CHEMISTRY 1AA3 TUTORIAL PROBLEM SET 7 WEEK OF MARCH 4, 2002

Transcription

1 EMISTRY 1AA3 TUTRIAL PRBLEM SET 7 WEEK F MAR 4, 2002 SLUTINS 1. The following reaction occurs in alkaline aqueous solution: I + l I + l and is first order in [I ] and in [l ]. A determination of dependence of rate on [ ] gives the following result: [ ] (mol/l) Rate (mol/l.s) (a) What is the order of reaction with respect to [ ]? onsidering the first and third experiments, as the [ ] is halved, the rate doubles, so rate is inversely proportional to [ ]. Thus, the order of reaction with respect to [ ] is 1. (b) What is the complete rate law? From the information given in the question, the reaction is first order in [I ] and in [l ], as well as the order of [ ] = 1 from part (a), and so: Rate = k [I ][l ][ ] 1 (c) The following mechanism has been suggested for this reaction: + + l l (fast equilibrium) I + l I + l + + (slow) The slow step is the rate-determining step, and the overall rate law mist be based on this step, and so we start with: Rate = k[i ][l] 1

2 owever, since [l] does not appear in the overall reaction, we must eliminate it from the rate law. We can do this be considering the fast equilibrium in step 1. Because the step is an equilibrium, rate of forward reaction = rate of reverse reaction, so we can isolate an expression for [l} and substitute in our rate law. k forward [ + ][l ] = k reverse [l] [l] = k forward [ + ][l ] k reverse And so: Rate = k[i ] k forward [ + ][l ] k reverse ollecting all the constants, we get: Rate = k [I ][ + ][l ] Is this proposed mechanism consistent with the above rate law? (int: Think about how the concentrations of + and are related in aqueous solution). In aqueous solution, the concentrations of + and are inversely related. Remember from K w = [ + ][ ], that as [ + ] increases, then [ ] decreases (and vice versa). And so, looking at our rate law, if we wish to use the concentration of [ ] instead of [ + ], in order to compare our result with part (b), we can substitute 1/[ ] for the [ + ] term: Rate = k [I ][l ] [ ] or Rate = k [I ][l ][ ] 1 This result from the mechanism is consistent with the experimental rate law, even though it does not appear to be at first glance. 2. The most common form of nylon (Nylon-6) is 63.68% carbon, 14.14% oxygen, 12.38% nitrogen and 9.80% hydrogen. Determine the empirical formula of Nylon-6. Assume we have g of compound. Then we can use the percent composition values given to determine the mole ratios of each of the elements present. 2

3 N g 9.80 g g g g/mol g/mol g/mol g/mol This gives the mole ratio : : : N of : : : Divide each quantity by the smallest number to get the ratio in terms of whole numbers: N N is the empirical formula. Nylon-6 is a polymer; the empirical formula represents the repeat unit of the polymer: N n 3. ydrocarbons ( x y ) and other organic molecules undergo combustion with oxygen to produce carbon dioxide and water. What volume of 2 (g) is produced when 96.1 g of propane gas undergoes combustion with oxygen at STP? The balanced reaction describing the combustion is: 3 8 (g) (g) 3 2 (g) (l) The conversion we need is: g 3 8 mol 3 8 mol 2 volume g mol mol 2 = 6.54 mol g mol 2 Now, use the ideal gas law to convert moles 2 to pressure: PV = nrt or V = nrt/p We are told that the 2 volume is measured at STP (273 K, 1 atm): V = (6.54 mol)( L atm/k mol)(273 K) / (1 atm) V = 147 L of 2 produced 3

4 4. Which of the following formulas could represent an alkyne, alkene or cycloalkene? (i) (ii) 6 12 (iii) 4 6 (iv) 8 18 alkyne general formula: n 2n-2 alkene general formula: n 2n cycloalkene general formula same as alkyne Formulas (i), (ii) could be alkenes, (iii) could be an alkyne or a cycloalkene. Formula (iv) represents an alkane (fully saturated with hydrogen atoms). 5. Draw line structures for all of the structural isomers that satisfy the molecular formula Name each structural isomer using the IUPA rules. hexane 2-methylpentane 3-methylpentane 2,2-dimethylbutane 2,3-dimethylbutane 6. For each of the given structural formulas, draw the Lewis structure. Remember that carbon is always tetravalent (has four bonds), and it can form double and triple bonds to many different atoms (e.g., N). Use these ideas to determine each Lewis structure, drawing multiple bonds where necessary to satisfy the valency at each atom. (i) 2 (ii) 3 N (iii) 3 (iv) 3 l 2 (v) 2 N 3 (i) (ii) (iii) N (iv) l l (v) N 4

5 7. Draw Newman projections for all of the different the staggered and eclipsed conformations of 1-bromobutane, looking along the 2-3 bond. Draw a graph showing energy on the y-axis and torsional (twist) angle on the x-axis. Show where each of these al isomers (rotamers) appears on a relative energy scale, as the twist angle around the 2-3 bond changes from Start with the lowest energy rotamer and work towards the highest energy conformation. The relative energies of the four possible eclipsed or staggered rotamers are not the same. Rotamer A is the lowest energy rotamer - the bulky groups are as far apart as possible (anti conformation). Rotamer D is the highest energy - the bulky groups eclipse each other. In B the bulky groups eclipse an - better than D but not as good as (gauche conformation), where the all groups are staggered, but and 3 are closer to each other than they are in A. The relative energy positions are shown on the sketched graph. A B D zero 60 degree 120 degree 180 degree 3 anti 3 3 gauche 3 same as same as 3 3 gauche 5