Page 1 of 8Chemistry 2080 - Spring 2011Prelim #1Tuesday March 1st, 7:30 to 9:00 pmName: ______KEY____________________________ 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 8 pages in this exam, and there is also a separateequation sheet.There are 6 questions; the points for each are indicated below.1. _____________/152. _____________/203. _____________/204. _____________/155. _____________/156. _____________/15Total _____________/100Page of 8Page 2 of 81) (15 points) Under certain conditions, the experimental rate law for the gas-phase reaction of molecularhydrogen with molecular bromine:H2 (g) + Br2 (g) –––––––> 2HBr (g)is given by: rate = k[H2][Br2]1/2Show that the unusual reaction order of one half for Br2 can be explained by the following mechanism: k1Step 1: Br2 ∏ 2Br. (rapid equilibrium) k-1 k2Step 2: Br. + H2 ––––––––> HBr + H. (slow) k3Step 3: H. + Br2 ––––––––> HBr + Br. (fast)____________________________________________________Assume an equilibrium in Step #1: k1[Br2] = k-1[Br.]2 . . . . . . (4)The RDS is Step #2 so Rate = k2[Br.][H2]But from equation #4, [Br.] = (k1[Br2]/k-1)1/2So, Rate = k2[H2] (k1[Br2]/k-1)1/2= kobs[H2][Br2]1/2Q.E.D.Page of 8Page 3 of 82) (20 points) Calculate the standard enthalpy of combustion at 298.15K per mole of a gaseous fuel that containsC3H8 and C4H10 in the mole fractions of 0.62 and 0.38, respectively.(Assume that the water produced in the combustion is liquid, not gas).Here are values of ΔH°f in kJ per mole (i.e. standard molar enthalpies of formation) at 298.15K, for somecompounds:CO2 (g) -393.5 CH4 (g) -74.81 C3H8 (g) -103.8H2O (l) -285.8 C2H6 (g) -84.68 C4H10 (g) -125.6Page of 8Page 4 of 83) (20 points) Given the following standard enthalpies of reaction:H2(g) + F2(g) –––––––> 2HF(g) ΔH° = –537 kJ . . . . . (a)C(s) + 2F2(g) –––––––> CF4(g) ΔH° = –680 kJ . . . . . (b)2C(s) + 2H2(g) –––––––> C2H4(g) ΔH° = +52.3 kJ . . . . . (c)Calculate the value of ΔH° for the reaction of ethylene (C2H4) with F2 :C2H4(g) + 6F2(g) –––––––> 2CF4(g) + 4HF(g)___________________________________________________________________________Take equation (a) x 2 2H2(g) + 2F2(g) –––––––> 4HF(g) ΔH° = –1074 kJPlus equation (b) x 2 2C(s) + 4F2(g) –––––––> 2CF4(g) ΔH° = –1360 kJand invert equation (c) C2H4(g) –––––––> 2C(s) + 2H2(g) ΔH° = –52.3 kJ _____________________________________________________Now add them all: C2H4(g) + 6F2(g) ––––> 2CF4(g) + 4HF(g) ΔH° = –2486 kJPage of 8Page 5 of 84) (15 points) Dimethyl ether (H3COCH3) decomposes at 450°C to give methane, hydrogen and carbonmonoxide according to the following balanced chemical equation: k H3COCH3 (g)––––––––> CH4 (g) + H2 (g) +CO (g)The reaction was carried out in a flask at constant volume. Initially only dimethyl ether was present and thepressure was 0.350 bar. The total pressure (Ptot) in the flask was measured as a function of time, and thefollowing data collected:Time (min) Ptot (bar)0 0.3507 0.43820 0.57332 0.67150 0.78260 0.829Determine graphically if the reaction is zero, first, or second order in dimethyl ether, and determine the rateconstant k. Assume ideal gas behavior for all gaseous components. Show your graphs below, and show thecalculation of the rate constant k. SEE ATTACHED PAGES SEE ATTACHED PAGESPage of 8Page 6 of 8(Extra space for answering question #4)To save time, we will call the components of this reaction mixture A (dimethyl ether), B, C, and D.We are given the total pressure Ptot at various times during the reaction, but we need to calculate the valueof PA (the partial pressure of A) at those times. Note that for every molecule of gaseous A that reacts, threemolecules of gaseous products result. Thus when the reaction is over, and all A has reacted, the pressure will be 3times the starting pressure of 0.35 bar (assuming that all four gaseous components behave as ideal gases). Thestarting pressure of A is 0.35 bar, so when x bar of A have reacted, the PA remaining will be (0.35 - x) bar, andPtot will be (0.35 – x + 3x) = (0.35 + 2x) bar. Rearranging this equation gives us:2x = (Ptot – 0.35) bar, and thus x = (Ptot – 0.35)/2 Substituting this value of x into the equation PA = (0.35 - x), shows us that PA = 0.35 – (Ptot – 0.35)/2 or simplifying this equation, PA = (1.05 – Ptot)/2We can now calculate the value of PA, ln(PA), and 1/(PA) and plot each versus time. The values of PA, ln(PA), and 1/(PA) are given in the table below, and the plots are on the attached pages. It is clear that only the plot of ln(PA) versus time is linear, and therefore the reaction is first order in A.The value of the rate constant can be got from the negative of the slope of the line in the plot of ln(PA)versus time. A least-squares fit performed on a computer (using Kaleidagraph) gave a value of 1.92x10-2 min-1for the first order rate constant. More realistically for the purposes of the exam, we can calculate the negative of theslope from the first and last points of the plot, since they do fit well on the straight line that can be drawn throughall the points.Thus the slope is: (– 1.0498 – (– 2.2073)) / (0 – 60) = – 1.92x10-2 min-1, and the first order rateconstant is therefore +1.92x10-2 min-1, in agreement with the least squares calculation.Page of 8Page 7 of 85) (15 points) The following data were collected for the reaction:A + B + C –––––––> products [A]