Page 1 of 8 Chemistry 2080 Spring 2011 Prelim 1 Tuesday March 1st 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 8 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 20 3 20 4 15 5 15 6 15 Total 100 Page 2 of 8 1 15 points Under certain conditions the experimental rate law for the gas phase reaction of molecular hydrogen with molecular bromine H2 g Br2 g is given by 2HBr g rate k H2 Br2 1 2 Show that the unusual reaction order of one half for Br2 can be explained by the following mechanism k1 Step 1 Br2 2Br rapid equilibrium k 1 k2 Step 2 Br H2 HBr H slow H Step 3 k3 Br2 HBr Br fast Assume an equilibrium in Step 1 k1 Br2 k 1 Br 2 The RDS is Step 2 so Rate k2 Br H2 But from equation 4 So Br k1 Br2 k 1 1 2 Rate k2 H2 k1 Br2 k 1 1 2 Q E D kobs H2 Br2 1 2 4 Page 3 of 8 2 20 points Calculate the standard enthalpy of combustion at 298 15K per mole of a gaseous fuel that contains C3H8 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 H f in kJ per mole i e standard molar enthalpies of formation at 298 15K for some compounds CO2 g 393 5 CH4 g 74 81 C3H8 g 103 8 H2O l 285 8 C2H6 g 84 68 C4H10 g 125 6 Page 4 of 8 3 20 points Given the following standard enthalpies of reaction H2 g F2 g 2HF g H 537 kJ a C s 2F2 g CF4 g H 680 kJ b 2C s 2H2 g C2H4 g H 52 3 kJ c Calculate the value of H 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 Plus equation b x 2 2C s 4F2 g 2CF4 g and invert equation c C2H4 g 2C s 2H2 g H 52 3 kJ Now add them all 4HF g C2H4 g 6F2 g 2CF4 g 4HF g H 1074 kJ H 1360 kJ H 2486 kJ Page 5 of 8 4 15 points Dimethyl ether H3COCH3 decomposes at 450 C to give methane hydrogen and carbon monoxide 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 the pressure was 0 350 bar The total pressure Ptot in the flask was measured as a function of time and the following data collected Time min Ptot bar 0 0 350 7 0 438 20 0 573 32 0 671 50 0 782 60 0 829 Determine graphically if the reaction is zero first or second order in dimethyl ether and determine the rate constant k Assume ideal gas behavior for all gaseous components Show your graphs below and show the calculation of the rate constant k SEE ATTACHED PAGES SEE ATTACHED PAGES Page 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 value of PA the partial pressure of A at those times Note that for every molecule of gaseous A that reacts three molecules of gaseous products result Thus when the reaction is over and all A has reacted the pressure will be 3 times the starting pressure of 0 35 bar assuming that all four gaseous components behave as ideal gases The starting pressure of A is 0 35 bar so when x bar of A have reacted the PA remaining will be 0 35 x bar and Ptot 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 2 We 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 1 for the first order rate constant More realistically for the purposes of the exam we can calculate the negative of the slope from the first and last points of the plot since they do fit well on the straight line that can be drawn through all the points Thus the slope is 1 0498 2 2073 0 60 1 92x10 2 min 1 and the first order rate constant is therefore 1 92x10 2 min 1 in agreement with the least squares calculation Page 7 of 8 5 15 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 0 150 0 042 0 036 6 50x10 6 2 0 050 0 084 0 036 1 44x10 6 3 0 150 0 021 0 036 3 25x10 6 4 0 300 0 036 0 072 7 88x10 6 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 0 100 M Comparing 1 and 3 we see that the only change is in B and in the initial rate The value of B in 1 is twice that in 3 and the initial rate in 1 is also exactly twice the value in 3 so the exponent b is clearly equal to 1 So the rate law can be written as Rate k A a B C c Next comparing 3 with 2 3 25x10 6 1 44x10 6 k k 0 150 0 050 a 0 021 0 084 0 …