test1s-1 - Chemistry 2000 Spring 2001 Section B Test 1...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chemistry 2000 Spring 2001 Section B Test 1 Solutions (a) We first compare experiments 1 and 2. At constant O2 , the concentration of NO is doubled, resulting in a (roughly) fourfold increase in the rate. Thus, the order with respect to [NO] is 2. Experiments 1 and 3 were carried out at identical NO concentrations but the oxygen concentration in experiment 3 was double that in experiment 1. The rate also doubled, so the order with respect to O2 is 1. The rate law is therefore ¡ k O2 NO 2 £¡ ¡ rate ¡ 1. ¢ (b) We should get similar values using any of the three experiments. From the first, we get ¤£ 70 ¢ 2 103 L2 mol 2 s ¥ 2 8 10 5 mol L 1s 1 0 0010 mol L 0 0020 mol L ¥ ¨§ ¥ ¤£ £ ¦¨ § ¥ rate O2 NO 2 1 £¥ £¦¢¡ ¡ ¢ k 2. 1 L of solution contains 9.919 mol of ethanol and weighs 913.9 g. The molar mass of ethanol is 46.068 g/mol. Thus £ 456 9 g ¢ ©¨ ¦ £ £¦¢ 9 919 mol 46 068 g mol £ § ¨ meth The rest (457.0 g) is water. The molar mass of water is 18.015 g/mol so the number of moles of water is 457 0 g 25 368 mol nH2 O 18 015 g mol The mole fraction of ethanol is therefore £ 0 2811 £ £ £ ¢ ¢ ¢ £ § £ 9 919 mol 9 919 25 368 mol £ £ £  ¢ Xeth 3. By stoichiometry, if we start with 0.9 atm of dimethyl ether and form 0.7 atm of methane, we have used 0.7 atm of dimethyl ether. We thus have 0.2 atm left. We need to find t such that x 0 2 atm with x0 0 9 atm: £  ¢ ¨  ¨§ ¦ ¨ £ ¦  1 3786 s   ¢ £ £   1733 s ln 0 22 ln 1 2 t 1 ln 1733 s 2 ¢  t 1733s 1 2  ¢  £ £ ln t 1733s  ¢ £ 0 22  02 09 ¦ ¢ ¢ ¨ t  £ ¢ 0 2 atm  £¦ ln 0 22  1 2 t 1733s 1 0 9 atm 2 1 h 3 min   ¢ 4. If the reaction is first order, a plot of ln A vs t should give a straight line. If the reaction is second order, a plot of 1 A vs t should give a straight line. Conversely, if we use the wrong plot, the data will not fit a line. ¡ ¡ § 5. We have a pair of equations in two unknowns: (477 C (727 C 750 K) 1000 K) ¢ # ! ! " ¢ ¢ £ ¥ 1 Ea 750R Ea 1000R " k∞ exp k∞ exp # 1 ¢ ¥ 0 085 s 140 s If we divide these two equations, we eliminate k∞ and obtain a single equation in one unknown: ¥ ¤ ¥ ¥ £ ¤ 104 K £ ¦¨ ¥ 2 22 ¤ 1 104 K ¢ ©¨ ¤ £ ¢ £ ¢ £¦ 8 314 510 J K 1mol ¥ (©¨ ¥ ¢ ¢ ¤ £¦    To get k∞ , we can go back to either of the original equations. For instance, ¢ ¢ ¥ ¥ ¢¡ ¡ ¡ ¡ ¢ ¥ 2 ¥ ¢¡ ¡ ¢¡ ¡  ¡  20 L mol 1 s 1 ¡ ¡ ¥ ¥ 10 K k 1 HBr Br HBr Br k1 H Br2 k1 ¡ ¥ ¤£ 14 ¢  ¢ k1 K 0 ¡ £ § ¢¥ 1 d HBr dt k1 H Br2 ¥ (c) k 10 ¡ ¤£ HBr Br  1 ¢ 63 1 35 ¤ k Ea 750R 11 1 10 s k∞ exp ¥ d HBr k1 H Br2 dt (b) At equilibrium, (a) k∞  6. 1 £ 0 085 s 13 k∞ 185 kJ mol § ¢ 2 22 ¥ Ea R Ea   7 41 1 1000 K £ £ ¢ 1 750 K Ea exp 3 33 10 4 K 1 R Ea 3 33 10 4 K 1 R % ¢ 4 Ea 1000R & ' £ 10 Ea R  ¤ ¢ exp Ea 750R  exp  exp  ¥ ln 6 07 Ea 750R Ea 1000R " exp 4 ¢ 10 ! $ 6 07 " 0 085 140 ...
View Full Document

Ask a homework question - tutors are online