Section 3 of 4

Section 3: Physical Chemistry

Physical chemistry notes covering energetics, reaction profiles, bond energy calculations, rates of reaction, and dynamic equilibria.

Edexcel IGCSE Hub › Section 3

Key Definitions

Exothermic Reaction
A reaction that releases thermal energy to the surroundings (temperature of surroundings increases).
Endothermic Reaction
A reaction that absorbs thermal energy from the surroundings (temperature of surroundings decreases).
Activation Energy
The minimum energy that colliding particles must have in order to react.
Catalyst
A substance that increases the rate of reaction by providing an alternative pathway with a lower activation energy, without being chemically changed or used up.
Reversible Reaction
A reaction where the products can react to reform the original reactants.
Dynamic Equilibrium
A state in a reversible reaction where the forward and reverse reactions happen at the same rate, and the concentrations of reactants and products remain constant.

Energetics

Exothermic & Endothermic Reactions

Exothermic Reactions

An exothermic reaction transfers energy to the surroundings, usually by heating. The temperature of the surroundings increases.

  • Combustion: burning fuels (CH₄ + 2O₂ → CO₂ + 2H₂O)
  • Neutralisation: acid + alkali → salt + water
  • Oxidation: metals reacting with oxygen

Everyday uses: self-heating cans, hand warmers.

Endothermic Reactions

An endothermic reaction takes in energy from the surroundings, usually by heating. The temperature of the surroundings decreases.

  • Thermal decomposition: CaCO₃ → CaO + CO₂
  • Dissolving ammonium nitrate: caused by breaking ionic bonds.
  • Citric acid + sodium hydrogencarbonate

Everyday uses: instant cold packs for sports injuries.

All reactions involve both bond breaking (endothermic - requires energy) and bond making (exothermic - releases energy). Whether the overall reaction is exothermic or endothermic depends on which process transfers more energy.

Core Practical: Temperature Changes Extended

Investigate the variables that affect temperature changes in reacting solutions, e.g. Acid + alkali neutralisation.

  1. Measure a fixed volume of dilute acid (e.g. 25 cm³ of HCl) into a polystyrene cup using a measuring cylinder.
  2. Record the starting temperature of the acid using a thermometer.
  3. Add a measured volume of alkali (e.g. 25 cm³ of NaOH) and stir.
  4. Record the highest temperature reached.
  5. Calculate the temperature change: ΔT = final temperature − initial temperature.
  6. Repeat with different concentrations of alkali to investigate how concentration affects the temperature change.
Thermometer Insulating Lid Polystyrene Cup (Good insulator) Reaction Mixture (e.g. Acid + Alkali) Glass Beaker

The polystyrene cup minimises heat loss to the surroundings, providing a more precise measurement of the maximum temperature reached.

A polystyrene cup is used because it is a good insulator - it minimises heat loss to the surroundings, making the temperature measurement more accurate.

Interpreting temperature data

A student adds NaOH to HCl in a polystyrene cup. The temperature rises from 21°C to 28°C. Is this exothermic or endothermic?

Step 1: ΔT = 28 − 21 = +7°C.

Step 2: The temperature of the surroundings (solution) increased.

Step 3: Energy was transferred to the surroundings → this is an exothermic reaction.

A common mistake is confusing the direction of energy transfer. If the temperature goes up, the reaction is exothermic (energy is released into the solution). If the temperature goes down, the reaction is endothermic (energy is taken from the solution).

Reaction Profiles

Reaction profiles (also known as energy level diagrams) represent the energy changes taking place during a chemical reaction. They show the relative energy levels of the reactants and products, the activation energy (Ea), and the overall enthalpy change (ΔH).

Exothermic Reaction Profile

Energy Progress of Reaction Reactants Products Eₐ ΔH (negative)

Endothermic Reaction Profile

Energy Progress of Reaction Reactants Products Eₐ ΔH (positive)

Bond Energy Calculations (HT)

Chemical reactions involve breaking old bonds (endothermic) and making new bonds (exothermic).

Overall energy = Energy to break bonds − Energy released making bonds
  • If more energy is released making bonds than is needed to break bonds → exothermic (negative value).
  • If more energy is needed to break bonds than is released making bonds → endothermic (positive value).

Worked Example 1: H₂ + Cl₂ → 2HCl (exothermic)

Bond energies: H-H = 436 kJ/mol, Cl-Cl = 242 kJ/mol, H-Cl = 431 kJ/mol

Breaking: 1 × H-H + 1 × Cl-Cl = 436 + 242 = 678 kJ

Making: 2 × H-Cl = 2 × 431 = 862 kJ

Overall: 678 − 862 = −184 kJ/mol (exothermic)

Worked Example 2: N₂ + O₂ → 2NO (endothermic)

Bond energies: N≡N = 941 kJ/mol, O=O = 498 kJ/mol, N=O = 587 kJ/mol

Breaking: 1 × N≡N + 1 × O=O = 941 + 498 = 1439 kJ

Making: 2 × N=O = 2 × 587 = 1174 kJ

Overall: 1439 − 1174 = +265 kJ/mol (endothermic - more energy needed to break bonds than is released making them)

Worked Example 3: Combustion of methane CH₄ + 2O₂ → CO₂ + 2H₂O

Bond energies: C-H = 413, O=O = 498, C=O = 805, O-H = 464 kJ/mol

Breaking: 4 × C-H + 2 × O=O = (4 × 413) + (2 × 498) = 1652 + 996 = 2648 kJ

Making: 2 × C=O + 4 × O-H = (2 × 805) + (4 × 464) = 1610 + 1856 = 3466 kJ

Overall: 2648 − 3466 = −818 kJ/mol (exothermic - combustion always releases energy)

A negative answer = exothermic (energy released). A positive answer = endothermic (energy absorbed). Always show your working clearly in the exam. The methane combustion calculation is one of the most commonly set exam questions.

Calorimetry Experiments & Energy Calculations

Calorimetry is the experimental measurement of heat energy changes during chemical reactions. We measure the temperature change of a known mass of water (or solution) and calculate the heat energy transferred (Q) using the formula:

Q = m × c × ΔT
  • Q: Heat energy change in Joules (J).
  • m: Mass of water or solution in grams (g) (1 cm³ of aqueous solution is assumed to have a mass of 1 g).
  • c: Specific heat capacity of water, which is approximately 4.18 J/g/°C (amount of heat required to raise the temperature of 1 g of water by 1°C).
  • ΔT: Change in temperature in °C (T_final - T_initial).

1. Simple Calorimetry for Dissolving Salts (Energy Change of Dissolution)

Used to measure the enthalpy change when a salt dissolves in water (e.g., ammonium nitrate or calcium chloride).

Procedure:
  1. Measure 50 cm³ of distilled water using a measuring cylinder and pour it into a polystyrene cup. The polystyrene cup acts as an insulator, minimizing heat loss to the surroundings.
  2. Place the polystyrene cup in a beaker for stability, insert a thermometer, and measure the initial temperature of the water.
  3. Weigh a known mass (e.g., 5.0 g) of the salt (e.g., ammonium nitrate, NH₄NO₃).
  4. Add the salt to the water and stir rapidly using the thermometer to ensure complete dissolution.
  5. Record the maximum or minimum temperature reached by the solution.

Worked Example: Dissolving Ammonium Nitrate

5.0 g of ammonium nitrate (NH₄NO₃) is dissolved in 50 g of water. The temperature falls from 21.0°C to 14.5°C. Calculate the molar enthalpy change (ΔH) in kJ/mol. (M_r of NH₄NO₃ = 80)

  1. Step 1: Calculate heat energy change (Q):
    ΔT = 21.0 - 14.5 = 6.5°C
    Q = m × c × ΔT = 50 g × 4.18 J/g/°C × 6.5°C = 1358.5 J = 1.3585 kJ
  2. Step 2: Calculate moles of salt (n):
    Moles = Mass ÷ M_r = 5.0 g ÷ 80 = 0.0625 mol
  3. Step 3: Calculate molar enthalpy change (ΔH):
    Since the temperature decreased, the reaction is endothermic, so ΔH must be positive.
    ΔH = +Q ÷ n = +1.3585 kJ ÷ 0.0625 mol = +21.7 kJ/mol

2. Simple Calorimetry for Displacement Reactions

Used to measure the temperature change during displacement reactions (e.g., zinc replacing copper(II) ions in solution).

Procedure:
  1. Measure 25 cm³ of 1.0 mol/dm³ copper(II) sulfate solution into a polystyrene cup.
  2. Record the temperature of the solution every minute for 3 minutes to establish a stable starting temperature.
  3. At the 4th minute, add a known mass (excess) of zinc powder and stir rapidly. Do not record the temperature at this minute.
  4. Record the temperature every minute from the 5th to the 10th minute.
  5. Plot a graph of temperature against time, and extrapolate the lines to the 4th minute to find the theoretical maximum temperature rise, correcting for heat loss.

Worked Example: Zinc & Copper(II) Sulfate Displacement

When excess zinc is added to 25 cm³ of 1.0 mol/dm³ CuSO₄ solution, the temperature rises by 18.5°C. Calculate the molar enthalpy change (ΔH) in kJ/mol. (Assume density of solution = 1.0 g/cm³, specific heat capacity = 4.18 J/g/°C)

  1. Step 1: Calculate heat energy change (Q):
    Q = m × c × ΔT = 25 g × 4.18 J/g/°C × 18.5°C = 1933.25 J = 1.933 kJ
  2. Step 2: Calculate moles of CuSO₄ reacted (n):
    Moles = Concentration × Volume (dm³) = 1.0 mol/dm³ × (25 ÷ 1000) dm³ = 0.025 mol
  3. Step 3: Calculate molar enthalpy change (ΔH):
    Since the temperature increased, the reaction is exothermic, so ΔH must be negative.
    ΔH = -Q ÷ n = -1.933 kJ ÷ 0.025 mol = -77.3 kJ/mol

3. Calorimetry for Combustion of Alcohols

Used to measure the heat energy released by burning fuels (e.g., ethanol, propanol) using a copper calorimeter.

Procedure:
  1. Measure 100 cm³ of cold water into a copper calorimeter (copper is a good thermal conductor, ensuring rapid heat transfer from the flame to the water). Record the initial temperature of the water.
  2. Weigh a spirit burner containing the alcohol (e.g., ethanol) with its cap on, and record its mass.
  3. Place the spirit burner under the copper calorimeter, remove the cap, and light the wick. Shield the apparatus with draft shields to minimize heat loss to the surrounding air.
  4. Stir the water constantly with the thermometer. Burn the alcohol until the temperature of the water has risen by about 20-30°C.
  5. Extinguish the flame using the burner cap, and record the final maximum temperature of the water.
  6. Re-weigh the spirit burner with its cap on to find the mass of alcohol burned.

Worked Example: Combustion of Ethanol

A student burns 0.75 g of ethanol (C₂H₅OH) to heat 100 g of water in a copper calorimeter. The temperature of the water rises from 18.0°C to 42.0°C. Calculate the molar enthalpy change of combustion (ΔH) in kJ/mol. (M_r of ethanol = 46)

  1. Step 1: Calculate heat energy change (Q):
    ΔT = 42.0 - 18.0 = 24.0°C
    Q = m × c × ΔT = 100 g × 4.18 J/g/°C × 24.0°C = 10032 J = 10.032 kJ
  2. Step 2: Calculate moles of ethanol burned (n):
    Moles = Mass ÷ M_r = 0.75 g ÷ 46 = 0.0163 mol
  3. Step 3: Calculate molar enthalpy change (ΔH):
    Combustion is always exothermic, so ΔH must be negative.
    ΔH = -Q ÷ n = -10.032 kJ ÷ 0.0163 mol = -615.5 kJ/mol
Sources of Error in Combustion Calorimetry:

The experimental value for the enthalpy of combustion is always significantly lower than the theoretical data book value. The main reasons for this discrepancy are:

  • Heat loss: Heat is lost to the surrounding air and the copper calorimeter itself instead of transferring to the water.
  • Incomplete combustion: Some of the alcohol burns in a limited oxygen supply, producing carbon (soot) and carbon monoxide instead of carbon dioxide, which releases less energy.
  • Evaporation of alcohol: Some alcohol evaporates from the wick of the burner while it is cooling or being weighed.

Rates of Reaction

Rate of Reaction

The rate of a chemical reaction is a measure of how quickly reactants are used up or how quickly products are formed.

rate = amount of product formed ÷ time

Some reactions are very fast (explosions), while others are very slow (rusting of iron).

Collision Theory

For a chemical reaction to occur, particles must:

  1. Collide with each other.
  2. Collide with sufficient energy - at least the activation energy (Eₐ).

Collisions that have enough energy to react are called successful collisions.

Successful vs Unsuccessful Collisions Successful Collision BAM! Fast-moving particles Energy ≥ Activation Energy (Eₐ) Reaction occurs Unsuccessful Collision Slow-moving particles Energy < Activation Energy (Eₐ) Particles just bounce off

Left: Particles collide with sufficient energy (≥ Eₐ) to react. Right: Particles collide with insufficient energy (< Eₐ) and simply bounce apart.

Not every collision leads to a reaction. Only those with energy ≥ Eₐ (the activation energy) are successful.

Factors Affecting Rate

Temperature

Increasing temperature increases rate. Particles move faster (more kinetic energy), so collisions are more frequent AND more energetic. A greater proportion of collisions exceed the activation energy.

Concentration (or Pressure for gases)

Increasing concentration increases rate. There are more particles in the same volume, so collisions are more frequent.

Surface Area

Increasing surface area increases rate. Using smaller pieces (or powders) exposes more reactant particles on the surface, so there are more opportunities for collisions.

Effect of Surface Area on Reaction Rate Large Solid Block Low Surface Area (Particles trapped inside cannot react) Broken into Powders High Surface Area (Many more exposed particles to collide)

Left: A large piece of reactant only exposes its outer faces to collisions. Right: Breaking it down exposes new surfaces, increasing the frequency of successful collisions.

When explaining a rate factor, always use collision theory: (1) state what happens to the frequency or energy of collisions, (2) explain how this affects the number of successful collisions.

Measuring Rate of Reaction

Methods for Measuring Reaction Rates Method 1: Gas Collection Gas Syringe Measure gas volume at regular intervals Method 2: Mass Loss 120.45g Cotton Wool Gas escapes, mass decreases Method 3: Disappearing Cross Precipitate forms, cross becomes invisible

Three common methods for measuring the rate of a reaction depending on the state of the products.

Core Practical 3.15: Investigate the Effect of Surface Area and Concentration on Rate

This investigation has two parts. The reaction used is between calcium carbonate (marble chips) and hydrochloric acid:

CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)

Part 1: Varying Surface Area of Marble Chips

  1. Measure 50 cm³ of 1.0 mol/dm³ hydrochloric acid into a conical flask.
  2. Place the conical flask on a digital mass balance.
  3. Weigh 5.0 g of large marble chips and add them to the flask.
  4. Immediately plug the neck of the flask with cotton wool. This is critical: it allows carbon dioxide gas to escape but prevents acid spray from spitting out of the flask (which would cause a false mass loss).
  5. Start a stopclock. Record the mass of the flask and its contents every 30 seconds for 5 minutes.
  6. Repeat the experiment using the same mass (5.0 g) of small marble chips, and then using 5.0 g of powdered calcium carbonate. Keep the volume, concentration, and temperature of the acid constant.
  7. Plot a graph of mass loss against time for each surface area.

Part 2: Varying Concentration of Acid

  1. Keep the surface area of the solid constant by using 5.0 g of small marble chips for each run.
  2. Perform the experiment first using 50 cm³ of 2.0 mol/dm³ hydrochloric acid.
  3. Repeat using 50 cm³ of 1.0 mol/dm³ acid, and then 0.5 mol/dm³ acid. Keep the mass of marble chips and the starting temperature constant.
  4. Alternative Method (Gas Collection): Instead of mass loss, connect the flask via a delivery tube to a gas syringe or an inverted measuring cylinder filled with water. Measure the volume of carbon dioxide gas collected at regular intervals.
  5. Plot a graph of gas volume against time for each concentration.

Analysis of Results:

  • The initial gradient of each curve shows the rate of reaction. A steeper curve indicates a faster rate.
  • Surface Area: Powdered calcium carbonate has the largest surface area, giving the steepest curve and fastest rate. Larger chips have the smallest surface area, giving the shallowest curve and slowest rate. More reactant particles are exposed on the surface, increasing the frequency of successful collisions.
  • Concentration: 2.0 mol/dm³ acid has the most acid particles per unit volume, producing the steepest curve. More particles in the same space leads to a higher frequency of successful collisions.

Core Practical 3.16: Investigate the Effect of Different Catalysts on the Decomposition of Hydrogen Peroxide

Hydrogen peroxide decomposes slowly at room temperature to form water and oxygen gas:

2H₂O₂(aq) → 2H₂O(l) + O₂(g)

Procedure:

  1. Measure 50 cm³ of 10-volume hydrogen peroxide solution into a conical flask.
  2. Set up a gas syringe and connect it to the flask via a delivery tube and a rubber bung.
  3. Weigh 0.5 g of manganese(IV) oxide (MnO₂) powder.
  4. Add the MnO₂ to the flask, immediately insert the bung to seal the flask, and start the stopclock.
  5. Record the volume of oxygen gas collected in the syringe every 10 seconds for 2 minutes (or until the reaction is complete).
  6. Repeat the experiment using 0.5 g of other solid catalysts, such as copper(II) oxide (CuO), iron(III) oxide (Fe₂O₃), and zinc oxide (ZnO). Keep the volume and concentration of hydrogen peroxide constant.
  7. Plot a graph of volume of oxygen gas against time for each catalyst.

Analysis of Results:

  • The solid that produces the steepest initial gradient is the most effective catalyst. Manganese(IV) oxide (MnO₂) is typically the most effective catalyst.
  • To prove that the solid acts as a catalyst (and is not consumed), the reaction mixture can be filtered to recover the catalyst. The recovered solid is washed with distilled water, dried in an oven, and re-weighed. Its mass should remain exactly 0.5 g.

Calculating mean rate of reaction

In an experiment, 48 cm³ of gas was collected in 60 seconds. Calculate the mean rate.

Formula: mean rate = volume of gas ÷ time

Calculation: 48 ÷ 60 = 0.80 cm³/s

Rate Graphs

The steeper the graph’s gradient, the faster the rate. The gradient decreases over time as reactants are used up. The graph eventually levels off when the reaction is complete.

Interpreting Changes on Graphs

  • Higher temperature or concentration: Steeper initial gradient, but same final amount of product (same amount of reactant).
  • Using a catalyst: Steeper gradient, same final product.
  • Using more reactant: Steeper gradient AND more total product.
Reaction Rate Graph Comparing Fast and Slow Reactions Time → Amount of Product → (or reactant used) Final amount of product (reaction is complete) Reaction A finishes Reaction B finishes Steeper initial gradient = Faster rate (More frequent successful collisions) Higher Temp / Conc. (Faster) Lower Temp / Conc. (Slower)

Rate graphs show how the amount of product changes over time. The steeper the line, the faster the reaction.

Interpreting two rate curves

Two experiments use the same mass of marble chips and volume of HCl. Experiment A is at 20°C; Experiment B is at 40°C. Both curves level off at the same volume of CO₂. Explain the differences.

Gradient: Curve B is steeper because at 40°C particles have more kinetic energy, so collisions are more frequent and more energetic → more successful collisions per second.

Final volume: Both level off at the same volume because the same amount of reactant was used - temperature does not change the total product, only how fast it forms.

Time to finish: Curve B levels off sooner because the reaction is faster at higher temperature.

The graph always starts steep and flattens out. The flat part shows the reaction has finished. A catalyst makes the graph steeper but doesn’t change where it levels off. If you are asked to calculate rate from a graph, draw a tangent at the required time and calculate gradient = Δy/Δx.

Catalysts

A catalyst is a substance that increases the rate of a chemical reaction without being used up in the process. Catalysts work by providing an alternative reaction pathway with a lower activation energy.

Collision Theory Link: By lowering the activation energy, a larger fraction of colliding particles possess energy equal to or greater than the new activation energy (E ≥ Ea). Consequently, a higher proportion of collisions are successful, leading to a higher frequency of successful collisions and thus a faster rate of reaction.

Energy Profile Diagram comparing catalysed and uncatalysed reactions Progress of Reaction → Energy → Reactants Products Eₐ without catalyst Eₐ with catalyst Uncatalysed Pathway Catalysed Pathway

A catalyst provides an alternative reaction pathway with a lower activation energy, meaning more particles have sufficient energy to react.

Biological catalysts are called enzymes. They are proteins that catalyse reactions in living organisms - for example, amylase breaks down starch.

Industrial Catalysts

  • Iron - used in the Haber process (N₂ + 3H₂ ⇌ 2NH₃) to manufacture ammonia.
  • Vanadium(V) oxide (V₂O₅) - used in the Contact process to make sulfuric acid.
  • Manganese dioxide (MnO₂) - catalyses the decomposition of hydrogen peroxide.
Catalysts reduce costs in industry because they can be reused, reduce the energy needed (lower activation energy), and speed up production. However, they must be kept clean - impurities can "poison" a catalyst and stop it working.

Reversible Reactions & Equilibria

Reversible Reactions

A reversible reaction is one that can proceed in both directions - products can re-form the reactants.

A + B ⇌ C + D

The ⇌ symbol indicates a reversible reaction.

Energy in Reversible Reactions

If the forward reaction is exothermic, the reverse reaction is endothermic - and they involve the exact same amount of energy.

Hydration of anhydrous copper sulfate:

CuSO₄ + 5H₂O ⇌ CuSO₄·5H₂O

Forward: white → blue (exothermic). Reverse: blue → white (endothermic, by heating).

Dynamic Equilibrium

In a closed system (nothing can enter or leave), a reversible reaction reaches dynamic equilibrium. At equilibrium:

  • The rate of the forward reaction equals the rate of the reverse reaction.
  • The concentrations of reactants and products remain constant (but are not necessarily equal).
At equilibrium, both reactions are still happening - it's "dynamic." But the overall concentrations don't change because the rates are balanced.
Dynamic Equilibrium tank analogy Reactants Products Forward Rate Reverse Rate Rates are EQUAL Constant Level Constant Level

In dynamic equilibrium, the forward and reverse reactions happen at the exact same rate. The amount of reactants and products remains constant because they are being formed as fast as they are used up.

Le Chatelier’s Principle (HT)

If a system at equilibrium is subjected to a change in conditions, the position of equilibrium will shift to oppose the change.

Effect of Temperature

  • Increase temperature: Equilibrium shifts in the endothermic direction (to absorb the extra heat).
  • Decrease temperature: Equilibrium shifts in the exothermic direction.

Effect of Pressure (for gas reactions)

  • Increase pressure: Equilibrium shifts to the side with fewer moles of gas.
  • Decrease pressure: Equilibrium shifts to the side with more moles of gas.

Effect of Concentration

  • Increase concentration of a reactant: Equilibrium shifts to the right (forward), producing more product.
  • Increase concentration of a product: Equilibrium shifts to the left (backward).
A catalyst does NOT affect the position of equilibrium - it speeds up both forward and reverse reactions equally. It only makes equilibrium reached faster.

Applying Le Chatelier’s Principle: The Haber Process

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)     (forward reaction is exothermic)

The Haber process is used to manufacture ammonia. Conditions are chosen as a compromise between yield and rate:

Worked Example: Predicting equilibrium shifts

For N₂ + 3H₂ ⇌ 2NH₃ (exothermic forward), predict what happens when:

1. Temperature is increased: Equilibrium shifts LEFT (endothermic direction) to absorb extra heat → less NH₃, lower yield. However, rate is faster.

2. Pressure is increased: Left side has 4 moles of gas (1 + 3), right side has 2 moles. Equilibrium shifts RIGHT (fewer moles) → more NH₃, higher yield.

3. More N₂ is added: Equilibrium shifts RIGHT to use up the extra N₂ → more NH₃ produced.

Compromise conditions: 450°C (moderate - low temp gives better yield but reaction is too slow), 200 atm (high pressure for better yield), iron catalyst (speeds up reaction without affecting yield). These give about 15% yield per pass - unreacted gases are recycled.

Thermal Decomposition of Ammonium Chloride

A classic example of a reversible reaction is the thermal decomposition of solid ammonium chloride (NH₄Cl) in a test tube.

NH₄Cl(s) ⇌ NH₃(g) + HCl(g)

Experimental Description:

  • When white solid ammonium chloride is heated in a test tube, it undergoes thermal decomposition (an endothermic reaction) to form two colourless gases: ammonia (NH₃) and hydrogen chloride (HCl).
  • As these gases rise up the test tube away from the heat source, they reach a cooler region at the top of the tube.
  • In this cooler region, the reverse reaction occurs: the ammonia and hydrogen chloride gases react together (an exothermic reaction) to reform the white solid ammonium chloride.
  • Observation: A white ring or deposit of ammonium chloride forms on the cooler upper walls of the test tube.