Consider a hypothetical reaction: A ——> B.
If this is a thermodynamically unfavorable reaction the ΔGo\' value will be positive. Let\'s assume it is +4.0 kcal/mol. In order to drive this reaction in the direction written it can be coupled to the hydrolysis of ATP. The free energy of ATP hydrolysis to ADP is shown:

ATP + H2O ——> ADP + Pi: ΔGo\' = –7.3 kcal/mol

Coupling the two reactions together gives the equation:

A + ATP + H2O ——> B + ADP + Pi + H+

The ΔGo\' for this coupled reaction is the sum of the ΔGo\' values of the two separate reactions, i.e. (-7.3kcal/mol) + (+4.0kcal/mol) = -3.3kcal/mol. This indicates that coupling ATP hydrolysis provides the energy necessary to make the conversion of A to B thermodynamically favorable.

Another useful example is to examine one of the reactions of glycolysis. In this case we will look at the oxidation of phosphoenolpyruvate to pyruvate catalyzed by the enzyme pyruvate kinase (PK).

phosphoenolpyruvate ——> pyruvate: ΔGo\' = –14.7 kcal/mol
This reaction releases sufficient energy to drive the synthesis of ATP from ADP and Pi which would normally be thermodynamically unfavorable with a ΔGo\' of +7.3kcal/mol. Note that this value is the reciprocal of the hydrolysis of ATP. This points out another fact that the ΔGo\' for a reaction in one direction is equal but mathematically opposite for the reciprocal direction. Coupling the two reactions together yields:

phosphoenolpyruvate + ADP + H+ ——> pyruvate + ATP: ΔGo\' = –7.4 kcal/mol
For the simple enzyme-catalyzed reaction:
S <==> P
S is substrate; and P is product.
The enzyme forms a complex with the substrate, in much the same way as a protein-ligand complex, and performs some chemical reaction/transformation of the bound substrate. The resultant product is released.

Two or more reactions in a cell sometimes can be coupled so that thermodynamically unfavorable reactions and favorable reactions are combined to drive the overall process in the favorable direction. In this circumstance the overall free energy is the sum of individual free energies of each reaction. This process of coupling reactions is carried out at all levels within cells. The predominant form of coupling is the use of compounds with high energy to drive unfavorable reactions.

The predominant form of high energy compounds in the cell are those which contain phosphate. Hydrolysis of the phosphate group can yield free energies in the range of –10 to –62 kJ/mol. These molecules contain energy in the phosphate bonds due to:

1. Resonance stabilization of the phosphate products

2. Increased hydration of the products

3. Electrostatic repulsion of the products

4. Resonance stabilization of products

5. Proton release in buffered solutions

The latter phenomenon indicates that the pH of the solution a reaction is performed in will influence the equilibrium of the reaction.

To effectively interpret the course of a reaction in the presence of a mixture of components, such as in the cell, one needs to account for the free energies of the contributing components. This is accomplished by calculating total free energy which is comprised of the individual free energies. In order to carry out these calculations one needs to have a reference state from which to calculate free energies. This reference state, termed the standard state, is chosen to be the condition where each component in a reaction is at 1M. Standard state free energies are given the symbol: G°.

The partial molar free energy of any component of the reaction is related to the standard free energy by the following:

G = G° + RTln[X]

From this equation one can see that when the component X, or any other component, is at 1M the ln[1] term will become zero and:

G = G°

The utility of free energy calculations can be demonstrated in a consideration of the diffusion of a substance across a membrane. The calculation needs to take into account the changes in the concentration of the substance on either side of the membrane. This means that there will be a ΔG term for both chambers and, therefore, the total free energy change is the sum of the ΔG values for each chamber:

Equation for total free energy change

This last equation indicates that if [A]2 is less than [A]1 the value of ΔG will be negative and transfer from region 1 to 2 is favored. Conversely if [A]2 is greater than [A]1 ΔG will be positive and transfer from region 1 to 2 is not favorable, the reverse direction will be.

One can expand upon this theme when dealing with chemical reactions. It is apparent from the derivation of ΔG values for a given reaction that one can utilize this value to determine the equilibrium constant, Keq. As for the example above dealing with transport across a membrane, calculation of the total free energy of a reaction includes the free energies of the reactants and products:

ΔG = G(products) - G(reactants)

Since this calculation involves partial molar free energies the ΔG° terms of all the reactants and products are included. The end result of the reduction of all the terms in the equation is:

Standard free energy relation to reactant and product concentrations

When the above equation is used for a reaction that is at equilibrium the concentration values of A, B, C and D will all be equilibrium concentrations and, therefore, will be equal to Keq. Also, when at equilibrium ΔG = 0 and therefore:

0 = ΔG° + RTlnKeq

Equation of equilibrium constant (Keq) relationship to free energy change

This demonstrates the relationship between the free energy values and the equilibrium constants for any reaction.

Biochemical Oxygen Demand

When biodegradable organic matter is released into a body of water, microorganisms especially bacteria feed on the waste, breaking it down into simpler organic and inorganic substances .when this decomposition takes place in an aerobic environment, i.e. in the presence of oxygen, the process produces non–objectionable, stable end product such as carbon dioxide, sulphates, ortho-phosphate and nitrate. A simplified representation of aerobic decomposition is given by following.

Organic Matter + O2 MICROORGANISMS CO2 + H2O + New cells + Stable Products

The amount of oxygen required by microorganism to oxidize organic waste aerobically is called as biochemical oxygen demand (BOD). It is generally expressed in mg of oxygen required per litre of waste.

MODELLING BOD AS FIRST ORDER REACTION:-

Rate of decomposition of organic wastes is proportional to the amount of waste available. If Lt represent the amount of oxygen demand left after time t, then, first order reaction equation will be,

dLt/dt = -k Lt (1)

Where k = BOD reaction rate constant 

Lt = Lo e –kt (2) 
Where Lo = Ultimate BOD.

The ultimate BOD is the sum of the amount of oxygen consumed by the waste in the first t days (BODt), plus the amount of oxygen remaining to be consumed after time t. i.e.

Lo = BODt + Lt (3)

Combining eq. (2) and (3)

BODt = Lo (1- e–kt) (4) 

Soap and Detergents

SOAPS

Ordinary soaps are derived from fats and oils by saponification with sodium hy-droxide. Saponification is a special case of hydrolysis in which an alkaline agent is present to neutralize the fatty acids as they are formed. Beef fat and cottonseed oil are used to pro-duce low-grade, heavy-duty soaps. Coconut and other oils are used in the produc-tion of toilet soaps.

All sodium and potassium soaps are soluble in water. If the water is hard, the calcium, magnesium, and any other ions causing hardness precipitate the soap in the form of metallic soaps. Soap must be added to precipitate all the ions causing hardness before it can act as a surfactant, usually indicated by the onset of frothing upon agitation The synthetic surfactants are of three major types: anionic, nonionic, and cationic. 

Cationic Detergents

The cationic detergents are salts of quaternary ammonium hydroxide. In quaternary ammonium hydroxide, the hydrogens of the ammonium ion have all been replaced with alkyl groups. The surface-active properties are contained in the cation.
The cationic detergents are noted for their disinfecting (bactericidal) properties. They are used as sanitizing agents for dishwashing where hot water is unavailable or undesirable.

The alkyl-benzene sulfonates de-rived from propene were highly resistant to biodegradation, and their persistence resulted in excessive foaming in rivers and groundwaters in the 1950s. This presented some of the first evidence of the potential harmful environmental consequences of synthetic organic chemicals. For this reason, the detergent manufacturing industry changed to the production of LAS surfactants. LAS is readily degradable under aerobic conditions, and its use has helped relieve the most serious problems of detergent foaming. However, unlike common soap, it is resistant