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: 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 641667 9.1 For a certain firstorder 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 ] e kt We have the data point that at t = 540 s, [ A ] [ A ] = . 325. Substituting, 0 . 325 = e k · 540 , so k = . 002081 · s 1 . b) What length of time would be required for 25% of the reactant to decompose. Solution Note that this is twentyfive percent of the reactant that decomposes , not twentyfive percent left, so we can’t just use twice the halflife. There is seventyfive percent remaining, so [ A ] [ A ] = . 75 = e . 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 were observed: p (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 Solution a) Although we are given only empirical data, we can use what we know about n thorder halflives to make the task easier. We know that in general, τ 1 / 2 ∝ p 1 n . So when we compare two halflives, τ ( a ) 1 / 2 τ ( b ) 1 / 2 = p ( a ) p ( b ) 1 n n = 1 ln τ ( a ) 1 / 2 ln τ ( b ) 1 / 2 ln p ( a ) ln p ( b ) We can perform this calculation three times for the given data as a safety check: 7 . 6 3 . 7 = 265 130 1 n n = . 01 ≈ 7 . 6 1 . 7 = 265 58 1 n n = . 01 ≈ 7 . 6 1 . 7 = 265 58 1 n n = . 03 ≈ To within reasonable error, all three equations show that this is a th order reation. b) Now that we have the rate law, we can measure the approximate rate constant from the halflife time. For a 0 th order reaction, k = p 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 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 ditbutyl 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) 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?...
View
Full
Document
This note was uploaded on 02/06/2009 for the course 5 5.60 taught by Professor Unknown during the Fall '08 term at MIT.
 Fall '08
 unknown

Click to edit the document details