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Engineering Life: Miracles in Biotechnology - Bio-Enzymes: Revolutionizing Household Cleaning-advanced

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Bio-enzymes are organic solutions produced by the fermentation of organic waste (like fruit peels) with sugar (jaggery) and water, using the catalytic activity of microorganisms.

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Enzymes are biological catalysts that speed up biochemical reactions. In household cleaning, four main types are used: Proteases (break down proteins), Lipases (break down fats/lipids), Amylases (break down starch), and Cellulases (soften fabrics and remove soil).

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The fermentation process involves the breakdown of glucose (C6H12O6C_{6}H_{12}O_{6}) into alcohol and carbon dioxide, eventually producing acetic acid and beneficial enzymes.

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The standard preparation ratio for bio-enzymes is 1:3:101:3:10, representing the proportions of Jaggery, Organic Waste (fruit peels), and Water respectively.

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Bio-enzymes help in reducing the Biochemical Oxygen Demand (BODBOD) of wastewater, making them eco-friendly alternatives to chemical detergents which increase water toxicity.

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The efficiency of bio-enzymes is highly dependent on pHpH levels. Most bio-enzymes are acidic, with a pHpH typically ranging between 3.03.0 and 4.04.0.

📐Formulae

C6H12O6→Fermentation2C2H5OH+2CO2C_{6}H_{12}O_{6} \xrightarrow{Fermentation} 2C_{2}H_{5}OH + 2CO_{2}

Ratio of Ingredients=1 (Jaggery):3 (Organic Waste):10 (Water)\text{Ratio of Ingredients} = 1 \text{ (Jaggery)} : 3 \text{ (Organic Waste)} : 10 \text{ (Water)}

Enzymatic Hydrolysis of Fats: Triglyceride+3H2O→LipaseGlycerol+3Fatty Acids\text{Enzymatic Hydrolysis of Fats: } \text{Triglyceride} + 3H_{2}O \xrightarrow{Lipase} \text{Glycerol} + 3\text{Fatty Acids}

Efficiency (%)=Initial Stain Mass−Final Stain MassInitial Stain Mass×100\text{Efficiency (\%)} = \frac{\text{Initial Stain Mass} - \text{Final Stain Mass}}{\text{Initial Stain Mass}} \times 100

💡Examples

Problem 1:

A student wants to prepare a total of 2.8 kg2.8\text{ kg} of bio-enzyme mixture following the standard 1:3:101:3:10 ratio. Calculate the mass of jaggery and fruit peels required.

Solution:

Total parts in the ratio = 1+3+10=141 + 3 + 10 = 14 parts. Total mass = 2.8 kg2.8\text{ kg}. Mass of 1 part = 2.814=0.2 kg\frac{2.8}{14} = 0.2\text{ kg}. Mass of Jaggery (11 part) = 1×0.2=0.2 kg1 \times 0.2 = 0.2\text{ kg}. Mass of Fruit Peels (33 parts) = 3×0.2=0.6 kg3 \times 0.2 = 0.6\text{ kg}. Mass of Water (1010 parts) = 10×0.2=2.0 kg10 \times 0.2 = 2.0\text{ kg}.

Explanation:

By dividing the total desired mass by the sum of the ratio parts, we find the value of a single 'unit' of the mixture. Multiplying this unit by the respective ratio parts gives the specific quantities needed for jaggery and fruit peels.

Problem 2:

During the fermentation of 180 g180\text{ g} of glucose (C6H12O6C_{6}H_{12}O_{6}), how many moles of CO2CO_{2} are released? (Molar mass of Glucose = 180 g/mol180\text{ g/mol})

Solution:

The chemical equation is: C6H12O6→2C2H5OH+2CO2C_{6}H_{12}O_{6} \rightarrow 2C_{2}H_{5}OH + 2CO_{2} Number of moles of Glucose = Given MassMolar Mass=180180=1 mole\frac{\text{Given Mass}}{\text{Molar Mass}} = \frac{180}{180} = 1\text{ mole}. From the balanced equation, 1 mole1\text{ mole} of Glucose produces 2 moles2\text{ moles} of CO2CO_{2}. Therefore, moles of CO2CO_{2} released = 2 moles2\text{ moles}.

Explanation:

Based on the stoichiometry of the anaerobic fermentation reaction, every mole of glucose consumed results in the production of two moles of carbon dioxide gas, which causes the pressure buildup in bio-enzyme containers.