Chemistry Course Specification - Grade 12 PDF

Course Specification Course Name: Chemistry Course Code: CHM71 Grade: 12 Aim: The Grade 12 Chemistry course provides students with a college-level foundation to support future advanced coursework in Chemistry. The cours...

Course Specification Course Name: Chemistry Course Code: CHM71 Grade: 12 Aim: The Grade 12 Chemistry course provides students with a college-level foundation to support future advanced coursework in Chemistry. The course will give students the opportunity to cultivate their understanding of Chemistry through inquiry-based investigations as they explore topics as Matter, Energy & Equilibrium, Oxidation & Reduction Reactions and Organic & Nuclear Chemistry. Course Outline: Unit Number of Periods Term 1 Unit 1: Matter, Energy & Equilibrium 45 1.1 – Energy and Chemical Change 12 1.2 – Reaction Rates 12 1.3 – Chemical Equilibrium 9 Term 2 1.4 – Acids and Bases 12 Unit 2: Oxidation & Reduction Reactions 18 2.1 – Redox Reactions 6 2.2 – Electrochemistry 12 Term 3 Unit 3: Organic & Nuclear Chemistry 24 3.1 – Hydrocarbons 12 3.2 – Substituted Hydrocarbons & Their Reactions 12 G12 Advanced Chemistry – CHM71 – Course Specification – Term 1 (Detailed KPIs) Page 1 of 52  Unit 1: Matter, Energy & Equilibrium 1.1 – Energy and Chemical Change Inspire Chemistry – Module 14 – Lesson 1: Energy CHM.5.5.01.001.01. Define Energy, Potential Energy and Kinetic Energy Energy is the ability to do work or produce heat Potential Energy is the energy due to the composition or position of an object Kinetic Energy is the energy of motion CHM.5.5.01.001.02. State the law of conservation of energy It states that in any chemical reaction or physical process, energy can be converted from one form to another, but it is neither created nor destroyed. CHM.5.5.01.001.03. Define chemical potential energy Chemical Potential Energy is the energy that is stored in a substance because of its composition. CHM.5.5.01.001.04. Describe how chemical energy is related to the heat lost or gained in chemical reactions (Exothermic and Endothermic reactions) In Exothermic Reactions: Chemical potential energy changes to heat, and the heat is released. In Endothermic Reactions: Heat is absorbed and changed to chemical potential energy. CHM.5.5.01.001.05. Compare and contrast temperature and heat Temperature is a measure of the average kinetic energy of the particles in a sample of matter Heat (q) is energy that is in the process of flowing from a warmer object to a cooler object. CHM.5.5.01.001.06. List the units of temperature and heat Units of Units of Heat Temperature Kelvin (K) Joule (J) Celsius (℃) calorie (cal) Calorie (Cal) calorie is the amount of heat required to raise the temperature of one gram of pure water one degree Celsius (℃). Joule is the SI unit for heat and energy G12 Advanced Chemistry – CHM71 – Course Specification – Term 1 (Detailed KPIs) Page 2 of 52  CHM.5.5.01.001.07. Perform interconversion between units of temperature and heat Interconversion between units of Interconversion between units of Temperature Heat 1J 0.2390 cal , 0.2390 cal 1J 𝐊 = ℃ + 𝟐𝟕𝟑 1 cal 4.184 J , ℃ = 𝐊 − 𝟐𝟕𝟑 4.184 J 1 cal 1 Cal 1000 cal , 1000 cal 1 Cal Examples: Interconversions between units of Temperature 1. Convert 39℃ to K K = ℃ + 273 = 39 + 273 = 312 K 2. Convert 962 K to ℃ ℃ = K − 273 = 962 − 273 = 689 ℃ 3. Convert 152 K to ℃ ℃ = K − 273 = 152 − 273 = − 121 ℃ Interconversions between units of Heat 1. Convert 457 Cal to J 1000 cal 4184 J 457 Cal × × = 1.91 × 10! J 1 Cal 1000 cal 2. Convert 326 kJ to cal 1000 J 1 cal 326 kJ × × = 7.79 × 10" cal 1 kJ 4.184 J 3. Convert 432 cal to kJ ".$%" ' $ +' 432cal × × = 1.81 kJ $ ()* $,,, ' CHM.5.5.01.001.08. Describe the relationship between the specific heat capacity of a substance and the resistance to change in temperature Specific heat capacity is the amount of heat required to raise the temperature of one gram of a given substance by one degree Celsius. - Unit: J/g. ℃ - It is a physical property that can be used to help identify the substance. - Since different substances have different compositions, each substance has its own specific heat capacity. - Water has the highest specific heat capacity. - The higher the heat capacity of a substance the more it resists the change in temperature when heated. G12 Advanced Chemistry – CHM71 – Course Specification – Term 1 (Detailed KPIs) Page 3 of 52  Specific Heat Capacities (J/g. ℃)of Some Common Substances (a 298 K or 25℃) Substance Specific heat Capacity Substance Specific heat Capacity Water(l) 4.184 Granite(s) (Solid) 0.803 Water(s) (Ice) 2.03 Calcium(s) 0.647 Water(aq) 2.01 Iron(s) 0.449 Ethanol(l) 2.44 Strontium(s) 0.301 Beryllium(s) 1.825 Silver(s) 0.235 Magnesium(s) 1.023 Barium(s) 0.204 Aluminum(s) 0.897 Lead(s) 0.129 Concrete(s) 0.84 Gold(s) 0.129 CHM.5.5.01.001.09. Describe how the same amount of heat affects the temperature of different objects of the same mass The same amount of heat affects the temperature of objects of the same mass with a high specific heat capacity much less than the temperature of those with low specific heat capacity. The substance that has the lower specific heat capacity will have a larger change in temperature because it requires less heat to raise the temperature of one gram of that substance by one degree Celsius. Example: Equal masses of two substances (X and Y) absorb the same amount of heat. The temperature of substance X increases as twice as much as the temperature of substance Y. Which substance (X or Y) has a higher specific heat capacity? Justify your answer. Substance Y has a higher specific heat capacity For equal masses, the substance with a greater specific heat capacity undergoes a smaller temperature change. CHM.5.5.01.002.01 Calculate the amount of heat absorbed by a substance as its temperature changes Using the formula: 𝐪 = 𝐦 × 𝐜 × ∆𝐓 q = Heat absorbed or released in Joules (J) m = mass of the substance in grams (g) c = specific heat capacity of the substance (J/g. ℃) ∆T = Change in temperature in ℃ = T final – T initial Example: How much heat is absorbed by a 34.0 g of water when heated from 30.0℃ to 60.0℃? q = m × c × ∆T = 34.0 × 4.184 × (60.0 − 30.0) = 4.27 × 10- J G12 Advanced Chemistry – CHM71 – Course Specification – Term 1 (Detailed KPIs) Page 4 of 52  CHM.5.5.01.002.02 Calculate the amount of heat released by a substance as its temperature changes Example: How much heat is released by a 50.0 g of water when cooled from 74.0℃ to 40.0℃? q = m × c × ∆T = 50.0 × 4.184 × 34.0 = 7.11 × 10- J CHM.5.5.01.002.03. Calculate the specific heat capacity of a sample given its mass and temperature change Example: A 1.55 g sample of stainless steel absorbs 141 J of heat when its temperature increases by 178℃. Calculate the specific heat capacity of stainless steel. q 141 c= = = 0.511 J/g. ℃ m × ∆T 1.55 × 178 Example: When a 3.4 g sample of olive oil absorbs 435 J of heat, its temperature increases from 21℃ to 85℃. Calculate the specific heat capacity of olive oil. q 435 c= = = 2.0 J/g. ℃ m × ∆T 3.4 × (85 − 21) Example: When a sample of water absorbs 5650 J of heat its temperature increases from 20.0℃ to 46.6℃. Calculate the mass of sample of water (Specific heat capacity of water is 4.184 J/g. ℃). q 5650 m= = = 50.8 g c × ∆T 4.184 × (46.6 − 20.0) Inspire Chemistry – Module 14 – Lesson 2: Heat CHM.5.5.01.002.04. Define calorimetry Calorimetry is the measurement of heat flow into or out of a system for chemical and physical processes. CHM.5.5.01.002.05. Define calorimeter A calorimeter is an insulated device used for measuring the amount of heat absorbed or released, (i.e. enthalpy change of a reaction), during a chemical or physical process. G12 Advanced Chemistry – CHM71 – Course Specification – Term 1 (Detailed KPIs) Page 5 of 52  CHM.5.5.01.002.06. Describe the two types of calorimeter (Bomb calorimeter & Coffee-cup calorimeter) Coffee-cup Calorimeter Bomb Calorimeter Constant-Pressure Calorimeter Constant-Volume Calorimeter - A coffee cup calorimeter is used for measuring - A bomb calorimeter is used to measure heat heat flow in a chemical solution. flow for solids with low to high temperature It cannot be used for reactions. It is used by food chemists to a) reactions which involve gases since the determine the caloric content of different food. gases would escape from the cup. - A sample is added in a steel inner chamber b) high temperature reactions, since these called the bomb, which is filled with oxygen at would melt the cup. a high pressure. - A coffee cup calorimeter is essentially a - Surrounding the bomb is a measured mass of polystyrene (Styrofoam) cup or two with a lid. water stirred by a stirrer to ensure uniform - The cup is partially filled with a known temperature. volume of water and a sensitive thermometer - The reaction is initiated by a spark, and the is inserted through the lid of the cup so that its temperature is recorded until it reaches its bulb is below the water surface. maximum. - When a chemical reaction occurs in the coffee - The heat that is released warms the water cup calorimeter, the water absorbs the heat of surrounding the chamber. the reaction. By measuring the temperature increase of the water, it is possible to calculate the quantity of heat released during the combustion reaction. qsystem = ΔH= – qsurrounding = – m´C´ΔT q metal = – q water - For an endothermic reaction, sign of ΔH is positive - For an exothermic reaction, sign of ΔH is negative - Sources of Error: 1) The foam cup will absorb heat 2) Some heat will be lost to the air 3) If the reactants are not completely mixed, temperature measurements will not be accurate G12 Advanced Chemistry – CHM71 – Course Specification – Term 1 (Detailed KPIs) Page 6 of 52  CHM.5.5.01.002.07. Perform different calculations in a calorimetry experiment (For Mixing aqueous solutions) Example: A student mixes 50. mL of 1.0 M HCl and 50. mL of 1.0 M NaOH in a calorimeter and notices that the temperature rises from 21.0℃ to 27.5℃. Calculate the enthalpy change, in kJ, if the total volume is 100 mL, the density is 1.0 g/mL, and the specific heat capacity of water is 4.184 J/g.℃. Mass of water = Density × Volume = 1.00 × 100 = 100 g ΔT = Tfinal – Tinitial = 27.5 – 21.0 = 6.5℃ ΔH = – qsurr = – (m ´ C ´ ΔT) = – (100 ´ 4.184 ´ 6.5) = – 2720 J = – 2.7 kJ CHM.5.5.01.002.08. Perform different calculations in a calorimetry experiment (For Adding a metal to water) Example: A 25 g block of an unknown metal X was heated to 95℃ then immersed in 13 mL of water (density = 1.0 g/mL) at 25℃. When the two substances reach equilibrium, the final temperature was 32℃. Calculate the specific heat capacity of metal X. (Density of water = 1.0 g/mL and specific heat capacity of water is 4.184 J/g.℃) q metal = – q water mmetal × Cmetal × ∆Tmetal = – (mwater × Cwater × ∆Twater) mass of water = density × volume = 1.0 × 13 = 13 g 25 × Cmetal × (32 – 95) = – (13 × 4.184 × (32 – 25)) Cmetal = 0.24 J/g.℃ Example: A 100. g block of gold heated to 78℃ is placed in a calorimeter containing 40.0 g of water at initial temperature of 21℃. If the specific heat capacity of gold is 0.129 J/g.℃, calculate the final temperature of the mixture. (specific heat capacity of water is 4.184 J/g.℃) q metal = – q water mmetal × Cmetal × ∆Tmetal=– (mwater × Cwater × ∆Twater) 100. × 0.129 × (Tfinal – 78) = − (40.0 × 4.184 × (Tfinal – 21)) Tfinal = 25.1℃ Example: A certain mass of a sample of lead is heated and placed in a coffee cup calorimeter containing 40.0 mL (density = 1.0 g/mL) of water at 18.0℃. The water reaches a temperature of 21.0℃. Calculate the amount of heat, in joules, released by the lead sample. (Density of water = 1.0 g/mL and specific heat capacity of water is 4.184 J/g.℃) q metal = – q water q metal =– (mwater × Cwater × ∆Twater)= − (40.0 × 4.184 × 3.0) = − 502 J 502 J are released G12 Advanced Chemistry – CHM71 – Course Specification – Term 1 (Detailed KPIs) Page 7 of 52  CHM.5.5.01.004.01 Define thermochemistry Thermochemistry is the study of heat changes that accompany chemical reactions and phase changes. CHM.5.5.01.004.02. Differentiate among the system, the surrounding and the universe The system is the specific part of the universe that contains the reaction or process to study. (Simply, the system is what you observe) The surrounding is everything in the universe other than the system. A universe is the system plus surroundings. Universe = System + Surroundings CHM.5.5.01.006.01. Define enthalpy (H) Enthalpy (H) is the heat content of a system at constant pressure. CHM.5.5.01.006.02. Define enthalpy change of reaction (∆Hrxn) while writing the equation used Enthalpy change of reaction (∆Hrxn) is the amount of energy released or absorbed during a chemical reaction. The value of ∆Hrxn can be determined by measuring the heat flow of the reaction at constant pressure. ∆𝑯𝒓𝒙𝒏 = 𝑯𝒇𝒊𝒏𝒂𝒍 − 𝑯𝒊𝒏𝒊𝒕𝒊𝒂𝒍 ∆𝑯𝒓𝒙𝒏 = 𝑯𝒑𝒓𝒐𝒅𝒖𝒄𝒕𝒔 − 𝑯𝒓𝒆𝒂𝒄𝒕𝒂𝒏𝒕𝒔 𝒒 = ∆𝑯𝒓𝒙𝒏 CHM.5.5.01.006.03. Perform and conduct an experiment to determine the enthalpy of a chemical reaction CHM.5.5.01.004.03. Identify direction of heat flow (For Exothermic and endothermic processes) The direction of heat flow is defined from the point of view of the system Endothermic Process Exothermic Process Heat flows into the system from its Heat flows from the system to its surroundings. surroundings. Heat flowing into a system from its Heat flowing out of a system into its surroundings is defined as positive. surroundings is defined as negative. G12 Advanced Chemistry – CHM71 – Course Specification – Term 1 (Detailed KPIs) Page 8 of 52  Example: a) Which diagram represents an exothermic process? Diagram (a) b) Which diagram represents an endothermic process? Diagram (b) c) Is heat flow positive or negative in diagram (a)? Negative d) Is heat flow positive or negative in diagram (b)? Positive e) What does a negative value for heat represent? Heat flowing out of a system into its surroundings f) What does a positive value for heat represent? Heat flowing into a system from its surroundings G12 Advanced Chemistry – CHM71 – Course Specification – Term 1 (Detailed KPIs) Page 9 of 52  CHM.5.5.01.004.04. Compare and contrast potential energy diagrams of exothermic and endothermic reactions in terms of general shape, enthalpy of reactants and products, activation energy of forward and backward reactions, and enthalpy of reaction and its sign Exothermic Reaction Endothermic Reaction - Enthalpy of Reactants > Enthalpy of Products - Enthalpy of Reactants < Enthalpy of Products - △H is negative - △H is positive - System gives off heat to the surroundings - System absorbs heat from its surroundings - System releases energy - System absorbs energy - Reaction feels warm to the touch - Reaction feels cold to the touch - It occurs when more energy is released during - It occurs when a greater amount of energy is product bond formation than is required to required to break the existing bonds in the break bonds in the reactants reactants than is released when new bonds form in the products CHM.5.5.01.004.05. Perform and conduct an experiment to differentiate between exothermic and endothermic reactions Inspire Chemistry – Module 14 – Lesson 3: Thermochemical equations CHM.5.5.01.006.04. Define thermochemical equation Thermochemical equation is a balanced chemical equation that includes the physical states of all reactants and products and the energy change, usually expressed as the change in the enthalpy, ∆𝐻. Thermochemical equation informs about the amount of heat released or absorbed in a chemical change at constant pressure. G12 Advanced Chemistry – CHM71 – Course Specification – Term 1 (Detailed KPIs) Page 10 of 52  CHM.5.5.01.006.05. Write thermochemical equation for exothermic and endothermic reactions Thermochemical equation for Exothermic Reactions A negative value for ∆𝐻 indicates that energy is released. 4 Fe(s) + 3 O2 (g) → 2 Fe2O3(s) + 1625 kJ or 4 Fe(s) + 3 O2 (g) → 2 Fe2O3(s) ∆𝐻= − 1625 kJ Thermochemical equation for Endothermic reactions A positive value for ∆𝐻 indicates that energy is absorbed. C(s) + 2 S(s) + 89.3 kJ → CS2 (l) or C(s) + 2 S(s) → CS2 (l) ∆𝐻 = + 89.3 kJ CHM.5.5.01.006.06. Calculate the enthalpy change for different thermochemical reactions Example: Consider the following thermochemical equation: 2NaHCO3(s) + 85 kJ → Na2CO3(s) + H2O(l) + CO2(g) Calculate the amount of heat (in kJ) required to decompose 4.48 mole NaHCO3(s). 2 NaHCO3 ∆𝐻 2 moles 85 kJ 4.48 moles ?? kJ Amount of heat = 190. kJ Example: Consider the following thermochemical equation: C(s) + 2 S(s) → CS2 (l) ∆𝐻 = 89.3 kJ Calculate the amount of heat (in kJ) absorbed when 8.80 g of CS2 are formed. CS2 ∆𝐻 1 mole 89.3 kJ 1 × 76 g 89.3 kJ 8.80 ?? kJ Amount of heat = 10.3 kJ CHM.5.5.01.006.07. Define enthalpy (heat) of combustion, ∆Hcomb, while determining on what basis it is defined Enthalpy (heat) of combustion (∆Hcomb) is the enthalpy change for the complete burning of one mole of the substance. Enthalpy of combustion is defined per one mole of reactant. Example: CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (l) + 890 kJ CHM.5.5.01.006.08. Write a thermochemical equation for the combustion of a compound, while including the sign of ΔH and identifying whether reaction is exothermic or endothermic Example: When 2 moles of solid calcium, Ca, combines with 1 mole of oxygen gas, O2, 2 moles of solid calcium oxide, CaO, is formed and 1200 kJ of heat is released. a) Write the thermochemical equation for the above combustion reaction. 2 Ca(s) + O2 (g) → 2 CaO(s) + 1200 kJ or 2 Ca(s) + O2 (g) → 2 CaO(s) ∆𝐻= − 1200 kJ b) Is the above reaction exothermic or endothermic? Justify your answer. Exothermic. ∆𝐻 has a negative value. G12 Advanced Chemistry – CHM71 – Course Specification – Term 1 (Detailed KPIs) Page 11 of 52  CHM.5.5.01.006.09. Perform calculations using enthalpy of combustion Example: The data below shows the thermochemical reaction of the combustion of ethene gas, C2H4. C2H4 (g) + 3 O2 (g) ® 2 CO2 (g) + 2 H2O (l) DH° = - 1411 kJ Deduce the value of enthalpies, in kJ, for each of the following thermochemical equations. a) 2CO2(g) + 2H2O(l) ® C2H4(g) +3O2(g) ________ ∆𝐻 = +1411 kJ $ - b) CO2(g) + H2O(l) ® = C2H4(g) + = O2(g) ________ ∆𝐻 = +705.5 kJ $ - c) C2H4(g) + = O2(g) ® CO2(g) + H2O(l) ________ ∆𝐻 = -705.5 kJ = Example: Consider the following thermochemical equation: C2H6O(l) + 3 O2(g) → 2CO2 (g) + 3H2O(l) ∆𝐻 = − 1368 kJ Calculate the amount of heat (in kJ) released when 11.5 g of C2H6O is burned. C2 H6 O ∆𝐻 1 mole 1368 kJ 1 × 46 g 1368 kJ 11.5g ?? kJ Amount of heat = 342 kJ CHM.5.5.01.003.01. Define molar enthalpy (heat) of vaporization, ∆Hvap Molar enthalpy (heat) of vaporization is the amount of heat required to vaporize one mole of a liquid at a constant temperature. It is an endothermic process. CHM.5.5.01.003.02. Define molar enthalpy (heat) of fusion, ∆Hfus Molar enthalpy (heat) of fusion is the amount of heat absorbed by one mole of solid substance as it melts to a liquid at constant temperature. It is an endothermic process. CHM.5.5.01.003.03. Define molar enthalpy (heat) of condensation, ∆Hcond Molar enthalpy (heat) of condensation is the amount of heat lost when one mole of a vapor condenses at a constant temperature. It is an exothermic process. G12 Advanced Chemistry – CHM71 – Course Specification – Term 1 (Detailed KPIs) Page 12 of 52  CHM.5.5.01.003.04. Define molar enthalpy (heat) of solidification, ∆Hsolid Molar enthalpy (heat) of solidification is the amount of heat lost when one mole of a liquid solidifies at a constant temperature. It is an exothermic process. CHM.5.5.01.006.10. Write thermochemical equation for the changes of state (vaporization, fusion, condensation and solidification) Vaporization H2O(l) → H2O(g) ΔHvap = 40.7 kJ/mol Endothermic Fusion H2O(s) → H2O(l) ΔHfus = 6.01 kJ/mol Endothermic Condensation H2O(g) → H2O(l) ΔHcond = − 40.7 kJ/mol Exothermic Solidification H2O(l) → H2O(s) ΔHsolid = − 6.01 kJ/mol Exothermic ∆𝑯𝒗𝒂𝒑 = − ∆𝑯𝒄𝒐𝒏𝒅 ∆𝑯𝒇𝒖𝒔 = − ∆𝑯𝒔𝒐𝒍𝒊𝒅 The amount of energy absorbed during The amount of energy absorbed during vaporization is numerically equal to the fusion is numerically equal to the amount of energy released during amount of energy released during condensation. solidification. CHM.5.5.01.003.05. Perform phase-change calculations using enthalpy of fusion, vaporization, condensation and solidification Example: Calculate the enthalpy change, in kJ, when 45.2 g of water vapor at 100℃ condenses to liquid water at the same temperature. (ΔHcond = - 40.7 kJ/mol) H2O(l) ΔHcond 1 mole 40.7 kJ 1 × 18 g 40.7 kJ 45.2 g ???/ kJ Amount of heat released = 102 kJ Enthalpy change = − 102 kJ G12 Advanced Chemistry – CHM71 – Course Specification – Term 1 (Detailed KPIs) Page 13 of 52  Example: Calculate the amount of heat absorbed when 360. g of liquid water vaporizes to water vapor at 100℃. (ΔHvap = 40.7 kJ/mol) H2O(l) ΔHvap 1 mole 40.7 kJ 1 × 18 g 40.7 kJ 360. g ??? kJ Amount of heat absorbed = 814 kJ CHM.5.5.01.003.06. Perform and conduct an experiment to determine the heat of fusion of ice CHM.5.5.02.004.01. Interpret the energy diagram of the phase changes for water (using heating and cooling curves of water) Heating Curve of Water Cooling Curve of Water - Heating curve is a plot of temperature versus - Cooling curve is a plot of temperature versus time for a process where energy is added at a time for a process where energy is released at constant pressure. a constant pressure. - It graphically describes the enthalpy changes - It graphically describes the enthalpy changes that take place during the phase changes. that take place during the phase changes. - The plateau at the boiling point is longer than the plateau at the melting point because it takes almost 7 times more energy (thus 7 times the heating time) to vaporize liquid water than to melt ice. - The slopes of other lines are different because the different states of water have different molar heat capacities. G12 Advanced Chemistry – CHM71 – Course Specification – Term 1 (Detailed KPIs) Page 14 of 52  Inspire Chemistry – Module 14 – Lesson 4: Calculating enthalpy change CHM.5.5.02.001.01. State Hess's Law Hess’s Law states that if you can add two or more thermochemical equations to produce a final equation for a reaction, then the sum of the enthalpy changes for the individual reactions is the enthalpy change for the final reaction. CHM.5.5.02.001.02. Describe the rules to be followed to apply Hess's Law Rules are to be followed: 1) Arrange equations so that reactants and products are on the correct sides of the arrows 2) If an equation is reversed, the sign of ∆H is also reversed 3) If an equation is multiplied to obtain a correct coefficient, the ∆H is also multiplied by the same number 4) When adding the equation, the individual ∆H are also added CHM.5.5.02.002.01. Calculate, using Hess's law, the ∆H of a reaction Example: Using the thermochemical reactions and their respective enthalpy values given below, calculate the enthalpy of the reaction below. 2 CH4 (g) + 2 NH3 (g) + 3 O2 (g) ® 2 HCN (g) + 6 H2O (l) Thermochemical equation Value of enthalpy in kJ Reaction 1: N2 (g) + 3 H2 (g) ® 2 NH3 (g) - 91.8 Reaction 2: C (s) + 2 H2 (g) ® CH4 (g) - 74.9 Reaction 3: 2 C (s) + H2 (g) + N2 (g) ® 2 HCN (g) + 270.3 Reaction 4: 2 H2 (g) + O2 (g) ® 2 H2O (l) - 579 Reaction 1 Reverse 2 NH3 (g) ® N2 (g) + 3 H2 (g) + 91.8 kJ Reaction 2 Reverse and multiply by 2 2 CH4 (g) ® 2C (s) + 4 H2 (g) + 149.8 kJ Reaction 3 Same 2 C (s) + H2 (g) + N2 (g) ® 2 HCN (g) + 270.3 kJ Reaction 4 Same and multiply by 3 6 H2 (g) + 3 O2 (g) ® 6 H2O (l) - 1737 kJ 2 CH4 (g) + 2 NH3 (g) + 3 O2 (g) ® 2 HCN (g) + 6 H2O (l) − 1225.1 kJ Enthalpy of the required reaction = − 1225.1 kJ G12 Advanced Chemistry – CHM71 – Course Specification – Term 1 (Detailed KPIs) Page 15 of 52  CHM.5.5.02.006.01. Define standard enthalpy (heat) of formation of a compound, ∆𝐻?@ , while determining on what basis it is defined Standard enthalpy (heat) of formation of a compound (∆𝐻AB )is the enthalpy change for the formation of one mole of the compound in its standard state from its constituent elements in their standard states at 25oC and 1 atm. Enthalpy of formation is defined in terms of one mole of product. Example: Pb (s) + 2 Cl2 (g) → PbCl4(l) ∆𝐻 = −329.2 kJ CHM.5.5.02.006.02. Identify the standard enthalpy (heat) of formation of a free element in its standard state The standard enthalpy for formation of a free element in its standard state, ∆𝐻?@ , is zero Example: ∆𝐻?@ for H2 (g), N2 (g), O2 (g), F2 (g), Cl2 (g), Br2 (l), I2 (s) Fe(s), and Na(s) is 0.0 kJ The standard state of an element is the normal physical state at 1 atm and 298 K. Example: Element Standard State Iron, Fe Solid Mercury, Hg Liquid Oxygen, O2 Gas Iodine, I2 Solid Bromine, Br2 Liquid Compound Formation Equation ∆𝑯𝒐𝒇 (kJ/mol) H2S (g) H2(g) + S(s) → H2S(g) −21 $ $ HF(g) H2(g) + = F2(g) → HF(g) −273 = - SO3(g) S(s) + = O2(g) → SO3(g) −396 SF6(g) S(s) + 3F2(g) → SF6(g) −1220 G12 Advanced Chemistry – CHM71 – Course Specification – Term 1 (Detailed KPIs) Page 16 of 52  CHM.5.5.02.006.03. Calculate, using the standard enthalpies of formation, the enthalpy of a reaction, ∆𝐻CDE ∆𝑯𝐨 = 𝚺 𝚫𝑯𝐨𝐟 (𝐏𝐫𝐨𝐝𝐮𝐜𝐭𝐬) − 𝚺 𝚫𝑯𝐨𝐟 (𝐑𝐞𝐚𝐜𝐭𝐚𝐧𝐭𝐬) Example: Use the following data that shows the standard enthalpies of formation of different substances to answer questions a and b. Substance Standard enthalpy of formation, ∆𝑯𝐨𝐟 (kJ/mol) NO2(g) +33.9 HNO3(aq) −173.2 NO(g) +90.2 H2O(l) −285.8 a) Calculate the enthalpy of the reaction below: 2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l) ∆𝐻B = Σ ∆𝐻AB (Products) − Σ ∆𝐻AB (Reactants) ∆𝐻AB (Products) = (3 × 𝐻A NO= ) + (𝐻A H= O) = (3 × 33.9) + (−285.8) = −184.1 kJ ∆𝐻AB (Reactants) = (2 × HA HNO- ) + (𝐻A NO) = (2 × −173.2) + (+90.2) = −256.2 kJ ∆𝐻B = +72.1 kJ b) Is the reaction above endothermic or exothermic? Explain your answer. Endothermic reaction. ∆𝐻 of the reaction is positive or greater than zero. CHM.5.5.02.006.04. Calculate enthalpy of reaction, ∆Hrxn, using standard enthalpies of formation of products and reactants then use sign of ∆H for determining if the reaction is exothermic or endothermic ∆𝑯𝐨 = 𝚺 𝚫𝑯𝐨𝐟 (𝐏𝐫𝐨𝐝𝐮𝐜𝐭𝐬) − 𝚺 𝚫𝑯𝐨𝐟 (𝐑𝐞𝐚𝐜𝐭𝐚𝐧𝐭𝐬) Example: P4O6 undergoes combustion according to the following balanced equation: P" OH(J) + 2O=(L) → P" O$,(J) a) Use the following equations to calculate the enthalpy change, ∆𝐻, for the combustion of P" OH(J). P"(J) + 5O=(L) → P" O$,(J) ∆𝐻 = −2940 kJ P"(J) + 3O=(L) → P" OH(J) ∆𝐻 = −1640 kJ Equation 1: Keep the same P"(J) + 5O=(L) → P" O$,(J) ∆𝐻 = −2940 kJ Equation 2: Reverse P" OH(J) → P"(J) + 3O=(L) ∆𝐻 = +1640 kJ Overall Equation: P" OH(J) + 2O=(L) → P" O$,(J) ∆𝐻CDE = −1300 kJ b) Based on your answer to part (a), is the combustion of P" OH(J) endothermic or exothermic? Justify your answer. Exothermic reaction. ∆𝐻 is negative or less than zero. G12 Advanced Chemistry – CHM71 – Course Specification – Term 1 (Detailed KPIs) Page 17 of 52  Inspire Chemistry – Module 14 – Lesson 5: Reaction spontaneity CHM.5.5.02.007.01. Define a spontaneous process A spontaneous process is a physical or chemical change that occurs without outside intervention and may require energy to be supplied to begin the process. CHM.5.5.02.008.01. Define entropy, S, while identifying its unit Entropy is a measure of the number of possible ways that the energy of a system can be distributed; related to the freedom of the system's particles to move and the number of ways they can be arranged. Entropy is a measure of the degree of disorder or randomness of a system. Unit J/K CHM.5.5.02.008.02. List the two driving factors for all chemical reactions Enthalpy and Entropy CHM.5.5.02.008.03. Recall the first, second and third laws of thermodynamics First law of thermodynamics Energy can neither be created nor destroyed Second law of thermodynamics It states that the disorder of the universe or entropy, ∆S, is constantly increasing Third law of thermodynamics The entropy of a perfect crystal is zero at 0K CHM.5.5.02.008.04. Predict the change in the entropy of a system, ∆𝑆MNMOPQ , (According to a set of rules) 1) The entropy of a substance always increases as it changes state from solid to liquid to gas. Ssolid > Sliquid > Sgas In solids, molecules have limited movement. In liquids, molecules have some freedom to move. In gases, molecules can move freely within their container. Example: H2O(l) → H2O(g) ∆𝑆JRJSTU > 0 (Positive) H2O(s) → H2O(g) ∆𝑆JRJSTU < 0 (Negative) 2) The dissolving of a gas in a solvent always results in a decrease in entropy Sgas > Saqueous Example: O2 (g) → O2 (aq) ∆𝑆JRJSTU < 0 (Negative) G12 Advanced Chemistry – CHM71 – Course Specification – Term 1 (Detailed KPIs) Page 18 of 52  3) Assuming no change in physical state occurs: The entropy of a system usually increases when the number of gaseous product particles is greater than the number of gaseous reactant particles Example: 2 NH3 (g) → N2 (g) + 3 H2 (g) ∆𝑆JRJSTU > 0 (Positive) The ∆𝑆JRJSTU s positive because two molecules of gas react and three molecules of gas are produced. 4) With some exceptions, entropy increases when a solid or a liquid dissolves in a solvent Example: NaCl(s) → NaW()V) + ClX ()V) ∆𝑆JRJSTU > 0 (Positive) 5) The random motion of the particles of a substance increases as its temperature increases. The kinetic energy of molecules increases with temperature. Increased kinetic energy means faster movement and more possible arrangements of particles. Hence ∆𝑆JRJSTU is Positive. CHM.5.5.02.008.05. Identify, from a list of given reactions, the reaction that shows an increase or decrease in entropy Example: For each of the following reactions, identify whether it shows an increase or decrease in entropy. Justify your answer. a) 2H2 (g) + O2 (g) ® 2H2O (l) Decrease in entropy. There are three moles of gaseous reactants and 2 moles of liquid products. Hence there is a decrease in disorder. b) 2NH4NO3 (s) ® 2N2 (g) + 4H2O (l) + O2 (g) Increase in entropy. There are two moles of solid reactants. In products, there are 3 moles of gaseous species and 4 moles of liquid species, hence there is an increase in disorder. Example: Predict the sign of ∆𝑆JRJSTU for each of the following changes. 2 N2O(g) → 2 N2(g) + O2(g) ∆𝑆JRJSTU is positive Mg(s) + Cl2(g) → MgCl2(s) ∆𝑆JRJSTU is negative HCl(g) + NH3(g) ) → NH4Cl(s) ∆𝑆JRJSTU is negative 2O3(g) → 3O2(g) ∆𝑆JRJSTU is positive 2H2S(g) + 3O2(g) → 2H2O(g) + 2SO2(g) ∆𝑆JRJSTU is negative C10H8(l) → C10H8(s) ∆𝑆JRJSTU is negative G12 Advanced Chemistry – CHM71 – Course Specification – Term 1 (Detailed KPIs) Page 19 of 52  CHM.5.5.02.008.06. Calculate entropy change using the table of standard values of entropies ∆ 𝑺𝐨𝒓𝒙𝒏 = 𝚺 𝑺𝐨𝒑𝒓𝒐𝒅𝒖𝒄𝒕𝒔 − 𝚺 𝑺𝐨𝒓𝒆𝒂𝒄𝒕𝒂𝒏𝒕𝒔 Example: B Calculate ∆𝑆YZ[ for the balanced chemical equation: 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) Species So (J/mol.K) NH3 (g) 192.8 O2 (g) 205.2 NO (g) 210.8 H2O (g) 188.8 B B B ∆ 𝑆YZ[ = Σ 𝑆\YB]^(SJ − Σ 𝑆YT)(S)[SJ B = n4 (𝑆_` ) + 6 o𝑆aB! ` pq − n4 o𝑆_a B " p + 5 o𝑆`B! pq = [(4×210.8) + (6×188.8)] – [(4×192.8) + (5×205.2)] = 1976.0 – 1797.2 = 178.8 J/K CHM.5.5.02.008.07. Describe how can entropy be used to determine whether a chemical reaction is more or less likely to be spontaneous - An increase in entropy favors a spontaneous reaction A decrease in entropy favors a non-spontaneous reaction - According to the second law of thermodynamics, the entropy of the universe must increase as a result of a spontaneous reaction or process. - Hence, for a spontaneous process: ∆𝑆^[bcTYJT > 0 ∆𝑺𝐮𝐧𝐢𝐯𝐞𝐫𝐬𝐞 = ∆𝑺𝐬𝐲𝐬𝐭𝐞𝐦 + ∆𝑺𝐬𝐮𝐫𝐫𝐨𝐮𝐧𝐝𝐢𝐧𝐠𝐬 - ∆𝑆^[bcTYJT tends to be positive for reactions and processes under the following conditions: 1) The reaction of process is exothermic, i.e. ∆𝐻JRSTU is negative. The heat released by an exothermic reaction raises the temperature of the surroundings, thus increase the entropy of the surrounding. Hence, ∆𝑆J^YYB^[]b[LJ is positive. 2) The entropy of the system increases, so ∆𝑆JRSTU is positive. - Exothermic chemical reactions accompanied by an increase in entropy are all spontaneous. CHM.5.5.02.009.01. Define free energy while identifying its unit Free energy, G, is the energy available to do work. Unit kJ G12 Advanced Chemistry – CHM71 – Course Specification – Term 1 (Detailed KPIs) Page 20 of 52  CHM.5.5.02.009.02. Describe free energy change and standard free energy change The free energy change, ∆Gsystem, is the difference between the system’s change in enthalpy, ∆Hsystem, and the product of the kelvin temperature and the change in entropy, T∆Ssystem. ∆Gsystem = ∆Hsystem − T∆Ssystem ∆Gsystem: The free energy change Unit: kJ/mol ∆Hsystem: The change in enthalpy Unit: kJ T: Temperature Unit: K (Kelvin) (K = ℃ + 273) ∆Ssystem: The change in entropy Unit: J/K Note: Convert the unit of ∆Ssystem from J/K to kJ/K by dividing by 1000. @ The standard free energy change, ∆𝐺MNMOPQ , is for a reaction or process that occurs under standard condition (298 K and 1 atm) ∆𝑮𝒐𝒔𝒚𝒔𝒕𝒆𝒎 = ∆𝑯𝒐𝒔𝒚𝒔𝒕𝒆𝒎 − 𝑻∆𝑺𝒐𝒔𝒚𝒔𝒕𝒆𝒎 The sign of Free Energy, ∆Gsystem If the sign of ∆Gsystem is negative, the reaction is spontaneous. If the sign of ∆Gsystem is positive, the reaction is non-spontaneous. CHM.5.5.02.009.03. Explain the relationship among the signs of ∆G, ∆H, ∆S that lead to spontaneous or non-spontaneous reaction while identifying the temperature conditions ∆Gsystem = ∆Hsystem − T∆Ssystem ∆Hsystem ∆Ssystem ∆Gsystem Reaction Spontaneity − + − Spontaneous at all temperatures − − −or + Spontaneous at lower temperatures + + −or + Spontaneous at higher temperatures + − + Never spontaneous at any temperature G12 Advanced Chemistry – CHM71 – Course Specification – Term 1 (Detailed KPIs) Page 21 of 52  CHM.5.5.02.009.04. Calculate free energy change, ∆G, when ∆H, ∆S, and temperature (in Kelvin or Celsius) are given while determining whether reaction is spontaneous or nonspontaneous Example: At 100℃, the reaction: 2 N2O (g) + 3 O2 (g) → 4 NO2 (g), ∆H =− 7.9 kJ and ∆S= + 40.0 J/K Calculate the Gibb’s free energy, ∆G, for the reaction. Is the reaction spontaneous? Justify your answer. Hint ∆G = ∆H − T∆S Divide ∆S by 1000 to have its 40.0 unit in kJ/K to be able to = (−7.9) − v(100 + 273) × w xy = −22.8 kJ proceed with the calculations 1000 and have the unit of ∆G in kJ. Reaction is spontaneous because ∆G is negative or < 0 CHM.5.5.02.009.05. Describe the significance of different values of Gibb’s free energy (> 0, < 0, = 0) ∆G Significance > 0 or Positive The reaction is nonspontaneous in the forward direction, but the reverse reaction will be spontaneous < 0 or Negative The reaction is spontaneous in the forward direction =0 The reaction is at equilibrium Note: For elements at standard state (Pure element at 25oC and 1 atm), the standard free energy is equal to zero, i.e. 𝐺AB = 0. CHM.5.5.02.009.06. Calculate standard Gibb’s free energy change using the table of standard values ∆ 𝑮𝐨𝒓𝒙𝒏 = z ∆ 𝑮𝐨𝒑𝒓𝒐𝒅𝒖𝒄𝒕𝒔 − z ∆ 𝑮𝐨𝒓𝒆𝒂𝒄𝒕𝒂𝒏𝒕𝒔 Example: Consider the reaction: CH4(g) + 8 O2(g) → CO2(g) + 2 H2O(g) + 4 O3(g) Use the standard free energies of formation to determine ΔG°rxn for this reaction at 25 °C. Species 𝐺AB (kJ/mol) CH4(g) − 50.5 O2 (g) 0.0 CO2(g) − 394.4 H2O(g) − 228.6 O3(g) 163.2 @ @ @ ∆ 𝐺CDE = 𝛴 ∆ 𝐺rC@stuOM − 𝛴 ∆ 𝐺CPvuOvEOM = no𝐺?@ 𝐶𝑂= p + 2 o𝐺?@ 𝐻= 𝑂p + 4o𝐺?@ 𝑂- pq − no𝐺?@ 𝐶𝐻" p + 8 o𝐺?@ 𝑂= pq = n(−394.4) + o2 × (−228.6)p + o4 × (163.2)pq − [(−50.5) + (8 × 0.0)] = (−198.8) − (−50.5) = −148.3 kJ G12 Advanced Chemistry – CHM71 – Course Specification – Term 1 (Detailed KPIs) Page 22 of 52  11.6– Reaction Rates Inspire Chemistry – Module 15 – Lesson 1: A model for reaction rates CHM.5.4.01.001.01. Define reaction rate while identifying its unit Reaction rate is the change in concentration of reactant and product per unit of time. Units mol/L.s or M/s The above figure represents Reaction Progress During a reaction, reactants are converted into products. The squares represent the reactants. The circles represent the products CHM.5.4.01.001.02. Explain what the reaction rate indicates about a particular chemical reaction The reaction rate indicates the rate of change of the concentration of reactant or product in mol/L.s CHM.5.4.01.020.01. Calculate the average reaction rate using the rate of consumption of reactants or the rate of production of products ∆ 𝐐𝐮𝐚𝐧𝐭𝐢𝐭𝐲 𝐀𝐯𝐞𝐫𝐚𝐠𝐞 𝐑𝐞𝐚𝐜𝐭𝐢𝐨𝐧 𝐑𝐚𝐭𝐞 = ∆𝐭 (∆[𝐑𝐞𝐚𝐜𝐭𝐚𝐧𝐭]) (∆[𝐏𝐫𝐨𝐝𝐮𝐜𝐭]) 𝐀𝐯𝐞𝐫𝐚𝐠𝐞 𝐑𝐞𝐚𝐜𝐭𝐢𝐨𝐧 𝐑𝐚𝐭𝐞 = − = ∆𝐭 ∆𝐭 The average reaction rate for the consumption of a reactant is the negative change in the concentration of the reactant divided by the elapsed time. o[𝐑𝐞𝐚𝐜𝐭𝐚𝐧𝐭]𝒂𝒕 𝒕𝒊𝒎𝒆 𝒕𝟐 − [𝐑𝐞𝐚𝐜𝐭𝐚𝐧𝐭]𝐚𝐭 𝐭𝐢𝐦𝐞 𝐭 𝟏 p 𝐀𝐯𝐞𝐫𝐚𝐠𝐞 𝐑𝐞𝐚𝐜𝐭𝐢𝐨𝐧 𝐑𝐚𝐭𝐞 = − 𝒕𝟐 − 𝒕𝟏 o[𝐏𝐫𝐨𝐝𝐮𝐜𝐭]𝒂𝒕 𝒕𝒊𝒎𝒆 𝒕𝟐 − [𝐏𝐫𝐨𝐝𝐮𝐜𝐭]𝐚𝐭 𝐭𝐢𝐦𝐞 𝐭 𝟏 p 𝐀𝐯𝐞𝐫𝐚𝐠𝐞 𝐑𝐞𝐚𝐜𝐭𝐢𝐨𝐧 𝐑𝐚𝐭𝐞 = 𝒕𝟐 − 𝒕𝟏 G12 Advanced Chemistry – CHM71 – Course Specification – Term 1 (Detailed KPIs) Page 23 of 52  Example: Excess of magnesium reacts with a 0.60 mol/L solution of sulfuric acid, H2SO4, according to the following chemical equation: Mg(s)+H2SO4(aq)→MgSO4(aq)+H2(g) The concentrations of sulfuric acid were recorded during the time of the reactions as shown in the table below. Time (min) 0.0 0.5 1.0 1.5 2.0 2.5 Concentration of H2SO4 (mol/L) 0.60 0.50 0.40 0.20 0.15 0.10 Calculate the rate of disappearance of H2SO4 between 1.0 minute and 2.5 minutes. ∆[a! |`% ] ( ,.$,X,.",) Rate = − =− = 0.20 mol/L. min ∆S (=.~X$.,) Example: Hydrochloric acid solution, HCl, reacts with excess of magnesium, Mg, according to the following chemical equation: Mg(s)+2HCl(aq)⟶MgCl2(aq)+H2(g) During the progress of the reaction, the concentration of MgCl2 ___ while that of HCl ______. increases, decreases CHM.5.4.01.020.02. Compare the concentrations of the reactants and products during the course of a chemical reaction (assuming no additional reactants are added) The concentrations of the reactants decrease, and the concentrations of the products increase at the same rate. CHM.5.4.01.020.03. Compare the average reaction rate measured over an initial, short time interval to one measured over a long time interval As the concentration of reactants decreases, the reaction rate decreases. Thus, the longer the time period, the smaller the average reaction rate will be. CHM.5.4.01.020.04. Perform and conduct an experiment to determine the reaction rate CHM.5.4.01.002.01. Describe collision theory The collision theory states that atoms, ions and molecules must collide in order to react. CHM.5.4.01.002.02. Explain why a collision between two particles is necessary for a reaction to occur Particles that are to react must come close to each other so that new bonds in the product can form, while old bonds in the reactants break. G12 Advanced Chemistry – CHM71 – Course Specification – Term 1 (Detailed KPIs) Page 24 of 52  CHM.5.4.01.002.03. List the two conditions, according to the collision theory, that must be met for a collision between reactant molecules to be effective in producing new chemical species The conditions are: 1) It must be energetic enough to supply the necessary activation energy 2) The colliding molecules must be oriented in a way that favors their efficient interaction G12 Advanced Chemistry – CHM71 – Course Specification – Term 1 (Detailed KPIs) Page 25 of 52  CHM.5.4.01.002.04. List the assumptions of the collision theory The assumptions are: 1) Reacting substances (atoms, ions or molecules) must collide 2) Reacting substances must collide in the correct orientation 3) Reacting substances must collide with sufficient energy to form an activated complex CHM.5.4.01.022.01. Define activated complex The activated complex is a temporary, unstable arrangement of atoms that may form products or may break apart to re-form the reactants. During the brief existence of the activated complex, bonds in the reactants are in the process of breaking, while new bonds are beginning to form to produce the products. CHM.5.4.01.022.02. Define activation energy, Ea The activation energy is the minimum amount of energy that reacting particles must have to form the activated complex and lead to a reaction. Activation energy is the barrier the reactants must overcome to form the products. CHM.5.4.01.022.03. Describe the relationship between activation energy and rate of the reaction The higher the activation energy, the slower the reaction rate A high activation energy means that few collisions will have the required energy to produce the activated complex. Hence, the reaction rate is slow. A low activation energy means more collisions will have the required energy to produce the activated complex. Hence, the reaction rate is faster. CHM.5.4.01.022.04. Interpret exothermic and endothermic reactions using the collision theory Exothermic Reaction Endothermic Reaction - In an exothermic reaction, the reactants collide - In an endothermic reaction, the reactants are with enough energy to overcome the activation at a lower energy than the products. energy barrier. - To react, the reactants must absorb enough - They form the activated complex, then release energy to overcome the activation energy energy and form products at a lower energy barrier and form higher-energy products. level. G12 Advanced Chemistry – CHM71 – Course Specification – Term 1 (Detailed KPIs) Page 26 of 52  CHM.5.4.01.022.05. Interpret how the speed of a chemical reaction is related to the spontaneity of the reaction The rate of a chemical reaction is unrelated to the spontaneity of the reaction. The reaction spontaneity, ∆𝐺, implies nothing about the speed of the overall reaction. ∆𝐺 indicates only the tendency for a reaction or process to proceed. CHM.5.4.01.001.03. Use a plot of concentration versus time to describe the progress of the reaction in terms of change in the concentration of reactants and products As the reaction proceeds, the concentration of reactants decrease and the concentration of products increase. Hence, the rate of reaction decreases and the reaction slows down. CHM.5.4.01.001.04. Describe the relationship between reactant concentration and reaction rate As concentration of reactant increases, rate of reaction increases. Inspire Chemistry – Module 15 – Lesson 3: Reaction rate laws CHM.5.4.01.005.01. Explain what the rate law for a chemical reaction tells about the reaction The rate law expresses the mathematical relationship between the rate of a chemical reaction and the concentration of reactants. CHM.5.4.01.005.02. Explain the function of the specific rate constant (k) in rate law equation The specific rate constant, k, relates reaction rate and concentration at a specific temperature. The specific rate constant does not change with concentration. The specific rate constant, k, describes the reaction rate and is determined experimentally. CHM.5.4.01.005.03. Identify under what circumstance the specific rate constant (k) is not a constant The specific rate constant, k, changes with temperature. G12 Advanced Chemistry – CHM71 – Course Specification – Term 1 (Detailed KPIs) Page 27 of 52  CHM.5.4.01.005.04. Identify the relation between the size of specific rate constant (k) and rate of reaction - The larger the value of the specific rate constant, k, the faster the reaction. - The value of the specific rate constant, k, in a rate law is large if the products form quickly. - The value of the specific rate constant, k, in a rate law is small if the products form slowly. CHM.5.4.01.005.05. Explain how the exponents in the rate equation for a chemical reaction relate to the coefficients in the chemical equation There is no relationship in general. In the relatively rare case of a single-step reaction with a single activated complex, the exponents are equal to the coefficients. CHM.5.4.01.005.06. Write the unit of specific rate constant (k) for different orders of reaction Order of reaction Unit of k M.time‒1 Zero order reaction mol/L.time First order reaction time‒1 M‒1.time‒1 Second order reaction L/mol.time M‒2.time‒1 Third order reaction L2/mol2.time CHM.5.4.01.005.07. Write the rate law for a first order reaction For a first order reaction, the rate law is: Rate = k[X]1 = k[X] Where: [X] = Concentration of reactant X Exponent 1 = Reaction order Reaction order defines how the rate is affected by the concentration of that reactant. CHM.5.4.01.005.08. Describe the graph for a first order reaction For a first order reaction, the reaction rate changes in the same proportion that the concentration of X changes. if the concertation of X is halved, the reaction rate is also halved. The rate constant, k, is equal to the slope of the line. G12 Advanced Chemistry – CHM71 – Course Specification – Term 1 (Detailed KPIs) Page 28 of 52  CHM.5.4.01.005.09. Write the rate law for a chemical reaction at a given temperature Rate= k[A]m[B]n where m and n are exponents to be determined experimentally Example: For each of the following statements, write the rate law that fits the description and predict the overall order of the reaction. a) For the reaction: Q + 2 R → P, the specific rate law constant, k, has a unit of min−1. Rate Law: Rate = k [Q] or Rate = k [R] Overall order of the reaction: 1 b) For the reaction: X + 2 Y → Z, reaction is first order in X, and the rate of reaction increases by a factor of four if the concentration of Y is doubled. Rate Law: Rate = k [X][Y]2 Overall order of the reaction: 3 c) For the reaction: 2 W + X → Q, when the initial concentrations of W and X each equals 0.20 M, the rate of the reaction is equal to the specific rate constant. Rate Law: Rate = k Overall order of the reaction: 0 d) For the reaction: S + T → W, the rate of the reaction is not affected when the concentration of S is doubled and the unit of the specific rate constant is L/mol.min Rate Law: Rate = k [T]2 Overall order of the reaction: 2 CHM.5.4.01.005.10. Describe the method of initial rate The method of initial rates determines the reaction order by comparing the initial rate of a reaction carried out with varying reactant concentrations. CHM.5.4.01.006.01. Use the method of initial rates to determine the order of a reaction with respect to each reactant CHM.5.4.01.006.02. Use the method of initial rates to write rate law of reaction CHM.5.4.01.006.03. Use the method of initial rates to calculate the specific rate constant and identify its unit G12 Advanced Chemistry – CHM71 – Course Specification – Term 1 (Detailed KPIs) Page 29 of 52  CHM.5.4.01.006.05. Use the method of initial rates to calculate the initial rate of a reaction Example: The initial rates listed in the following table were determined for the reaction: CO (g)+ Cl2(g)→COCl2(g) Initial concentration of CO Initial concentration of Cl2 Initial rate Trial in 𝐦𝐨𝐥/𝐋 in 𝐦𝐨𝐥/𝐋 in 𝐦𝐨𝐥/(𝐋. 𝐬𝐞𝐜) 1 0.120 0.200 2.00 × 10X- 2 0.060 0.200 1.00 × 10X- 3 0.060 0.400 4.00 × 10X- 4 0.120 0.400 ????????? 1) Find the order of the reaction in respect to CO. Rate = 𝑘[CO]m[Cl2]n Consider trials 1 and 2 where [Cl2] is kept constant, when [CO] is doubled, rate doubles Reaction is first order in CO and m = 1 2) Find the order of the reaction in respect to Cl2. Rate = 𝑘[CO]m[Cl2]n Consider trials 2 and 3 where [CO] is kept constant, when [Cl2] is doubled, rate quadruples or increases by a factor of 4 Reaction is second order in Cl2 and n = 2 3) Write the rate law of this reaction. Rate = 𝑘[CO]1[Cl2]2 4) Calculate the specific rate constant of this reaction. Choose any trial to find the value of k. Trial 1: Rate = 𝑘[CO]1[Cl2]2 2.00 × 10X- = 𝑘 (0.120)(0.200)= 𝑘 = 4.17 × 10X$ M−2.sec−1 or L2/mol2.sec 5) Calculate the value of the initial rate in trial 4. Rate = 𝑘[CO] [Cl2]2 = 4.17 × 10X$ × 0.120 × (0.400)= = 8.00 × 10X- mol/L.s G12 Advanced Chemistry – CHM71 – Course Specification – Term 1 (Detailed KPIs) Page 30 of 52  Example: The initial rates listed in the following table were determined for the reaction: 2 NO (g) + Cl2 (g) → 2 NOCl (g) 1) What is the order of the reaction with respect to NO? Second order 2) What is the order of the reaction with respect to Cl2? First order 3) What is the rate law of the above reaction? Rate = k [NO]2 [Cl2] 4) What is the value of the rate constant? Using experiment 1 Rate = k [NO]2 [Cl2] 1.0 × 10 –2 = k (0.13)2 (0.20) k = 3.0 M–2.s–1 Inspire Chemistry – Module 15 – Lesson 4: Instantaneous reaction rates and reaction mechanisms CHM.5.4.01.020.05. Define the instantaneous rate of a reaction Instantaneous rate is the slope of the straight line tangent to the curve at a specific time in a chemical reaction. Instantaneous rate is the reaction rate at a specific time. CHM.5.4.01.020.06. Interpret the graph for the instantaneous rate of a reaction The variables that are plotted: x-axis: Time y-axis: Concentration of a reactant (in mol/L) The instantaneous rate for a specific point in the reaction progress can be determined from the tangent to the curve that passes through that point. G12 Advanced Chemistry – CHM71 – Course Specification – Term 1 (Detailed KPIs) Page 31 of 52  CHM.5.4.01.020.07. Explain how the rate law for a chemical reaction can be used to determine the instantaneous rate of a reaction By using the actual experimental rate constant and reactant concentrations into the rate law. CHM.5.4.01.020.08. Calculate the instantaneous rate of a reaction from experimental data Example: Consider the reaction: 2N2O5(g) → 4NO2(g) + O2(g) The experimentally rate law is: Rate = k[N2O5] Calculate the instantons rate law when [N2O5] = 0.450 M and k = 1.0 × 10‒5 s‒1. Rate = k[N2O5] = (1.0 × 10‒5)(0.450) = 4.5 × 10‒6 mol/L.s Example: Consider a reaction that occurs between X and Y to form XY. The reaction is first order in X and first order in Y. The rate constant, k, is 0.600 mol/(L.s) Calculate the rate of reaction when [X] = 2.20 × 10‒2 M and [Y] = 1.4 × 10‒2 M. Rate = k[X][Y] = (0.600)(2.20 × 10‒2)( 1.4 × 10‒2) = 1.85 × 10‒4 mol/L.s CHM.5.4.01.022.06. Define reaction mechanism Reaction mechanism is the complete sequence of elementary steps that make up a complex reaction. CHM.5.4.01.022.07. Compare and contrast elementary chemical reaction and complex reaction Elementary chemical reaction: A reaction that occurs in a single step. Complex reaction: A reaction that consists of two or more elementary steps. CHM.5.4.01.022.08. Distinguish among an intermediate, a catalyst and an activated complex An intermediate An intermediate in a complex reaction is a substance produced in one elementary and consumed in a subsequent elementary step A catalyst A substance that is a reactant in one step and a product in another step An activated complex In an activated complex, bonds are partially broken and partially formed, so the complex represents a maximum on the energy profile CHM.5.4.01.022.09. Define the rate-determining step for a chemical reaction The rate-determining step is the slowest of the elementary steps in a complex reaction. It is the step that has the largest activation energy, Ea. G12 Advanced Chemistry – CHM71 – Course Specification – Term 1 (Detailed KPIs) Page 32 of 52  CHM.5.4.01.022.10. Determine, using a given reaction mechanism, the rate determining step (the slowest step) CHM.5.4.01.022.11. Determine, using a given reaction mechanism, the reactant and product CHM.5.4.01.022.12. Determine, using a given reaction mechanism, the intermediate and catalyst CHM.5.4.01.022.13. Determine, using a given reaction mechanism, the overall reaction Example: A generic reaction occurs in a two-step mechanism: Step 1: X ⟶ Y +Z slow Step 2: Y+W⟶V fast Identify each of the following: a) Reactant(s): X and W b) Product(s): Z and V c) Catalyst: The reaction does not have a catalyst d) Intermediate(s): Y e) Rate-determining step: Step 1 (The slower step) f) Overall reaction: X+W→Z+V Example: Consider the following three-steps reaction mechanism. Step 1: X + Q → R= Slow Step 2: R = + W → WR = Fast Step 3: X + WR = → S + Q Fast Identify each of the following: a) Reactant(s): X and W b) Product(s): S c) Catalyst: Q d) Intermediate(s): R2 and WR2 e) Rate-determining step: Step 1 (It is the slowest step) f) Overall reaction: 2X + W → S G12 Advanced Chemistry – CHM71 – Course Specification – Term 1 (Detailed KPIs) Page 33 of 52  Example: Consider the following three-steps reaction mechanism. Step 1: Ce"W =W -W -W ()V) + Mn()V) → Ce()V) + Mn()V) Fast Step 2: Ce"W -W -W "W ()V) + Mn()V) → Ce()V) + Mn()V) Slow Step 3: TlW "W -W =W ()V) + Mn()V) → Tl()V) + Mn()V) fast Identify each of the following: a) Reactant(s): Ce"W W ()V) and Tl()V) b) Product(s): Tl-W -W ()V) and Ce()V) c) Catalyst: Mn=W ()V) d) Intermediate(s): Mn-W "W ()V) and Mn()V) e) Rate-determining step: Step 2 (It is the slowest step) f) Overall reaction: 2Ce"W W -W -W ()V) + Tl()V) → Tl()V) + 2Ce()V) CHM.5.4.01.022.14. Determine, using the energy of reaction graph for a given mechanism, the rate determining step (the slowest step) CHM.5.4.01.022.15. Determine, using the energy of reaction graph for a given mechanism, the reactant and product CHM.5.4.01.022.16. Determine, using the energy of reaction graph for a given mechanism, the intermediate and the activated complex Example: The diagram below represents the energy profile of a given reaction. Identify each of the following: a) Reactant(s): X b) Product(s): Y c) Intermediate(s): Z d) Activated complex(es): Y and T e) Overall reaction: X→W G12 Advanced Chemistry – CHM71 – Course Specification – Term 1 (Detailed KPIs) Page 34 of 52  Example: The diagram below represents the energy profile of a given reaction. Identify each of the following: a) Reactant(s): X and R b) Product(s): W and S c) Intermediate(s): Z d) Activated complex(es): Y and T e) Overall reaction: X+R→W+S f) Rate-determining step: Arrow 1 (It has a larger activation energy, Ea) 11.7– Chemical Equilibrium Inspire Chemistry – Module 16 – Lesson 1: A state of dynamic balance CHM.5.4.02.001.01. Define chemical equilibrium A chemical equilibrium is a state in which the forward and reverse reactions balance each other because they take place at equal rates. Rate of forward reaction = Rate of backward reaction G12 Advanced Chemistry – CHM71 – Course Specification – Term 1 (Detailed KPIs) Page 35 of 52