This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: CHAPTER 14 SOLUTIONS 10th Edition (9th Edition in brackets, where different) 10. This question is a nice combination of gas law concepts and rate law. a) Use the given mass and formula to work out the number of moles, then use this to get the pressure. The Frst ʻ partial pressure ʼ will be the initial total pressure. 1/ (2x 14.01 + 5 x 16)= 9.26 x 103 moles of gas. P= nRT/V P= 1.71 x 10 2 atm = 13.0 mmHg. b) Just such that you remember, the formula for working with halflives will be: ln( 1/2) = kt 1/2 But since this is a single halflife, you know that there will be exactly half as much of the reactant left. Half as much gas means half as much pressure due to it: 6.5 mmHg. c) So as the reaction proceeds, it makes 2.5 moles of product for every mole of N 2 O 5 reacted. Half of the reactant (4.83 x 10 3 moles) has become 1.21 x 102 moles. You can use this to calculate a partial pressure due to this product, or realize that 2.5 more moles means 2.5 times the pressure: 6.5 mmHg x 2.5= 16.3 mmHg, plus the 6.5 mmHg from the remaining unreacted gas, gives 22.8 mmHg. 13 . Questions like this are simple to work out by examining the initial rates and looking for the key ratios, i.e. when you double the Cl 2 , you double the rate, therefore it ʼ s Frst order with respect to chlorine, while when you double the NO, the rate quadruples, meaning that it ʼ s 2nd order with respect to NO, therefore 3rd order overall. However, make sure that you can work through the process as outlined on page 579 in the text. Brie¡y: Rate 1= k x NO m x (Cl 2 ) n Rate 2= k x NO m x (2Cl 2 ) n Rate2/Rate1= 2 = k/k x NO m /NO m x (2Cl 2 ) n /(Cl 2 ) n = 2 n 2= 2 n n=1 Do that again for Rate1 and Rate 3, to get 4=2 m , so m=2. Overall, the Rate= k x NO 2 x Cl 2 . 19. a) After every halflife (which are intrinsically 1st order, so don ʼ t let the intro throw you) the amount drops by a factor of two. So 1.6 g to .4 grams in 38 minutes is a factor of 4, or two halflives. So the halflife must be 19 minutes . b) Using this, run it through the basic formula for half lives to obtain the decay constant, or k, in terms of minutes (usually this would be in seconds, but no reason to bother, since we ʼ re working in minutes): ln (1/2)= k (19 min) k= 0.0365 Then for 1st order reactions: ln (A t / 1.60 g)= k (60) A t / 1.60= e k(60) A t = 0.18 grams. 25. a) First order reactions conform to the basic formula where rate=k[A], but this is hard to see or prove, and we don ʼ t know the rate anyway. Expressing the same formula in its integrated form: ln (A t /A o )= kt ln (A t ) = kt + ln (A o ) Makes it easy to see; in this form, the natural log of the concentration at any time, when plotted against that time, will give a straight line (of slope k). If you have patience or a computer, you can do so if the line is straight, the reaction is ¡rst order....
View
Full
Document
This note was uploaded on 04/04/2011 for the course SCIENCE CHEM 120 taught by Professor Fenster during the Winter '11 term at McGill.
 Winter '11
 Fenster
 Chemistry, Mole, Mass, Work

Click to edit the document details