NAME KEY FRANCIS SCOTT CHEMISTRY 419 SPRING 2010 2103 Final Examination May 22 2010 Answer each question in the space provided use back of page if extra space is needed Answer questions so the grader can READILY understand your work only work on the exam sheet will be considered Write answers where appropriate with reasonable numbers of significant figures You may use only the Student Handbook a calculator and a straight edge 1 10 points The diffusion coefficient of lysozyme MW 14 1 kg mol is 0 104 10 9 m2s 1 How long does it take this protein on average to diffuse an rms distance of exactly 1 m in solution The root mean square distance traveled in three dimensional diffusion is r2 DO NOT WRITE IN THIS SPACE 1 10 2 10 6 Dt 3 10 Substitution into this equation gives the result t r2 6D 4 10 1 10 6 m 2 6 0 104 10 9 m 2 s 1 5 10 6 10 3 1 60 10 s 1 6 ms 7 10 8 10 9 10 10 10 11 10 Extra credit 12 10 TOTAL PTS 110 PERCENTAGE NAME CHEM 419 Final Exam Spring 2010 page 2 2 10 points Iodine reacts with a ketone in aqueous solution to form an iodoketone by the following reaction I2 ketone iodoketone I H Birk and Walters J Chem Ed 69 585 587 1992 report the following initial rate data for the iodination of acetone measured at 23 C H mol dm 3 0 0404 0 0809 0 323 0 323 0 323 Acetone mol dm 3 1 33 1 33 0 333 0 667 0 333 I2 mol dm 3 0 00665 0 00665 0 00665 0 00665 0 00332 106 Initial rate mol dm 3 s 1 1 85 3 89 3 76 7 55 3 75 Write the rate law for this reaction showing the orders with respect to the reactants and the hydrogen ion Show your work and explain how you got the answers HINT Remember that orders are generally whole numbers or perhaps half order So a slight difference from integral orders may be considered to be due to experimental error The various orders are found by taking the appropriate ratios For example since the iodine and acetone are constant in the first two experiments it can be used to determine the dependence on hydrogen ion concentration 0 404 0 0809 n 1 85 3 89 Taking the logarithm of both sides gives the following equation 0 69438n 0 74322 Solving this for n gives n 0 74322 0 69438 1 07 which is close to 1 Similarly one may use the third and fourth experiments to determine the order with respect to acetone 0 333 0 667 n 3 76 7 55 Taking the logarithm of both sides gives the following equation 0 69465n 0 6971 Solving this for n gives n 0 6971 0 69465 1 00 Finally one may use the third and fifth experiments to determine the order with respect to iodine 0 00665 0 00332 n 3 76 3 75 Taking the logarithm of both sides gives the following equation 0 69465n 0 00266 Solving this for n gives n 0 00266 0 69465 0 004 which is close to 0 So the rate equation looks like this v k Acetone 1 H 1 I 2 0 k Acetone H Interestingly it does not depend on the concentration of iodine Score for Page NAME CHEM 419 Final Exam Spring 2010 page 3 3 10 points Viscosity like many other dynamic processes can be considered to be thermally activated If that is the case the viscosity obeys the equation T 0 e E RT where 0 is a pre exponential factor and E is an activation energy The viscosity of liquid benzene has been measured over a relatively wide range of temperature as given in the table From these data determine the values of 0 and E for benzene Show your work clearly HINT Be careful Note that the formula is subtly different from the Arrhenius formula for the temperature dependence of a rate constant cP t C 5 0 836 40 0 492 80 0 318 120 0 219 160 0 156 T K 278 15 313 15 353 15 393 15 433 15 1 T 1 K 0 00359 5 0 00319 3 0 00283 2 0 00254 4 0 00230 9 ln cP 0 179127 0 709277 1 145704 1 518684 1 857899 0 000000 0 200000 0 400000 ln cP 0 600000 y 1290 5x 4 8176 R2 0 9994 0 800000 1 000000 1 200000 1 400000 1 600000 1 800000 2 000000 0 002 0 0022 0 0024 0 0026 0 0028 0 003 0 0032 0 0034 0 0036 0 0038 1 T 1 K The plot of the logarithm of the viscosity versus 1 T should be linear The intercept is logarithm of 0 and E can be determined from the slope Using the data on the plot one finds that 0 The activation energy is exp 4 8176 cP 0 008086 cP E slope R 1290 5 K 8 3144349 J K 1 mol 1 10 73 kJ mol 1 Score for Page NAME CHEM 419 Final Exam Spring 2010 page 4 4 10 points For each of the phrases in column A match it with the appropriate statement or definition in column B by putting the proper number in the blank Column A 14 Activated complex theory Column B 1 Approximation that the concentrations of reactive intermediates do not change in time 15 Electrophoresis 2 Chemical process that occurs in a single step 2 Elementary step 3 Constant in the Arrhenius equation 10 Initial rate method 4 Correction for the friction coefficient of nonspherical molecules 8 Laminar flow 5 Describes how the reaction rate depends on concentration 4 Perrin factor 6 Diffusion limits the reaction rate 12 Random walk 7 Equation for the rms velocity 5 Reaction order 8 Flow may be decomposed into layers of constant speed 13 Sedimentation coefficient 9 Graph of potential energy of configurations 1 Steady state approximation 10 Measurement of change at the earliest times of a reaction 11 Moving through a gel in a worm like fashion 12 Series of steps each of which is not correlated with the previous or subsequent step in direction 13 Terminal velocity divided by the acceleration 14 Transition state theory 15 Transport that occurs when a charged molecule is placed in an electric field Score for Page NAME CHEM 419 Final Exam Spring 2010 page 5 5 10 points In the stratosphere ozone is converted to molecular oxygen by reaction with atomic chlorine Cl O3 ClO O2 1 1 260 K T The rate constant for this second order reaction is thermally activated and is k 1 7 10 dm mol s e where the temperature is in kelvins At 20 km above the surface of the Earth the temperature may be assumed to be 50 C a What is the rate constant k for this reaction 20 km above the surface of the Earth 10 …