HOTS-Chem_Equi

LECTURER’S GUIDE Based on Matriculation syllabus ANALYZING EVALUATING CREATING Nurul Hanani Rusli Nor Hasniza Ibrahim A game-based Learning Module to Enhance HOTS HOTS Challenge in Chemical Equilibrium

Time Allocation 6 Module Component 4 Theme 1: Dynamic Equilibrium 8 Theme 2: Equilibrium Constant 24 Theme 3: Le Chatelier’s Principle 40 References 64 Reward and Punishment 7 Introduction 1 TABLE OF CONTENT Higher-order Thinking Skills (HOTS) 2 Chemical Equilibrium 3 Module Objectives 4 How to play? 5 Fun Fact 56 HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium Chemequi Community 63

INTRODUCTION The ability to think critically using Higher-Order Thinking Skills (HOTS) has garnered special attention within our educational system since the introduction of the Malaysian Education Blueprint (MEB) (2013-2025). The development of HOTS is crucial for equipping students with essential 21st-century skills. For matriculation students, these thinking skills are particularly important, as they lay the foundation for higher educational pursuits. The HOTS-Chem_Equi Module has been designed as a platform for matriculation students to practice and enhance their Higher-Order Thinking Skills, including the abilities to analyze, evaluate, and create, with a specific focus on the concept of Chemical Equilibrium. This module adopts a game-based learning approach, encouraging students to employ higher-order thinking skills to solve problems presented within the game. The interactive activities foster a collaborative environment, enabling students to work in groups and fostering selfmotivation while facilitating meaningful learning in the area of Chemical Equilibrium. The content of this module aligns with the Chemistry Curriculum Specification of SK015, encompassing the matriculation syllabus for Chemical Equilibrium topics. The module comprises a game-board set, 'Fun Facts' related to Chemical Equilibrium, Chemical Equilibrium tasks catering to three levels of HOTS (analyzing, evaluating, and creating), and QR codes linked to the Chemequi Padlet, facilitating communication between students and lecturers. This module has been developed with the intent of providing matriculation students with an engaging and interactive learning experience, while simultaneously honing their critical thinking skills during the exploration of Chemical Equilibrium. Matriculation graduates equipped with HOTS are poised to excel, particularly when confronted with the challenges of advanced educational levels in the future HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium 1

HIGHER-ORDER THINKING SKILLS (HOTS) According to Revised Taxonomy Bloom (Anderson & Krathwohl, 2001), the higher-order thinking skills include the skills of analyzing, evaluating, and creating, which when these skills are implemented into the chemistry subject, can create connections to chemistry's unique aspects to increase students' engagement with the subject. This is because, the analyzing, evaluating, and creating skills in chemistry call for students to do beyond simply remembering and comprehending factual information; rather, they necessitate extensive cognitive operations Analyzing in the context of chemistry, is the practice of separating chemical reactions or concepts into their parts and distinguishing how those parts relate to one another. Evaluating skills allow students to judge the value or quality of information or arguments in chemistryrelated phenomena or situations. Creating thinking skills is students’ ability to generate new ideas or concepts while learning chemistry by applying the concepts of chemistry that they learn, as a new strategy in solving problem. Considering the importance of Higher-Order Thinking Skills (HOTS) in the field of chemistry, this module provides interactive game-based activities to assist matriculation students in enhancing their HOTS, with a particular focus on the Chemical Equilibrium topic. HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium 2

CHEMICAL EQUILIBRIUM As indicated in the matriculation syllabus, the Chemical Equilibrium topic consists of three subtopics: Dynamic equilibrium, Equilibrium constant, and Le Chatelier’s Principle. According to Mai et al. (2021), the Chemical Equilibrium topic consists of complex conceptual understanding. Students need to predict and explain observations involving specific terms that have a different meaning to everyday words. Oztay et al. (2023) found that students often struggle to grasp the abstract knowledge of dynamic equilibrium, which involves the simultaneous balance between forward and reverse reactions. This topic also contains mathematical problem-solving involving the equilibrium constant and equilibrium expressions. This section of the topic requires students to analyze the given information and evaluate the appropriate equations or formulas to be used while performing mathematical operations like quadratic equations to grasp the underlying principles of the topic (Jusniar et al., 2020). Additionally, applying Le Chatelier's principle poses significant challenges for students in learning chemical equilibrium. Students may have limited critical thinking skills using the underlying principles and concepts related to Le Chatelier's principle, leading to incorrect predictions and generating ideas to encounter problems in the equilibrium system (Kraska, 2022) Therefore, addressing these challenges emphasizes the importance of appropriate learning approaches in chemical equilibrium at the matriculation level. This module targets these difficulties, providing additional support and guidance to help students overcome their challenges and empowering their HOTS in this topic. HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium 3

MODULE OBJECTIVES HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium The HOTS-Chem_Equi Module aims to: Provide a learning platform for students to practice HOTS 15 of Dynamic Equilibrium 15 of Equilibrium Constant 15 of Le Chatelier’s Principle COMPONENT: 1 gameboard 50cm x 50cm 4 toggles and 1 dice 1 student Chemequi Booklet containing 15 Fun Fact of Chemical Equilibrium and 45 Challenges containing : Padlets: Chemequi Comunity (Scan via QR Code on gameboard/ in booklet) MODULE COMPONENTS HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium 4 Create an active learning among students Enhance students’ HOTS in Chemistry Create a meaningful and interactive play learning in Chemical Equilibrium **Students need at least 1 smartphone in each group

9. REWARDS AND PUNISHMENT Correct : Reward (Points according to types of Challenge Scores + board space on your turn) Wrong: Punishment (Accrding to the board space on your turn) 10. RECORD THE POINTS Teacher will record the points in Equibrilliant Padlet . 5. ON YOUR TURN Roll the dice and move according to the dice number 6. CHALLENGE Make selection (Awesome, Clever or Brilliant) as the group’s challenge and go to the challenge page in the booklet. (suggested time: 1 minutes) 7. RESPONSE Discuss and solve the task within the time limit (suggested time: 4 minutes) 8. FEEDBACK Present the solution. Teacher mark and give quick feedback (suggested time: 3 minutes) 1. SELECT THEME (Theme is selected based on subtopic: Dynamic Equilibrium, Equilibrium Constant, and Le Chatelier’s Principle) 2. FIND PLAYERS (Minimum 2 groups, Maximum 4 groups) Upload the group photos with group's name into the Chemequi Community padlet. 3. PREPARE THE GAME-BOARD & CHEMEQUI BOOKLET Get ready with the Chemequi Booklet. Put the toggles at 'Start here' 4. DECIDE TURN HOW TO PLAY? HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium 5

ROUND CHALLENGE RESPONSE FEEDBACK TOTAL TIME ROUND 1 1 MINUTE 4 MINUTES 3 MINUTES 8 MINUTES ROUND 2 1 MINUTE 4 MINUTES 3 MINUTES 8 MINUTES ROUND 3 1 MINUTE 4 MINUTES 3 MINUTES 8 MINUTES Lecturer facilitates the game cycles. Lecturer can modify the time allocated for the activities accordingly Lecturer records the points according to types of challenge and marks the solution presented by students. (Lecturer can refer to the answer scheme provided) Instruction for teacher: TIME SUGGESTION: 24 MINUTES PER GROUP TO COMPLETE 3 ROUNDS TIME ALLOCATION HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium 6

REWARD AND PUNISHMENT REWARDS Double points Move forward 3 steps Bonus 5 points Bonus 10 points PUNISHMENT Move back 1 step Deduct 2 points Deduct 5 points Lost a turn Points according to solution presented, marked by teacher HOW TO WIN? HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium Each Challenge selection will get specific points (if the answer is correct): Awesome: 10 points Clever: 30 points Brilliant: 50 points 7 Cumulative points Group with highest points will be the WINNER Cumulative points ADDITIONAL POINTS (Quickest to finish the round) 1st : 5 points 2nd: 3 points 3rd: 1 points On the board space Rewards on board space

HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium 8 THEME 1: DYNAMIC EQUILIBRIUM

AWESOME : CHALLENGE 1 The Haber-Bosch process is an industrial chemical reaction used to synthesize ammonia (NH₃) from nitrogen (N₂) and hydrogen (H₂) gases. The balanced chemical equation for this reaction is: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) Point out the concept of reversible reactions using the HaberBosch process as an example. Clue: Dynamic Equilibrium DYNAMIC EQUILIBRIUM In the case of the Haber process, the reaction proceeds in both directions: the reactants N₂ and H₂ combine to form NH₃, and NH₃ can decompose to reform N₂ and H₂. At the beginning of the reaction, when only N₂ and H₂ are present, the forward reaction dominates. As ammonia forms, the concentration of NH₃ increases, and the reverse reaction gains strength. Eventually, the rates of the forward and reverse reactions become equal, leading to a state of dynamic equilibrium. At this point, the concentrations of N₂, H₂, and NH₃ remain constant over time. AWESOME : CHALLENGE 1 5 MARKS HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium 9

The mixture of nitrogen dioxide (NO₂) and dinitrogen tetroxide (N₂O₄) give brown color and can cause irritation of the eyes and nose. The reversible reaction between nitrogen dioxide (NO₂) and dinitrogen tetroxide (N₂O₄) is shown below: 2NO₂(g) ⇌ N₂O₄(g) Differentiate the characteristics of nitrogen dioxide (NO₂) and dinitrogen tetroxide (N₂O₄) in term of concentration and rate of the system before and at the equilibrium state. Clue: characteristic of Dynamic Equilibrium AWESOME : CHALLENGE 2 DYNAMIC EQUILIBRIUM HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium Constant Concentrations: At equilibrium, the concentrations of reactants and products remain constant over time. In the case of the NO₂ and N₂O₄ reaction, initially, only NO₂ is present. As the reaction proceeds, the concentration of N₂O₄ increases, while NO₂ concentration decreases. Eventually, the concentrations of both NO₂ and N₂O₄ reach a point where they do not change, indicating the system has reached equilibrium. Reaction Rates: In a system at equilibrium, the forward and reverse reactions occur at equal rates. As reactants convert to products, products also convert back to reactants at the same rate. This balance of opposing reaction rates leads to a steady concentration of reactants and products. AWESOME : CHALLENGE 2 5 MARKS 10

AWESOME : CHALLENGE 3 Consider the equilibrium between molecular Phosphorus pentachloride (PCl₅), Phosphorus trichloride (PCl₃), and Chlorine gas (Cl₂). Illustrate how the law of mass action applies in the reaction between PCl₅ to formPCl₃ and Cl₂ Clue: Law of mass action DYNAMIC EQUILIBRIUM Answer: Law of Mass Action: The law of mass action states that the rate of a chemical reaction is directly proportional to the product of the concentrations of the reactants, each raised to the power of its stoichiometric coefficient, at a given temperature. Mathematically, for the reaction PCl₅ ⇌ PCl₃ + Cl₂, the law of mass action is expressed as: AWESOME : CHALLENGE 3 5 MARKS HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium 11

AWESOME : CHALLENGE 4 Consider the reversible reaction of hydrogen iodide (HI) decomposing into hydrogen (H₂) and iodine (I₂): 2HI(g) ⇌ H₂(g) + I₂(g) By using the reversible decomposition reaction of hydrogen iodide (HI) to hydrogen (H₂) and iodine (I₂), illustrate how the characteristics that bring the system to an equilibrium state. Clue: concentration and rates DYNAMIC EQUILIBRIUM i) Constant Concentrations of Reactants and Products: initially, only hydrogen iodide (HI) is present. As the reaction proceeds, HI decomposes to form hydrogen (H₂) and iodine (I₂). Eventually, the concentrations of HI, H₂, and I₂ do not change, indicating that the forward and reverse reactions occur at the same rate. At equilibrium, the concentrations of reactants and products remain constant over time. ii) Dynamic Equilibrium and Opposing Reaction Rates: In the case of the hydrogen iodide decomposition reaction, the forward reaction (2HI → H₂ + I₂) results in the formation of H₂ and I₂, while the reverse reaction (H₂ + I₂ → 2HI) regenerates hydrogen iodide. A system at equilibrium is in dynamic equilibrium, meaning that both the forward and reverse reactions continue to occur at the same rate AWESOME : CHALLENGE 4 5 MARKS HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium 12

AWESOME : CHALLENGE 5 Consider the reversible reaction between nitrogen dioxide (NO2) and dinitrogen tetroxide (N₂O₄): 2NO₂(g) ⇌ N₂O₄(g) Examine the curve of concentration of reactant and product against time for the reversible reaction between nitrogen dioxide (NO₂) and dinitrogen tetroxide (N₂O₄). Based on your analysis, distinguish the factors influencing the observed behavior of the curve. Clue: Concentration of reactant and product Rate of reactant and product DYNAMIC EQUILIBRIUM NO₂ At the beginning of the reaction, only nitrogen dioxide (NO₂) is present, so the concentration of NO₂ is relatively high. As the reaction proceeds, NO₂ molecules combine to form dinitrogen tetroxide (N₂O₄). As a result, the concentration of NO₂ decreases over time, while the concentration of N₂O₄ increases. Eventually, the concentrations of both NO₂ and N₂O₄ remain constant, leading to a curve that levels off over time. Changes in Concentrations of Reactant and Product: Initially, the rate of the forward reaction (2NO₂ → N₂O₄) is high as there is an abundance of NO₂. As N₂O₄ forms, the concentration of NO₂ decreases, and the rate of the reverse reaction (N₂O₄ → 2NO₂) increases. Eventually, the rates of the forward and reverse reactions become equal, and the concentrations of NO₂ and N₂O₄ reach a constant value Shape of the Curve: AWESOME : CHALLENGE 5 5 MARKS HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium 13

In ocean water, carbon dioxide (CO₂) from atmosphere reacts with water (H₂O) to form carbonic acid (H₂CO₃) in a reversible reaction: CO₂(g) + H₂O(l) ⇌ H₂CO₃(aq) A continuous formation of carbonic acid (H₂CO₃) in the ocean leading to a decrease in pH and ocean acidification. Thus, the exchange of carbon dioxide between the ocean waters in a reversible way ensures that the system remains dynamic. Justify how the nature of this reversible reaction benefits the ocean life. DYNAMIC EQUILIBRIUM Clue: Dynamic Equilibrium CLEVER : CHALLENGE 1 CO₂(g) + H₂O(l) ⇌ H₂CO₃(aq) The formation of carbonic acid (H₂CO₃) from carbon dioxide and water is the forward reaction, while the decomposition of carbonic acid back into carbon dioxide and water is the reverse reaction. As carbon dioxide is continuously exchanged between the atmosphere and ocean, the concentrations of CO₂, H₂O, and H₂CO₃ remain constant As the rate forward and reverse reactions are equal The system reach an equilibrium state over time and benefits the ocean life. CLEVER : CHALLENGE 1 5 MARKS HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium 14

Long-term exposure to high levels of nitrogen dioxide can cause chronic lung disease. The equation below shows the reversible reaction of the production of NO₂ gas. N₂O₄(g) ⇌ 2NO₂(g) Millions of tons of Nitrogen dioxide gas (NO₂) are produced each year for use primarily in the production of fertilizers. But as long as the dynamic equilibrium is established in the reaction system, the pollution of NO₂ is still under controlled. By using the knowledge of dynamic equilibrium characteristic, justify the equilibrium system in the reaction produced the NO₂ gas. DYNAMIC EQUILIBRIUM CLEVER : CHALLENGE 2 Clue: concentration of reactant and product, rate of reaction Constant Concentrations: In the case of the N₂O₄ and NO₂ reaction, initially, only N₂O₄ is present. As the reaction proceeds, the concentration of NO₂ increases, while N₂O₄ concentration decreases. Eventually, the concentrations of both NO₂ and N₂O₄ reach a point where they do not change, indicating the system has reached equilibrium. Reaction Rates: In a system at equilibrium, the forward and reverse reactions occur at equal rates. As reactants convert to products, products also convert back to reactants at the same rate. This balance of opposing reaction rates leads to a steady concentration of reactants and products. CLEVER : CHALLENGE 2 5 MARKS HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium 15

Hemoglobin will combine with oxygen in the blood and an equilibrium will be established from the attachment and dissociation of oxygen and hemoglobin. The reaction occur according to the equation below. Hb (aq) + 4O ₂ (g) ⇋ Hb(O₂) ₄ (aq) The curve below show that the system in the body of a healthy person. Based on the curve, conclude how the system is in equilibrium? Clue: concentration of reactant and product, rate of reaction DYNAMIC EQUILIBRIUM CLEVER : CHALLENGE 3 Hemoglobin As the reaction proceeds, O₂ molecules combine with hemoglobin to form oxyhemoglobin Hb(O₂)₄. Eventually, the concentrations of both O₂ and Hb(O₂) ₄ stabilize, leading to a curve that levels maintain over time. Changes in Concentrations of Reactant and Product: Initially, the rate of the forward reaction is high as there is an abundance of O₂. As Hb(O₂) ₄ forms, the concentration of O₂ decreases, and the rate of the reverse reaction increases. Eventually, the rates of the forward and reverse reactions become equal, and the concentrations of O₂ and Hb(O₂) ₄ reach a constant value Shape of the Curve: As a result, the concentration of O₂ decreases over time, while the concentration of Hb(O₂) ₄ increases. CLEVER : CHALLENGE 3 5 MARKS HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium 16

∆H = +ve Photochromic compound Darken colour Clue: Dynamic equilibrium Photochromism is a chemical process in which photochromic compound undergoes a reversible reaction with a different color. The photochromic compound is used in the production of sun glass to protect eyes. During sunny day, justify how the darken color of the sun glass maintain by using your knowledge of dynamic equilibrium. DYNAMIC EQUILIBRIUM CLEVER : CHALLENGE 4 During a sunny day, the light is continuously exposed so the forward reaction occur produce the darken colour product. As the reaction proceeds, product of darken colour increase while photochromic decrease. At equilibrium, the concentrations of reactants and products remain constant over time. Eventually, the concentrations of both photochromic and darken product reach a point where they do not change, The rate of forward and reverse reaction become equal indicating the system has reached equilibrium. So the dark colour of sun glasses maintain. CLEVER : CHALLENGE 4 5 MARKS HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium 17

Mr Johnson who owns an ammonia production factory has found an optimum way to produce the product in optimum amount. According to the equation of reversible reaction, Can you conclude how the system maintains equilibrium after a new condition is applied in increasing the production of Ammonia? N₂(g) + 3H₂(g) ⇌ 2NH₃(g) Clue: Dynamic equilibrium DYNAMIC EQUILIBRIUM CLEVER : CHALLENGE 5 When a new condition is applied to maximize the yield of the product, the equilibrium proceeds to the right to form more product. Initially, only reactants (N₂ and H₂) is present. As the reaction proceeds, the concentration of NH₃ increases, while the concentration of N₂ and H₂ decrease. Eventually, the concentrations of N₂ , H₂ ,and NH₃ reach a point where they do not change, indicating the system has reached equilibrium. The rate of forward and reverse reaction become equal So the system reach the equilibrium state CLEVER : CHALLENGE 5 5 MARKS HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium 18

BRILLIANT : CHALLENGE 1 reaction between zinc (Zn) and manganese dioxide (MnO₂)facilitates the storage and release of electrical energy. When electricity is supplied, the reaction occur as below: Chemical Reaction: Zn(s) + MnO₂(s) + H₂O(l) ⇌ ZnO(s) + Mn(OH)₂(s) By using your knowledge of dynamic equilibrium, suggest how this reaction can be used as rechargeable battery? Electric supply Clue: Reversible reaction DYNAMIC EQUILIBRIUM During charging (discharging), the battery undergoes reversible reaction: Zn(s) + MnO₂(s) + H₂O(l) ⇌ ZnO(s) + Mn(OH)₂(s) Forward Reaction (Charging): Zinc metal (Zn) and manganese dioxide (MnO₂) react with water (H₂O) to produce zinc oxide (ZnO) and manganese hydroxide (Mn(OH)₂). Reverse Reaction (Discharging): During discharge, the battery provides electrical energy, and the reverse reaction occurs, converting zinc oxide and manganese hydroxide back into zinc metal and manganese dioxide. The battery operates in dynamic equilibrium, where the charging and discharging reactions occur simultaneously at the same rate. The forward and reverse reactions maintain a stable state within the battery. BRILLIANT : CHALLENGE 1 5 MARKS HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium 19

BRILLIANT : CHALLENGE 2 Dinitrogen tetroxide, N₂O₄, dissociates into two NO₂ molecules in an endothermic reaction as shown in the reversible equation and the diagram below. As NO₂is usually considered as a primary pollutant gas, what should a responsible company suggest to reduce the formation of NO₂ in the industry? N₂O₄(g) ⇌ 2NO₂(g) ∆H = +ve t₁ Answer: to reduce the formation of NO₂ in the industry, the company should: prepare high pressure conditions, lower the temperature, and reduce the volume of the container. BRILLIANT : CHALLENGE 2 5 MARKS HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium 20 DYNAMIC EQUILIBRIUM

BRILLIANT :CHALLENGE 3 Carbon monoxide, CO, "fools" hemoglobin (Hb) into mistaking it for oxygen because it also bonds to hemoglobin, and the equilibrium expression is as follows: Hb (aq) + 4CO (g) ⇋ Hb(CO)₄ (aq) Oxygen reacts with the carboxyhemoglobin to produce properly oxygenated hemoglobin, along with carbon monoxide: Hb(CO)₄ (aq) + 4O₂ (g) ⇋ Hb(O₂)₄ (aq) + 4CO₂ (g). carboxyhemoglobin carboxyhemoglobin Oxygenated hemoglobin According to the reversible equation stated above, how would you suggest for treatment if someone has been poisoned by Carbon monoxide gas? To save his life, you want the oxygenated hemoglobin to be produced in his body back immediately. To save someone’s life who has been poisoned by Carbon monoxide is by helping his body to produce oxygenated hemoglobin. The best way is pure oxygen gas must be introduced to the body to reverse the effects of carbon monoxide. It will react with the carboxyhemoglobin to produce properly oxygenated hemoglobin that is needed to supply oxygen to the whole body. BRILLIANT : CHALLENGE 3 5 MARKS HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium 21 DYNAMIC EQUILIBRIUM

Phosphorus Pentachloride is a white to pale yellow, crystalline solid with a pungent odor. It is used in the manufacture of other chemicals, in aluminum metallurgy, and in the pharmaceutical industry. The reaction of Phosphorus Pentachloride production is shown in the equation below: PCl₃(g) + Cl₂(g) ⇌ PCl₅(g) Can you design an optimum way to produce PCl₅ as an industrial product? BRILLIANT : CHALLENGE 4 Phosphorus Pentachloride PCl₃(g) + Cl₂(g) ⇌ PCl₅(g) Increase the rate of forward reaction Increase the concentration of reactants (PCl₃ and Cl₂) Remove products to decrease the concentration of the product Phosphorus Pentachloride BRILLIANT : CHALLENGE 4 5 MARKS HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium 22 DYNAMIC EQUILIBRIUM

BRILLIANT : CHALLENGE 5 Clue: Reversible reaction, concentration, rate of reaction When you go hiking, the higher level at the top of mountain gives lower pressure that decrease the amount of oxygen gas in the body. Lack of oxygen make it difficult for the production of oxyhaemoglobin resulting in short of breath. With the knowledge of dynamic equilibrium, suggest a way to maintain the body system so that oxyhaemoglobin will continuously produced in the body? The reaction occur according to the equation below. Hb (aq) + 4O ₂ (g) ⇋ Hb(O₂) ₄ (aq) Hemoglobin Oxyhaemoglobin First, supply the oxygen to the body to increase the concentration of oxygen. When the reaction proceeds, the concentration of oxyhemoglobin is increased. Eventually, the backward reaction occurs and the concentration of reactants and product get stable and maintain over time as the body system achieved the chemical equilibrium. When reach the equilibrium, the rate of reverse and forward is equal and maintain in the body. BRILLIANT : CHALLENGE 5 5 MARKS HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium 23 DYNAMIC EQUILIBRIUM

HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium THEME 2: EQUILIBRIUM CONSTANT 24

AWESOME : CHALLENGE 1 Kc for the reaction is 0.497 at 500 K. In certain experiment, 4.0 x 10⁻² mole of PCl₅, 2.5 x 10⁻³ mole of PCl₃, and 6.8 x 10⁻² moles of Cl₂ are mixed in a 2.0 L flask. In which direction will the system proceed to reach equilibrium? PCl₅(g) ⇌ PCl₃(g) + Cl₂ EQUILIBRIUM CONSTANT [PCl₅ ] = 4.0 x 10⁻² mol/ 2.0 L = 0.020 mol/L [PCl₃ ] = 2.5 x 10⁻³mol/ 2.0 L = 1.25 x 10⁻³ mol/L [Cl₂ ] = 6.8 10⁻² mol/ 2.0 L = 0.034 mol/L Qc < Kc. The equilibrium position proceed to the right until equilibrium is established. PCl₅(g) ⇌ PCl₃(g) + Cl₂ AWESOME : CHALLENGE 1 Kc = 0.497 5 MARKS HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium 25

AWESOME : CHALLENGE 2 A sample of solid NH₄HS is placed in a 2.58 L flask containing 0.100 mol NH₃(g) at equilibrium. NH₄HS(s) ⇌ NH₃(g) + H₂S(g) What will be the total gas pressure when equilibrium is established at 25ᵒC. Given Kp = 0.108 at 25ᵒC . (R= 0.08206 L.atm.mol⁻¹.K⁻¹) EQUILIBRIUM CONSTANT NH₄HS(s) ⇌ NH₃(g) H₂S(g) P equilibrium (atm) - 0.948 x PV = nRT Pressuse of NH₃ = nRT/V = 0.948 atm Kp = (PNH₃) (PH₂S) = 0.108 (0.948) (x) = 0.108 x = 0.114 At equilibrium, PNH₃ = 0.948atm ; PH₂S = 0.114 atm Ptotal = 0.948 + 0.114 = 1.062 atm NH₄HS(s) ⇌ NH₃(g) + H₂S(g) AWESOME : CHALLENGE 2 5 MARKS HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium 26

AWESOME : CHALLENGE 3 Kc for the formation of nitrosyl chloride, an orange-yellow compound, from nitric oxide and molecular chlorine is 6.5 x 10⁴ at 35ᵒC. In a certain experiment, 2.0 x 10⁻² mole of NO, 8.3 x 10⁻³ mole of Cl₂, and 6.8 moles of NOCl are mixed in a 2.0–L flask. In which direction will the system proceed to reach equilibrium? 2NO(g) + Cl₂(g) ⇌ 2NOCl(g) EQUILIBRIUM CONSTANT AWESOME : CHALLENGE 3 5 MARKS 2NO(g) + Cl₂(g) ⇌ 2NOCl(g) [ NO ] = 2.0 x 10⁻² mol/ 2.0 L = 0.010 mol/L [Cl₂ ] = 8.3 x 10⁻³ mol/ 2.0 L = 4.15 x 10⁻³ mol/L [NOCl ] = 6.8 mol/ 2.0 L = 3.4 mol/L Qc > Kc. The equilibrium position proceed to the left until equilibrium is established. Kc= 6.5 x 10⁴ at 35ᵒC HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium 27

AWESOME : CHALLENGE 4 The Nitrogen dioxide is a poisonous gas that contribute to the gas pollutant. In a country, the polluted air was studied and the atmospheric oxidation of nitrogen dioxide, (NO₂) was recorded at 184ᵒC with a pressure of 1.000 atm of NO and 1.000 atm of O₂. At equilibrium, PO₂ = 0.506 atm. Calculate Kp. 2NO(g) + O₂(g) ⇌ 2NO₂(g) EQUILIBRIUM CONSTANT 2NO(g) + O₂(g) ⇌ 2NO₂(g) Initial(atm) 1 1 0 Change (atm) -2x -x +2x Equilibrium(atm) 1-2x 1-x = 0.506 2x 2NO(g) + O₂(g) ⇌ 2NO₂(g) AWESOME : CHALLENGE 4 5 MARKS 1-x = 0.506 x = 0.494 At equilibrium, P(NO) = 1 – 2(0.494) = 1- 0.988 = 0.012 atm PO₂ = 0.506 atm PNO₂ = 2(0.494) atm = 0.988 atm = 1.34 x 10⁴ HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium 28

1.50 mol H₂(g) and 2.00 mol I₂(g) are introduced into a 15.0–L flask at 445ᵒC and allowed to come to equilibrium. At equilibrium, concentration of HI(g) is 0.05 mol/L. What will be the value of Kc? = 50.2 at 445ᵒC H₂(g) + I₂(g) ⇌ 2HI(g) AWESOME : CHALLENGE 5 EQUILIBRIUM CONSTANT H₂(g) + I₂(g) ⇌ 2HI(g) H₂ (g) ⇌ I₂ (g) 2HI (g) Initial (M) 0.1 1.3 0 Change (M) -x -x +2x Equilibrium (M) 0.1-x 1.3-x 2x = 0.05 AWESOME : CHALLENGE 5 5 MARKS [ H₂ ] = 1.5mol/ 15.0 L = 0.1 mol/L [ I₂ ] = 2.0 mol/ 15.0 L = 1.3 mol/L 2x = 0.05 x= 0.025 At equlibrium, [ H₂ ] = 0.1 mol/L - 0.025 = 0.075 mol/L [ I₂ ] = 1.3 - 0.025 = 1.275 mol/L = = 0.0261 HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium 29

You have an unopened soda can containing carbonated water, which is a mixture of water and carbon dioxide gas under pressure. The fizzing process involves the dissolution of carbon dioxide (CO₂) gas into the liquid phase and the formation of carbon dioxide gas bubbles in the air phase. Dissolution of CO₂ gas in liquid: CO₂(g) ⇌ CO₂(aq) Note: CO₂(g) represents carbon dioxide gas, and CO₂(aq) represents dissolved carbon dioxide in the liquid. With the knowledge of homogenous and heterogeneous equilibria, figure out the type of equilibria in the unopened soda can. EQUILIBRIUM CONSTANT CLEVER : CHALLENGE 1 CLEVER : CHALLENGE 1 5 MARKS HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium 30 In the unopened soda can, there exist CO₂ gas that dissolved in the liquid drink of CO₂. Thus, the equilibrium is said to be heterogeneous equilibrium.

The Haber process is a crucial industrial reaction for the production of ammonia (NH₃) from nitrogen gas (N₂) and hydrogen gas (H₂) under high pressure and temperature. The balanced equation for the reaction is: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) The equilibrium constant Kc for the Haber process at a temperature of 300K is 0.500. Calculate the corresponding Kp at the same temperature. At different temperature, the value of Kc increasing. What would happen to the value of Kp by relating both Kc and Kc? (R = 0.0821 L.atm.mol⁻¹.K⁻¹) Clue: Use Kp = Kc (RT) ∆n CLEVER : CHALLENGE 2 EQUILIBRIUM CONSTANT Equilibrium Constant Expression Kc: Calculating Kp from Kc: Given: Kc = 0.500 We know Kp = Kc (RT) ∆n, where n = (2 - (1 + 3)) = -2 The temperature (T) = 300 K. R = 0.0821 L.atm/(mol.K) Kp = 0.500 * (0.0821 L.atm/(mol.K) * 300 K)^(-2) Kp = 0.500 * (24.63)^(-2) Kp ≈ 0.500 * 0.0016 Kp ≈ 0.0008 CLEVER : CHALLENGE 2 5 MARKS HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium 31

The decomposition of ammonium hydrogen Sulfide, NH₄HS, on heating in a sealed flask, is an endothermic process. If Kp for this reaction is 0.11 at 25°C, find out the partial pressure of NH₃ in the flask (atm) at equilibrium. Justify the change in the Kp value if the partial pressure of NH₃ gas is increased in the system. NH₄NH(g) ⇌ H₂S(g) + NH₃(g) Clue: Use I.C.E table CLEVER : CHALLENGE 3 EQUILIBRIUM CONSTANT NH₄NH (s) ⇌ H₂S(g) NH₃ (g) Initial(atm) - 0 0 Change (atm) - +x +x Equilibrium(at m) - x x Kp = (PH₂S)(PNH₃) = 0.11 (x)(x) = 0.11 x = (PH₂S)=(PNH₃) = 0.332 mol/dm³ Increase the partial pressure of NH₃ will not effect the Kp value, but it will change the direction of equilibrium system to achieve new equilibrium. The Kp value remain constant. NH₄NH(g) ⇌ H₂S(g) + NH₃(g) CLEVER : CHALLENGE 3 5 MARKS HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium 32

Clue: Compare Qc of both labs with Kc Phosphorus pentachloride (PCl₅) is a greenish-yellow crystalline solid with an irritating odor. KC for the reaction is 0.497 at 500 K. The reaction equation is shown below: PCl₅(g) ⇌ PCl₃(g) Cl₂(g) There are two experiments of the decomposition of PCl₅ are established in different laboratories, which are lab A and Lab B. Which lab will produce more PCl₃? EQUILIBRIUM CONSTANT CLEVER : CHALLENGE 4 Lab A 0.020 mol/L of PCl₅, 0.1 mol/L of PCl₃ 0.034 mol/L of Cl₂ Lab B 0.120 mol/L of PCl₅, 0.25 mol/L of PCl₃ 0.4 mol/L of Cl₂ Qc < Kc. The equilibrium position proceed to the right until equilibrium is established. [PCl₅ ] = 0.020 mol/L [PCl₃ ] = 0.1 mol/L [Cl₂ ] = 0.034 mol/L Lab A [PCl₅ ] = 0.120 mol/L [PCl₃ ] = 0.25 mol/L [Cl₂ ] = 0.40 mol/L Lab B Qc > Kc. The equilibrium position proceed to the left until equilibrium is established. PCl₅(g) ⇌ PCl₃(g) Cl₂(g) Kc= 0.497 Therefore, Lab A will produce more PCl₃ CLEVER : CHALLENGE 4 5 MARKS HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium 33

Hydrogen Bromide (HBr) decomposed into Hydrogen gas (H₂) and Bromine gas (Br₂). The equation of the reaction is shown below: 2HBr(g) ⇌ H₂(g) + Br₂(g) Initially, 2 mole of HBr is introduced into a 2.0 L flask and is allowed to decompose to reach equilibrium. At equilibrium, the concentration of Br₂ is 0.01M. Find the degree of dissociation of HBr in the reaction. Clue: Use I.C.E. table and find the degree of dissociation, α EQUILIBRIUM CONSTANT CLEVER : CHALLENGE 5 2HBr(g) ⇌ H₂(g) Br₂(g) Initial(M) 1.0 0 0 Change (M) -2x +x +x Equilibrium(M) 1.0-x x x= 0.01 CLEVER : CHALLENGE 5 5 MARKS At equilibrium, [Br₂] = 0.01M = x Degree of dissociation, α = 2HBr(g) ⇌ H₂(g) Br₂(g) HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium 34

BRILLIANT : CHALLENGE 1 A student is conducting an experiment in a closed container to study the reaction between nitrogen gas (N₂) and hydrogen gas (H₂) to form ammonia gas (NH₃) using the following balanced chemical equation: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) Kc= 3.7x10⁸(at 298 K) Consider the moles of N₂, H₂, and NH₃ are 0.2 mole initially. The flask of 1.0 L is used in the experiment. Suggest a way to predict the direction of the reaction in the experiment. Clue: Reaction Quotient EQUILIBRIUM CONSTANT BRILLIANT : CHALLENGE 1 5 MARKS [ N₂ ] = 0.2 mol/L [ H₂ ] = 0.2 mol/L [ NH₃ ] = 0.2 mol/L N₂(g) + 3H₂(g) ⇌ 2NH₃(g) Kc = 3.7x10⁸(at 298 K) Compare Qc and Kc Qc < Kc. The system is not yet in equilibrium. The equilibrium position shift to the right until the equilibrium is achieved. HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium 35

BRILLIANT : CHALLENGE 2 Kc for the formation of nitrosyl chloride, an orange-yellow compound, from nitric oxide and molecular chlorine is 6.5 x 10⁴ at 35ᵒC. In a certain experiment, 0.010 mol/L of NO, 4.15 x 10⁻³ mol/L of Cl₂, and 3.4 mol/L of NOCl are mixed in a 2.0–L flask. Prove that the system is not in equilibrium. Suggest the direction of the reaction to reach the equilibrium. 2NO(g) + Cl₂(g) ⇌ 2NOCl(g) Clue: Reaction Quotient EQUILIBRIUM CONSTANT BRILLIANT : CHALLENGE 2 5 MARKS [ NO ] = 0.010 mol/L [Cl₂ ] = 4.15 x 10⁻³ mol/L [NOCl ] = 3.4 mol/L Qc ≠ Kc. So the system is not in equilibrium. Since Qc >Kc, The equilibrium position proceed to the left to establish the equilibrium. 2NO(g) + Cl₂(g) ⇌ 2NOCl(g) Kc = 6.5 x 10⁴ HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium 36

BRILLIANT : CHALLENGE 3 Pure SO₂Cl₂ at 0.75 atm and 298K is allowed to reach equilibrium in a closed container. When the system reached the equilibrium, the total pressure is found out to be 1.36 atm. SO₂Cl₂(g) ⇌ SO₂(g) + Cl₂(g) What will be the value of Kp for the reaction? suggest how to increase the value of Kp in the equilibrium system. (R = 0.0821 L.atm.mol⁻¹.K⁻¹) Clue: : Use I.C.E table and Kp = Kc (RT) ∆n EQUILIBRIUM CONSTANT SO₂Cl₂(g) ⇌ SO₂(g) + Cl₂(g) Initial(M) 0.75 0 0 Change (M) -x +x +x Equilibrium(M) 0.75-x x x BRILLIANT : CHALLENGE 3 5 MARKS The value of Kp can be increased by increasing the Temperature of the system. This can be proven by the relationship in the equation which increasing in T will increase the value of Kp Ptotal = 1.36 atm 1.36 = 0.75-x + x + x x = 0.610 PSO₂Cl₂ = 0.140 atm PSO₂ = PCl₂ = 0.610 atm HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium 37

BRILLIANT : CHALLENGE 4 In a closed vessel, Nitrogen dioxide (NO₂) and Hydrogen gas (H₂) are allowed to reach equilibrium and the reaction is shown in the equation below. 2NO₂(g) + 7H₂(g) ⇌ 2NH₃(g) + 4H₂O (l) Kp= 6.7 x 10⁻² at 298K Write the Qp expression for this system. If thhe pressure of all gaseous involved in the reaction is 1 atm. How to change the Qp value for this reaction so that the position change to the right? Clue: : Reaction Quotient EQUILIBRIUM CONSTANT BRILLIANT : CHALLENGE 4 5 MARKS 2NO₂(g) + 7H₂(g) ⇌ 2NH₃(g) + 4H₂O (l) Kp= 6.7 x 10⁻² at 298K Since Qp > Kp, the equilibrium position shift to the left. the pressure of NO₂ and H₂ increase until the value of Qp < 6.7 x 10⁻² (OR) the pressure of NO₂ increase until the value of Qp < 6.7 x 10⁻² (OR) the pressure of H₂ increase until the value of Qp < 6.7 x 10⁻² For the direction to change to the right, the Qp value should be less than Kp (6.7 x 10⁻²) According to Qp expression, the value will decrease when: 1. 2. 3. HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium 38

BRILLIANT : CHALLENGE 5 In a closed container, 0.76M of Nitrogen gas (N₂), 0.60M of Hydrogen gas (H₂), and 0.48M of ammonia gas (NH₃) are allowed to mix at the Temperature of 375⁰C. The Kc value is for the equilibrium system is 1.2. N₂(g) + 3H₂(g) ⇌ 2NH₃(g) Does the system already reach the equilibrium? If it is not, suggest either the amount of product is increasing or decreasing. Clue: : Reaction Quotient (Q) EQUILIBRIUM CONSTANT BRILLIANT : CHALLENGE 5 5 MARKS Qc > Kc. The system is not yet in equilibrium. The equilibrium position shift to the left to reach the equilibrium. So the amount of product (NH₃) is decreasing. N₂(g) + 3H₂(g) ⇌ 2NH₃(g) Kc = 1.2 at 375⁰C HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium 39

HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium THEME 3: LE CHATELIER’S PRINCIPLE 40

AWESOME : CHALLENGE 1 Consider the equilibrium between molecular oxygen and ozone. What will happen to the amount of the equilibrium components when the Helium is added by maintaining the pressure, based on Le Chatelier’s principle. 3O₂(g) ⇌ 2O₃(g) ∆H = 284kJ LE CHATELIER’S PRINCIPLE 3O₂(g) ⇌ 2O₃(g) ∆H = 284kJ Adding inert gas (Helium) at constant pressure, the partial pressure of O₂ and O₃ will decrease. System will increase the partial pressure by shift the equilibrium position to the direction with more number of moles. So equilibrium position shift to the left. more oxygen gas is produced. Production of ozone is decrease. AWESOME : CHALLENGE 1 5 MARKS HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium 41

AWESOME : CHALLENGE 2 How does an increase in temperature affect the equilibrium concentration of the underlined substance. CaO(s) + H₂O(l) ⇌ Ca(OH)₂(g) ∆H = – 82 kJ LE CHATELIER’S PRINCIPLE Answer: CaO(s)  +  H₂O(l)  ⇌   Ca(OH)₂(g)    ∆H = – 82 kJ When Temperature increases, the system will decrease the Temperature by absorbing the added heat. System favors endothermic reaction to reattain equilibrium. Equilibrium position shift to the left. The formation of Ca(OH)₂ decrease. The concentration of Ca(OH)₂ will decrease. AWESOME : CHALLENGE 2 5 MARKS HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium 42

AWESOME : CHALLENGE 3 Consider the following equilibrium: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) ∆H < 0 Differentiate the concentration of the mixture when the temperature is decreased in the equilibrium system? LE CHATELIER’S PRINCIPLE 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) ∆H < 0 When the Temperature decreases, the system will increase the Temperature by releasing heat. The system favors exothermic reactions. The equilibrium position shifts to the right to reattain equilibrium. The concentration of SO₃ will increase. The concentration of SO₂ and O₂ will decrease. AWESOME : CHALLENGE 3 5 MARKS HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium 43

Consider the following equilibrium: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) ∆H < 0 Point out the equilibrium direction when adding Xenon gas at constant pressure. AWESOME : CHALLENGE 4 LE CHATELIER’S PRINCIPLE Answer: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) ∆H < 0 Adding inert gas (Xenon) at constant pressure, partial pressure of SO₂, O₂, and SO₃ will decrease. System will increase the partial pressure by shift the equilibrium position to the direction with more number of moles. So equilibrium position shift to the left. SO₂ and O₂ will increase. SO₃ will decrease. AWESOME : CHALLENGE 4 5 MARKS HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium 44

How does increasing in pressure affect the equilibrium position of the following reaction? C(graphite) + 2H₂(g) ⇌ CH₄(g) ∆H < 0 AWESOME : CHALLENGE 5 LE CHATELIER’S PRINCIPLE C(graphite) + 2H₂(g) ⇌ CH₄(g) ∆H < 0 When the pressure increases, the system will reduce the disturbance by reducing the pressure of the system. Increase in pressure will change the equilibrium position to the side with fewer number of moles to reduce the pressure. The equilibrium position will shift to the right until a new equilibrium is established. More CH₄ will be produced. Amount of H₂ will decrease. AWESOME : CHALLENGE 5 5 MARKS HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium 45

When Fatimah Gently shakes the carbonated drinks in a bottle at room temperature, she observes bubbles formed in the bottle. Then, she opens the bottle cap slightly while adding 1 teaspoon of vinegar into the bottle. Surprisingly, she observes that there are more bubbles in the bottles after the addition of the vinegar. What actually happens in the bottle that results in more bubbles forming? Help Fatimah to investigate this phenomenon using Le chatelier’s principle and the equation given. (i) CO₂(g) ⇌ CO₂ (aq) + Heat (ii) CO₂(aq) + H₂O(l) ⇌ H₂CO₃(aq) (iii) H₂CO₃(aq) ⇌ H⁺ (aq) + HCO₃ ⁻ (aq) CLEVER : CHALLENGE 1 LE CHATELIER’S PRINCIPLE Clue: the bubble gas is carbon dioxide gas. When vinegar is added, the concentration of H⁺ increase making the concentration of the H₂CO₃ increase. By referring to eq (ii), Increasing H₂CO₃ will make the equilibrium position shift to the left and the concentration of CO₂ (aq) in water increase. Increasing the CO₂ (aq) change the equilibrium position in equation (i) to the left and lead to the production of more CO₂ gas as bubbling in the bottle. (i) CO₂(g) ⇌ CO₂ (aq) + Heat (ii) CO₂(aq) + H₂O(l) ⇌ H₂CO₃(aq) (iii) H₂CO₃(aq) ⇌ H⁺ (aq) + HCO₃⁻ (aq) CLEVER : CHALLENGE 1 5 MARKS HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium 46

Ali has bottles of carbonated drink which are left at room temperature. He then put some bottles into a container with ice cubes and other bottles into a warm container. He was amazed at the difference in the amount of bubbles in the bottles accordingly. The hot container bottles contain more bubbles than the cold ones. By applying Le chatelier’s principle with the help of the equation given, justify the phenomena in Ali’s situation. CO₂(g) ⇌ CO₂(aq) + Heat CLEVER : CHALLENGE 2 LE CHATELIER’S PRINCIPLE Hot container: More bubbles in the bottles Cold container:fewer bubbles in the bottles CO₂(g) ⇌ CO₂ (aq) + Heat When in the hot water, the Temperature increases. The system will decrease the temperature by absorbing the added heat. The system favors endothermic reactions. So equilibrium position shifts to the left resulting in more bubbles forms in the bottle that is put in the hot container. While in the ice cubes, the Temperature decrease. The system will increase the Temperature by releasing heat. The system favors exothermic reaction. The equilibrium position shift to the right resulting less bubble forms in the bottle CLEVER : CHALLENGE 2 5 MARKS HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium 47

In carbonated beverages (soft drinks), carbon dioxide is dissolved in water, forming carbonic acid. CO₂(g) ⇌ CO₂ (aq) + Heat CO₂(aq) + H₂O(aq) ⇌ H₂CO₃(aq) In a birthday party, Umar gently shake the carbonated drink in a bottle before he poured the drink into a clear glass. When he poured the drink into the glass, he was amazed to see there was more bubble gas evolved in the drink. He thought of Le Chatelier’s principle regarding this situation. Can you justify the phenomenon? Use the equations given. Bubble Carbonic acid CLEVER : CHALLENGE 3 LE CHATELIER’S PRINCIPLE Clue 1:Pressure Clue 2: When the cap of bottle is removed, the volume increase CO₂(g) ⇌ CO₂ (aq) + Heat CO₂(aq) + H₂O(aq) ⇌ H₂CO₃(aq) When he shaked the closed bottle, the pressure is at high state. The system will reduce the pressure by shift the equilibrium position to the right. More H₂CO₃ formed in the bottle. But once the cap of the bottle is removed, the pressure is lower. The system will increase the pressure by shift to the side to produce more moles of substances. Thus, resulting the equilibrium position shift to the left. more CO₂ (aq) formed and shift the equilibrium position in equation 1 to the left and produce more CO₂(g) as bubbles. CLEVER : CHALLENGE 3 5 MARKS HOTS-CHEM_EQUI HOTS Challenge in Chemical Equilibrium 48