Search Results

< Back Chapter 1: Equilibrium Explore the fascinating world of chemical equilibrium and learn how it impacts various chemical reactions and processes. Chapter 1: Equilibrium - This chapter explores the concept of chemical equilibrium, which is the state in which the concentrations of the reactants and products in a chemical reaction remain constant over time. Students will learn about the equilibrium constant, Le Chatelier's principle, and how to calculate equilibrium concentrations. Previous Next Study Material Notes Part 1 Notes Part 2 Notes Part 3 Notes Part 4 Excercises 1 Excercises 2 Excercises 3 Excercises 4 As any reaction proceeds, [reactants] decreases and [products] increases. Reversible reaction: a reaction which can go both ways. Equilibrium: is the point when the [reactants] and [products] becomes constant. Equilibrium: is the point when the [reactants] and [products] becomes constant. Equilibrium is recognized by constancy of macroscopic properties in a closed system at constant temperature. Macroscopic properties are observable properties or measurable properties like pressure, concentration, color, size, volume and mass. Macroscopic properties are observable properties or measurable properties like pressure, concentration, color, size, volume and mass. Each set of equilibrium concentrations is called an equilibrium position. Steady state: indicates a situation where macroscopic properties are constant but equilibrium does not exist as the system is not closed e.g a blue Busen burner flame. Equilibrium in physical changes: Solubility of iodine Vapor pressure of water Equilibrium in chemical reactions: NO2-N2O4 system Equilibrium is dynamic in nature since at equilibrium two microscopic processes are occurring in opposite direction at the same rate resulting in no observable macroscopic changes. Equilibrium is dynamic in nature since at equilibrium two microscopic processes are occurring in opposite direction at the same rate resulting in no observable macroscopic changes. Equilibrium is dynamic in nature since at equilibrium two microscopic processes are occurring in opposite direction at the same rate resulting in no observable macroscopic changes. Adding a catalyst or a noble gas does not alter the state of equilibrium. Equilibrium may not reached in certain equilibrium system because the activation energy of the forward reaction is too high. Le Chatelier's Principle: if an equilibrium system is subjected to a change processes will occur to partially counteract the imposed change. Low temperature is required in the Haber Process for a desirable good yield and high temperature is necessary for a satisfactory rate. The compromise used industrially involves an intermediate temperature around 450°C and even then the success of the process depends upon the presence of a suitable catalyst to achieve a reasonable reaction rate. High pressures are required in the Haber Process for a good yield and a satisfactory high rate. It is expensive and dangerous to build up a pressure. A pressure of 200 atm is actually used as a compromise. Mass Action Expression: For a general reaction: αA + βB ⇆ ΥC + ΔD Mass action expression = Q = [C]γ[D]Δ[A]/α[B]β At equilibrium: mass action expression is constant and is given a special name, equilibrium constant, Keq Concentrations of solids and liquids are NOT included in the equilibrium expression. These values are constant and are incorporate in the value of Keq directly. Temperature and the nature of solvent are the only values which determine the value of the equilibrium constant. There is only one equilibrium constant for a particular system at a particular temperature but there are an infinite number of equilibrium positions Significance of value of K If reactants and products are mixed are mixed, three things may occur:  An equilibrium is established  A reaction occurs in the forward direction  A reaction occurs in the backward direction. To find out what is occurring one needs to find the value of the mass action expression, Q and compare it to Keq. a. If Q >Keq reaction is NOT at equilibrium the backward reaction is taking place (system shifts to the left). b. If Q = Keq reaction is at equilibrium No shift occurs. c. If Q

Heating water from 20°C to 80°C Grade 10 SABIS SABIS Endothermic

< Back A level Hydrocarbons ​ ​ Previous Next

< Back A level Nitrogen compounds ​ ​ Previous Next

< Back Sabis Grade 12 Chemistry Concise content for Grade 12 SABIS Curriculum Course content click here https://www.k-chemistry.com/all-sabis-chapters/chapter-6-sabis-grade-12-part-3 Previous Next

< Back A level Chemistry of transition elements ​ ​ Previous Next

Phase Change ​ ​ The transition of a substance from one state of matter to another due to changes in temperature and/or pressure.

Sodium chloride is a very stable compound because Na+ ion has 10 electrons around it (like noble gas before it, Ne) and the Cl- has 18 electrons around it (like noble gas after it. Ar). Grade 10 SABIS ​

Avogadro's Hypothesis ​ ​ Avogadro's hypothesis states that equal volumes of different gases, at the same temperature and pressure, contain an equal number of particles. This means that regardless of the type of gas, the number of molecules or atoms in a given volume is the same. ✨ Lesson: Avogadro's Hypothesis ✨ 🔬 Introduction: Avogadro's hypothesis is a fundamental concept in chemistry that helps us understand the relationship between the number of particles and the amount of substance. It provides a link between the macroscopic world we observe and the microscopic world of atoms and molecules. Let's delve into Avogadro's hypothesis and explore its implications.💡 Avogadro's Hypothesis: 🔹 Definition: Avogadro's hypothesis states that equal volumes of different gases, at the same temperature and pressure, contain an equal number of particles. 🌡️🧪🔒🧪 Implications of Avogadro's Hypothesis: ✅ Equal Volumes: Regardless of the gas, equal volumes of different gases contain the same number of particles. 📊✅ Molar Volume: The concept of molar volume is established by Avogadro's hypothesis. At standard temperature and pressure (STP), the molar volume is approximately 22.4 liters. 📏✅ Moles and Particles: Avogadro's hypothesis allows us to relate the number of moles to the number of particles in a substance. One mole of any substance contains 6.02 × 10^23 particles, known as Avogadro's number. 🧪🧪🧪🔍 Example: Consider oxygen gas (O2) and nitrogen gas (N2) at the same temperature and pressure. According to Avogadro's hypothesis, equal volumes of these gases will contain the same number of particles. If we have 1 liter of oxygen gas, it will contain the same number of molecules as 1 liter of nitrogen gas. ⚖️🌬️🧪 Quiz (Basic Understanding): 1️⃣ What does Avogadro's hypothesis state? a) Equal volumes of different gases contain an equal number of particles. b) The mass of a substance is proportional to the number of particles. c) The volume of a gas is inversely proportional to its pressure. 2️⃣ What is the molar volume at STP? a) 6.02 × 10^23 liters b) 22.4 liters c) 1 liter 3️⃣ How many particles are there in one mole of a substance? a) 6.02 × 10^23 particles b) 1 particle c) 10 particles 4️⃣ According to Avogadro's hypothesis, what happens to the number of particles when comparing equal volumes of different gases? a) The number of particles is different. b) The number of particles is the same. c) The number of particles depends on the temperature.🔍 Answers: 1️⃣ a) Equal volumes of different gases contain an equal number of particles. 2️⃣ b) 22.4 liters 3️⃣ a) 6.02 × 10^23 particles 4️⃣ b) The number of particles is the same. 🌟 Well done! You've gained a basic understanding of Avogadro's hypothesis and its significance in chemistry. Keep exploring the fascinating world of atoms and molecules to uncover more exciting concepts! 🧪🔬✨

The use of bond energy data A Level Chemistry CIE Carrying out calculations using cycles and relevant energy terms is an essential aspect of thermochemistry. These calculations involve utilizing energy cycles, such as Hess's Law cycles, and incorporating relevant energy terms, including bond energy data, to determine enthalpy changes and other thermodynamic quantities. When using cycles to perform calculations, we start by constructing an energy cycle that relates the desired reaction to known reactions with known enthalpy changes. This involves breaking down the target reaction into a series of intermediate reactions for which we have the corresponding enthalpy changes. In the case of Hess's Law cycles, we manipulate the intermediate reactions by multiplying, reversing, or combining them to obtain the target reaction and its associated enthalpy change. By summing up the enthalpy changes of the intermediate reactions, taking into account their stoichiometric coefficients, we arrive at the overall enthalpy change for the target reaction. To carry out calculations using bond energy data, we utilize the concept that the enthalpy change of a reaction is related to the difference in bond energies between the bonds broken and the bonds formed during the reaction. Bond energy data provides information about the average energy required to break specific types of bonds. To calculate the enthalpy change using bond energy data, we start by identifying the bonds broken and formed in the reaction. We then sum up the bond energies for the bonds broken, subtract the sum of the bond energies for the bonds formed, and account for the stoichiometry of the reaction. For example, if we want to calculate the enthalpy change for the combustion of methane (CH4), we can use bond energy data to determine the energy changes associated with breaking the C-H bonds in methane and forming the bonds in the combustion products (CO2 and H2O). By subtracting the sum of the bond energies for the reactant bonds from the sum of the bond energies for the product bonds, we obtain the enthalpy change for the combustion reaction. It's important to note that bond energy data represents average values and can vary depending on the specific molecular environment and conditions. Additionally, bond energy calculations assume that all bonds in a molecule have equal energy, neglecting any effects of neighboring atoms or functional groups. Carrying out calculations using cycles and bond energy data allows us to determine enthalpy changes and make predictions about energy transformations in chemical reactions. These calculations provide valuable insights into the thermodynamic behavior of systems and assist in the design and optimization of chemical processes. In summary, performing calculations using cycles and relevant energy terms involves constructing energy cycles, such as Hess's Law cycles, to relate desired reactions to known reactions with enthalpy changes. Bond energy data is used to calculate enthalpy changes based on the energy differences between bonds broken and formed. These calculations enhance our understanding of energy transformations in chemical systems and aid in predicting thermodynamic behavior.

Given the heats of formation calculate the heat of reaction (simple application) Grade 10 SABIS ​ Given the heats of formation calculate the heat of reaction (simple application) Given: C(diamond) + O2(g) → CO2(g) ΔH = −395.4 kJ C(graphite) + O2(g) → CO2(g) ΔH = −393.5 kJ a) Find ΔH for the manufacture of diamond from graphite: C(graphite) → C(diamond) H = -393.5 + 395.4 = + 1.9 kJ b) Is heat absorbed or evolved as graphite is converted to diamond? Absorbed When given the heats of formation, we can calculate the heat of reaction for a specific process by applying the concept of Hess's law. Hess's law states that the overall heat of a reaction is independent of the pathway taken. This allows us to use known heats of formation to determine the heat of reaction for a desired process. Let's consider the given example of the reaction between graphite (C(graphite)) and oxygen gas (O2) to form carbon dioxide (CO2): C(graphite) + O2(g) → CO2(g) ΔH = -393.5 kJ To find the heat of reaction for the conversion of graphite to diamond, we need to compare the heats of formation of diamond (C(diamond)) and graphite (C(graphite)). By using the concept of Hess's law, we can subtract the heat of formation of graphite from the heat of formation of diamond. a) Find ΔH for the manufacture of diamond from graphite: C(graphite) → C(diamond) ΔH = -393.5 kJ + 395.4 kJ = +1.9 kJ By adding the heats of formation of the reactants and products, we find that the heat of reaction for the manufacture of diamond from graphite is +1.9 kJ. This positive value indicates that heat is absorbed during the conversion process, as the energy required to form the diamond is greater than the energy released during the formation of graphite. b) Is heat absorbed or evolved as graphite is converted to diamond? Heat is absorbed. Based on the positive value of ΔH (+1.9 kJ) for the conversion of graphite to diamond, we can conclude that heat is absorbed during this process. This means that energy is taken in from the surroundings to convert graphite into diamond. To summarize, when given the heats of formation, we can calculate the heat of reaction by applying Hess's law. In the example provided, we determined the heat of reaction for the conversion of graphite to diamond by subtracting the heats of formation of the reactants and products. The positive value of ΔH indicates that heat is absorbed during this process, indicating an energy input required for the conversion. Understanding this concept helps us analyze the energy changes and transformations that occur during chemical reactions.

Boiling water until it evaporates, then condensing the steam Grade 10 SABIS SABIS Physical