Order Of Reaction And Half Life : Chemical Kinetics

📘 Introduction

Chemical reactions proceed at different rates depending on the concentration of reactants. This relationship is described using the concept of Order of Reaction. Understanding zero, first, and second order reactions is critical in predicting product formation, controlling reaction speed, and interpreting real-world chemical behavior. An equally important concept is Half-Life, which tells us how long it takes for half the reactant to be consumed.

🟢 Zero Order Reaction

In zero-order reactions, the rate of reaction does not depend on the concentration of the reactant.
Rate law: Rate = k
Unit of rate constant (k): mol L^-1 s^-1

Example: Decomposition of NH₃ on platinum surface.
Graph: [A] vs Time is linear with negative slope.

🔵 First Order Reaction

In a first-order reaction, the rate depends linearly on the concentration of one reactant.
Rate law: Rate = k[A]
Unit of k: s^-1

Example: Radioactive decay, hydrolysis of ester.
Graph: ln[A] vs Time gives a straight line.

🟡 Second Order Reaction

The rate of a second-order reaction depends on the square of one reactant or the product of two reactants.
Rate law: Rate = k[A]² or Rate = k[A][B]
Unit of k: L mol^-1 s^-1

Example: 2NO + O₂ → 2NO₂
Graph: 1/[A] vs Time gives a straight line.

🧭 Half-Life (t_½) Concept

Half-life is the time required to reduce the concentration of a reactant to half its initial value.

• Zero order: t_½ = [A]₀/2k
• First order: t_½ = 0.693/k (independent of concentration)
• Second order: t_½ = 1/k[A]₀

Zero Order Reaction

In zero order reactions, the rate of reaction is independent of the concentration of reactants. The concentration decreases linearly with time.
Example: Photochemical decomposition of HI on a gold surface.

First Order Reaction

In first order reactions, the rate of reaction is directly proportional to the concentration of one reactant. The graph shows an exponential decrease in concentration.
Example: Decomposition of N₂O₅.

Second Order Reaction

In second order reactions, the rate depends on the square of the concentration of the reactants. The graph of [R] vs. time shows a curve that decreases steeply.
Example: Reaction between NO₂ and CO or 2A → Products.

⚙️ Applications in Real Life

• 📺 Radioactive decay in carbon dating is a first-order reaction.
• 💊 Drug elimination from the body often follows first-order kinetics.
• 🧪 Enzyme-catalyzed reactions exhibit zero-order behavior at saturation.
• 🚗 Pollution breakdown reactions can follow second-order kinetics.

🔟 IIT-JEE Style Conceptual Questions

1. A reaction is zero-order in [A]. What happens to rate if [A] is doubled?

2. Half-life of a first-order reaction is 10 mins. How long for 75% to decay?

3. For a reaction: Rate = k[A]². What is the unit of k?

4. For a first-order reaction, ln[A] vs time gives a straight line. What does the slope represent?

5. If [A]₀ = 0.5 M and after one t_½ it becomes 0.25 M, what is the order?

6. In which order reaction, t_½ is inversely proportional to initial concentration?

7. Rate = k. What type of reaction is this?

8. For a second-order reaction, what is the shape of [A] vs time graph?

9. What is the dimension of rate constant for first-order reaction?

10. Which reaction order has a constant half-life?

MCQ for IIT JEE

🧪 IIT-JEE Quiz: Reaction Order & Half-Life

1. For a zero-order reaction, the rate of reaction:

A. Increases with concentration
B. Decreases with concentration
C. Remains constant
D. Depends on catalyst only

✔ Answer: C. Remains constant

2. The unit of rate constant for a first-order reaction is:

A. mol L⁻¹ s⁻¹
B. s⁻¹
C. L mol⁻¹ s⁻¹
D. No units

✔ Answer: B. s⁻¹

3. For a second-order reaction, half-life is:

A. Directly proportional to [A]₀
B. Inversely proportional to [A]₀
C. Independent of [A]₀
D. Proportional to square of [A]₀

✔ Answer: B. Inversely proportional to [A]₀

4. In a first-order reaction, time to complete 75% is:

A. 2t½
B. t½
C. 3t½
D. 4t½

✔ Answer: C. 3t½

5. A reactant decreases linearly with time. The reaction is:

A. First-order
B. Second-order
C. Zero-order
D. Third-order

✔ Answer: C. Zero-order

6. t½ = 10 min for a 1st order reaction. Time for 87.5% reaction:

A. 20 min
B. 30 min
C. 40 min
D. 50 min

✔ Answer: B. 30 min

7. Integrated rate law for 2nd-order A → Product:

A. [A] = [A]₀ - kt
B. ln[A] = ln[A]₀ - kt
C. 1/[A] = 1/[A]₀ + kt
D. [A] = [A]₀e⁻ᵏᵗ

✔ Answer: C. 1/[A] = 1/[A]₀ + kt

8. ln[A] vs time gives a straight line for:

A. Zero-order
B. First-order
C. Second-order
D. All of these

✔ Answer: B. First-order

9. For k = 0.693 s⁻¹ in a 1st-order reaction, t½ is:

A. 1 s
B. 0.5 s
C. 10 s
D. 100 s

✔ Answer: A. 1 s

10. Which is incorrect for zero-order reactions?

A. Rate is independent of [A]
B. [A] vs time is a straight line
C. Half-life increases with [A]₀
D. Half-life is constant

✔ Answer: D. Half-life is constant

🔗 Internal Links

🎮 Reaction Order Game
📊 Rate vs Concentration Graph
🧠 Colligative Property Game

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JJ Thomson and Rutherfords Experiments

🔬 Life and Experiments of J.J. Thomson and Ernest Rutherford

🌟 Introduction (परिचय)

Atomic science started to shine in the early 20th century with the contributions of two great physicists: J.J. Thomson and Ernest Rutherford. Their groundbreaking experiments led to models that changed how we understand atoms today. These foundations are now seen in technologies like television displays, mobile processors, and even nuclear medicine.

👨‍🔬 J.J. Thomson: Life & Achievements (जीवन और उपलब्धियाँ)

Sir Joseph John Thomson was born in 1856 in Manchester, UK. He studied engineering but became more inclined towards physics. He later became the director of the famous Cavendish Laboratory. In 1897, he discovered the electron — a fundamental negatively charged particle — for which he received the Nobel Prize in Physics (1906).

📎 Thomson's Cathode Ray Experiment

Thomson used a cathode ray tube — a glass tube from which air is removed — and observed a stream of rays coming from the cathode. These rays:

• Were negatively charged
• Could be deflected by electric and magnetic fields
• Had mass-to-charge ratio (e/m) lower than hydrogen ion

Conclusion: He proposed the "Plum Pudding Model" — atom is a positively charged sphere with electrons embedded in it like raisins in pudding.

⚠️ Drawbacks of Thomson Model

- Could not explain the nuclear structure
- No explanation for alpha scattering or spectral lines

🔬 Rutherford: Life & Achievements (जीवन और प्रयोग)

Ernest Rutherford, born in New Zealand in 1871, is known as the "Father of Nuclear Physics." He studied at Cambridge and worked with J.J. Thomson. In 1909, he and his assistants Geiger and Marsden performed the famous Gold Foil Experiment. He received the Nobel Prize in Chemistry (1908).

💡 Rutherford's Alpha Scattering Experiment

A thin gold foil was bombarded with alpha particles (positively charged helium nuclei). Observations:

• Most passed through undeflected
• Some deflected at small angles
• Few bounced back

Conclusion: Atom has a small, dense, positively charged center called the nucleus. Electrons revolve around it like planets around the sun — called the Rutherford Model.

❌ Limitations of Rutherford's Model

- Failed to explain electron stability in orbits
- Did not justify atomic spectra
- Classical physics predicted electrons would spiral into the nucleus

📱 Applications in Modern Devices (आज के उपकरणों में प्रयोग)

The experiments of Thomson and Rutherford have deep influence in:

• Television Screens: CRTs use cathode rays (based on Thomson’s work)
• Mobile Phones: Atomic models help in transistor-level design
• Nuclear Reactors: Based on understanding of nuclear structure
• X-rays and Imaging: Use behavior of atomic particles

These devices wouldn't have been possible without the pioneering work on atomic structure.

🔗 Internal Blog Links

• 🧠 Play Colligative Memory Game
• 🎯 Limiting Reagent Game
• 📚 Atomic Structure Related Articles

📌 Hindi Summary (हिन्दी में सारांश)

जे.जे. थॉमसन ने इलेक्ट्रॉन की खोज की और 'प्लम पुडिंग मॉडल' दिया।
अर्नेस्ट रदरफोर्ड ने न्यूक्लियस की खोज की और परमाणु का सौरमंडल जैसा मॉडल प्रस्तुत किया।
इनके मॉडल की सीमाएं भी थीं, लेकिन आधुनिक विज्ञान और तकनीक — जैसे मोबाइल, टीवी, रिएक्टर — इन्हीं खोजों पर आधारित हैं।

🔍 Final Thoughts

The evolution from Thomson’s cathode rays to Rutherford’s nucleus marked a revolution in atomic theory. While modern quantum models like Bohr’s or Schrödinger’s have advanced further, the legacy of these early experiments remains strong. Every time you unlock your smartphone, watch TV, or undergo an MRI, you’re witnessing the legacy of these brilliant scientists.

🧪 Rutherford & J.J. Thomson Quiz

📚 Empirical formula

© 2025 STEMFACT.IN — All Rights Reserved. Do not reproduce without permission.

Chemistry Solution With All Terms And Memory Game

💡 Solution Chemistry: Complete Guide with Advanced Topics & Memory Game

🔍 What is a Solution?

A solution is a homogeneous mixture composed of two or more substances. The substance present in the larger amount is usually the solvent, and the one in lesser amount is the solute.

• Solvent: The component that dissolves the solute (e.g., water).
• Solute: The component that gets dissolved (e.g., salt).

🧪 Henry’s Law

Henry’s Law states that the solubility of a gas in a liquid is directly proportional to the pressure of that gas above the liquid at constant temperature.

Mathematical Expression: C = kP, where:

• C is the concentration of gas in the liquid.
• P is the partial pressure of the gas.
• k is the Henry’s law constant.

Applications: Carbonated beverages, deep-sea diving, and anesthesia.

🧪 Types of Solutions

Depending on the physical states of solute and solvent, solutions can be:

• Solid in liquid (e.g., sugar in water)
• Gas in liquid (e.g., CO₂ in soda)
• Liquid in gas (e.g., water vapor in air)

🔬 Non-Ideal Solutions

Non-ideal solutions deviate from Raoult’s Law due to differing interactions between solute and solvent molecules.

✅ Positive Deviation

• Weaker solute-solvent interactions
• Increased vapor pressure
• Example: Ethanol and Acetone

✅ Negative Deviation

• Stronger solute-solvent interactions
• Decreased vapor pressure
• Example: Acetone and Chloroform

♻️ Azeotropes

An azeotrope is a mixture of two or more liquids that boils at a constant temperature and composition.

• Minimum Boiling Azeotrope: Ethanol + Water
• Maximum Boiling Azeotrope: Hydrochloric acid + Water

They cannot be separated by simple distillation.

📏 Colligative Properties

• Relative Lowering of Vapor Pressure
• Boiling Point Elevation
• Freezing Point Depression
• Osmotic Pressure

These depend on the number of solute particles, not their identity.

🧮 van’t Hoff Factor (i)

Formula: i =

observed value⁄calculated value

It adjusts for dissociation/association of solutes. For example, NaCl dissociates into Na⁺ and Cl⁻, so i ≈ 2.

Graphical view of Positive and negative Deviation

Positive & Negative Deviations

Graphical view of Depression in Freezing Point

Depression in Freezing Point

Graphical view of Elevation in Boiling Point

Elevation in Boiling Point

🎮 Memory Puzzle Game

Match the concepts and their terms by clicking below:

🎮 Play Colligative Properties Memory Game

✅ Conclusion | निष्कर्ष

This article includes basic to advanced level content from solubility concepts to memory games.

📌 Bookmark this article to revise before exams and Olympiads!

Colligative Properties IIT - JEE level Problems

Question 1

A solution contains 5.85 g of NaCl in 100 g of water. Calculate the depression in freezing point. (K_f = 1.86 K kg/mol)

Answer: ΔT_f = 3.72 K (Using van't Hoff factor i = 2)

Question 2

A 0.1 molal solution of urea (non-electrolyte) causes a depression in freezing point of 0.186 K. What is the molal depression constant of the solvent?

Answer: K_f = ΔT_f/m = 0.186 / 0.1 = 1.86 K kg/mol

Question 3

Which of the following 0.01 molal aqueous solution will have the lowest freezing point?
A. NaCl
B. Glucose
C. BaCl₂
D. Urea

Answer: C. BaCl₂ (i = 3, highest van't Hoff factor)

Question 4

Calculate the osmotic pressure of a 0.1 M solution of glucose at 27°C. (R = 0.0821 L atm/mol K)

Answer: π = nRT/V = 0.1 × 0.0821 × 300 = 2.46 atm

Question 5

A solution containing 1 g of solute in 100 g of water gave a boiling point elevation of 0.52°C. If K_b = 0.52 K kg/mol, find the molar mass of the solute.

Answer: M = (1000 × 1 × 0.52) / (100 × 0.52) = 100 g/mol

Question 6

1.8 g of glucose (C₆H₁₂O₆) is dissolved in 100 g of water. Calculate the depression in freezing point. (K_f = 1.86 K kg/mol)

Answer: ΔT_f = 0.186 K

Question 7

Which of the following shows the highest boiling point elevation?
A. 0.1 M NaCl
B. 0.1 M BaCl₂
C. 0.1 M Glucose
D. 0.1 M AlCl₃

Answer: D. AlCl₃ (i = 4, highest number of particles)

Question 8

A 1 molal solution of a non-volatile solute has a freezing point depression of 1.86°C. What will be the freezing point of this solution?

Answer: 0 - 1.86 = -1.86°C

Question 9

Osmotic pressure of a solution at 300 K is 2.46 atm. Find the concentration of solute (R = 0.0821 L atm/mol K).

Answer: C = π / RT = 2.46 / (0.0821 × 300) = 0.1 M

Question 10

Which property is used to determine molar mass of a solute in dilute solution?
A. Surface tension
B. Colligative property
C. Viscosity
D. Refractive index

Answer: B. Colligative property

Question 11

3 g of a non-volatile solute dissolved in 100 g water produces a boiling point elevation of 0.3°C. Molar mass of solute is 60. What is K_b?

Answer: K_b = ΔT_b × M × 1000 / (W × m) = 0.3 × 60 × 1000 / (3 × 100) = 6 K kg/mol

Question 12

Identify the colligative property which is not affected by ionization:
A. Relative lowering of vapor pressure
B. Boiling point elevation
C. Freezing point depression
D. None of these

Answer: D. None of these (all are affected by van’t Hoff factor)

Question 13

A 0.1 M solution of MgSO₄ behaves as 0.15 M in osmotic pressure calculation. What is the van’t Hoff factor?

Answer: i = 0.15 / 0.1 = 1.5

Question 14

Freezing point of a solution of 10 g urea in 180 g of water is?
(K_f = 1.86 K kg/mol, Molar mass of urea = 60 g/mol)

Answer: ΔT_f = 1.86 × (10/60) / (0.18) ≈ 1.72 K; T_f = -1.72°C

Question 15

Which factor decides magnitude of colligative properties?
A. Nature of solute
B. Number of solute particles
C. Volume of solvent
D. Temperature

Answer: B. Number of solute particles

🔗 Related Links

• 🧪 More Chemistry Experiments
• 🎲 Play Solubility Game
• 📘 Class 12 Chemistry Notes

Solubility test of iron sulphate , copper sulphate , pottasium nitrate

🧪 Lab Experiment: Color and Solubility of Common Salts

In this experiment, we observed the solubility and color changes of three different salts: Iron(II) sulfate (FeSO₄), Copper(II) sulfate (CuSO₄), and Potassium nitrate (KNO₃). These compounds were mixed with cold water in separate conical flasks, and real-life reactions were documented with photographs. This hands-on activity helped students understand how chemical substances behave in aqueous solutions.

📚 Background: What is Solubility?

Solubility is the ability of a substance (solute) to dissolve in a solvent (usually water) and form a solution. When a salt dissolves, its ions break apart and disperse in water. This process is influenced by factors like temperature, nature of the solute and solvent, and particle size.

Color changes often happen due to hydration or chemical interactions with water or air. Transition metals like copper and iron show vivid colors due to d-d electronic transitions.

🟢 Iron(II) Sulfate – Surprising Color Shift

When I took a sample of Iron(II) sulfate, it appeared light green in its dry form. I carefully poured it into a conical flask containing cold water. Surprisingly, the color changed — it became more intense, slightly murky green. This was fascinating for all of us.

Possible reason: FeSO₄ undergoes slight oxidation in air, forming brown Fe³⁺ ions or hydrated iron complexes. This is a classic example of transition metal sensitivity.

🔵 Copper(II) Sulfate – The Blue Beauty

Copper sulfate was deep blue even before mixing. After adding it to water, it dissolved completely, forming a clear blue solution. This is due to the formation of [Cu(H₂O)₆]²⁺ ions – a hydrated complex responsible for the blue color.

This part of the experiment was very visual and enjoyed by students. It confirmed that CuSO₄ is highly soluble and reacts immediately in water.

⚪ Potassium Nitrate – Totally Transparent

Potassium nitrate (KNO₃) is a colorless crystalline compound. When added to water, it dissolved instantly, forming a transparent solution. There was no color change observed. This is expected as KNO₃ is a neutral salt formed from a strong acid and base (HNO₃ + KOH).

Its dissociation was complete, releasing K⁺ and NO₃⁻ ions into solution. Students found this salt visually boring but scientifically important.

📊 Scientific Concepts

To understand ionic behavior in solution, chemists use the molar conductivity formula:

Λ_m = Λ_m⁰ − A√c

Where:

• Λ_m = molar conductivity at concentration c
• Λ_m⁰ = molar conductivity at infinite dilution
• A = constant depending on ion interaction
• c = concentration in mol/L

This formula is useful in comparing how easily different salts conduct electricity in solution.

👩‍🔬 Student Observations & Reactions

Students were excited to see visible changes in the lab. Some comments:

• "I didn’t know iron salt could change color so fast!"
• "Copper salt is so beautiful, like blue ink."
• "Potassium nitrate looks like nothing happened, but it’s all inside the solution!"

The use of real lab tools and flasks made this experiment memorable. Even students who were afraid of chemistry found it approachable.

🌱 Real-Life Applications of These Salts

• FeSO₄: Used in iron supplements and soil treatment for iron deficiency in plants.
• CuSO₄: Used as a fungicide in agriculture and to clean aquarium water.
• KNO₃: Used in fertilizers, food preservatives, and fireworks due to its oxidizing properties.

Understanding how these behave in water helps us see their importance in industry and daily life.

🧠 Extension Activity

You can extend this experiment by:

• Measuring temperature changes during dissolution
• Adding acids or bases to the solutions and observing reactions
• Testing electrical conductivity using a simple circuit

This builds deeper understanding of ionic chemistry and thermodynamics.

🔗 Related Articles

• Empirical Formula & Calculations
• 🎮 Limiting Reagent Spin Game
• How Fireflies Create Light (Bioluminescence)

✅ Conclusion

This lab experiment was a great success. It visually demonstrated how different salts dissolve, react, and behave in water. From the vibrant blue of copper sulfate to the neutral clarity of potassium nitrate, students witnessed chemistry in action.

Taking pictures, noting color changes, and discussing applications made the lesson more than just theory — it became an experience. Such hands-on activities should be a regular part of learning.

Oxalic Acid Experiment to Prepare 1/20 M

Preparation of 0.1 M Oxalic Acid Solution (250 mL)
with RDKit & Real Lab Activity

Preparing standard solutions is one of the most basic yet crucial lab activities in chemistry. Recently, I conducted a practical experiment to prepare a 0.1 M (1/10 M) solution of oxalic acid using 250 mL of distilled water. This post documents every step of the process — from selecting glassware and using the measuring toolbox to understanding the calculations and integrating RDKit for molecular insights.

🔬 1. Why Use Oxalic Acid?

Oxalic acid (H₂C₂O₄·2H₂O) is a dibasic acid often used in titrations and standard solution preparation because of its stable molecular mass and reliable properties. It's widely used to standardize KMnO₄ solutions in redox reactions.

⚖️ 2. Tools and Chemicals Used

• Oxalic Acid (white crystalline form)
• Measuring Tool Box (electronic or manual balance, beaker, glass rod, funnel)
• 250 mL volumetric flask
• Distilled Water
• Funnel, glass rod, wash bottle

I used a special lab toolbox with a mini weighing balance, which allowed accurate measurement of small masses in grams.

📐 3. Understanding the Calculation

To prepare a 0.1 M solution, we use the molarity formula:

Molarity (M) = moles / volume (L)

We rearrange this to calculate the mass needed:

Required Mass = Molarity × Molar Mass × Volume (L)

For oxalic acid dihydrate (H₂C₂O₄·2H₂O), the molar mass = 126 g/mol

Volume = 250 mL = 0.250 L Molarity = 0.1 M → Mass = 0.1 × 126 × 0.250 = 3.15 g

🧪 4. Procedure I Followed in the Lab

1. Weighed exactly 3.15 g of oxalic acid crystals using the digital balance.
2. Dissolved them in about 150 mL of distilled water in a beaker using a glass rod.
3. Transferred the solution into a 250 mL volumetric flask using a funnel.
4. Washed the beaker and funnel with distilled water and poured that into the same flask.
5. Added more distilled water slowly till the bottom of the meniscus touched the 250 mL mark.
6. Closed the flask and inverted it several times to mix evenly.

📸 5. Observations & Practical Notes

• Oxalic acid crystals appeared as white, danedar powder
• They dissolved easily in water with slight cooling effect
• Final solution appeared transparent, indicating full dissolution

🧠 6. Why Precision is Critical

Even a small error in weighing or water level can lead to incorrect molarity. That’s why I used a toolbox with proper measuring glassware and an accurate digital balance.

🧬 7. RDKit: Chemistry Meets Code

RDKit is an open-source cheminformatics tool that allows us to visualize and manipulate chemical structures. It helps verify molecular properties like molar mass, bonding, and atom count.

💡 You can explore oxalic acid using RDKit tools here: Explore Oxalic Acid on RDKit

SMILES for Oxalic Acid: OC(=O)C(=O)O

🔗 8. Real-World Uses of Oxalic Acid

• Used in removing rust and stains
• Standardization of permanganate solutions
• Cleaning agent in textile and leather industries
• Found naturally in rhubarb and spinach

📚 9. Related Concepts

• Molarity vs Normality
• Ostwald's Law
• RDKit Documentation

📌 10. Conclusion

Preparing a 0.1 M solution of oxalic acid may seem basic, but it involves real precision and understanding of chemistry concepts like molarity, molecular weight, and solubility. With tools like RDKit, we now have the ability to digitally verify and simulate chemical behaviors alongside real lab work.

This experiment reinforced my understanding of lab measurements and the importance of accurate calculations in solution preparation.

🔗 Related Read: Understand Mole Concept with Real Lab Activities

How Fireflies Glow | Firefly Bioluminescence Explained

🔥 How Fireflies Glow? | फायरफ्लाई रौशनी कैसे पैदा करती है?

🧪 What is Bioluminescence?

Bioluminescence is a natural process in which living organisms produce light through a chemical reaction inside their bodies. This is not like fire or bulb heat — it is a cold light (no heat loss).

बायोल्यूमिनेसेंस वह प्रक्रिया है जिसमें जीव रासायनिक अभिक्रिया से प्रकाश उत्पन्न करते हैं। यह बिना गर्मी के होता है — इसलिए इसे कोल्ड लाइट कहते हैं।

🐞 How Do Fireflies Produce Light?

Fireflies produce light using two key chemicals: luciferin and luciferase. Here's the process:

• Luciferin + Oxygen + ATP + Luciferase → Oxyluciferin + Light

This reaction takes place in the abdomen of the firefly, where special light-producing cells are located.

फायरफ्लाई के पेट में विशेष कोशिकाएं होती हैं, जो लुसिफरीन नामक रसायन को ऑक्सीजन और एटीपी की मदद से लाइट में बदलती हैं।

🌌 Why Fireflies Glow?

• 🔸 To attract mates (प्रजनन के लिए)
• 🔸 To warn predators (शिकारियों को चेतावनी)
• 🔸 Communication between species (प्रजातियों के बीच संवाद)

🔬 Other Insects & Creatures That Glow

Bioluminescence is not only seen in fireflies. Other glowing insects or organisms include:

• 🌊 Marine worms – found deep in ocean
• 🐛 Glow-worms – similar to fireflies
• 🪳 Click beetles – glow from thorax
• 🧬 Jellyfish & plankton – produce glowing effect in water

समुद्र के कीड़े, जेलिफ़िश, और प्लवक भी बायोल्यूमिनेसेंस दिखाते हैं। यह उन्हें गहरे पानी में जीवित रहने में मदद करता है।

💡 Fun Fact

The light produced by fireflies is nearly 100% efficient — unlike electric bulbs that lose energy as heat.

फायरफ्लाई की रौशनी पूरी तरह से ऊर्जा कुशल होती है, जबकि बल्ब ऊर्जा को गर्मी में भी गंवाते हैं।

🔥 Try This Quick Quiz:

Q: Which enzyme helps fireflies produce light?

🧠 Science in Daily Life

Scientists are studying fireflies to design bio-inspired lighting and even medical imaging tools. Their light-producing ability has huge potential in **biotech** and **healthcare**.

वैज्ञानिक फायरफ्लाई की रोशनी का प्रयोग नई तकनीकों जैसे बायो लाइटिंग और शरीर के अंदर की जांच के लिए कर रहे हैं।

🔒 All rights reserved © 2025 STEMFact.in | Do not copy content without permission.

🎮 Firefly Catching Game

Click the glowing fireflies before they disappear!

Score: 0 | Time: 30s

🔙 Read Previous Post: Unsaturation Test with Bromine in Lab 🔥

🔬 Go to RDKit Chemistry Application

Unsaturation Test with Bromine in Lab

Lab Activity: Unsaturation Test using Bromine

🧪 English: Today in the lab, I conducted an unsaturation test. I started with 10 ml of ethyl alcohol and added a few drops of concentrated sulfuric acid. This mixture helped me generate ethene gas via a dehydration reaction. Thos time the solution was transparent in color. I means it was colorless. To verify the presence of unsaturation, I added carbon tetrachloride, which led to the formation of an oily layer, typical of hydrocarbon presence.

I then carefully broke a sealed glass ampoule of bromine (orange liquid). The vapor spread rapidly due to bromine's volatile nature. I immediately covered my face with a handkerchief due to the pungent and irritating odor. Thankfully, the lab window was open, which provided ventilation. Upon adding a few drops of bromine to the solution, a red gel-like layer appeared with an orange layer floating on top.

Since the standard tripod stand could not support my tall candle, I placed the candle above the inverted beakers and filter paper above the tripod stand and thereafter I began slow heating.

After about an hour, the orange and red colors completely disappeared. This confirmed the presence of unsaturation, as bromine reacts with double bonds in alkenes, causing decolorization.

🔥 Hindi: आज की प्रयोगशाला में मैंने असंतृप्तता की पहचान की। एथिल अल्कोहल में कुछ बूँदें गाढ़ा सल्फ्यूरिक अम्ल मिलाया, जिससे एथीन गैस बनी। फिर कार्बन टेट्राक्लोराइड मिलाया, जिससे तैलीय परत दिखी। मैंने नारंगी रंग के ब्रोमीन को तोड़ा, जिसकी गंध तेज़ थी। उसे सावधानी से मिलाने पर लाल जैल जैसी परत बनी, ऊपर नारंगी रंग तैरता रहा। गर्म करने पर रंग समाप्त हो गया, जिससे असंतृप्तता सिद्ध हुई।

Chemistry Behind the Reaction

Ethene is an alkene, containing a carbon-carbon double bond (C=C). When bromine is added, it reacts with this double bond to form a dibromo compound. The disappearance of bromine's orange color is a qualitative indicator of unsaturation:

CH2=CH2 + Br2 → CH2Br-CH2Br

Why Bromine is Used

• Bromine is highly reactive with alkenes and alkynes.
• Its orange color makes it easy to observe changes.
• It doesn’t react with saturated hydrocarbons, so it's selective.

Safety Measures

• Always wear gloves and goggles when handling bromine.
• Work in a well-ventilated area or fume hood.
• Never inhale bromine vapors — they are toxic and corrosive.
• Use a handkerchief or mask in emergency exposure.

Applications of Unsaturation Test

This test is commonly used in organic chemistry labs and industries for:

• Detecting unsaturated fats and oils
• Verifying the presence of alkenes in synthetic reactions
• Educational demonstrations in chemistry classes

MCQs for Practice

Q1. What color change occurs when bromine reacts with an unsaturated compound?

• A) Red to blue
• B) Orange to colorless ✅
• C) Yellow to green
• D) No change

Q2. Which gas is evolved by dehydrating ethyl alcohol with sulfuric acid?

• A) Ethene ✅
• B) Methane
• C) Acetylene
• D) Propane

Bromine Properties

Bromine Properties

• Symbol: Br
• Atomic Number: 35
• Appearance: Reddish-orange liquid
• Boiling Point: 58.8°C
• Highly volatile, toxic, and corrosive

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