# Wright CHM 1220 - CHM 1220_Exam2_Fall2017_answers (3 pages)

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## CHM 1220_Exam2_Fall2017_answers

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- Chm 1220 - General Chemistry II

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Exam 2 Multiple Choice Answers 1 2 3 4 5 6 7 8 9 10 A A C B D B B C E D Exam 2 Written Problem Solutions 11 7 points The following initial rate data were obtained for the reaction A B C products Experiment 1 2 3 4 5 A mM 1 25 2 5 1 25 1 25 3 01 B mM 1 25 1 25 3 02 3 02 1 00 C mM 1 25 1 25 1 25 3 75 1 15 Rate mM s 8 7 17 4 50 8 457 a Write the rate law for this reaction What is the overall order of the reaction From Experiments 1 2 B C constants From Experiments 3 4 A B constants From Experiments 3 4 A C constants rate doubles A doubles rate times 9 C tripled rate times 5 8 B times 2 4 first order with respect to A second order with respect to C second order with respect to B 50 8 log 8 7 0 766 2 Rate k A B C reaction order B 1 2 2 5 3 02 0 383 b Determine the value of the rate constant Use this information to predict the reaction rate for Experiment 5 log 1 25 Rate law 2 Using data from any experiment Using k just determined k overall reaction order is the sum of the exponents 2 Rate 8 7mM s 1 2 85 mM 4 s 1 2 2 2 2 A B C 1 25 mM 1 25 mM 1 25 mM Rate 2 85 mM 4 s 1 A B 2 C 2 2 85 mM 4 s 1 3 01 mM 1 00 mM 1 15 mM 11 3 mM 2 2 s 12 5 points The following reaction was monitored as a function of time AB A B A plot of 1 AB versus time yields a straight line with slope 0 055 M 1 s 1 a What is the value of the rate constant for this reaction at this temperature Since the plot of 1 AB versus time is a straight line reaction is second order 1 1 kt AB AB 0 indicates slope k 0 055 M 1 s 1 b Write the rate law for the reaction 2 Reaction is second order Rate k AB c If the initial concentration of AB is 0 250 M and the reaction mixture initially contains no products what are the concentrations of A Plug in values 1 and B after 75 s 1 1 1 1 integrated rate law 0 055 M s AB AB 0 123 M 1 1 kt AB AB 0 75 s 0 250 M 8 125 M A B AB consumed This is the amount that A B 0 250 0 123 0 127 M remains of AB 13 8 points The decomposition of urea NH2 2CO in 0 10 M HCl is a first order process with the rate only dependent upon the urea concentration The following equation describes the reaction NH2 2CO aq 2H aq H2O 2NH4 aq CO2 g The reaction has an activation energy of 131 kJ mol and at 75 C k 2 25 10 5 min 1 If this reaction is to be run at 85 C starting with a urea concentration of 0 0020 M how many minutes will it take for the urea to drop to 0 0012 M Knowns Unknowns Arrhenius equation Need to first figure T1 85 C 358 K k1 k1 Ea 1 1 relating temperature ln out k at 85 C k2 2 25 10 5 min 1 T2 75 C 348 K to rate constant k2 R T2 T1 Plug in values solve for k Ea 131 103 J mol 131 103 J Ea 1 1 mol 1 1 10 7 1 27 9 43 ln 2 25 10 5 min 1 R T2 T1 8 314 J 348 K 358 K mol K ln k 10 7 1 27 k e 9 43 8 00 10 5 min 1 ln k1 ln k2 Finally the reaction is first order so can use a first order integrated rate law to calculate time ln NH2 2 CO kt ln NH2 2 CO 0 1 NH2 2 CO 1 0 0012 M t ln ln 5 1 k NH2 2 CO 0 8 00 10 min 0 0020 M t 6381 min 106 h 14 12 points Cyclohexane C and methylcyclopentane M are isomers with the chemical formula C 6H12 The equilibrium constant for their rearrangement shown at the right is 0 140 at 25 C C aq M aq a A solution of 0 0200 M cyclohexane and 0 100 M methylcyclopentane is prepared Is the system at equilibrium If not will it form more reactants or products M 0 1 Qc Kc the reaction will shift left and To determine if at equilibrium 5 0 140 Qc 5 make more reactant molecules calculate Q and compare to K C 0 02 b What are the concentrations of cyclohexane and methylcyclopentane at equilibrium Equilibrium expression KC M 0 140 C 0 1 x 0 140 0 02 x Substitute for equilibrium quantities C M Initial Conc M 0 0200 0 100 Change in Conc M x x Equilibrium Conc M 0 0200 x 0 100 x 0 100 x 0 140 0 02 x 1 14 x 0 0972 x 0 0853 M Multiple through solve for x Finally solve for equilibrium concentrations C 0 02 x 0 02 0 0853 0 105 M M 0 1 x 0 1 0 0853 0 0147M c If the temperature is raised to 50 C the concentration of cyclohexane becomes 0 1 M when equilibrium is reestablished Calculate the new equilibrium constant C M Initial Conc M 0 105 0 0147 Change in Conc M x x Equilibrium Conc M 0 1 Can use a table to relate the changes in concentration Can figure out x from the first column 0 105 M x 0 1 x 0 005 M Because the concentration of C has decreased the reaction must have shifted to the right Can figure out equilibrium concentration of M knowing x Finally calculate new Kc KC M 0 0147 M x 0 0147 M 0 005 M 0 0197 M M 0 0197 0 197 C 0 1 d Is the reaction exothermic or endothermic at 25 C Explain your conclusion Because Kc increased when the temperature was raised this implies that the heat term of the reaction is on the reactant side This would be an endothermic reaction because energy is required to be inputted to form products 15 8 points Carbon monoxide replaces oxygen in oxygenated hemoglobin according to the reaction HbO2 aq CO aq HbCO aq O2 aq a Use the reactions and associated equilibrium constants at body temperature below to find the equilibrium constant for the reaction shown above 2Hb aq 2O2 aq 2HbO2 aq Hb aq CO aq HbCO aq flipped and divided by 2 no change add together Kc 3 24 Kc 306 HbCO2 aq Hb aq …

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