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< Back Reaction kinetics ​ ​ Previous Next 🔬 Chapter 9: Rates of Reaction 🔬 Learning Outcomes 🎯: Understand reaction kinetics and the factors affecting the rates of chemical reactions. Recognize the role of surface area, concentration, temperature, and catalysts in reaction rates. Understand the concept of activation energy and its role in determining the rate of reaction. Differentiate between homogeneous and heterogeneous catalysts. Understand the Boltzmann distribution of molecular energies and how it changes with temperature. Factors Affecting Rate of Reaction 📈: Surface Area : Finely divided solids have a larger surface area, leading to more frequent collisions and a faster reaction rate. Concentration and Pressure : Higher concentration or pressure leads to more frequent collisions between reactant molecules, increasing the reaction rate. Temperature : At higher temperatures, molecules have more kinetic energy, leading to more frequent and successful collisions. Catalysts : Catalysts increase the rate of reaction by providing an alternative reaction pathway with a lower activation energy. Activation Energy ⚡: Activation energy is the minimum energy required by colliding particles for a reaction to occur. It acts as a barrier to reaction, and only particles with energy greater than the activation energy can react. Boltzmann Distribution 📊: The Boltzmann distribution represents the number of molecules in a sample with particular energies. At higher temperatures, the distribution changes, showing that more molecules have energy greater than the activation energy, leading to an increase in reaction rate. Catalysis 🧪: Catalysts lower the activation energy, allowing a greater proportion of molecules to have sufficient energy to react. Homogeneous catalysts are in the same phase as the reactants, while heterogeneous catalysts are in a different phase. Enzymes are biological catalysts that provide an alternative reaction pathway of lower activation energy.

Volume at STP Grade 10 SABIS SABIS 1.00 mole of ANY gas occupies 22.4 dm3

Conservation of Atoms Grade 10 SABIS SABIS In chemical reactions, the number of atoms of each element in the reactants is equal to the number of atoms of the same element in the products.

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Application on Hess’s Law medium Grade 10 SABIS ​ Question 1: Given the following reactions and their respective enthalpy changes: C(s) + O2(g) → CO2(g) ΔH1 = -393.5 kJ/mol H2(g) + 1/2O2(g) → H2O(l) ΔH2 = -286.0 kJ/mol C(s) + H2(g) → CH4(g) ΔH3 = -74.8 kJ/mol Calculate the enthalpy change for the reaction: CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) Answer 1: To calculate the enthalpy change for the given reaction, we can use Hess's Law. By manipulating the given reactions, we can cancel out the common compounds and add the enthalpy changes. Multiplying reaction 1 by 2 gives: 2C(s) + 2O2(g) → 2CO2(g) 2ΔH1 = 2(-393.5 kJ/mol) = -787.0 kJ/mol Multiplying reaction 2 by 2 gives: 2H2(g) + O2(g) → 2H2O(l) 2ΔH2 = 2(-286.0 kJ/mol) = -572.0 kJ/mol Adding reactions 3, 2, and 1 gives: C(s) + H2(g) + 2H2(g) + O2(g) + 2O2(g) → CH4(g) + 2H2O(l) + 2CO2(g) ΔH3 + 2ΔH2 + 2ΔH1 = -74.8 kJ/mol + (-572.0 kJ/mol) + (-787.0 kJ/mol) = -1433.8 kJ/mol Since the given reaction is the reverse of the calculated reaction, the enthalpy change for the given reaction is the negative of the calculated value. ΔH = -(-1433.8 kJ/mol) = 1433.8 kJ/mol Question 2: Given the following reactions and their respective enthalpy changes: 2SO2(g) + O2(g) → 2SO3(g) ΔH1 = -198.2 kJ/mol S(s) + O2(g) → SO2(g) ΔH2 = -296.8 kJ/mol 2S(s) + 3O2(g) → 2SO3(g) ΔH3 = -792.0 kJ/mol Calculate the enthalpy change for the reaction: 2SO2(g) + O2(g) → 2SO3(g) + 198.2 kJ Answer 2: To calculate the enthalpy change for the given reaction, we can use Hess's Law. By manipulating the given reactions, we can cancel out the common compounds and add the enthalpy changes. Multiplying reaction 2 by 2 gives: 2S(s) + 2O2(g) → 2SO2(g) 2ΔH2 = 2(-296.8 kJ/mol) = -593.6 kJ/mol Adding reactions 1 and 2 gives: 2SO2(g) + O2(g) + 2S(s) + 2O2(g) → 2SO3(g) + 2

Solving Problems Grade 10 SABIS SABIS To determine the energy released or required

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< Back Chapter 7: Electrons and the Periodic Table Learn about the electron configurations of atoms and how they determine the chemical and physical properties of elements. Chapter 7: Electrons and the Periodic Table - This chapter covers the behavior of electrons in atoms and their relationship to the periodic table. Students will learn about electron configurations, the periodic trends in atomic properties, and chemical bonding. Previous Next

< Back Previous Next 🎯 Solubility of Ionic Compounds 🎯 The solubility of ionic compounds can be predicted with some general rules, which you already listed. These rules are not absolute, but they offer a good starting point for understanding solubility. You'll be able to predict whether certain compounds will dissolve in water and under what conditions. 🌊💡🔋 Electrolysis of Aqueous Solutions 🔋💡🌊 An aqueous solution contains dissolved ions and also ions from the water itself. When it comes to electrolysis: 👉 At the cathode: The less reactive cation (positive ion) will be discharged. In general, this is often a metal ion or a hydrogen ion (H+). 👉 At the anode: The less reactive anion (negative ion) will be discharged. Usually, this is the hydroxide ion (OH-) or a halide ion if the solution is concentrated. 🎨🌈 pH Scale and Indicators 🌈🎨 The pH scale ranges from 0 (strongly acidic) to 14 (strongly basic). A pH of 7 is neutral. Indicators are chemicals that change color based on the pH of the solution. Common indicators include: Litmus: Red in acid, blue in base. Methyl orange: Red in acid, yellow in base. Thymolphthalein: Colourless in acid, blue in base. Universal indicator: Displays a range of colors across the pH scale. 💡⚡ Electrolysis Examples ⚡💡 You've provided multiple examples for the electrolysis of different substances, including: Copper sulfate (CuSO4) : Copper is deposited at the cathode, and oxygen is released at the anode. Sulfuric acid (H2SO4) : Hydrogen is released at the cathode, and oxygen at the anode. Dilute NaCl : Hydrogen is released at the cathode, and oxygen at the anode. Concentrated HCl : Hydrogen is released at the cathode, and chlorine at the anode. Concentrated NaCl (brine) : Hydrogen is released at the cathode, and chlorine at the anode. Note that in all these reactions, the less reactive ions are the ones getting discharged at the electrodes. In each case, the electrolyte concentration changes during the electrolysis process, often getting more concentrated. 🏭🔋 Applications of Electrolysis 🔋🏭 Electrolysis has many important industrial applications, including the production of chlorine, hydrogen, and sodium hydroxide from brine. These products have a wide range of uses, such as in the production of plastics, fuels, soaps, detergents, and more. And there you have it! A journey through the intricate world of solubility, pH, electrolysis, and their applications. Remember, the more you understand these processes, the easier it becomes to understand how the world of chemistry works around us. Keep exploring! 🧪🎆🚀 Go To lesson 3

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Relative magnitude of heat involved in chemical and nuclear changes Grade 10 SABIS ​ On the other hand, the heat involved in nuclear changes is orders of magnitude larger than in chemical changes. Nuclear reactions involve changes in the nucleus of an atom, such as nuclear fission or nuclear fusion. These reactions release or absorb an enormous amount of energy due to the conversion of mass into energy, as described by Einstein's famous equation, E = mc^2. The energy released in nuclear changes is millions or billions of times greater than that released in chemical reactions. The heat involved in nuclear reactions is typically measured in millions of electron volts (MeV) or joules (J). The energy released in nuclear fission or fusion reactions can be in the range of millions or billions of joules per mole of reactants or products. For example, the energy released in a typical chemical combustion reaction, such as the burning of a hydrocarbon fuel, is on the order of tens or hundreds of kilojoules per mole. In contrast, the energy released in a nuclear fission reaction, such as the splitting of a uranium nucleus, is on the order of millions of electron volts per nucleus. It's important to note that while nuclear changes involve much larger energy releases, they are also associated with unique challenges and considerations, including the potential for radioactive materials and the requirement for precise control and safety measures. In summary, the relative magnitude of heat involved in chemical and nuclear changes differs significantly. Chemical changes involve relatively small energy changes associated with the breaking and formation of chemical bonds, while nuclear changes involve much larger energy releases due to the conversion of mass into energy. Understanding and quantifying these energy changes are crucial in various scientific, technological, and energy-related applications.

Find energy supplied by electric current using W = IVt Grade 10 SABIS ​ To find the energy supplied by an electric current, the equation W = IVt is utilized, where W represents the energy supplied, I is the current flowing through the circuit, V is the potential difference (voltage) across the circuit, and t is the time for which the current flows. The equation W = IVt is derived from the fundamental relationship between electrical power, current, voltage, and time. Power (P) is defined as the rate at which energy is transferred or consumed, and it can be calculated as the product of current and voltage, P = IV. Multiplying this power by time (t), we obtain the energy supplied or consumed, which is given by the equation W = IVt. The unit of current (I) is measured in amperes (A), the unit of voltage (V) is measured in volts (V), and the unit of time (t) is measured in seconds (s). For example, let's consider a scenario where a circuit has a constant current of 2 amperes (A) flowing through it, a voltage of 12 volts (V) across the circuit, and the current flows for a duration of 10 seconds (s). Using the equation W = IVt, we can calculate the energy supplied as follows: W = (2 A) * (12 V) * (10 s) = 240 joules (J) Therefore, in this case, the energy supplied by the electric current is 240 joules (J). It's important to note that this equation assumes that the current and voltage remain constant during the entire time period. In real-world scenarios, the current and voltage may vary over time, requiring more advanced calculations to determine the total energy supplied. The equation W = IVt is widely applicable in various electrical systems, such as household circuits, electronic devices, and power grids. It allows for the measurement and calculation of energy consumption or supply, enabling us to understand and analyze the energy usage and requirements of electrical systems. By utilizing the equation W = IVt, we can quantitatively assess the energy consumed or supplied by an electric current. This information is essential for managing energy resources, estimating costs, and optimizing energy efficiency in various applications. In summary, finding the energy supplied by an electric current involves using the equation W = IVt, where W represents the energy supplied, I is the current, V is the voltage, and t is the time. By multiplying the current, voltage, and time, we can determine the energy transferred or consumed. Understanding and calculating the energy supplied by electric current are essential in various fields, including electrical engineering, energy management, and sustainable technology.