sodium salt of metham C S -, Na+ S C S - H 2

Transcription

1 Fechner - Levy Vapam Spill - Sacramento River - July 14, 1991 On July 14, 1991 a train derailment resulted in the spill of approximately 13,000 gal of the soil fumigant sodium metham (Vapam) to the Sacramento River, 70 Km upstream from Lake Shasta, California largest reservoir. The spill killed almost all aquatic life in the river. In an aqueous environment, metham, which is highly soluble, hydrolyzes to form methyl isocyanate and H 2 S as major products: sodium salt of metham H 3 C H N S C S -, Na+ H 3 C H N S C S - H 2 O H 3 C N C S + H 2 S metham methyl isothiocyanate 1 Rate constants for this reaction are k a := 300 and k' n :=. Metham also undegoes indirect molesec sec photolysis to methyl isocyanate, with a pseudo-first-order rate constant of approximately k p := sec The Calfornia Regional Water Quality Control Board urgently needs your advice, both as to the maximum concentrations of metham that will be encountered at different points along the Sacramento River, and as to the total load that will enter Lake Shasta. a. Calculate the maximum concentration, in moles per, of sodium, which behaves in a conservative fashion, as well as Metham that will be encountered, both at the town of Dunsmuir, 10 km downstream, and at the point where the Sacramento River reaches Shasta Lake (70 Km downstream from the spill). b. How important are the processes of longitudinal dispersion, hydrolysis, and indirect photolysis in attenuating the peak concentration of metham between Dunsmuir and Shasta Lake? Quantify your answer. c. What are the half lives for hydrolysis and indirect photolysis? d. How important are these two processes in collectively attenuating the total load of Metham between Dunsmuir and Shasta Lake. Fechner Levy 2-27.mcd 1 of 11

2 e. Data are currently not available as to the Henry's Law constant for Metham. Would you recommend measuring this parameter, or do you feel it is reasonable to neglect air-water exchange of Metham? Indicate your logic. f. Estimate the total mass of methyl isocyanate (in moles) that will be transported to Lake Shasta, ignoring loses from any process (such as air-water interchange) that might remove this compound during its transport in the Sacramento River. Consider some additional useful information: ph := 7.8 ph of river water Q Sacramento := flow in Sacramento River min width Sacramento := 1.2m width of Sacramento River d Sacramento := 0.3m mean depth of Sacramento river Q Sacramento V := mean velocity in Sacramento River V = m width Sacramento d Sacramento s q spill := 13000gal volume of spill conc := 33% concentration by weight of the sodium salt of metham in solution. NOTE: Is not the mass of sodium nor the mass of metham r := 1 gm desity of metham solution ml D L m 2 := longitudinal dispersion coefficient min MW metham := gm molecular weight of sodium salt of metham mole MW sodium := 23 gm mole MW metham := 106 gm mole MW miso := 73 gm mole Mass := q spill r conc Mass = kg mass of sodium salt of metham We must solve for the mass of sodium and that of metham separately using stoichiometry. 1 mole of sodium is associated with 1 mole metham. Fechner Levy 2-27.mcd 2 of 11

3 XMW sodium + XMW metham = Mass X := Mass ( ) MW sodium + MW metham X = mol moles of sodium and metham Mass sodium := XMW sodium Mass sodium = kg Mass metham := XMW metham Mass metham = kg Mass sodium M sodium := mass per unit channel area, of sodium width Sacramento d Sacramento Mass metham M metham := mass per unit channel area, of metham width Sacramento d Sacramento Solution The spill may be considered as a pulse injection. Na + may considered conservative while metham is not. The concentration of sodium at any point in the river is given by: C( x = M 4pD L exp We will use this equation to predict the sodium concentration at 10 km and 70 km over time. In order to do this we first must compute the mass of contaminant from the volume of the spill and the liquid characteristics t := 0.001, C sodium ( x M sodium := exp This equation treats the compound as 4 conservative, note there are no terms for decay. It would be appropriate for Na + C metham ( x := M metham exp Fechner Levy 2-27.mcd 3 of 11

4 In order to compute the total load passing a specified point we integrate the [concentration*flow velocity] with respect to time. The limits of integration bracket the time span during which the concentration is > 0.0 load conservative ( x) 6 := 0 M sodium exp load conservative ( 10km) = kg Q Sacramento dt load conservative ( 70km) = kg note that the load values are the same as the mass originally discharged and does not change between 10 km and 70 km. This would be expected for a conservative substance. Part a C sodium ( 10km C metham ( 10km Conservative Concentration - 10 km sodium metham t Maximum sodium concentration at 10 km is 25,742 / Maximum metham concentration at 10 km is 118,640 / Fechner Levy 2-27.mcd 4 of 11

5 Sodium Concentration - 70 km C sodium ( 70km C metham ( 70km Maximum sodium concentration at 70 km is 9777 / Maximum metham concentration at 70 km is 44,266 / t Comments : Dispersion reduces the Na + concentration by a factor of = C( x = M exp 4 exp ( k ) Here "k" is the sum of the first order rate constants for the various removal processes. In this problem we have hydrolysis and indirect photolysis. The overall first order rate constant for hydrolysis is ph dependent and given by: t := k' T := 0, k' n sec k' T = k a mole sec10 ph 1 sec.693 The half life for metham hydrolysis is: t H := t H = day k' T.693 The half life for metham indirect photolysis is: t p := t p = 0.08 day k p Fechner Levy 2-27.mcd 5 of 11

6 Obviously photolysis is the more rapid of the two removal mechanisms C M ( x M metham := exp 4 exp ( k' T ) compute the concentration resulting from dispersion and hydrolysis Now compute the resulting load passing 10 km and 70 km accounting for hydrolysis load disp_hyd ( x) 6 := 0 M metham exp ( ) Q Sacramento exp k' T dt load disp_hyd ( 10km) = kg load disp_hyd ( 70km) = kg The analogous equation incorporating only dispersion and indirect photolysis is: C M2 ( x := M metham exp 4 exp k p ( ) load disp_pho ( x) 6 := 0 M metham exp load disp_pho ( 10km) = kg ( ) Q Sacramento exp k p dt load disp_pho ( 70km) = kg The analogous equation incorporating dispersion, hydrolysis and photolysis is: C M3 ( x := M metham exp 4 exp ( k p + k' ) T Fechner Levy 2-27.mcd 6 of 11

7 load disp_pho_hyd ( x) 6 := 0 M metham exp ( ) Q Sacramento exp k p + k' T dt load disp_pho_hyd ( 10km) = kg load disp_pho_hyd ( 70km) = kg Note that dispersion serves to spread out the chemical and reduce the concentration but does not reduce the load. Hydrolysis and indirect photolysis actually reduce the load. At 10 km: The loss due to hydrolysis is given by: load disp_pho ( 10km) load disp_pho_hyd ( 10km) = kg load disp_pho ( 10km) load disp_pho_hyd ( 10km) km Mass = 0.84 % fraction of total mass lost to hydrolysis at 10 The loss due to I. photolysis is given by: load disp_hyd ( 10km) load disp_pho_hyd ( 10km) = kg load disp_hyd ( 10km) load disp_pho_hyd ( 10km) photolysis at 10 km Mass = % fraction of total mass lost to indirect At 70 km: The loss due to hydrolysis is given by: load disp_pho ( 70km) load disp_pho_hyd ( 70km) = kg load disp_pho ( 70km) load disp_pho_hyd ( 70km) at 70 Km Mass = % fraction of total mass lost to hydrolysis Fechner Levy 2-27.mcd 7 of 11

8 The loss due to indirect photolysis is given by: load disp_hyd ( 70km) load disp_pho_hyd ( 70km) = kg load disp_hyd ( 70km) load disp_pho_hyd ( 70km) Mass indirect photolysis at 70 km = % fraction of total metham mass lost to The result at 70 km implies that, by the time the slug has reached 70 km approximately 163 kg of metham, approximately 1.0% of original mass, has been hydrolyzed to methyl isothiocyanate and approximately kg lost due to photolysis, approximately 64% of total mass. The number of moles of metham hydrolyzed is : mole hydrolyzed := load disp_pho 70km ( ) load disp_pho_hyd ( 70km) MW metham mole hydrolyzed start of the problem. There is a 1 to 1 molar ratio of metham to methyl isocyanate, see the reaction at the Thus, the amount of methyl isocyanate getting to Lake Shasta is : mole hydrolyzed MW miso = kg Below we have plotted the concentration of metham assuming 1. dispersion+removal by indirect photolysis, 2. dispersion+hydrolysis and 3. dispersion+both processes simultaneously. Fechner Levy 2-27.mcd 8 of 11

9 C M2 ( 10km C M ( 10km C M3 ( 10km Metham Concentration - 10 km t dispersion + indirect photolysis dispersion + hydrolysis dispersion+hydrolysis+indirect photolysis C M2 ( 70km C M ( 70km C M3 ( 70km Metham Concentration - 70 km dispersion+ indirect photolysis dispersion+ hydrolysis dispersion+hydrolysis+indirect photolysis t Fechner Levy 2-27.mcd 9 of 11

10 Additional Conclusions The maximum sodium concentration at 10 km is 25,742 /. The maximum metham concentration at 10 km is 118,640 /. At 70 km the maximum sodium concentration is 9,777 /. The maximum metham concentration is 44,266 / Surprisingly then the concentration is decreased mostly because of dispersion, although dispersion does not actually remove metham. 2. Indirect photolysis is the primary removal mechanism rather than hydrolysis. The drop in metham concentration due only to dispersion is : The drop in metham concentration between 10 and 70 km due to dispersion and hydrolysis is : Thus, the drop due only to hydrolysis is: The drop due to dispersion + photolysis between 10 and 70 km is: Thus the drop due only to indirect photolysis is: The drop due to dispersion+hydrolysis+indirect photolysis between 10 and 70 km is : The percentage of the total drop in concentration between 10 and 70 km resulting from photolysis and hydrolysis is : the total drop in concentration, the remainder is due to dispersion. = % thus, loss by these 2 processes account for about 15% of Fechner Levy 2-27.mcd 10 of 11

11 Should the Henry's Law constant be estimated and the loss due to air water transfer be estimated? Answer - I say yes for the following reasons: 1. Since metham is a soil fumigant it is probably quite volatile. 2. The concentration of metham in the overlying atmosphere is probably very close to zero. 3. The resulting metham concentration after the spill is quite far from the equilibrium concentration in water. All of these would result in substantial water/air transfer. Fechner Levy 2-27.mcd 11 of 11