Page 1 of 6 Chemistry 2080 Spring 2010 Prelim 1 Tuesday March 2nd 7 30 to 9 00 pm Name Last First M I Lab Day Time CU ID Lab TA Name Instructions Answer all questions in the spaces provided You have 90 minutes You must show all work for full credit There are 6 pages in this exam and there is also a separate equation sheet There are 5 questions the points for each are indicated below 1 20 2 20 3 20 4 20 5 20 Total 100 Page 2 of 6 1 20 points a Write a balanced equation for the combustion of liquid cyclohexane C6H12 in oxygen to give gaseous carbon dioxide and liquid water b The standard enthalpy change for this combustion reaction at 25 C and 1 atm pressure is 3 920 kJ mol of cyclohexane Use this figure to calculate the standard enthalpy of formation H f of liquid cyclohexane at 25 C and 1 atm pressure Note that H f for H2O l 285 8 kJ mol H f for H2O g 241 8 kJ mol and H f for CO2 g 393 5 kJ mol all at 25 C a C6H12 l 9O2 g 6CO2 g 6H2O l H rxn 3 920 kJ b H rxn p H f products r H f reactants 3 920 kJ i e H rxn 6x H f CO2 g 6x H f H2O l 1x H f C6H12 l 9x H f O2 g So H rxn 3 920 kJ 6molx 393 5kJ mol 6molx 285 8kJ mol 1molx H f C6H12 l 9x0 Thus 3 920 kJ 2 361 kJ 1 714 8 kJ H f C6H12 l Hence H f C6H12 l 3 920 kJ 4 076 kJ 156 kJ for 1 mol of cyclohexane i e H f C6H12 l 156 kJ mol Page 3 of 6 2 20 points Acetoacetic acid CH3COCH2COOH a reagent used in organic synthesis decomposes in acidic solution producing acetone and CO2 g CH3COCH2COOH aq CH3COCH3 aq CO2 g This first order decomposition has a half life of 144 minutes a How long will it take for a sample of acetoacetic acid to be 65 decomposed b How many liters of CO2 g measured at 24 5 C and 748 Torr are produced as a 10 0 g sample of CH3COCH2COOH decomposes for 575 min Ignore the aqueous solubility of CO2 g Page 4 of 6 3 20 points Imagine a slab of water ice of surface area 1 203 square meters and uniform thickness of 15 00 cm The temperature of this ice slab is a uniform 12 00 C How many hours of sunshine would be required to a raise the temperature of the entire slab to 0 00 C and then b uniformly melt enough of the ice so that the thickness is reduced to 10 00 cm The sunshine irradiates the ice with 195 watts meter2 energy Assume that all of the energy of the sunshine is absorbed by the ice and goes to heat or melt it Further you can assume that the heating and melting is uniform and that the only source of warming and melting of the ice is the direct radiation by the sun You may take the density of ice as 0 917 g mL Note that it is not necessary to know the area of the ice slab since each square meter of its surface behaves like any other so we can do the calculation for just one square meter Some students assumed that the area of 1 203 square meters represented the total area of the slab including the top surface the bottom surface and the edges The dimensions of this slab depend on what shape you assume it is and the answers vary accordingly a The area of one square meter is 100 x 100 cm2 i e 104 cm2 The volume of a piece of ice that has a top surface area of 1 00x104 cm2 and is 15 0 cm thick is 1 50x105 mL The mass of a piece of ice of volume 1 50x105 mL is 1 50x105 mL x 0 917 g mL 1 376 x105 g To raise the temperature of 1 376 x105 g of ice from 12 00 C to 0 00 C would take 1 376 x105 g x 2 01 J g C x 12 00 C 3 319 x 106 Joules Remembering that 1 watt is equivalent to 1 J sec and the sun s irradiation corresponds to 195 watts per square meter we can see that 3 319 x 106 J will be supplied by the sun in 3 319 x 106 J 195 J sec 1 702 x 104 sec 1 702 x 104 sec is equivalent to 1 702 x 104 sec 3 60 x 103 sec hr 4 73 hours b We want to melt enough ice so that the thickness is reduced by 5 00 cm to 10 00 cm The volume of a piece of ice that has a top surface area of 1 00x104 cm2 and is 5 00 cm thick is 5 00 x104 mL The mass of a piece of ice of volume 5 00 x104 mL is 5 00 x104 mL x 0 917 g mL 4 585 x104 g To melt a piece of ice of mass 4 585 x104 g and leave it as liquid water still at 0 C would take 4 585 x104 g x 6 01 kJ mol 18 015 g mol kJ of energy i e 1 530 x 104 kJ This corresponds to 1 530 x 107 J of energy and would take 1 530 x 107 J 195 J sec 7 846 x 104 sec of sunshine 7 846 x 104 seconds is equivalent to 7 846 x 104 sec 3 60 x 103 sec hr 21 8 hours The total which was not actually asked for is 26 5 hrs of sunshine Page 5 of 6 4 20 points Suppose that we have the following reaction A B C D The initial rate law for this reaction is as follows Initial Rate k A a B b C c The following initial rate data were obtained for this reaction as a function of the given initial concentrations of A B and C the initial rates are the initial rates of formation of D Experiment 1 2 3 4 5 A M B M C M 0 100 0 200 0 100 0 300 0 100 0 200 0 200 0 400 0 200 0 600 0 050 0 100 0 050 0 050 0 100 Initial Rate M sec 1 1 59 6 36 3 18 14 31 4 77 From this set of initial concentrations and rates determine 1 the rate law i e find the numerical values of the exponents a b and c 2 the value of the rate constant k and 3 the initial rate of the reaction when A 0 400 M B 0 300 M and C 1 00 M Be sure to give the units of the …