5.60_PS_9_2008_solutions

# 5.60_PS_9_2008_solutions - MASSACHUSETTS INSTITUTE OF...

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MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Chemistry 5.60 Physical Chemistry Spring 2008 Problem Set #9 Solutions Posted: May 9, 2008 Readings: Silbey Alberty Bawendi, Pages 641-667 9.1 For a certain first-order reaction, it is observed that 32 . 5% of the reactant remains after 540 s. a) Calculate the rate constant for the reaction Solution The reaction is first order, so [ A ]( t ) = [ A ] 0 e - kt We have the data point that at t = 540 s, [ A ] [ A ] 0 = 0 . 325. Substituting, 0 . 325 = e - k · 540 , so k = 0 . 002081 · s - 1 . b) What length of time would be required for 25% of the reactant to decompose. Solution Note that this is twenty-five percent of the reactant that decomposes , not twenty-five percent left, so we can’t just use twice the half-life. There is seventy-five percent remaining, so [ A ] [ A ] 0 = 0 . 75 = e - 0 . 002081 · t and t = 138 . 2 s 9.2 It is possible to follow the rate of decomposition of NH 3 on a surface of W by noting the increase in pressure of gases at constant volume and temperature. The reaction is 2NH 3 N 2 + 3H 2 . At 1 , 100 C the following half times and initial pressures p 0 were observed: p 0 (mm Hg) 265 130 58 τ 1 / 2 (min) 7.6 3.7 1.7 a) Determine the order of the reaction and the rate constant at 1 , 100 C. b) Predict the total pressure 3 . 0 min after NH 3 at 200 mm Hg is admitted to a rigid vessel containing W at 1 , 100 C. 1

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Solution a) Although we are given only empirical data, we can use what we know about n th -order half-lives to make the task easier. We know that in general, τ 1 / 2 p 0 1 - n . So when we compare two half-lives, τ ( a ) 1 / 2 τ ( b ) 1 / 2 = p ( a ) 0 p ( b ) 0 1 - n n = 1 - ln τ ( a ) 1 / 2 - ln τ ( b ) 1 / 2 ln p ( a ) 0 - ln p ( b ) 0 We can perform this calculation three times for the given data as a safety check: 7 . 6 3 . 7 = 265 130 1 - n n = - 0 . 01 0 7 . 6 1 . 7 = 265 58 1 - n n = 0 . 01 0 7 . 6 1 . 7 = 265 58 1 - n n = - 0 . 03 0 To within reasonable error, all three equations show that this is a 0 th order reation. b) Now that we have the rate law, we can measure the approximate rate constant from the half-life time. For a 0 th order reaction, k = p 0 2 τ 1 / 2 = 265 mm Hg 2 × 7 . 6 min = 17 mm Hg · min - 1 The rate law for a 0 th order reaction is p NH 3 = p 0 - kt = 200 mm Hg - 17 mm Hg · min - 1 × t min . For t = 3 min, p NH 3 = 148 mm Hg. Each mole of ammonia is converted into one mole of nitrogen and one mole of hydrogen. Thus, p N 2 = p H 2 = (200 - 148) mm Hg = 52 mm Hg p = p NH 3 + p N 2 + p H 2 = (148 + 2 × 52) mm Hg = 252 mm Hg 2
9.3 For the gas phase decomposition of di-t-butyl peroxide (CH 3 )COOC(CH 3 ) 3 ( g ) -→ 2(CH 3 ) 2 CO ( g ) + C 2 H 6 ( g ) The following table gives the data for the total pressure of the system vs. time (assume pure reactant at t = 0): t (min) 0 3 6 9 12 15 18 21 p (Torr) 173.5 286.5 344.1 378.8 402.2 418.9 431.5 441.3 a) What is the order of this reaction? b) What is the rate constant for this reaction? c) Calculate the fraction of di-t-butyl peroxide that will have decomposed after 36 minutes. Solution For gaseous kinetics, partial pressures take the role of the observable variable, rather than concentrations. Thus, if we had a table of the partial pressure of the reagent versus time, we could plot this data in many forms, only one of which would give a straight line. The uniqueness of this plot determines the order of the reaction.

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