a2s - Chemistry 2000B Spring 2002 Assignment 2 Solutions t...

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Unformatted text preview: Chemistry 2000B Spring 2002 Assignment 2 Solutions t t1 ¥ ¤ £ 1 t1 2 2 ln x x0 t1 2 ln 1 2 ln 0 05 5 3s ln 1 2 ¢ ¤ ln 2 t1 2 . The results at the ¡ © ¢ ¤ ¡ ¥ ¨ ¨ ¢ © ¨ ¨ £ § § ¡ (a) We can calculate the rate constant from the half-life by k two temperatures are as follows: ¥ © ¨ © ¡ ¥ © 23 s ¡ ln ¢ ¡ t 2 ¢ £ T ( C) T (K) k (min 1 ) 140 413.15 0.0154 180 453.15 0.0347   ¡ Ea RT ¢  ¢  ¢ £  ¤ ¢ ¤  ¡ ¢  ¥ ¢  ¢ ¢ ¢ ¡ ¤© ¢ ¨©   ¢ ¡ £ § § ¡ £ ¢ ¢  1 150 min  ¨  31 6 103 J mol 8 314 510 J K 1mol 1 453 15 K 1 ¢ ¤ ¨ exp ¡ ¡ 1  ¥ ¡  £  0 0347 min keEa RT   ¥  ¢  © ¢    To get the preexponential factor, we rearrange the Arrhenius equation to k∞ We can use either of the data points. For instance, ¢ ¡ k∞ ¤ e Ea 453 15R Ea Ea exp 453 15R 413 15R e Ea 413 15R Ea 1 1 R 413 15 K 453 15 K 0 0347 min 1 8 314 510 J K 1mol 1 ln 0 0154 min 1 1 1 413 15 K 453 15 K 31 6 kJ mol ¢ Ea . Therefore  £ k180 k140  ln k180 k140 k∞ e  ¥ According to the Arrhenius equation, k  2. 1 2 t ¦ x x0 x x0 ¡ ln 0 05. For a first-order reaction, ¢ 1. If 95% has decayed, 5% remains, i.e. x x0 . (b) This is a very similar exercise to question 1: © ¥ © ln 1 2 74 min ¡ ¥ ¨ 20 min ¢ 1 mg kg 13mg kg ln ! ¨ © ¡ ¨ © ¡ ¥ ¨ t ln x x0 t1 2 ln 1 2 Meat fried at 180 C for 74 min would look, smell, feel and taste like shoe leather, so this isn’t a really practical way to get the benzylpenicillin levels down to acceptable levels.  3. At equilibrium for this reaction, k B2 10   10 12 L mol 1s  ¢ 13 s 1 1 3 1013 mol L   ¡  ¡  " ¡ ¢ # $" % ¡ k K 1 ¢ k k " K ¢ % ¡  % # § # ¡ § k # B2 A ¢% kA  4. First, we have to calculate the pressures of all the reactants and products using the ideal gas law. For instance, 15  ¢ © ¡ ¨© ¢  106 Pa 15 atm 15. Proceeding similarly, we get ¡ ¨© ¢  ¢   15 atm 1 atm 673 15 K & ¡ ¡ ¨ ¢ ¡ ¡ § The activity of HCl is therefore aHCl 1 ¢ ¢ PHCl 20 g 0 55 mol 36 4606 g mol nRT 0 55 mol 8 314 510 J K 1mol V 2 10 3 m3 ¢ nHCl ¡ n (mol) P (Pa) O2 0.25 7 0 105 Cl2 0.85 2 4 106 H2 O 0.67 1 9 106 a 6.9 23 18  ¢  ¢  ¢ The equilibrium constant is therefore 18 2 69 ©¢ ¨ 4 ¨ 2 © 23 15 0 51 ¡ ¨ © © ¨ 2 ¢ © ¡ © ¨ ¨ ¨ © ¡ ¨ © (a) Under these conditions, 0 3002 0 400 ¢ ¢ ¡ a2 HI aH2 ¡ ¡ © a2 HI aH2 aI2 0 225 ¢ ¨© ¨ ¡ Q ¢ Since Q K , the reaction must proceed from left to right (hydrogen and iodine combine into hydrogen iodide) to reach equilibrium. ' 5. aCl2 2 aH2O aHCl 4 aO2 ¢ K 2 (b) Let x the amount of H2 used, expressed as an activity (proportional to pressure in atm). aH2 aHI Initial 0.400 0.300 Final 0 400 x 0 300 2x  ¢  4x2 1 20x ¢   ¨ ¡ ¢ ¢ ¡ ¡ 0 300 2x 2 0 0900 4x2 1 552x 0 0508 2 ¢ ¢  ¢ ¡ © ¢ ¢ ¢ ¢  ¨  x 0 0 300 2x 0 400 x  ¡ (© ¡  § 0 352 0 400 a2 HI aH2 © 0 352 K ¢ ¢ The equilibrium condition becomes ¢ § ¨ ¢ This last equation is solved using the quadratic equation: 1 552 1 795 8 ¢ ) ¢ © ¡ 0 0508 ¢ ¢  ¨© 1 5522 4 4 24  ¨ ¢ 0 1) 1 552   ¢ ¡ ¨ © x We know that x, the amount of H2 used, has to be positive. Therefore, we must take the positive sign in the above equation: 0 030 ¢ 1 552 1 795 8 ¢  ¢ ¡ ¢  ¡ x The final pressures of the two gases are therefore  0 030 atm 0 370 atm 2 0 030 atm 0 361 atm ¢ ¢ ¢ 04 03 2 ¢ ¡ ¡ ¢ © ¢ ¨ ¡ ¢  PH2 and PHI ¡ 6. The reaction is N2 O4 g 2NO2 g . The equilibrium constant is therefore K a2 2 aN2 O4 . NO We need to find the pressures of the reactant and product to solve this problem. We know the initial number of moles of N2 O4 :  2 33 g 0 0253 mol 92 0110 g mol 0 0253 mol 8 314 510 J K 1mol 1296 cm3 100 cm m 3 ¢  4 81 104 Pa & ¢ ¡ © 296 05 K 0 475 atm ¢ ¢ ¨© 1  ¢  © ¡ ¢ ¨© ¢ ¡ ¢ ¢ ¨ ¡  § ¨ The (total) equilibrium pressure (P) is 441 mm Hg, which is 0.580 atm. The total pressure is the sum of the partial pressures: PN2 O4 PNO2 ¢  ¡ P Now let x be the amount (expressed as a a pressure in atm) of N2 O4 reacted, and consider the following table: 3 ¢ 3 4  i PN2 O4 ¡ nN2O4 PN2 O4 (atm) PNO2 (atm) Initial 0.475 0 Final 0 475 x 2x  ¢ The final pressure (in atm) is therefore ¡ ¢ 2  ¢ ¢ ¡ ¢ ©  ¢ ¨ ¢ ¢ ¢ ¡ ¡ ¡ ¨ ¢ ¢ § § ¢ ¡ 5© ¢ ¡ ¢  ¢ ¡ 4 ¢ PN2 O4 and PNO2 ¡ ¡ K 0 475 x 2x 0 475 x 0 580 atm 0 106 atm 0 475 0 106 atm 0 369 atm 2 0 106 atm 0 211 atm 2 0 211 0 121 0 369 ¢  P x § ...
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This note was uploaded on 03/03/2012 for the course CHEM 2000 taught by Professor Roussel during the Fall '06 term at Lethbridge College.

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