Calculate the theoretical oxygen demand (ThOD, mg\/L), chemical oxygen demand (COD, mg\/L), and nitrogenous oxygen demand (NOD, mg\/L) for the wastewater from a dairy operation which contains 2000 mg\/L lactose (C12H22O11), 200 mg\/L of protein (represented by glycine, an amino acid, NH2CH2COOH) and 50 mg\/L of urea ((NH2)2CO). (Note: ThoD = COD + NOD) You will first need to determine the COD\/wt, NOD\/wt and ThOD\/wt ratios (g OD\/g compound) for each of the compounds, calculate the associated COD\u2019, NOD\u2019 and ThOD (mg\/L), for each compound, then for the combined mixture of the three. Please summarize your results using the following table.<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n To calculate the Theoretical Oxygen Demand (ThOD)<\/strong>, Chemical Oxygen Demand (COD)<\/strong>, and Nitrogenous Oxygen Demand (NOD)<\/strong> for the wastewater mixture, we need to calculate the oxygen demand for each compound based on their oxidation reactions, convert them to g O\u2082\/g compound, and then multiply by the concentration in the wastewater (mg\/L).<\/p>\n\n\n\n Oxidation: COD reaction<\/strong> (carbon oxidation): NOD reaction: To assess the oxygen demand of a dairy wastewater mixture, we analyze its content of lactose, glycine, and urea. Each compound contributes to oxygen demand in different ways based on its chemical structure.<\/p>\n\n\n\n Lactose is a carbohydrate that undergoes complete oxidation to carbon dioxide and water. This process requires molecular oxygen and contributes solely to the Chemical Oxygen Demand (COD)<\/strong>. Based on the stoichiometry of its oxidation, lactose has a COD\/wt of 1.123 g O\u2082\/g. With a concentration of 2000 mg\/L, its oxygen demand equals 2246 mg\/L.<\/p>\n\n\n\n Glycine, representing proteins, contributes to both COD<\/strong> (oxidation of the carbon portion) and Nitrogenous Oxygen Demand (NOD)<\/strong> (nitrification of the amino group). Its oxidation yields a COD\/wt of 0.64 and NOD\/wt of 0.853. At 200 mg\/L, glycine contributes 128 mg\/L to COD and 171 mg\/L to NOD, totaling 299 mg\/L as Theoretical Oxygen Demand (ThOD)<\/strong>.<\/p>\n\n\n\n Urea, common in dairy waste from urine or metabolism, contributes only to NOD. Upon hydrolysis, it produces ammonia, which is then oxidized to nitrate in biological treatment, demanding oxygen. Its NOD\/wt is high (2.133), and at 50 mg\/L concentration, it contributes 107 mg\/L to oxygen demand.<\/p>\n\n\n\n The combined ThOD<\/strong> of the mixture (sum of COD and NOD) is 2652 mg\/L<\/strong>, which indicates the total potential oxygen consumption during biological treatment. This value is critical for designing and managing wastewater treatment systems, especially in ensuring sufficient aeration and preventing oxygen depletion in receiving waters.<\/p>\n","protected":false},"excerpt":{"rendered":" Calculate the theoretical oxygen demand (ThOD, mg\/L), chemical oxygen demand (COD, mg\/L), and nitrogenous oxygen demand (NOD, mg\/L) for the wastewater from a dairy operation which contains 2000 mg\/L lactose (C12H22O11), 200 mg\/L of protein (represented by glycine, an amino acid, NH2CH2COOH) and 50 mg\/L of urea ((NH2)2CO). (Note: ThoD = COD + NOD) You […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-207517","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/207517","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=207517"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/207517\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=207517"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=207517"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=207517"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\nSTEP 1: Balanced Reactions & OD\/Weight Ratios<\/strong><\/h3>\n\n\n\n
1. Lactose (C\u2081\u2082H\u2082\u2082O\u2081\u2081) \u2013 COD only<\/strong><\/h4>\n\n\n\n
[
C_{12}H_{22}O_{11} + 12O_2 \\rightarrow 12CO_2 + 11H_2O
]
Molecular weight = 342 g\/mol
O\u2082 required = 12 mol \u00d7 32 g\/mol = 384 g O\u2082\/mol<\/strong>
COD\/wt = 384 \/ 342 = 1.123 g O\u2082\/g<\/strong><\/p>\n\n\n\n2. Glycine (NH\u2082CH\u2082COOH) \u2013 COD + NOD<\/strong><\/h4>\n\n\n\n
[
NH_2CH_2COOH + 1.5O_2 \\rightarrow 2CO_2 + 2H_2O + NH_3
]
NOD reaction<\/strong> (ammonia nitrification):
[
NH_3 + 2O_2 \\rightarrow NO_3^- + H_2O + H^+
]<\/p>\n\n\n\n\n
Total = 112 g O\u2082\/mol
Molecular weight = 75 g\/mol
COD\/wt = 48 \/ 75 = 0.64 g O\u2082\/g<\/strong>
NOD\/wt = 64 \/ 75 = 0.853 g O\u2082\/g<\/strong>
ThOD\/wt = 1.493 g O\u2082\/g<\/strong><\/li>\n<\/ul>\n\n\n\n3. Urea ((NH\u2082)\u2082CO) \u2013 NOD only<\/strong><\/h4>\n\n\n\n
[
(NH_2)_2CO + H_2O \\rightarrow 2NH_3 + CO_2 \\
2NH_3 + 4O_2 \\rightarrow 2NO_3^- + H_2O + H^+
]
4 mol O\u2082 per mol urea = 128 g O\u2082
Molecular weight = 60 g\/mol
NOD\/wt = 128 \/ 60 = 2.133 g O\u2082\/g<\/strong><\/p>\n\n\n\n
\n\n\n\nSTEP 2: Calculate COD\u2019, NOD\u2019, ThOD (mg\/L)<\/strong><\/h3>\n\n\n\n
Compound<\/th> COD\/wt (g\/g)<\/th> NOD\/wt (g\/g)<\/th> ThOD\/wt (g\/g)<\/th> Conc. (mg\/L)<\/th> COD’ (mg\/L)<\/th> NOD’ (mg\/L)<\/th> ThOD (mg\/L)<\/th><\/tr><\/thead> Lactose<\/td> 1.123<\/td> 0<\/td> 1.123<\/td> 2000<\/td> 2246<\/td> 0<\/td> 2246<\/td><\/tr> Glycine<\/td> 0.64<\/td> 0.853<\/td> 1.493<\/td> 200<\/td> 128<\/td> 171<\/td> 299<\/td><\/tr> Urea<\/td> 0<\/td> 2.133<\/td> 2.133<\/td> 50<\/td> 0<\/td> 107<\/td> 107<\/td><\/tr> Total<\/strong><\/td> <\/td> <\/td> <\/td> <\/td> 2374<\/strong><\/td> 278<\/strong><\/td> 2652<\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
\n\n\n\nSummary<\/strong><\/h3>\n\n\n\n
\n
\n\n\n\nExplanation (300 words)<\/strong><\/h3>\n\n\n\n