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Chapter-5- Energy and Energy Changes: Thermochemistry الطاقة وتغيرات الطاقة: الكيمياء الحرارية

General Chemistry I ( Chem 1010) 1437/1438 (Semester 1). Chapter-5- Energy and Energy Changes: Thermochemistry الطاقة وتغيرات الطاقة: الكيمياء الحرارية. Dr. El Hassane ANOUAR

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Chapter-5- Energy and Energy Changes: Thermochemistry الطاقة وتغيرات الطاقة: الكيمياء الحرارية

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  1. General Chemistry I (Chem 1010) 1437/1438 (Semester 1) Chapter-5-Energy and Energy Changes: Thermochemistryالطاقة وتغيرات الطاقة: الكيمياء الحرارية Dr. El Hassane ANOUAR Chemistry Department, College of Sciences and Humanities, Prince Sattam bin Abdulaziz University, P.O. Box 83, Al-Kharij 11942, Saudi Arabia. Faculty of Applied Sciences, UiTM

  2. 5. 1 Measuring energy in chemical reactions • Almost, all chemical reactions involve either the release or the absorption of heat. Heat is released • Burning of coal and gasoline • Dynamite and TNT decomposition (detonation) Heat is released • The combustion of hydrogen • gas in oxygen Heat is released • Heat is the transfer of thermal energy between two bodies that are at different temperatures. • Thermochemistry is the study of heat change in • chemical reactions.

  3. 5. 1 Measuring energy in chemical reactions • Chemical reactions or physical changes are classified as:. • Exothermic process in which heat is evolved (q is negative) • Endothermic process in which heat is absorbed (q is positive)

  4. 5. 1 Measuring energy in chemical reactions • To analyze energy changes associated with chemical reactions we must first define the system and its surroundings. For chemists, systems include substances involved in chemical and physical changes; and the surroundings are the rest of the universe outside the system

  5. 5. 1 Measuring energy in chemical reactions • The heat (q) absorbed or evolved by a reaction depends on the conditions under which the reaction occurs.. • During a chemical reaction, the heat transfer ismeasuredby a calorimeter device. = + The system is isolated: = 0 = + = Calorimeter constant where: A constant-pressure calorimeter A constant-volume calorimeter

  6. 5. 1 Measuring energy in chemical reactions • The energy of a system increases when its temperature is raised. The increase depends on the conditions under which the heating takes place. • At constant pressure, Cp is the heat capacity at constant pressure. The heat capacity is an extensive property which is defined as the amount of heat needed to change the temperature of a substance by 1°C. Its unit is J/°C; kJ/°C or Cal/°C or J/K or Cal/K.. 1 cal = 4.18 J For some applications, it is useful to know the specific heat capacity, (more informally, the ‘specific heat’) of a substance, which is the heat capacity of the sample divided by the mass, usually in grams:

  7. 5. 1 Measuring energy in chemical reactions Exercise 5. 1 How many kilojoules of heat will raise the temperature of 3.0 Kg of water from 22.0°C to 25.0°C, if c (H2O) = 4.184 J g-1 °C-1. Solution

  8. 5. 1 Measuring energy in chemical reactions Exercise 5. 2 A 10.40 g sample of silver was heated to 100.0°C. It was then added to 28.0 g of water in an insulated container. The water temperature rise from 25.0°C to 26.48°C. What is the specific heat of silver? Solution

  9. 5. 2 Heat of reactions and Thermochemistry • The heat absorbed or evolved by a reaction depends on the conditions under which the reaction occurs. • Usually, a reaction takes place in a vessel open to the atmosphere and therefore at the constant pressure of the atmosphere. The heat of reaction as qp, the subscript p indicating that the process occurs at constant pressure. • Enthalpy (denoted H) is a property of a substance that is related to the heat of reaction qp. It is an extensive property of a substance that can be used to obtain the heat absorbed or evolved in a chemical reaction. Note: An extensive property is a property that depends on the amount of substance (e.g. mass and volume.)

  10. 5. 2 Heat of reactions and Thermochemistry • Enthalpy is a state function. This means The change in enthalpy does not depend on how the change was made, but only on the initial state and final state of the system. The difference in altitude is independent of the path taken between the campsites. Note: A state function is a property of a system that depends only on its present state, which is determined by variables such as temperature and pressure, and is independent of any previous history of the system.

  11. 5. 2 Heat of reactions and Thermochemistry • The change in enthalpy , ∆H for a reaction at a given temperature and pressure (called the enthalpy of reaction) is obtained by subtracting the enthalpy of the reactants from the enthalpy of the products. Thus , • ∆H = Hfinal - Hinitial • Since, we start from reactants and end with products. Thus, the,∆H is • ∆H= H(products) - H(reactants) • ∆Hcan be positive or negative, depending on the process. • Endothermic process : Heat absorbed by the system from the surroundings => ∆H > 0 • Exothermic process; Heat released by the system to the surroundings =>∆H< 0

  12. 5. 2 Heat of reactions and Thermochemistry • ∆Hcan be positive or negative, depending on the process. • Endothermic process : Heat absorbed by the system from the surroundings => ∆H > 0 • Exothermic process: Heat released by the system to the surroundings • =>∆H< 0

  13. 5.3 Hess’s law of heat summation • In 1840, the Russian chemist Germain Henri Hess, a professor at the University of St. Petersburg, discovered by experiment that the enthalpy change for a chemical reaction is independent of the path by which the products are obtained. • Hess’s law of heat summation states that “for a chemical equation that can be written as the sum of two or more steps, the enthalpy change for the overall equation equals the sum of the enthalpy changes for the individual steps”.

  14. 5.3 Hess’s law of heat summation • 5.3.1 Manipulating thermochemical equations • A thermochemical equation: The chemical equation for a reaction (including phase labels), in which the equation is given a molar interpretation, and the enthalpy of reaction for these molar amounts is written directly after the equation. • The following guidelines are helpful in writing and interpreting thermochemical equations: • If we multiply both sides of a thermochemical equation by a factor n, then ∆H must also change by the same factor. x 2 =

  15. 5.3 Hess’s law of heat summation • 5.3.1 Manipulating thermochemical equations • The following guidelines are helpful in writing and interpreting thermochemical equations: • When we reverse a thermochemical equation, we change the roles of reactants and products. Consequently, the magnitude of ∆H for the equation remains the same, but its sign changes. Reverse reaction

  16. 5.3 Hess’s law of heat summation • Exercise 5. 3 • Giving the following thermochemical equations: • Find the for the reaction: • (4) • Solution

  17. 5.4 Standard states • The term standard state refers to the standard thermodynamic conditions chosen • for substances when listing or comparing thermodynamic data: • 1 atm pressure • Specified temperature (usually 25°C) • An allotrope is one of two or more distinct forms of an element in the same physical state

  18. 5.4 Standard states • The enthalpy change for a reaction in which reactants in their standard states yield products in their standard states is called the standard enthalpy of reaction(∆H°). • The standard enthalpy of formation, (also called the standard heat of formation) of a substance, is the enthalpy change for the formation of one mole of the substance in its standard state fromits elements in their reference form and in their standard states. • By convention, the standard enthalpy of formation of any element in its most stable form is zero. (C, graphite) = 0 (O2) = 0

  19. 5.4 Standard states • One we know values, we can calculate the standard enthalpy of reaction, defined as the enthalpy of a reaction carried out at 1 atm. • For example, consider the reaction is given by In general where m and n denote the stoichiometric coefficients for the reactants and products

  20. 5.4 Standard states Example 5.5 Consider the reaction: Calculate the standard heat of reaction, ∆H° = +31.4 kJ/mol; = -285. 9 kJ/mol; = -1088.7 kJ/mol Solution

  21. 5.4 Standard states Example 5.6 When 1.00 g of oxygen difluoride reacts with water at 1.0 atm and 25°C, 5.98 kJ of heat is liberated. Calculate ∆H° for this reaction in kJ: Solution

  22. 5.5 Conservation of energy and first law of themodynamics • Thermodynamics is the study of the interconversion of heat and other kinds of energy between a system and its surrounding (i.e., changes in the state of the system). • The state of a systemis defined by the values of all relevant macroscopic properties (e.g., composition, energy, temperature, pressure, and volume). • Energy, pressure, volume, and temperature are state functions. • State functions are properties that are determined by the state of the system, regardless of how that condition was achieved.

  23. 5.5 Conservation of energy and first law of themodynamics 5.5.1 The first law of thermodynamics • Theinternal energy, E, is the sum of the kinetic and potential energies of the particles making up the system. • The kinetic energy includes the energy of motion of electrons, nuclei, and molecules. • The potential energy results from the chemical bonding of atoms and from the attractions between molecules. • Internal energy is a state function, i.e., a property of a system that depends only on its present state, which is completely determined by variables such as temperature and pressure. • When a system changes from one state to another, its internal energy changes from one definite value to another.

  24. 5.5 Conservation of energy and first law of themodynamics 5.5.1 The first law of thermodynamics • The change in internal energy, ∆E, from the initial state value of the internal energy, Ei, and the final state value of the internal energy, Ef. ∆E = Ef - Ei • Ef < Ei (∆E < 0) => Energy is transferred from the system to the surrounding. • Ef > Ei(∆E > 0) => Energy is absorbed by the system from the surrounding. • If Ef =Ei (∆E = 0) => No change in energy between the system and the surrounding (isothermal process).

  25. 5.5 Conservation of energy and first law of themodynamics 5.5.1 The first law of thermodynamics • The exchange of energy between the system and its surrounding are of two kinds of energy heatand work. • Heat , q is energy that moves into or out of the system because of a temperature difference between the system and its surroundings. • Work, wis the energy exchange that results when a force F moves an object through a distance d:w = F ×d • The first law of thermodynamics, which is based on the law of conservation of energy, states that energy can be converted from one form to another, but cannot be created or destroyed. It states that the change in internal energy of a system, E, equals q + w ∆E = q + w

  26. 5.5 Conservation of energy and first law of themodynamics 5.5.1 The first law of thermodynamics • The transfer of energy from the system to the surroundingsdoes not change the total energy of the universe (energy can be converted from one form to another, but cannot be created or destroyed). That is, the sum of the energy changes must be zero: ∆Esys+ ∆Esur = 0 • Energy gained in one place must have been lost somewhere else. • Energy can be changed from one form to another, i.e., the energy lost by one system can be gained by another system in a different form.

  27. 5.5 Conservation of energy and first law of themodynamics 5.5.2 Work in physical and chemical systems • In thermodynamics, work has a broader meaning that includes: • Mechanical work(e.g, example, a crane lifting a steel beam) • Electrical work (e.g., a battery supplying electrons to light the bulb of a flashlight). • Surface work (blowing up a soap bubble). • One way to illustrate mechanical work is to study the expansion or compression • of a gas. The work done by the gas on the surroundings is For gas expansion (work done by the system, w < 0), ∆V > 0. For gas compression (work done on the system, w > 0), ∆V < 0. Change in volume, = Vf – Vi

  28. 5.5 Conservation of energy and first law of themodynamics • 5.5 Conservation of energy and first law of themodynamics 5.5.2 Work in physical and chemical systems 5.5.2 Work in physical and chemical systems For gas expansion (work done by the system, w < 0), ∆V > 0. For gas compression (work done on the system, w > 0), ∆V < 0. Change in volume, = Vf – Vi < 0 => wsur > 0 Thus, > 0 => wsur < 0 Thus,

  29. 5.5 Conservation of energy and first law of themodynamics 5.5.2 Work in physical and chemical systems Note that: 1 atm. L = 101.325 J = 101.325 × 10-3 kJ 1 atm. L = 24.2 cal = 24.2 × 10-3 kJ In J or atm. L

  30. 5.5 Conservation of energy and first law of themodynamics 5.5.2 Work in physical and chemical systems Example 5.6 Calculate ∆E when a system (a) Absorbs 60.0 kJ of heat and does 55.0 kJ of work (b) Evolves 75.0 kJ of heat and does 45.0 kJ of work (c) Evolves 45.0 kJ of heat and has 45.0 kJ of work done on it Solution

  31. 5.5 Conservation of energy and first law of themodynamics 5.5.2 Work in physical and chemical systems Example 5.7 If 1000 mL of gas is compressed to 500 mL under a constant external pressure of 6 atm, and if the gas also absorbs 25.20 kJ of heat, what are the values of q w ∆E for the gas ∆E for the surrounding express in kJ? Solution

  32. 5.6 Heat of reactions 5.6.1 The heat of reactions at constant volume At constant volume, ∆V = 0 => W = -P∆V = 0 Thus, qv < 0 => Exothermic process ∆E = q = qv qv > 0 => Endothermic process 5.6.2 The heat of reactions at constant pressure The enthalpy, H of a system is defined as: H = E + PV If only the expansion work is considered for the system, then qp< 0 => Exothermic process ∆H = qp qp> 0 => Endothermic process

  33. 5.6 Heat of reactions • 5.6 Heat of reactions 5.6.3 Reaction involving gases • To calculate the enthalpy change of a gaseous reaction, we assume ideal gas behavior and constant temperature. In this case, ∆H = ∆E + P∆V ∆H = ∆E + ∆n(g) RT Where At standard state (T = 298 K, and P = 1 atm) ∆H° = ∆E° + ∆n(g)R (298)

  34. 5.6 Heat of reactions Example 5.8 Calculate ∆E° for the combustion of octane C8H18(l), to CO2 (g) and H2O (l). If the heat of combustion at constant pressure of octane is -5470.71 kJ/mol. Solution

  35. 5.6 Heat of reactions Example 5.9 What is ∆E when 1.00 mol of liquid water vaporizes at 100 °C? The heat of vaporization of water at 100°C is 44.66 kJ/mol Solution

  36. 5.6 Heat of reactions Example 5.10 For the reaction NH4Cl(aq) + NaNO2(aq)  N2(g) + 2H2O(l) + NaCl(aq) The change in internal energy is -333 kJ and the change in volume to produce 1.00 mol of N2 increases by 22.4 L at 1 atm. Calculate the change in enthalpy for the above reaction. Solution

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