a1s-1 - Chemistry 2000 Spring 2001 Section B Assignment 1...

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Unformatted text preview: Chemistry 2000 Spring 2001 Section B Assignment 1 Solutions 1. 4 ¡ nH2 O 0 0036 ¡ ¦ ¥ ¥ (a) Suppose that we take 1 L of this solution. This liter of solution contains 1.697 mol of acetone and weighs 0.9849 kg (from the concentration and density, respectively). The acetone weighs ¡ £¤ ¡ © £ macetone ¡ ¡ ¤ ¢ 2. ¡ § ¡ §¡ ¨ §¦ § ¦ ¡ ¡¡ ¢¡ ¡ ¤ ¢ ¡ £¤ £ ¢ ¡¡ ¡ mH2 O XK 8 0g 0 051 mol 158 032 g mol 250 mL 0 998 23 g mL 250 g 250 g 13 9 mol 18 015 g mol nK 0 051 mol nH2 O nK nMnO 13 9 0 051 0 051 mol ¡ nKMnO4 1 697 mol 58 079 g mol 98 56 g The water contained in our liter of solution therefore weighs ¡ ¡ 1 697 mol 0 8863 kg 98 56 g 886 3 g ¡¢ ¡ c ˜ ¡ ¡ The molality is therefore 984 9 ¡¡ mwater 1 915 mol kg (b) Consider again our liter of solution. We know how many moles of acetone it contains. The number of moles of water can be determined from the mass: ¡ ¡  ¡ ¢ ¡  ¡   ¡§¡ ¡ ¡ ¥ ¡ ¡ ¢¡ ¡ Xacetone 3. (a) ∆A ∆t 0 382 886 3 g 49 20 mol 18 015 g mol 1 697 mol 0 03334 1 697 49 20 mol ¡ nwater 0 400 mol L 0 32 s 1 0 056 mol L 1s 1 (b) ∆A ∆B 2 0 113 mol L 1s 1 ∆t ∆t Notes: I am using ∆’s instead of d ’s because these are average rates. However, it would not be wrong to use derivative notation. It’s a matter of taste, and I’m not completely consistent about it myself. Also note that I got the answer to this question using more digits than are shown in my answer to question 3a. This is the correct procedure, even if the two numbers are then not quite in the stoichiometric ratio of 2:1. This is fine since the last significant figure shown in any calculation is supposed to be somewhat uncertain. (That’s why we stop at that digit rather than some other.) ¡      ¡  ¢ 4. The rate law is assumed to be of a simple form, e.g. dx dt kxn . If we increase x by a factor of 3, dx dt increases by a factor of 9 32 . Therefore, the order of the reaction with respect to the reactant (n) is 2. ¢ (a) Experiments 1, 2 and 3 were performed at identical nitrogen monoxide concentrations. The concentration of ozone was doubled from experiment 1 to experiment 2 and the rate increased by a factor of 2. Comparing experiments 1 and 3, the ratio of the ozone concentrations is 3 and the ratio of the rates is also 3. The rate is therefore proportional to the ozone concentration. Experiments 3, 4 and 5 were performed at identical ozone concentrations. The NO concentration was doubled in experiment 4 compared to experiment 3 and the rate doubled. Comparing experiments 3 and 5, we see that both the NO concentrations and rates are in ratios of 1:3. Thus, the rate is also proportional to the NO concentration. The rate law is therefore d NO2 k NO O3 dt ¡     (b) I’ll use data from experiment 1, but we could in principle use any of the data points: k  ¤ £ ¡ ¢¡ ¡  ¡ ¤ ¢    ¡ £ ¤  ¢  ¡   ¡ £    ¢  d NO2 dt NO O3 66 0 10 6 mol L 1s 1 1 00 10 6 mol L 3 00 10 6 mol L 2 20 107 L mol 1s 1 Note that I got rid of the SI prefixes on the quantities appearing in this calculation to avoid getting an answer in awkward units (22 0 L µmol 1s 1 or 22 0 L µM). Strange units like these are sometimes useful, but it is generally considered bad style to have a prefix in the denominator of units in the SI system. The prefixes should really only be used as prefixes, i.e. at the beginning of a unit, whether it is a simple unit (g, mol, L, etc.) or a compound unit (such as µmol L 1s 1 ).  5. 2 (c) ¡    ¡ £¤    ¡£      k NO O3 3 1 1 30 10 6   ¡ £¤ ¢ 2 20 107 L mol 1 s 143 µmol L 1s 1 mol L 5 00 10 6 ¤¢ d NO2 dt mol L ...
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