Exam #2 Multiple Choice Answers 1. A 2. A 3. C 4. B 5. D 6. B 7. B 8. C 9. E 10. DExam #2 Written Problem Solutions 11. (7 points) The following initial rate data were obtained for the reaction: Experiment [A] (mM) [B] (mM) [C] (mM) Rate (mM/s) 1 1.25 1.25 1.25 8.7 2 2.5 1.25 1.25 17.4 3 1.25 3.02 1.25 50.8 4 1.25 3.02 3.75 457 5 3.01 1.00 1.15 ? (a) Write the rate law for this reaction. What is the overall order of the reaction? (b) Determine the value of the rate constant. Use this information to predict the reaction rate for Experiment 5. 12. (5 points) The following reaction was monitored as a function of time: 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? (b) Write the rate law for the reaction. (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 and B after 75 s? 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: 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? Reaction is second-order: integrated rate law Arrhenius equation relating temperature to rate constant: 122111lnTTREkkaNeed to first figure out k at 85 °C: A + B + C → products     1422122 85.2 .251 .251 .2517.8][C]A][B[Rate smMmMmMmMsmMk22][C]A][B[Rate kUsing data from any experiment: Rate law: From Experiments 1 & 2: [B], [C] = constants rate doubles, [A] doubles first order with respect to [A] From Experiments 3 & 4: [A], [C] = constants rate times 5.8, [B] times 2.4 second order with respect to [B] From Experiments 3 & 4: [A], [B] = constants rate times 9, [C] tripled second order with respect to [C] 2383.0766.025.102.3 log7.88.50log][B order,reaction overall reaction order is the sum of the exponents 5221 Using k just determined:      smMmMmMmMsmMsmM 3.11 .151 .001 01.3 85.2][C]A][B[ 85.2Rate22142214AB → A + B  2ABRate kSince the plot of 1/[AB] versus time is a straight-line reaction is second-order    0AB1AB1 ktindicates 11 0.055 sMkslope   0AB1AB1 ktPlug-in values:    111 125.8 0.2501 75 055.0AB1 MMssM M 123.0AB (NH2)2CO(aq) + 2H+(aq) + H2O → 2NH4+(aq) + CO2(g) Knowns & Unknowns: k1 = ? T1 = 85 °C = 358 K T2 = 75 °C = 348 K Ea = 131×103 J/mol k2 = 2.25×10-5 min-1 Plug-in values, solve for k,   43.927.17.10K 5831K 4831KmolJ314.8molJ10131min 1025.2ln31-51543.9min1000.8 ekFinally, the reaction is first-order, so can use a first-order integrated rate law to calculate time:    02222CO)(NHlnCO)(NHln  kt  MMkt 0020.0 0012.0lnmin 1000.81CO)(NHCO)(NHln11-502222h 106min 6381 t122111lnlnTTREkka27.17.10ln kThis is the amount that remains of [AB]    M 127.0 123.0250.0BA      consumedABBA Equilibrium expression: 14. (12 points) Cyclohexane (C) and methylcyclopentane (M) are isomers with the chemical formula C6H12. The equilibrium constant for their rearrangement (shown at the right) is 0.140 at 25 °C. (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? (b) What are the concentrations of cyclohexane and methylcyclopentane at equilibrium? (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. (d) Is the reaction exothermic or endothermic at 25 °C? Explain your conclusion. 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. (b) Suppose that an air mixture becomes polluted with carbon monoxide at a level of 0.10%. Assuming the air contains 20.0% oxygen and that the oxygen and carbon monoxide ratios that dissolve in the blood are identical to the ratios in the air, what is the ratio of HbCO to HbO2 in the blood stream? Comment as to what this means as far as relative amounts of HbCO and HbO2 and/or toxicity. [C] [M] Initial Conc. (M) 0.0200 0.100 Change in Conc. (M) +x -x Equilibrium Conc. (M) 0.0200 + x 0.100 - x [C] [M] Initial Conc. (M) 0.105 0.0147 Change in Conc. (M) -x +x Equilibrium Conc. (M) 0.1 ?   140.0CMCKFinally, solve for equilibrium concentrations:   MxMx0147.00853.01.01.0M 105.00853.002.002.0C140.002.01.0xxSubstitute for equilibrium quantities: Multiple through & solve for x: ccKQ   502.01.0CMcQC(aq) ⇌ M(aq) To determine if at equilibrium, calculate Q and compare to K: the reaction will shift left and make more reactant molecules 140.05 xx  02.0140.0100.00972.014.1 xMx 0853.0Can use a table to relate the changes in concentration Because the concentration of [C] has decreased the reaction must have shifted to the right. Can figure out x from the first column: MxxM 005.0 1.0 105.0Can figure out equilibrium concentration of [M] knowing x:   0197.0 005.0 0147.0 0147.0MMMxMMFinally, calculate new Kc:   197.00.10197.0CMCKBecause 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