Determinacion de la eficiencia de conversion por el metodo de cloratos

Páginas: 6 (1285 palabras) Publicado: 11 de marzo de 2011
DETERMINATION OF SODIUM HYPOCHLORITE SODIUM EFFICIENCY CONVERTIONBASED ON CHLORATE AND CHLORIDE CONCENTRATION IN HYPOCHLORITE The principle of these procedures is that the only secondary reaction on the sodium hypochlorite absorption column leads to the formation of sodium chlorate and sodium chloride as follows:

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 (23/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 + (23/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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