2 • ν most probable = 2 k B T m 1 / 2 • ν rms = 3 k B T m 1 / 2 2 • R = N Avogadro k B • R = 8 . 3144349 J K 1 mol 1 • N Avogadro = 6 . 022 x 10 23 mol 1 • M ( molar mass ) of Argon = 39 . 948 gram mol 1 • M ( molar mass ) of Krypton = 83 . 798 gram mol 1 2 Quiz 2 2.1 Problem 1 What is the order of the reaction for each species and the total reaction order for each rate expression given below: Rate = k [ ClO ] [ BrO ] 1 , 1 , overall = 2 Rate = k [ NO ] 2 [ O 2 ] 2 , 1 , overall = 3 Rate = k [ HI ] 2 [ O 2 ] [ H + ] 1 / 2 2 , 1 , 1 / 2 , overall = 5 / 2 2.2 Problem 2 What is the overall order of the reaction corresponding to the following rate constants? k = 1 . 63 x 10 4 M 1 s 1 order = 2 k = 1 . 63 x 10 4 M 2 s 1 order = 3 k = 1 . 63 x 10 4 M 1 / 2 s 1 order = 3 / 2 3 2.3 Problem 3 Consider the following firstorder reaction, the decomposition of cyclobutane at 438 C at constant volume: C 4 H 8 ( g ) → 2 C 2 H 4 ( g ) Express the rate of this reaction in terms of the change in total pressure as a function of time. Initially there is no product, and thus the partial pressure of product initially is 0. We will also call the initial pressure, which is due only to reactant, as P in the following discussion. The rate for this reaction is: Rate = d [ C 4 H 8 ] dt = 1 2 d [ C 2 H 4 ] dt We can rewrite this in terms of the partial pressures of each species as suming ideal gas behavior: Rate = 1 RT d P C 4 H 8 dt = 1 2 RT d P C 2 H 4 dt Now let’s consider the pressure issue, since we were asked to determine the rate in terms of the total pressure. Keep in mind that we are consideringthe rate in terms of the total pressure....
View
Full
Document
This note was uploaded on 02/02/2012 for the course CHEM 444 taught by Professor Dybowski,c during the Fall '08 term at University of Delaware.
 Fall '08
 Dybowski,C
 Physical chemistry, pH

Click to edit the document details