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Chemistry Form 5 KSSM-122-149
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Thermochemistrythermochemistry.
3 Heat change in reactions 3 Heat of reaction 3 Application of endothermic and exothermic reactions in daily life
- Endothermic reaction
- Energy level diagram
- Exothermic reaction
- Heat of combustion
- Heat of displacement
- Heat of neutralisation
- Heat of precipitation
- Heat of reaction
- Thermochemical equation
KeyKeyKey WdsWds
What will you learn?What will you learn?
“Affordable, ready to drink and on the go”. These are the important concepts that Datuk Kenneth Warren Kolb had in mind when he invented Hot Can, a smart self-heating can that contains beverages and soups. These self-heating cans allow users to enjoy hot food or drinks without heating. The can works through an exothermic reaction between two chemicals. There are two chambers surrounding each other. The outer chamber stores food or drinks while the interior chamber stores the chemicals that react when mixed.
Common chemicals used are aluminium and silica, calcium oxide and water, and copper sulphate and zinc. When both materials combine, a chemical reaction generates enough heat to raise the temperature of the can.
####### Bulletin
Do all chemical reactions
release heat?
How do you calculate heat change?
Which is the most suitable fuel for daily usage?
Laboratory Activity Determining Exothermic and Endothermic Reactions
3A Aim: To study the types of reactions based on heat change and the changes in thermometer readings when substances dissolve in water. Materials: Solid sodium hydroxide, NaOH, solid anhydrous calcium chloride, CaCl 2 , solid ammonium nitrate, NH 4 NO 3 , solid sodium thiosulphate, Na 2 S 2 O 3 and distilled water. Apparatus: Polystyrene cups with lids, spatula, measuring cylinder and thermometer. Procedure:
- Measure 20 cm 3 of distilled water and pour it into a polystyrene cup.
- Put a thermometer in the polystyrene cup and leave it for two minutes. Record the initial temperature of the distilled water.
- Add one spatula of solid sodium hydroxide, NaOH into the polystyrene cup.
- Stir the mixture carefully with the thermometer.
- Record the highest or the lowest temperature of the mixture.
- Repeat steps 1 to 5 by replacing solid sodium hydroxide, NaOH with solid ammonium nitrate, NH 4 NO 3 , solid sodium thiosulphate, Na 2 S 2 O 3 and solid anhydrous calcium chloride, CaCl 2. Results: Construct a table to record your observations. Discussion:
- Based on your results, determine: (a) which reaction absorbs heat. (b) which reaction releases heat.
- With reference to 1(a) and 1(b), classify the reactions as endothermic or exothermic reactions.
Learning Science PAKPAK 2121 Through Inquiry
Prepare a complete report after carrying out this laboratory activity.
CAUTIONCAUTION Sodium hydroxide corrodes. Handle with care.
kubupublication.com/Kimia/Tingkatan5/Video28.htm Endothermic Reaction bit/kpkt5v
Energy Level Diagram
- During a chemical reaction, heat is absorbed or released. This heat is called heat of reaction, and is given the symbol ΔH. The unit for heat of reaction is kJ mol−1.
- In a chemical reaction, when heat is released to the surroundings, ΔH is given a negative sign. When heat is absorbed from the surroundings, ΔH is given a positive sign.
- The energy change in a chemical reaction can be represented using an energy level diagram. The energy level diagram shows the difference in the heat energy content between the reactants and the products. The symbol Δ, delta is the fourth letter in the ∆H = H products – H reactants Greek alphabet that denotes difference or change.
Heat of reaction, ΔH is the heat change of one mole of reactant that reacts or one mole of product that is formed.
Thermochemistry Chapter 3
The reaction between magnesium, Mg and sulphuric acid, H 2 SO 4 forming magnesium sulphate, MgSO 4 and hydrogen gas, H 2 is an exothermic reaction.
When 1 mole of Mg reacts with 1 mole of H 2 SO 4 to form 1 mole of MgSO 4 and 1 mole of H 2 gas, 467 kJ of heat is released to the surroundings.
During the reaction, the temperature of the mixture increases.
The total energy content of the products (MgSO 4 and H 2 ) is lower than the total energy content of the reactants (Mg and H 2 SO 4 ). Therefore, ΔH is negative.
The reaction between nitrogen gas, N 2 and oxygen gas, O 2 forming nitrogen monoxide gas, NO is an endothermic reaction.
When 1 mole of N 2 gas reacts with 1 mole of O 2 gas to form 2 moles of NO gas, 180 kJ heat energy is absorbed from the surroundings.
During the reaction, the temperature of the mixture decreases.
The total energy content of the product (NO) is higher than the total energy content of the reactants (N 2 and O 2 ). Therefore, ΔH is positive.
Figure 3 Energy level diagram for an exothermic reaction
Figure 3 Energy level diagram for an endothermic reaction
Draw an arrow from the energy level of the reactant to the energy level of the product
Label “Energy” Draw two energy levels
Write the value of ΔH
Write the reactant and the product of the reaction at the correct energy level
Draw an arrow upwards
- The following thermochemical equation shows an exothermic reaction, while Figure 3 shows the energy level diagram for the reaction.
Mg(s) + H 2 SO 4 (aq) → MgSO 4 (aq) + H 2 (g) ΔH = - 467 kJ mol−
The following bridge map shows how the energy level diagram is constructed.
The following thermochemical equation shows an example of an endothermic reaction, while, Figure 3 shows the energy level diagram for the reaction.
N 2 (g) + O 2 (g) → 2NO(g) ΔH = +180 kJ mol−
Mg(s) + H 2 SO 4 (aq)
N 2 (g) + O 2 (g)
∆H = - 467 kJ mol−
∆H = +180 kJ mol−
MgSO 4 (aq) + H 2 (g)
Theme 3 Heat
Figure 3 Types of heat of reactions
Figure 3 General steps to calculate heat of reactions
- The heat of reaction can be determined through experiments by establishing the temperature change when the reaction occurs. The value of temperature change obtained is used to calculate heat of reaction.
- Figure 3 shows a flow chart of the steps to calculate heat of reaction.
3 HEAT OF REACTION
Pupils are able to: 3.2 determine heat of precipitation through activity. 3.2 determine heat of displacement through activity. 3.2 compare heat of neutralisation through experiments for reactions between item: (a) strong acid and strong alkali, (b) weak acid and strong alkali, (c) strong acid and weak alkali, (d) weak acid and weak alkali. 3.2 compare heat of combustion for various types of alcohol through experiment.
- The heat of reaction is normally named according to the types of reactions that occur. Figure 3 shows the types of heat of reaction.
Physics Link: Q = mcθ θ = change in temperature
The Joule symbol J is the SI issued unit for energy, work and heat that is used in honour of James Prescott Joule (1818-1889).
Heat of Reaction
Heat of Precipitation
Heat of Combustion
Heat of Displacement
Heat of Neutralisation
- The number of moles, n is according to the types of heat of reactions as follows:
Step 1 Step 2 Step 3 Step 4
State the heat of reaction, ∆H with +/- signs and the correct units: ∆H = +/-X kJ mol−
Calculate the heat change for 1 mole of reactant or 1 mole of product formed in proportion.
Calculate the heat change in the reaction: Q = mcθ θ = change in temperature
Determine the number of moles of the reactants and products formed, n mole
Mole of precipitate formed for heat of precipitation
Mole of metal displaced for heat of displacement
Mole of water formed for heat of neutralisation
Mole of fuel burnt for heat of combustion
Number of moles, n represents
Form 4 Chemistry: Preparation of insoluble salts.
Figure 3 Usage of barium sulphate in medicine
Heat of precipitation is the heat change when 1 mole of precipitate is formed from their ions in an aqueous solution.
The thermochemical equation for the formation of barium sulphate, BaSO 4 precipitate is:
Based on the thermochemical equation, 42 kJ of heat is released when 1 mole of barium sulphate, BaSO 4 precipitate is formed. Therefore, the heat of precipitation of barium sulphate, BaSO 4 is -42 kJ mol−1.
Based on the conversation above, do you know that barium sulphate, BaSO 4 is an insoluble salt that is used in the fields of medicine? Try to reflect on the insoluble salts learned in form 4. All insoluble salts are precipitates.
Ba2+(aq) + SO 4 2−(aq) → BaSO 4 (s) ∆H = - 42 kJ mol−
Why do I have to drink the contrast media liquid?
- Most chemical reactions carried out to determine the heat of reaction involve aqueous solutions. Some assumptions are made during calculations: ◆ The density of any aqueous solution is equal to the density of water, 1 g cm−3. ◆ The specific heat capacity of any aqueous solution is equal to the specific heat capacity of water, which is 4 J g−1 °C−1. ◆ No heat is lost to the surroundings. ◆ No heat is absorbed by the apparatus of the experiment.
The contrast media liquid contains barium sulphate. Therefore, the digestive tract, stomach and intestines can be easily seen on the X-ray.
Photograph 3 White precipitate of barium sulphate, BaSO 4
- Insoluble salts are formed by a double decomposition reaction that involves heat change, which is known as heat of precipitation. What is meant by heat of precipitation?
Madam, you have to drink the contrast media liquid before taking the X-ray.
Determining Heat of Precipitation
Laboratory Activity 3B
Aim: To determine the heat of precipitation of silver chloride, AgCl and magnesium carbonate, MgCO 3. Materials: 0 mol dm−3 of silver nitrate, AgNO 3 solution, 0 mol dm−3 of sodium chloride, NaCl solution, 0 mol dm−3 of magnesium nitrate, Mg(NO 3 ) 2 solution and 0 mol dm−3 of sodium carbonate, Na 2 CO 3 solution. Apparatus: Two polystyrene cups with lids, measuring cylinder and thermometer. Operational definition - Heat of Precipitation: when sodium chloride, NaCl solution is added to silver nitrate, AgNO 3 solution to produce 1 mole of silver chloride, AgCl precipitate, the thermometer reading increases. Procedure:
Measure 25 cm 3 of 0 mol dm−3 of silver nitrate, AgNO 3 solution and pour it into a polystyrene cup.
Put a thermometer into the solution and leave aside for two minutes.
Record the temperature of the solution.
Measure 25 cm 3 of 0 mol dm−3 of sodium chloride, NaCl solution, and pour it into another polystyrene cup.
Put a thermometer into the solution and leave it aside for two minutes. Record the temperature of the solution.
Pour the sodium chloride, NaCl solution quickly and carefully into the polystyrene cup containing the silver nitrate, AgNO 3 solution.
Cover the polystyrene cup and stir the mixture using the thermometer as shown in Figure 3.
Record the highest temperature of the mixture.
Repeat steps 1 to 8 by replacing silver nitrate, AgNO 3 solution with magnesium nitrate, Mg(NO 3 ) 2 solution, and sodium chloride, NaCl solution with sodium carbonate, Na 2 CO 3 solution. Results: Construct a suitable table to record your results and observations. Discussion:
State the type of reaction that occurred.
Calculate the heat of precipitation of silver chloride, AgCl and magnesium carbonate, MgCO 3 [Use the heat change formula, Q = mcθ] [Given: Specific heat capacity of solution, c = 4 J g−1 °C−1; density of solution = 1 g cm−3]
25 cm 3 of 0 mol dm− of silver nitrate, AgNO 3 solution
25 cm 3 of 0 mol dm− of sodium chloride, NaCl solution
Mixed solution
Polystyrene cups
Thermometer Stir
CAUTIONCAUTION Silver nitrate solution may cause dark stains on your skin and clothes.
- Write the thermochemical equation for the precipitation of silver chloride, AgCl and magnesium carbonate, MgCO 3.
- Construct the energy level diagram for the precipitation of silver chloride, AgCl and magnesium carbonate, MgCO 3.
- The theoretical value for the heat of precipitation of silver chloride, AgCl is -65 kJ mol−1. Is this value the same as the value obtained in this experiment? Explain your answer. Conclusion: What is the conclusion that can be drawn from this experiment?
Solution: Step 1: Calculate the number of moles of silver chloride, AgCl formed.
Number of moles of potassium chloride, KCl = = 0 mol
Number of moles of silver nitrate, AgNO 3 = = 0 mol
From the equation, 1 mole of potassium chloride, KCl reacts with 1 mole of silver nitrate, AgNO 3 to produce 1 mole of silver chloride, AgCl. Therefore, 0 mole of potassium chloride, KCl reacts with 0 mole of silver nitrate, AgNO 3 to produce 0 mole of silver chloride, AgCl.
Step 2: Calculate the heat change. Given ∆H = - 65 kJ mol− When 1 mole of silver chloride, AgCl is formed, 65 kJ of heat is released.
Therefore, 0 mole of silver chloride, AgCl is formed, = 0 kJ heat is released.
Step 3: Calculate the rise in temperature. Mass of solution = (20 + 20) cm 3 1 g cm−3 = 40 g Heat released, Q = 655 J 40 g 4 J g−1 °C−1 θ = 655 J
Rise in temperature, θ = = 3 °C
Example: The following equation shows the precipitation of silver chloride, AgCl. AgNO 3 (aq) + KCl(aq) → AgCl(s) + KNO 3 (aq) ∆H = - 65 kJ mol−
If 20 cm 3 of 0 mol dm−3 of silver nitrate, AgNO 3 solution is added into 20 cm 3 of 0 mol dm− of potassium chloride, KCl solution, calculate the rise in temperature of the mixture. [Given: Specific heat capacity of solution, c = 4 J g−1 °C−1; density of solution = 1 g cm−3 ]
0 20 1000
0 mol 65 kJ 1 mol
655 J 40 g 4 J g−1 °C−
Laboratory Activity 3C Determining Heat of Displacement
Aim: To determine and compare the heat of displacement of copper from copper(II) sulphate, CuSO 4 solution with zinc metal, Zn and magnesium metal, Mg. Materials: 0 mol dm−3 of copper(II) sulphate, CuSO 4 solution, magnesium powder, Mg and zinc powder, Zn. Apparatus: Polystyrene cups with lids, thermometer, measuring cylinder and spatula. Procedure:
- Based on Figure 3, list down the suggested apparatus and materials, plan the procedure for the experiment and consider any precautionary steps that need to be taken to calculate the heat of displacement of copper, Cu.
- Discuss the planned procedures with your teacher before you proceed. Results: Construct a suitable table to record your readings. Discussion:
- Based on the experiment: (a) write the chemical and the ionic equations for both reactions. (b) calculate the heat of displacement, ∆H of copper, Cu by magnesium, Mg and zinc, Zn. [Given: Specific heat capacity of solution: c = 4 J g−1 °C−1; density of solution = 1 g cm−3] (c) explain why the value of heat of displacement of copper, Cu by magnesium metal, Mg and zinc metal, Zn is different. (d) draw the energy level diagram for both reactions.
- Why are both metals used in excess in this experiment?
- Apart from the change in temperature, state other observations that can be found during the experiment.
Step 3: Calculate the heat change. Mass of solution, m = 50 cm 3 1 g cm− = 50 g Heat change, Q = 50 g 4 J g−1 °C−1 4 °C = 840 J = 0 kJ
Step 4: Calculate the heat change for 1 mole of iron, Fe displaced. Displacement of 0 mole of iron, Fe releases 0 kJ of heat. Therefore, the displacement of 1 mole of iron, Fe can release = 67 kJ of heat.
Step 5: Write the heat of displacement, ∆H. Heat of displacement of iron, Fe by magnesium, Mg ∆H = - 67 kJ mol−
The negative sign (-) shows that this is an exothermic reaction.
Figure 3 Determining the heat of displacement of copper
Copper(II) sulphate, CuSO 4 solution Precipitate ofcopper, Cu
Metal powder Lid
Thermometer
Form 4 Chemistry: Neutralisation
A student carried out an experiment to determine the heat of displacement for the reaction between copper, Cu and silver nitrate, AgNO 3 solution. In this experiment, excess copper powder, Cu is added into 100 cm 3 of 0 mol dm−3 of silver nitrate, AgNO 3 solution. The heat of displacement in this experiment is -105 kJ mol−1. [Specific heat capacity of solution, c = 4 J g−1 °C−1; density of solution = 1 g cm−3] (a) Write the thermochemical equation for this reaction. (b) Calculate the heat released in this experiment. (c) Calculate the change in temperature. (d) Draw the energy level diagram for this reaction. (e) The experiment is repeated using 100 cm 3 of 1 mol dm−3 of silver nitrate, AgNO 3 solution and excess copper powder, Cu. Calculate the change in temperature in this experiment. Explain why the change in temperature is different from the calculated value in (c).
PAKPAK 2121
- Neutralisation is a reaction between an acid and an alkali to produce only salt and water.
- Neutralisation reaction is an exothermic reaction.
- Hydrogen ions, H+ from acid reacts with hydroxide ions, OH− from alkali to form water, H 2 O molecules. The ionic equation is as follows: H+(aq) + OH−(aq) → H 2 O(l)
The heat of neutralisation is the heat change when one mole of water is formed from the reaction between an acid and an alkali.
- What is the operational definition of the heat of displacement in this experiment?
- Will the heat of displacement, ∆H of copper, Cu be the same if copper(II) sulphate, CuSO 4 solution is replaced with copper(II) nitrate, Cu(NO 3 ) 2 solution? Explain your answer. Conclusion: Write the conclusion for this experiment.
Figure 3 Experiment of neutralisation
The reaction that takes place between acid and alkali is exothermic. Heat is released to the surroundings.
Madam, why does the conical flask feel hot when I titrate hydrochloric acid into the sodium hydroxide solution?
B. Reactions between a strong acid and a weak alkali, and between a weak acid and a weak alkali
Problem statement: Does the reaction between a strong acid and a weak alkali produce a higher heat of neutralisation compared to the reaction between a weak acid and a weak alkali? Hypothesis: Construct a hypothesis that involves the reactions of a strong acid, a weak acid with a weak alkali and the heat of neutralisation.
Variables: State all variables.
Materials: 1 mol dm−3 of hydrochloric acid, HCl, 1 mol dm−3 of ethanoic acid, CH 3 COOH and 1 mol dm−3 of ammonia, NH 3 solution.
Apparatus: Polystyrene cups with lids, thermometer and measuring cylinder.
Procedure: Using the materials and apparatus provided, plan and carry out an experiment to investigate the effects of strengths of acids on the heat of neutralisation with weak alkalis.
Results: Copy and complete Table 3 to record the initial temperature of the acids and alkalis, average temperature of acids and alkalis, highest temperature and increase in temperature.
Discussion:
- Write the chemical equation for each neutralisation reaction that takes place.
- Calculate the heat of neutralisation, ∆H for each reaction. [Given: Specific heat capacity of solution: c = 4 J g−1 °C−1; density of solution = 1 g cm−3]
- Construct the energy level diagram for each neutralisation reaction.
- Compare the value of the heat of neutralisation for each neutralisation reaction in this experiment.
- Explain the reason for the difference in the value of the heat of neutralisation.
- The theoretical value of the heat of neutralisation between a strong acid and a strong alkali is -57 kJ mol−1. Compare this value with the heat of neutralisation obtained from this experiment. Suggest a reason for this difference.
Conclusion: Can the hypothesis be accepted? What is the conclusion of this experiment?
Prepare a complete report after carrying out this experiment.
Reacting mixture
Hydrochloric acid, HCl and sodium hydroxide, NaOH solution
Ethanoic acid, CH 3 COOH and sodium hydroxide, NaOH solution
Hydrochloric acid, HCl and ammonia, NH 3 solution
Ethanoic acid, CH 3 COOH and ammonia, NH 3 solution
Initial temperature of acid (°C) Initial temperature of alkali (°C) Average temperature of acid and alkali (°C) Highest temperature of mixture (°C) Temperature rise (°C)
- The theoretical value of the heat of neutralisation between a strong acid and a strong alkali is -57 kJ mol−1.
- Table 3 shows the heat of neutralisation of various neutralisation reactions.
Table 3 Heat of neutralisaton of various neutralisation reactions Example Heat of neutralisation, ∆H (kJ mol−1) Strong acid + Strong Alkali → Salt + Water - Weak acid + Strong Alkali → Salt + Water - Strong acid + Weak alkali → Salt + Water - Weak acid + Weak alkali → Salt + Water -
Note that there is an influence of acid and alkali strengths over the heat of neutralisation. The value of the heat of neutralisation is lower when weak acids or weak alkalis are used. This is explained in the flow chart of Figure 3.
The heat of neutralisation between weak acids and weak alkalis is the lowest: (a) More heat energy is needed to completely ionise both weak acids and weak alkalis. (b) Therefore, hydrogen ions, H+ and hydroxide ions, OH− produced can completely react to form 1 mole of water.
Hydrochloric acid, HCl is a monoprotic acid while sulphuric acid, H 2 SO 4 is a diprotic acid.
Complete neutralisation between a strong diprotic acid and a strong alkali produces twice the quantity of heat compared to a strong monoprotic acid.
Form 4 Chemistry:
- Strength of acids and alkalis
- Basicity of acids
HCl(aq) + NaOH(aq) → NaCl(aq) + H 2 O(l) ∆H = -57 kJ mol−
H 2 SO 4 (aq) + 2NaOH(aq) → Na 2 SO 4 (aq) + 2H 2 O(l) ∆H = -114 kJ
How about the neutralisation reaction between sulphuric acid, H 2 SO 4 , a diprotic acid, with sodium hydroxide, NaOH solution?
Weak acids or weak alkalis ionise partially in water and some remain as molecules.
Some of the heat released during neutralisation is absorbed and used to completely ionise the weak acid or weak alkali in water.
erefore, the heat released is lower.
Figure 3 Explanation related to heat of neutralisation that involves weak acids or weak alkalis
- Based on the conversation in Figure 3, burning of different fuels will produce different energy value. Combustion of fuel is an exothermic reaction.
Figure 3 Cooking using wood
Step 2: Calculate the heat change. Heat released in the reaction, Q = mcθ = (60 + 60) g 4 J g−1 °C−1 (40 − 28) °C = 6300 J = 6 kJ
Step 3: Calculate the heat change for the formation of 1 mole of water. Formation of 0 mole of water, H 2 O releases 6 kJ of heat.
Therefore, the formation of 0 mole of water, H 2 O releases of heat, which is 52 kJ mol−1.
Step 4: Write the heat of neutralisation, ∆H. Heat of neutralisation, ∆H = -52 kJ mol−
When 100 cm 3 of 2 mol dm−3 of dilute hydrochloric acid, HCl is added to 100 cm 3 of 2 mol dm−3 of sodium hydroxide, NaOH solution, the temperature of the reaction increases from 30 °C to 43 °C. [Specific heat capacity of solution, c = 4 J g−1 °C−1; density of solution = 1 g cm−3] (a) Calculate the heat of neutralisation. (b) Write the thermochemical equation. (c) Construct an energy level diagram for this reaction. (d) Predict the temperature change if the hydrochloric acid, HCl is replaced with nitric acid, HNO 3 with the same volume and concentration. Explain your answer.
A ctivity 3B
Each substance releases different heat of combustion.
Madam, why does cooking using wood take a longer time than cooking using a gas stove?
Consider the complete combustion of methane, CH 4 in oxygen, O 2.
The thermochemical equation shows that when one mole of methane, CH 4 is completely burnt in oxygen, O 2 , heat released is 394 kJ.
The heat released is known as heat of combustion.
The heat of combustion is the heat released when 1 mole of a substance is completely burnt in excess oxygen, O 2.
CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O(l) ∆H = -394 kJ mol−
3B Determining the Heat of Combustion of Various Alcohols.
Aim: To determine the heat of combustion of various alcohols.
Problem statement: Does the number of carbon atoms per molecule in alcohol affect the heat of combustion?
Hypothesis: The higher the number of carbon atoms per molecule of alcohol, the higher the heat of combustion.
Variables: (a) Manipulated variable : Types of alcohol (b) Responding variable : Heat of combustion (c) Fixed variable : Volume of water
Materials: Methanol, CH 3 OH, ethanol, C 2 H 5 OH, propanol, C 3 H 7 OH, butanol, C 4 H 9 OH and water, H 2 O.
Apparatus: Copper can, tripod stand, pipeclay triangle, thermometer (0 °C − 100 °C), measuring cylinder, spirit lamp, electronic weighing balance, wind shield and wooden block.
- Using a measuring cylinder, measure 200 cm 3 of water and pour it into a copper can.
- Place the copper can on a tripod stand as shown in Figure 3.
- Measure and record the initial temperature of the water.
- Fill the spirit lamp with methanol, CH 3 OH until about three quarter full.
- Weigh and record the mass of the lamp with its lid and its content.
- Place the lamp under the copper can and light up the wick of the lamp.
Figure 3 Apparatus to determine the heat of combustion
Wind shield Wind shield
Water Pipeclay triangle
Alcohol Wooden block
Spirit lamp
Make sure the flame from the spirit lamp touches the bottom of the copper can.
Precautionary steps
Carbon compound: Combustion of alcohol on page 86.
CAUTIONCAUTION Be cautious when handling alcohols. They are very volatile and highly flammable.
- Multiple Choice
Subject : Bahasa melayu
School : sekolah menengah kebangsaan maxwell, kuala lumpur.
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2023 kssm f5 chemistry complete syllabus.
This online learning platform will show you the road map and step-by-step guidance on how to score well in Chemistry subject. In General : Structured Lessons + Suitable Examples + SPM Format Exercise + A little Hardwork from YOU = Good Understanding
Course curriculum
F5 chapter 1 redox part 1 (2023 version).
Lesson 1 Introduction to Redox Reaction, Oxygen Method, Hydrogen Method, Oxidation Number Method (V1)
Lesson 2 Oxidation Number Method (V2)
Lesson 3 Conversion of Chemical equation to Ionic equation and Half equation (V3)
Lesson 4 Conversion of Fe2+ to Fe3+ Part 1 (V4)
Lesson 5 Conversion of Fe2+ to Fe3+ Part 2 (V5)
F5 Chapter 1 Redox Part 2 (2023 version)
Lesson 6 Conversion of Fe3+ to Fe2+ (V6)
Lesson 7 U-tube (V7)
Lesson 8 Metal Displacement (V8)
Lesson 9 Halide Displacement Part 1 (V9)
Lesson 10 Halide Displacement Part 2 (V10)
F5 Chapter 1 Redox Part 3 (2023 version)
Lesson 11 Standard Electrode Potential (V11)
Lesson 12 Voltaic Cell Part 1 (V12)
Lesson 13 Voltaic Cell Part 2 (V13)
Lesson 14 Introduction to Electrolytic Cell and Concept of Electrolytes (V14)
F5 Chapter 1 Redox Part 4 (2023 version)
Lesson 15 Electrolysis of Molten Compounds (V15)
Lesson 16 Electrolysis of Aqueous Solution (Factor 1 : E Values) (V16)
Lesson 17 Electrolysis of Aqueous Solution (Factor 2 : Concentration) (V17)
Lesson 18 Tutorial for Electrolysis of Aqueous Solution (V18)
Lesson 19 Electrolysis of Aqueous Solution (Factor 3 : Type of Electrode) (V19)
Lesson 20 Application of Electrolysis in Industry (Electroplating and Purification of Impure Copper) (V20)
Lesson 21 Extraction of Metals (V21)
Tutorial : Extraction of Metals (V22)
Lesson 22 Concept of Rusting, Atmospheric Corrosion, Electrochemical Corrosion (V23)
Lesson 23 Experiments for Rusting (V24)
Tutorial : Concept of Rusting, Atmospheric Corrosion, Electrochemical Corrosion (V25)
F5 Chapter 2 Carbon Compounds Part 1
Lesson 1 Introduction to Carbon Compounds (V26)
Tutorial 1 (V27)
Lesson 2 Alkanes (Drawing and Naming) (V28)
Self-Test 1 (V29)
Lesson 3 Isomers of Alkanes (V30)
Self-Test 2 (V31)
Lesson 4 Physical and Chemical Properties of Alkanes (V32)
Tutorial 2 (V33)
Lesson 5 Alkenes (Drawing and Naming) (V34)
Self-Test 3 (V35)
Lesson 6 Isomer of Alkenes (V36)
Lesson 7 Physical and Chemical Properties of Alkenes (V37)
Lesson 8 Oxidation and Polymerisation of Alkenes (V38)
Tutorial 3 (V39)
Tutorial 4 (V40)
F5 Chapter 2 Carbon Compounds Part 2
Lesson 9 Alkynes (Drawing and Naming) (V41)
Self-Test 4 (V42)
Lesson 10 Isomers of Alkynes (V43)
Lesson 11 Alcohols (Drawing, Naming and Isomers) (V44)
Self-Test 5 (V45)
Lesson 12 Physical and Chemical Properties of Alcohols (V46)
Tutorial 5 (V47)
Tutorial 6 (V48)
Lesson 13 Carboxylic Acids and it's Acid Properties (V49)
Tutorial 7 (V50)
Lesson 14 Esters (V51)
Tutorial 8 (V52)
Tutorial 9 (V53)
F5 Chapter 3 Thermochemistry
Lesson 1 Introduction to Thermochemistry (V54)
Tutorial (V55)
Lesson 2 Heat of Displacement (V56)
Lesson 3 Heat of Neutralisation Part 1 (V57)
Extra Question on Limiting and Excess Reagent (V58)
Lesson 4 Heat of Neutralisation Part 2 (V59)
Lesson 5 Heat of Combustion (V60)
Lesson 6 Heat of Precipitation (V61)
Lesson 7 Predict Change in Temperature (V62)
F5 Chapter 4 Polymers
Lesson 1 Basic Concept of Polymers, Classification of Polymers (V63)
Tutorial 1 (V64)
Lesson 2 Addition Polymerisation (V65)
Lesson 3 Condensation Polymerisation, Effect of Polymers on Environment (V66)
Tutorial 2 (V67)
Tutorial 3 (V68)
Lesson 4 Natural Rubber, Vulcanised Rubber (V69)
Tutorial 4 (V70)
Lesson 5 Coagulation of Latex (V71)
Tutorial 5 (V72)
F5 Chapter 5 Consumer and Industrial Chemistry
Lesson 1 Oils and Fats (V73)
Tutorial 1 (V74)
Lesson 2 Soap (V75)
Tutorial 2 (V76)
Lesson 3 Detergent (V77)
Tutorial 3 (V78)
Lesson 4 Food Additives (V79)
Tutorial 4 (V80)
Lesson 5 Medicines (V81)
Tutorial 5 (V82)
Lesson 6 Cosmetic and Nanotechnology (V83)
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Introduction to redox reaction, oxygen method, hydrogen method.
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Tutorial 1 (Basic of Redox, Oxidation Number)
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Chemistry Specialist Mr Martin Teo
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This notebook is written by the Chemistry Specialists, Mr. Martin and Ms. Su, based on the latest form 4 Chemistry KSSM syllabus to help students save time and revise more efficiently. The notes given are simplified for the ease of student's understanding. This book would help students get a better grasp of the topic in Chemistry and definitely be helpful for students revising for the examinations.
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[BUNDLE] Form 4 & 5 KSSM Chemistry Online Lessons
FORM 5 KSSM BIOLOGY COMPLETE SYLLABUS
| KSSM F5 Add Math Complete syllabus |
This course will guide you through the entire KSSM syllabus with the focus on accurate answering techniques for SPM. The course will highlight the important and popular areas of the chapter to prepare you well for exams!
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The heat of neutralisation of a strong acid with a strong alkali is almost the same for all acids and alkalis. This is because the same reaction always takes places. The reaction is H +(aq) + OH -(aq) → H 2 O (l) Heat change of neutralization reaction is affected by 3 factors: Quantity of acid and alkali. Basicity of the acid and alkali.
The value of temperature change obtained is used to calculate heat of reaction. General Steps to Calculate Heat of Reactions. Step 1. Determine the number of moles of the reactants and products formed, n mole. Step 2. Calculate the heat change in the reaction: 𝑄=𝑚𝑐𝜃. Q = m c θ. Q = mc\theta Q = mcθ. Step 3.
Heat of Neutralization in J: heat gained by solution + heat gained by calorimeter 2 7, 00 J 293 + J 2, 94 = 9 J. Heat of Neutralization: (total joules released / # of moles water produced) ( 92, 94 J / .02 moles) 9= 5 This is a neutralization reaction therefore energy is lost and the H should be a negative value making it -1( 5 9) = - 5 9 kJ/mol.
Explain the reason for the difference in the value of the heat of neutralisation. The theoretical value of the heat of neutralisation between a strong acid and a strong alkali is -57 kJ mol−1. Compare this value with the heat of neutralisation obtained from this experiment. Suggest a reason for this difference.
Pandai videos is an initiative to aggregate educational videos that are already publicly available and categorise them based on the Malaysian national school curriculum KSSR & KSSM Chemistry SPM: Heat of neutralisation
neutralization, and for the case of HCl and NaOH is the following reaction: HCl + NaOH → NaCl + H 2 O ΔH = −52 kJ / mole The heat involved in this process is -52 kJ / mole (per mole means with the coefficients shown in the balanced equation). The negative sign means that the heat is being released. Reactions that release heat are
Energy changes and formations of chemical bonds. Applications of exothermic and endothermic reactions. Heat of Reaction. Calculating Heat Change. Finding Heat of Reaction. Heat of Precipitation. Heat of Displacement. Heat of Neutralisation. Heat of Combustion.
The Heat of Reaction is the heat absorbed in a reaction at standard state condition between the numbers of moles of reactants shown in the equation for the reaction. The Heat of reaction is represented by the symbol ∆H. The unit of ∆H is kJmol -1. If the reaction is exothernic, ∆H shows a value of negative. If the reaction is endothernic ...
Heat of Neutralization - Part 2 | ThermochemistryForm 5 Chemistry kssm Chapter 3 ThermochemistryThis video is created by http://www.onlinetuition.com.my/More...
2023 KSSM F5 Chemistry COMPLETE syllabus . ... Lesson 23 Experiments for Rusting (V24) Tutorial : Concept of Rusting, Atmospheric Corrosion, Electrochemical Corrosion (V25) ... Lesson 4 Heat of Neutralisation Part 2 (V59) Lesson 5 Heat of Combustion (V60) Lesson 6 Heat of Precipitation (V61)