Energy Changes in Physical and Chemical Processes PDF

Created by Turbolearn AI Energy Changes in Physical and Chemical Processes Enthalpy and Enthalpy Change All physical and chemical changes involve energy changes, primarily in the form of heat. Reactions either absorb or release heat energy. While some, like combustion, are obvious, others, like rus...

Created by Turbolearn AI Energy Changes in Physical and Chemical Processes Enthalpy and Enthalpy Change All physical and chemical changes involve energy changes, primarily in the form of heat. Reactions either absorb or release heat energy. While some, like combustion, are obvious, others, like rusting, are less so. Enthalpy (H): The heat content of a chemical substance. It represents the total internal energy, but its absolute value cannot be measured directly. We can, however, measure changes in enthalpy. We can measure the change in enthalpy, which is always accompanied by a temperature change. For example, burning coal releases heat (a temperature increase) because the enthalpy of the products (carbon dioxide and other materials) is less than the enthalpy of the reactants (coal). The difference is the released heat. ΔH = Hproducts − Hreactants If ΔH is negative, the reaction is exothermic (heat is released). If ΔH is positive, the reaction is endothermic (heat is absorbed). This chapter focuses on ΔH , not absolute enthalpy values. The Law of Conservation of Energy Energy cannot be created or destroyed, only transformed. This is the Law of Conservation of Energy. The universe's total energy remains constant. Physical and chemical changes involve energy transfer between: System: The specific matter being studied. Surroundings: Everything else. We measure the energy change of the surroundings (which is measurable) to determine the energy change of the system (which is not directly measurable). Because energy is conserved: ΔEsystem = −ΔEsurroundings Page 1  Created by Turbolearn AI Endothermic Processes A process is endothermic if the system absorbs heat from the surroundings. The system's energy increases. The surroundings' energy decreases. The surroundings' temperature decreases. ΔH is positive. Example: Melting ice. Heat from the surroundings is absorbed by the ice, causing the water molecules to overcome attractive forces and become liquid. Liquid water has higher energy than ice. Exothermic Processes A process is exothermic if the system releases heat to the surroundings. The system's energy decreases. The surroundings' energy increases. The surroundings' temperature increases. ΔH is negative. Example: Freezing water. Water molecules lose kinetic energy (released as heat to the surroundings), slowing down enough for stronger attractive forces to form ice. Energy Level Diagrams These diagrams show energy changes. The x-axis represents reaction progress, and the y-axis represents energy (kJ). Melting: The energy level of liquid water is higher than that of ice, indicating a positive ΔH. Freezing: The energy level of ice is lower than that of liquid water, indicating a negative ΔH. Combustion: This is exothermic; ΔH is negative. Heat of Solution Dissolving a compound in water involves heat exchange (heat of solution): Page 2  Created by Turbolearn AI Exothermic dissolution: Heat is released, increasing water temperature. Endothermic dissolution: Heat is absorbed, decreasing water temperature. Example: Experiments with sodium hydroxide (exothermic), ammonium nitrate (endothermic), and sodium chloride (minimal change). The results are summarized in a table (shown in lecture). Thermodynamics Study Guide Energy Level Diagrams and Bond Energies Endothermic and exothermic processes can be represented using energy level diagrams. However, these diagrams are simplified; all reactions involve both energy input and output. A reaction is classified as exothermic or endothermic based on the net energy change. To understand this, consider bond breaking and bond formation: Bond breaking: Requires energy input (endothermic). Bond formation: Releases energy (exothermic). Example: Formation of Water The formation of water (H2O) from hydrogen (H2) and oxygen (O2) involves: 1. Breaking bonds in the reactants (H2 and O2). 2. Rearranging atoms. 3. Forming new bonds in the product (H2O). The overall enthalpy change depends on which process (bond breaking or bond formation) has a higher energy. Bond Bond Energy (kJ/mol) H-H 436 O=O 495 O-H 460 Calculation for Water Formation: Page 3  Created by Turbolearn AI Bond breaking (endothermic): 2 mol H-H bonds: 2 * 436 kJ/mol = 872 kJ 1 mol O=O bond: 1 * 495 kJ/mol = 495 kJ Total: 872 kJ + 495 kJ = 1367 kJ Bond formation (exothermic): 4 mol O-H bonds: 4 * 460 kJ/mol = 1840 kJ Enthalpy change (ΔH ): ΔH = Bond breaking energy - Bond formation energy ΔH = 1367 kJ - 1840 kJ = -473 kJ Since ΔH is negative, the formation of water is exothermic. Activation Energy Most reactions require activation energy to start. This is the initial energy needed to break bonds and initiate the reaction. Example: Methane and Oxygen Methane (CH4) and oxygen (O2) don't react until a spark (activation energy) breaks bonds, allowing new bonds to form and releasing energy. Calculating Enthalpy Change The enthalpy change (ΔH ) can be calculated using average bond energies: ΔH = Bond breaking energy - Bond formation energy Example: Consider a reaction where bond breaking energies sum to 2642 kJ and bond formation energies sum to 3450 kJ. ΔH = 2642 kJ - 3450 kJ = -808 kJ This is an exothermic reaction. Endothermic vs. Exothermic Reactions Page 4  Created by Turbolearn AI Endothermic reactions: Absorb heat from surroundings; temperature decreases. Examples include electrolysis, the reaction of sodium carbonate with ethanoic acid, and photosynthesis. The products have higher energy than the reactants. Exothermic reactions: Release heat to surroundings; temperature increases. Examples include acid-alkali neutralization, the reaction of calcium oxide with water, and respiration. The products have lower energy than the reactants. ΔH is negative. Latent Heat Latent heat of fusion: Heat required to change a solid to a liquid at constant temperature. Molar heat of fusion is the latent heat for one mole of substance. Latent heat of vaporization: Heat required to change a liquid to a gas at constant temperature. Molar heat of vaporization is the latent heat for one mole of substance. Determining Enthalpy Change: Experiment and Calculations Enthalpy change can be calculated using: ΔH = mcΔT where: m = mass of solution c = specific heat capacity ΔT = temperature change Example: Molar Heat of Neutralization The enthalpy change when an acid and alkali react to form one mole of water. An experiment measuring the temperature change during the reaction of hydrochloric acid and sodium hydroxide would be used to calculate the molar heat of neutralization. Thermochemistry Study Guide Page 5  Created by Turbolearn AI Heat of Reaction Scenario: 50 cubic centimeters of 2 molar hydrochloric acid reacts with 50 cubic centimeters of water, resulting in a temperature change of 13.5 degrees Celsius (or Kelvin). Note: A change in temperature has the same value in Celsius and Kelvin. Calculations: Total volume: 50 cm³ + 50 cm³ = 100 cm³ Mass of water: 100 cm³ * (1 g/cm³) = 100 g = 0.1 kg Enthalpy change (ΔH): ΔH = mcΔT = (0.1 kg)(4.2 kJ/kg·K)(13.5 K) = 5.67 kJ Since the reaction is exothermic, ΔH = -5.67 kJ. Moles of water formed: 1 mole (from the reaction's stoichiometry). Molar enthalpy change: -5.67 kJ/ 1 mol = -5.67 kJ/mol The equation ΔH = mcΔT calculates enthalpy change, where m is mass, c is specific heat capacity, and ΔT is the change in temperature. Thermochemical Equation: The reaction is represented as: Reaction → -5.67 kJ/mol Molar Heat of Solution Page 6  Created by Turbolearn AI Scenario: 4 grams of ammonium nitrate dissolved in 50 cubic centimeters of water, causing a temperature drop from 20°C to 8°C. Calculations: Mass of water: 50 cm³ * (1 g/cm³) = 50 g = 0.05 kg ΔT = 20°C - 8°C = 12°C = 12 K Enthalpy change (ΔH): ΔH = mcΔT = (0.05 kg)(4.2 kJ/kg·K)(12 K) = 2.52 kJ The reaction is endothermic, so ΔH = +2.52 kJ. Moles of ammonium nitrate: (4 g) / (80 g/mol) = 0.05 mol Molar enthalpy change: (2.52 kJ) / (0.05 mol) = 50.4 kJ/mol Thermochemical Equation: The reaction is represented as: Reaction → +50.4 kJ/mol Molar Heat of Displacement Scenario: 4 grams of zinc powder added to 100 cubic centimeters of copper(II) sulfate solution; temperature rose from 21°C to 46°C. Calculations: Mass of solution (approximated as water): 100 cm³ * (1 g/cm³) = 100 g = 0.1 kg ΔT = 46°C - 21°C = 25°C = 25 K Enthalpy change (ΔH): ΔH = mcΔT = (0.1 kg)(4.2 kJ/kg·K)(25 K) = 10.5 kJ The reaction is exothermic, so ΔH = -10.5 kJ. Moles of zinc: (4 g) / (65 g/mol) = 0.062 mol (approximately 0.06 mol) Molar enthalpy change: (-10.5 kJ) / (0.06 mol) ≈ -175 kJ/mol Thermochemical Equation: The reaction is represented as: Reaction → -175 kJ/mol Molar Heat of Combustion Page 7  Created by Turbolearn AI Scenario: Combustion of ethanol; measured heat transferred to 100 cm³ of water. Experiment: Initial mass of ethanol: 24 g Final mass of ethanol: 23.75 g Mass of ethanol used: 0.25 g Initial water temperature: 24°C Final water temperature: 44°C ΔT = 44°C - 24°C = 20°C = 20 K Mass of water: 100 cm³ * (1 g/cm³) = 100 g = 0.1 kg Calculations: Heat absorbed by water: ΔH = mcΔT = (0.1 kg)(4.2 kJ/kg·K)(20 K) = 8.4 kJ Moles of ethanol: (0.25 g) / (46 g/mol) = 0.0054 mol (approximately 0.005 mol) Molar enthalpy change: (-8.4 kJ) / (0.005 mol) ≈ -1680 kJ/mol (Note: Negative because it's exothermic) Thermochemical Equation: The reaction is represented as: Reaction → -1680 kJ/mol Hess's Law Statement: The heat change in a reaction is the same whether it occurs in one step or several steps. This is a consequence of the law of conservation of energy. Equation: ΔHtotal = ΔH1 + ΔH2+... Example: Formation of carbon dioxide (CO2) from carbon (C) and oxygen (O2) can occur through various reaction pathways, but the overall enthalpy change remains constant. Thermochemistry Study Guide Hess's Law Page 8  Created by Turbolearn AI Hess's Law states that the total enthalpy change for a reaction is independent of the pathway taken. This means that the total enthalpy change is the sum of the enthalpy changes for each step in a multi-step process. Example 1: Conversion of Carbon to Carbon Dioxide There are two pathways to convert carbon to carbon dioxide: A. Direct conversion: C(s) + O2(g) → CO2(g) ΔH1 = −393.5kJ B. Two-step conversion: 1. C(s) + 12 O2(g) → CO(g) ΔH2 = −110.5kJ 2. CO(g) + 12 O2(g) → CO2(g) ΔH3 = −283.0kJ According to Hess's Law: ΔHtotal = ΔH2 + ΔH3 = −110.5kJ + (−283.0kJ) = −393.5kJ. This is equal to ΔH1, demonstrating the law. Example 2: Formation of Sulfur(VI) Oxide Direct formation of SO3(g) is not possible, but we can use a two-step process: 1. S(s) + O2(g) → SO2(g) ΔH1 = −297.4kJ 2. SO2(g) + 12 O2(g) → SO3(g) ΔH2 = −97.9kJ ΔHSO3 = ΔH1 + ΔH2 = −297.4kJ + (−97.9kJ) = −395.3kJ Example 3: Formation of Methane Direct formation of methane (CH4) from its elements is difficult to measure. We can use the combustion of methane and its constituent elements to calculate its heat of formation: 1. Theoretical formation: C(s) + 2H2(g) → CH4(g) ΔHf (unknown) 2. Combustion of methane: CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ΔH = −890kJ 3. Combustion of carbon: C(s) + O2(g) → CO2(g) ΔH = −393.3kJ 4. Combustion of hydrogen: H2(g) + 12 O2(g) → H2O(l) ΔH = −286kJ Using Hess's Law: ΔHf + (−890kJ) = −393.3kJ + 2(−286kJ) which solves to ΔHf = −75kJ/mol Page 9  Created by Turbolearn AI Enthalpy of Solution, Hydration, and Lattice Energy Enthalpy of Solution: The heat change when one mole of a solute dissolves in a solvent. Hydration Energy: The energy released when one mole of gaseous ions is hydrated (surrounded by water molecules). Lattice Energy: The energy required to completely separate one mole of a solid ionic compound into its gaseous ions. The enthalpy of solution is the sum of the lattice energy and hydration energy. For sodium chloride (NaCl): Lattice energy = +781 kJ/mol (endothermic) Hydration energy of Na+ = −390kJ/mol (exothermic) Hydration energy of Cl− = −384kJ/mol (exothermic) ΔHsolution = 781kJ/mol + (−390kJ/mol) + (−384kJ/mol) = +7kJ/mol Standard Enthalpy Changes When experiments are carried out at 25°C and 1 atm pressure, the enthalpy changes measured are called standard enthalpy changes. They are denoted as ΔH θ, with a subscript indicating the type of enthalpy change. Page 10