Page 1 of 7Chemistry 2080 - Spring 2012Prelim #1Tuesday February 28th, 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 7 pages in this exam, and there is also a separateequation sheet.There are 6 questions; the points for each are indicated below.1. _____________/152. _____________/153. _____________/204. _____________/205. _____________/156. _____________/15Total _____________/100Page of 7Page 2 of 71) (15 points) Suppose we have 0.100 mol of helium gas at 25.0°C in a cylinder that is fitted with a frictionlesspiston of zero mass. The internal and external pressure at the start of our experiment is 2.40 atmospheres. (a) The external pressure is suddenly reduced to 1.20 atmospheres and the gas in the cylinder thereforeexpands against this new pressure. How much work (in joules) is done by the system, and is the sign of “w”positive or negative?(b) Instead of the process (a) as just described, suppose that the expansion is carried out in two steps. Theexternal pressure is first suddenly reduced from 2.40 to 1.80 atmosphere, and then in a second step is suddenlyreduced from 1.80 to 1.20 atmospheres. How much total work (in joules) is done by the system this time?__________________________________________________________________________________(a) First we need to calculate the starting volume of gas in the cylinder. For this we can use the ideal gas equation:PV = nRT, where P = 2.40 atm, n = 0.100 mol, R = the gas constant 0.0820574 L atm mol-1K-1and T = (25.0 + 273.2)K = 298.2 K. So V = nRT/P = 1.020 LThe final volume after the expansion can be calculated similarly (or we can just note that since the finalpressure is half the initial pressure, the volume will be doubled to 2.040 LSo the work done in the expansion (a), is –PΔV = –(1.20 atm)(2.040 - 1.020 L) = –1.224 L atmThe conversion factor from (L atm) to joules can be got from the values of the gas constant on the equationsheet thus: (8.314472 J mol-1K-1)/(0.0820574 L atm mol-1K-1) = 101.3 J L-1 atm-1So the work done in (a) is –(1.224 L atm) x (101.3 J L-1 atm-1) = –1.24 x 102 JThe work leaves the system, so its sign is negative(b) We start again with 1.020 L of helium at 298.2 K and 2.40 atm pressure. The gas then expands against anexternal pressure of 1.80 atm, so the final volume will be given by V = nRT/P where P = 1.80 atm.After this expansion, the volume will be (0.100 mol)(0.0820574 L atm mol-1K-1)(298.2 K)/(1.80 atm) = 1.359 LSo the work done in this expansion is: –PΔV = –(1.80 atm)(1.359 - 1.020 L) = –0.610 L atmNow the external pressure is reduced again, down to 1.20 atm, so there will be a second expansion, andthe final volume will be (as in (a)) = 2.040 L.The work done in this second expansion is –PΔV = –(1.20 atm)(2.040 - 1.359 L) = –0.817 L atmSo the total work done in the double expansion (b) is –0.610 – 0.817 L atm = – 1.427 L atmi.e. –(1.427 L atm) x (101.3 J L-1 atm-1) = –1.45 x 102 JPage of 7Page 3 of 72) (15 points) Ozone in the upper atmosphere is decomposed by nitric oxide through the reaction:O3 + NO –––––> O2 + NO2The experimental rate expression for this reaction is:rate = k[O3][NO]Explain which, if any, of the following mechanisms are consistent with the observed rate expression.(a) O3 + NO ––––> O + NO3(slow)O + O3 –––––> 2O2(fast)NO3 + NO –––––> 2NO2(fast)(b) O3 + NO ––––> O2 + NO2(slow)(c) NO + NO ∏ N2O2(fast equilibrium)N2O2 + O3 –––––> 2NO2 + O (slow)O + O3 ––––––> 2O2(fast)__________________________________________________________________________________(a) The rate is determined by the slow step, which is the first step, so rate = k[O3][NO] which is as foundexperimentally. So (a) is consistent with the observed rate expression.(b) Is clearly consistent with the observed rate expression, since there is just the one step and the two reactantsare O3 and NO, so the rate = k[O3][NO](c) The rate for this mechanism is determined by the rate of the slow step (let its rate constant be k2):Rate = k2[N2O2][O3]But we cannot have the reactive intermediate N2O2 in the rate expression, so we make use of the prior fastequilibrium to express [N2O2] in terms of [NO], which is one of the reactants.As it is an equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction. Let therate constants be k1 and k-1 for the forward and reverse reactions respectively, so we have:k1[NO]2 = k-1[N2O2]whence [N2O2] = k1[NO]2/k-1So rate = (k2k1/k-1)[NO]2[O3], which is NOT consistent with the observed rate expression.Page of 7Page 4 of 73) (20 points) The following data were collected for the reaction:A + B + C –––––––> products [A] (M) [B] (M) [C] (M) Initial Rate (M sec-1)(1) 0.24 0.30 0.040 20.4(2) 0.24 0.20 0.080 13.6(3) 0.12 0.40 0.060 38.47(4) 0.24 0.10 0.040 6.8Determine the rate law for this reaction, i.e. determine the values of a, b and c in the rate expression:Rate = k [A]a[B]b[C]calso calculate the rate constant k. What would be the Initial Rate of reaction if all three reactants were present at a concentration of 0.15 M?__________________________________________________________________________________If we compare experiments #1 and #4:Rate1/Rate4 = {k (0.24)a(0.30)b(0.04)c}/{k (0.24)a(0.10)b(0.04)c} = 20.4/6.8 = 3.0So, after cancelling-out identical factors on the top and bottom:We get: (0.30)b/(0.10)b = (0.30/0.10)b = 3.0b = 3.0 so by inspection b = 1Next, comparing experiments #2 and #4:{k (0.24)a(0.20)(0.08)c}/{k (0.24)a(0.10)(0.04)c} = 13.6/6.8 = 2.0After cancelling-out identical factors on the top and bottom:We get: (0.20/0.10)(0.08/0.04)c = 2.0i.e. (0.08/0.04)c = 1.0 and the only way that can be true is if c = 0Next, comparing experiments #3 and #4:{k (0.12)a(0.40)(0.06)0}/{k (0.24)a(0.10)(0.04)0} = 38.47/6.8 = 5.657and simplifying this we get: (0.12/0.24)a(4.0) = 5.657 or 0.5a x 4.0 = 5.657i.e. 0.5a = 1.414 and taking