Lake silbersee is located in the German city ofNuernberg. The lak\'s water quali
ID: 676717 • Letter: L
Question
Lake silbersee is located in the German city ofNuernberg. The lak's water quality has been diminishedbecause of high hydrogen sulfide concentrations that originate froma nearby leaking landfill. To combat the problem the citydecided to aerate the lake in an attempt to oxidize the odorousH2S to non-odourous sulfate ion according to thefollowing oxidation reaction: H2S+2O2-->SO42-+2H It has been determined experimentally that the reactionfollows frst order kinetics with respect to both oxygen andhydrogen sulfide concentrations: d[H2S]/dt =-k[H2S][O2] The present rate of aeration maintains that the oxygenconcentration in the lake at 2 mg/L. The rate constant k forthe reaction was determined to be 1000L/mole-day. If theaeration completly inhibits the production of hydrogen sulfide, howlong would it take to reduce H2S concentration fomm500M to 1M? Lake silbersee is located in the German city ofNuernberg. The lak's water quality has been diminishedbecause of high hydrogen sulfide concentrations that originate froma nearby leaking landfill. To combat the problem the citydecided to aerate the lake in an attempt to oxidize the odorousH2S to non-odourous sulfate ion according to thefollowing oxidation reaction: H2S+2O2-->SO42-+2H It has been determined experimentally that the reactionfollows frst order kinetics with respect to both oxygen andhydrogen sulfide concentrations: d[H2S]/dt =-k[H2S][O2] The present rate of aeration maintains that the oxygenconcentration in the lake at 2 mg/L. The rate constant k forthe reaction was determined to be 1000L/mole-day. If theaeration completly inhibits the production of hydrogen sulfide, howlong would it take to reduce H2S concentration fomm500M to 1M?Explanation / Answer
From the rate constant units it is clear that the reactionis following second order kinetics. Formula : 1/[A] = kt + 1/ [A]0 Given that k = 1000L/mole-day [A] = 1M [A]0= 500 M Uponsubstituting in the above formula , 1/(1 * 10 ^ -6 ) M = (1000L/mole-day) t + 1 /500 *10 ^ -6 M ( 1 * 10 ^ 6 - 0.002 * 10 ^ 6 ) M / (1000L/mole-day) =t t = 0.998 * 10 ^ 3 day = 998 day So it will take 998 days to reduce from 500M to1M. Given that k = 1000L/mole-day [A] = 1M [A]0= 500 M Uponsubstituting in the above formula , 1/(1 * 10 ^ -6 ) M = (1000L/mole-day) t + 1 /500 *10 ^ -6 M ( 1 * 10 ^ 6 - 0.002 * 10 ^ 6 ) M / (1000L/mole-day) =t t = 0.998 * 10 ^ 3 day = 998 day So it will take 998 days to reduce from 500M to1M.Related Questions
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