Search Results

Products Grade 10 SABIS SABIS The substances that are formed during a chemical reaction.

Effect of changing concentration on rate of reaction: Grade 10 SABIS ​ Increasing the concentration of a reactant increases the number of particles in a given volume thus the reacting particles will collide more frequently so the number of collisions will increase per unit time, thus rate of reaction increases.

< Back Chapter 5: Stoichiometry Discover the quantitative relationships between reactants and products in chemical reactions and learn how to apply stoichiometric calculations. Chapter 5: Stoichiometry - This chapter covers the calculations involved in chemical reactions. Students will learn how to balance chemical equations, calculate limiting reactants, and determine percent yield. The chapter also covers the mole concept and how to use it in stoichiometry problems. Previous Next

< Back Previous Next

< Back Polymerisation ​ ​ Previous Next

< 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.

< Back Previous Next

Graphite is a solid non-metal element which is brittle yet conducts electricity Grade 10 SABIS ​

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.

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

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