Page 1 of 6Chemistry 2080 - Spring 2010Prelim #1Tuesday March 2nd, 7:30 to 9:00 pmName: _____________________________________ CU ID: ____________________ Last First M.I.Lab: _______________________________ Lab TA Name: __________________ Day TimeInstructions: Answer all questions in the spaces provided. You have 90 minutes. You mustshow all work for full credit. There are 6 pages in this exam, and there is also a separateequation sheet.There are 5 questions; the points for each are indicated below.1. _____________/202. _____________/203. _____________/204. _____________/205. _____________/20Total _____________/100Page of 6Page 2 of 61) (20 points) (a) Write a balanced equation for the combustion of liquid cyclohexane (C6H12) in oxygen togive gaseous carbon dioxide and liquid water. (b) The standard enthalpy change for this combustion reaction at25°C and 1 atm pressure is – 3,920 kJ/mol of cyclohexane. Use this figure to calculate the standard enthalpy offormation (ΔH°f) of liquid cyclohexane at 25°C and 1 atm pressure. Note that ΔH°f for H2O(l) = –285.8kJ/mol, ΔH°f for H2O(g) = –241.8 kJ/mol, and ΔH°f for CO2(g) = –393.5 kJ/mol, all at 25°C._______________________________________________________________________________(a) C6H12(l) + 9O2(g) –––––––> 6CO2(g) + 6H2O(l) ΔH°rxn = – 3,920 kJ (b) ΔH°rxn = Σ νpΔH°f (products) – Σ νrΔH°f (reactants) = – 3,920 kJ i.e. ΔH°rxn = [6xΔH°f (CO2(g)) + 6xΔH°f (H2O(l))] - [1xΔH°f (C6H12(l)) + 9xΔH°f (O2(g))] So: ΔH°rxn = – 3,920 kJ = [6molx(-393.5kJ/mol) + 6molx(-285.8kJ/mol)] - [1molxΔH°f (C6H12(l)) + 9x0]Thus: – 3,920 kJ = [-2,361 kJ - 1,714.8 kJ] – [ΔH°f (C6H12(l))]Hence: ΔH°f (C6H12(l)) = 3,920 kJ – 4,076 kJ = – 156 kJ for 1 mol of cyclohexane.i.e. ΔH°f (C6H12(l)) = – 156 kJ/molPage of 6Page 3 of 62) (20 points) Acetoacetic acid, CH3COCH2COOH, a reagent used in organic synthesis, decomposes in acidicsolution, 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 ofCH3COCH2COOH decomposes for 575 min? [Ignore the aqueous solubility of CO2(g).]_______________________________________________________________________________Page of 6Page 4 of 63) (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 thethickness is reduced to 10.00 cm. The sunshine irradiates the ice with 195 watts/meter2 energy. Assume that all ofthe energy of the sunshine is absorbed by the ice, and goes to heat or melt it. Further you can assume that theheating and melting is uniform, and that the only source of warming and melting of the ice is the direct radiation bythe 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 likeany other, so we can do the calculation for just one square meter. (Some students assumed that the area of 1,203square 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 JoulesRemembering that 1 watt is equivalent to 1 J/sec, and the sun’s irradiation corresponds to 195 watts per squaremeter, 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 gTo 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 ofsunshine.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 of 6Page 5 of 64) (20 points) Suppose that we have the following reaction: A + B + C –––––––––> DThe initial rate law for this reaction is as follows: Initial Rate = k [A]a[B]b[C]cThe following initial rate data were obtained for this reaction, as a function of the given initialconcentrations of A, B, and C (the initial rates are the initial rates of formation of D):Experiment [A] (M) [B] (M) [C] (M) Initial Rate (M.sec-1) 1 0.100 0.200 0.050 1.592 0.200 0.200 0.100 6.363 0.100 0.400 0.050 3.18 4 0.300 0.200 0.050 14.315 0.100 0.600 0.100 4.77From this set of initial concentrations and rates, determine (1) the rate law (i.e. find the numerical valuesof 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 rate and of “k”._______________________________________________________________________________(1) Comparing Expt 4 and Expt 1: the only change