Determinacion de la eficiencia de conversion por el metodo de cloratos
NaClO = 1/3 NaClO3 + 2/3 NaCl
Procedure #1 : Based on the determination ofsodium chlorate and sodium chloride in sodium hypochlorite: The reaction efficiency is determined by the following formula:
Eff = {[100 x (mol/l active Cl2)] / [(mol/l active Cl2) + (3 x mol/l NaClO3)]}
As example: • • Active chlorine by weight = 10.8% Density: 1.1676
Then, the active chlorine concentration in mol/l is: 0.108 x 1.1676 x 1000 = 126.1 gpl Cl2 = 1.7761 mol/lhypochloritereactorefficiencydetermination.VictorCattaneo/ag/cavs/forANT/110110
•
Chlorate concentration as NaCl03 = 887 ppm (mg/kg)
Then, the chlorate concentration in mol/l is: 887 x 1.1676 / 1000 = 1.03566 gr/l = 0.00973 mol / l Therefore, the reaction efficiency will be:
Eff = 100 x 1.7761 / (1.7761 + 3 X 0.00973) = 98.38 %
The reaction efficiency could be also calculated based on the chlorideconcentration as per the following procedure: Procedure # 2: Based on the determination of total sodium chloride and active chlorine in sodium hypochlorite: In this case, the active chlorine and the sodium chloride numbers are enter in the table listed in the document below (page 12 and page 13).
hypochloritereactorefficiencydetermination.VictorCattaneo/ag/cavs/forANT/110110
CONVE & AVS,INC.
Summary of Decomposition Pathways of NaOCl:
2 NaOH+Cl2 = NaOCl + NaCl+H2O + 350 Kcal / kg Cl2 f ( pH, Temperature) f ( Light, Impurity Concentrations)
3 NaOCl = NaClO3 + 2NaCl
2 NaOCl = 2NaCl + O2
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Reaction Efficiency Losses During Bleach Production Inert 2 Mol NaOH
Main Reaction: 2NaOH + Cl2 = NaOCl + NaCl + H2O
1 Mol Cl2 Bleach
MainSid R M i Side Reaction: ti NaOCl = 1/3NaClO3 + 2/3NaCl 2nd Side Reaction : 2 NaOCl = 2NaCl + O2 Assumption: does not occur
Reaction Efficiency E ff = 1 1- NaOCl + /3 NaClO3 + 2/3 NaCl + NaCl => NaOCl N OCl + /3 N ClO3 + (2 3/3 N Cl /3 NaClO3 (23/3 NaCl
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Calculation of Reaction Efficiency by total Chloride Content 2 Mol NaOHReaction Effi i R i Efficiency E ff = 1 1-
% Efficiency 100 x 2(1- η)NaOCl ( ) (6 - η)/3 NaCl η /3NaClO3
1 Mol Cl2
NaOCl NaOCl + (23/3 NaCl + /3 NaClO3 NaOCl => NaCl (by reduction with H2O2) NaCl NaCl + NaClO3
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Calculation of Reaction Efficiency with Unknown Content of NaClO3, Solving Equation (4)Inert 2 Mol NaOH
200 x Available Cl2, mol/l (1) Total NaCl, mol/l NaClO3, mol/l 100 x 2 (1 - η), NaOCl (2) % Efficiency (6 - η)/3 NaCl η /3 NaClO3 )/3, /3,
(3) (4)
% Efficiency
1 Mol Cl2
% Efficiency 100(1 - η)
Bleach
(3B 6) (7 6 B ) η (3B 1) η 2 0
where B
Total NaCl, mol/l Available Cl2, mol/l
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FormulaCalculation of Reaction Efficiency by total Chloride Content in Bleach
(1) Concentrati A ti Cl2 A , mol/l C t tion Active l/l
(10)
(11)
(2) 2NaOH Cl2 NaCl NaOCl H2O (3) NaOCl H2O2 NaCl H2O O2 (4) Total concentration Cl - S , mol/l
Note : value of S corrected for NaCl content in NaOH
η X (1- η) 3 A E 2 (1- η) (1- η) 100 B(1- η) η /3
(5) Concentration total NaClO3 X, mol/l (6) S / A B 200 x A (7) Eficiency %, E 1 n S X 100 x 2 (1- η) (8) Eficiency %, E (6 - η)/3 η /3 A Dividing Equation # 7 i t Di idi E ti into (1- n) E 2 (1- η) (9) 100 B(1- η) X(1 η) /A ( (
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(12) B(1 η)2 (1 η)η/3 2(1 η) B(1- ) (1- ) /3 2(1- )
(13) (3B 6) (7 6 B) η (3B 1) η2 0 (14) (3B 1) a (15) (7 6B) b ( ) ( (16) (3B 6) c )...
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