Page 1 of 7 Chemistry 2080 Spring 2012 Prelim 1 Tuesday February 28th 7 30 to 9 00 pm Name KEY 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 7 pages in this exam and there is also a separate equation sheet There are 6 questions the points for each are indicated below 1 15 2 15 3 20 4 20 5 15 6 15 Total 100 Page 2 of 7 1 15 points Suppose we have 0 100 mol of helium gas at 25 0 C in a cylinder that is fitted with a frictionless piston 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 therefore expands 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 The external pressure is first suddenly reduced from 2 40 to 1 80 atmosphere and then in a second step is suddenly reduced 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 1 and T 25 0 273 2 K 298 2 K So V nRT P 1 020 L The final volume after the expansion can be calculated similarly or we can just note that since the final pressure is half the initial pressure the volume will be doubled to 2 040 L So the work done in the expansion a is P V 1 20 atm 2 040 1 020 L 1 224 L atm The conversion factor from L atm to joules can be got from the values of the gas constant on the equation sheet thus 8 314472 J mol 1K 1 0 0820574 L atm mol 1K 1 101 3 J L 1 atm 1 So the work done in a is 1 224 L atm x 101 3 J L 1 atm 1 1 24 x 102 J The 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 an external 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 L So the work done in this expansion is P V 1 80 atm 1 359 1 020 L 0 610 L atm Now the external pressure is reduced again down to 1 20 atm so there will be a second expansion and the final volume will be as in a 2 040 L The work done in this second expansion is P V 1 20 atm 2 040 1 359 L 0 817 L atm So the total work done in the double expansion b is 0 610 0 817 L atm 1 427 L atm i e 1 427 L atm x 101 3 J L 1 atm 1 1 45 x 102 J Page 3 of 7 2 15 points Ozone in the upper atmosphere is decomposed by nitric oxide through the reaction O3 NO O2 NO2 The 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 NO NO N2O2 fast equilibrium N2O2 O3 2NO2 O slow O O3 2O2 fast c a The rate is determined by the slow step which is the first step so rate k O3 NO which is as found experimentally 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 reactants are 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 fast equilibrium 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 the rate 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 1 So rate k2k1 k 1 NO 2 O3 which is NOT consistent with the observed rate expression Page 4 of 7 3 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 8 Determine 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 c also 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 0 So 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 1 Next 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 0 After cancelling out identical factors on the top and bottom We get 0 20 0 10 0 08 0 04 c 2 0 i e 0 08 0 04 c 1 0 and the only way that can be true is if c 0 Next 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 5 657 …