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John E. McMurry • Robert C. Fay General Chemistry: Atoms First Chapter 8 Thermochemistry: Chemical Energy Lecture Notes Alan D. Earhart Southeast Community College • Lincoln, NE Copyright © 2010 Pearson Prentice Hall, Inc. Energy and Its Conservation Energy: The capacity to supply heat or do work. Kinetic Energy (EK): The energy of motion. Potential Energy (EP): Stored energy. Copyright © 2010 Pearson Prentice Hall, Inc. Chapter 8/2 Energy and Its Conservation Law of Conservation of Energy: Energy cannot be created or destroyed; it can only be converted from one form to another. Copyright © 2010 Pearson Prentice Hall, Inc. Chapter 8/3 Energy and Its Conservation Total Energy = EK + EP Energy and Its Conservation Thermal Energy: The kinetic energy of molecular motion, measured by finding the temperature of an object. Heat: The amount of thermal energy transferred from one object to another as the result of a temperature difference between the two. Copyright © 2010 Pearson Prentice Hall, Inc. Chapter 8/5 Energy and Its Conservation First Law of Thermodynamics: Energy cannot be created or destroyed; it can only be converted from one form to another. Copyright © 2010 Pearson Prentice Hall, Inc. Chapter 8/6 Internal Energy and State Functions First Law of Thermodynamics: The total internal energy of an isolated system is constant. ∆E = Efinal - Einitial Internal Energy and State Functions First Law of Thermodynamics: The total internal energy of an isolated system is constant. ∆E = Efinal - Einitial Internal Energy and State Functions First Law of Thermodynamics: The total internal energy of an isolated system is constant. ∆E = Efinal - Einitial Internal Energy and State Functions CH4(g) + 2O2(g) CO2(g) + 2H2O(g) + 802 kJ energy ∆E = Efinal - Einitial = -802 kJ 802 kJ is released when 1 mole of methane, CH4, reacts with 2 moles of oxygen to produce 1 mole of carbon dioxide and two moles of water. Copyright © 2010 Pearson Prentice Hall, Inc. Chapter 8/10 Internal Energy and State Functions State Function: A function or property whose value depends only on the present state, or condition, of the system, not on the path used to arrive at that state. Expansion Work Work = Force x Distance w=Fxd Expansion Work C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g) 6 mol of gas 7 mol of gas Expansion Work Expansion Work: Work done as the result of a volume change in the system. Also called pressure-volume or PV work. Energy and Enthalpy ∆E = q + w q = heat transferred w = work = -P∆V q = ∆E + P∆V Constant Volume (∆V = 0): qV = ∆E Constant Pressure: qP = ∆E + P∆V Copyright © 2010 Pearson Prentice Hall, Inc. Chapter 8/15 Energy and Enthalpy qP = ∆E + P∆V = ∆H Enthalpy change or Heat of reaction (at constant pressure) Enthalpy is a state function whose value depends only on the current state of the system, not on the path taken to arrive at that state. ∆H = Hfinal - Hinitial = Hproducts - Hreactants Copyright © 2010 Pearson Prentice Hall, Inc. Chapter 8/17 The Thermodynamic Standard State C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g) ∆H = -2044 kJ C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) ∆H = -2220 kJ Thermodynamic Standard State: Most stable form of a substance at 1 atm pressure and at a specified temperature, usually 25 °C; 1 M concentration for all substances in solution. C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g) ∆H° = -2044 kJ Copyright © 2010 Pearson Prentice Hall, Inc. Chapter 8/18 Enthalpies of Physical and Chemical Change Enthalpy of Fusion (∆Hfusion): The amount of heat necessary to melt a substance without changing its temperature. Enthalpy of Vaporization (∆Hvap): The amount of heat required to vaporize a substance without changing its temperature. Enthalpy of Sublimation (∆Hsubl): The amount of heat required to convert a substance from a solid to a gas without going through a liquid phase. Copyright © 2010 Pearson Prentice Hall, Inc. Chapter 8/19 Enthalpies of Physical and Chemical Change Chapter 8/20 Enthalpies of Physical and Chemical Change 2Al(s) + Fe2O3(s) 2Fe(s) + Al2O3(s) ∆H° = -852 kJ Exothermic: Heat (enthalpy) flows from the system to the surroundings. Ba(OH)2 •8H2O(s) + 2NH4Cl(s) BaCl2(aq) + 2NH3(aq) + 10H2O(l) ∆H° = +80.3 kJ Endothermic: Heat (enthalpy) flows from the surroundings to the system. Copyright © 2010 Pearson Prentice Hall, Inc. Chapter 8/21 Enthalpies of Physical and Chemical Change 2Al(s) + Fe2O3(s) 2Fe(s) + Al2O3(s) ∆H° = -852 kJ Exothermic: Heat (enthalpy) flows from the system to the surroundings. 2Fe(s) + Al2O3(s) 2Al(s) + Fe2O3(s) ∆H° = +852 kJ Endothermic: Heat (enthalpy) flows from the surroundings to the system. Copyright © 2010 Pearson Prentice Hall, Inc. Chapter 8/22 Calorimetry and Heat Capacity Measure the heat flow at constant pressure (∆H). Calorimetry and Heat Capacity Measure the heat flow at constant volume (∆E). Calorimetry and Heat Capacity Heat Capacity (C): The amount of heat required to raise the temperature of an object or substance a given amount. q = C x ∆T Molar Heat Capacity (Cm): The amount of heat required to raise the temperature of 1 mol of a substance by 1 °C. q = (Cm) x (moles of substance) x ∆T Specific Heat: The amount of heat required to raise the temperature of 1 g of a substance by 1 °C. q = (specific heat) x (mass of substance) x ∆T Copyright © 2010 Pearson Prentice Hall, Inc. Chapter 8/25 Calorimetry and Heat Capacity Calorimetry and Heat Capacity Assuming that a can of soda has the same specific heat as water, calculate the amount of heat (in kilojoules) transferred when one can (about 350 g) is cooled from 25 °C to 3 °C. q = (specific heat) x (mass of substance) x ∆T J Specific heat = 4.18 g °C Mass = 350 g Temperature change = 25 °C - 3 °C = 22 °C Copyright © 2010 Pearson Prentice Hall, Inc. Chapter 8/27 Calorimetry and Heat Capacity Calculate the amount of heat transferred. 4.18 J x 350 g x 22 °C Heat evolved = = 32 000 J g °C 32 000 J x 1 kJ = 32 kJ 1000 J Copyright © 2010 Pearson Prentice Hall, Inc. Chapter 8/28 Hess’s Law Hess’s Law: The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for the individual steps in the reaction. Haber Process: 3H2(g) + N2(g) 2NH3(g) ∆H° = -92.2 kJ Multiple-Step Process 2H2(g) + N2(g) N2H4(g) ∆H°1 = ? N2H4(g) + H2(g) 2NH3(g) ∆H°2 = -187.6 kJ 3H2(g) + N2(g) 2NH3(g) ∆H°1+2 = -92.2 kJ Copyright © 2010 Pearson Prentice Hall, Inc. Chapter 8/29 Hess’s Law ∆H°1 + ∆H°2 = ∆H°1+2 ∆H°1 = ∆H°1+2 - ∆H°2 = -92.2 kJ - (-187.6 kJ) = 95.4 kJ Standard Heats of Formation Standard Heat of Formation (∆H°f ): The enthalpy change for the formation of 1 mol of a substance in its standard state from its constituent elements in their standard states. Standard states C(s) + 2H2(g) CH4(g) ∆H°f = -74.8 kJ 1 mol of 1 substance Copyright © 2010 Pearson Prentice Hall, Inc. Chapter 8/31 Standard Heats of Formation Chapter 8/32 Standard Heats of Formation ∆H° = ∆H°f (Products) - ∆H°f (Reactants) aA + bB cC + dD ∆H° = [c ∆H°f (C) + d ∆H°f (D)] - [a ∆H°f (A) + b ∆H°f (B)] Products Reactants Copyright © 2010 Pearson Prentice Hall, Inc. Chapter 8/33 Standard Heats of Formation Using standard heats of formation, calculate the standard enthalpy of reaction for the photosynthesis of glucose (C6H12O6) and O2 from CO2 and liquid H2O. 6CO2(g) + 6H2O(l) C6H12O6(s) + 6O2(g) ∆H° = ? H° = [∆H°f (C6H12O6(s))] - [6 ∆H°f (CO2(g)) + 6 ∆H°f (H2O(l))] ∆H° = [(1 mol)(-1273.3 kJ/mol)] [(6 mol)(-393.5 kJ/mol) + (6 mol)(-285.8 kJ/mol)] = 2802.5 kJ Copyright © 2010 Pearson Prentice Hall, Inc. Chapter 8/34 Standard Heats of Formation 6CO2(g) + 6H2O(l) C6H12O6(s) + 6O2(g) ∆H° = 2802.5 kJ (1) (2) (3) Why does the calculation “work”? 1. Reverse the “reaction” and reverse the sign on the standard enthalpy change. C(s) + O2(g) CO2(g) ∆H° = -393.5 kJ C(s) + O2(g) ∆H° = 393.5 kJ Becomes CO2(g) Copyright © 2010 Pearson Prentice Hall, Inc. Chapter 8/35 Standard Heats of Formation 6CO2(g) + 6H2O(l) C6H12O6(s) + 6O2(g) ∆H° = 2802.5 kJ (1) (2) (3) Why does the calculation “work”? 1. Multiply the coefficients by a factor and multiply the standard enthalpy change by the same factor. CO2(g) C(s) + O2(g) ∆H° = -393.5 kJ Becomes 6CO2(g) 6C(s) + 6O2(g) ∆H° = 6(393.5 kJ) = 2361.0 kJ Copyright © 2010 Pearson Prentice Hall, Inc. Chapter 8/36 Standard Heats of Formation 6CO2(g) + 6H2O(l) C6H12O6(s) + 6O2(g) ∆H° = 2802.5 kJ (1) (2) (3) Why does the calculation “work”? 2. Reverse the “reaction” and reverse the sign on the standard enthalpy change. 1 H2(g) + O2(g) 2 H2O(l) ∆H° = -285.8 kJ Becomes H2O(l) 1 H2(g) + O2(g) ∆H° = 285.8 kJ 2 Copyright © 2010 Pearson Prentice Hall, Inc. Chapter 8/37 Standard Heats of Formation 6CO2(g) + 6H2O(l) C6H12O6(s) + 6O2(g) ∆H° = 2802.5 kJ (1) (2) (3) Why does the calculation “work”? 2. Multiply the coefficients by a factor and multiply the standard enthalpy change by the same factor. H2O(l) 1 H2(g) + O2(g) 2 ∆H° = 285.8 kJ 6H2(g) + 3O2(g) ∆H° = 6(285.8 kJ) = 1714.8 kJ Becomes 6H2O(l) Copyright © 2010 Pearson Prentice Hall, Inc. Chapter 8/38 Standard Heats of Formation 6CO2(g) + 6H2O(l) C6H12O6(s) + 6O2(g) ∆H° = 2802.5 kJ (1) (2) (3) Why does the calculation “work”? C6H12O6(s) ∆H° = -1273.3 kJ 6CO2(g) 6C(s) + 6O2(g) ∆H° = 2361.0 kJ 6H2O(l) 6H2(g) + 3O2(g) ∆H° = 1714.8 kJ 6C(s) + 6H2(g) + 3O2(g) 6CO2(g) + 6H2O(l) C6H12O6(s) + 6O2(g) Copyright © 2010 Pearson Prentice Hall, Inc. ∆H° = 2802.5 kJ Chapter 8/39 Bond Dissociation Energies Bond Dissociation Energy: • The amount of energy that must be supplied to break a chemical bond in an isolated molecule in the gaseous state and is thus the amount of energy released when the bond forms. or • Standard enthalpy changes for the corresponding bond-breaking reactions. Copyright © 2010 Pearson Prentice Hall, Inc. Chapter 8/40 Bond Dissociation Energies Bond Dissociation Energies H2(g) + Cl2(g) 2HCl(g) ∆H° = D(Reactant bonds) - D(Product bonds) ∆H° = (DH-H + DCl-Cl) - (2DH-Cl) ∆H° = [(1 mol)(436 kJ/mol) + (1mol)(243 kJ/mol)] - (2mol)(432 kJ/mol) = -185 kJ Copyright © 2010 Pearson Prentice Hall, Inc. Chapter 8/42 Fossil Fuels, Fuel Efficiency, and Heats of Combustion CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ∆H°c = [∆H°f (CO2(g)) + 2 ∆H°f (H2O(l))] - [∆H°f (CH4(g))] = [(1 mol)(-393.5 kJ/mol) + (2 mol)(-285.8 kJ/mol)] [(1 mol)(-74.8 kJ/mol)] = -890.3 kJ Copyright © 2010 Pearson Prentice Hall, Inc. Chapter 8/43 Fossil Fuels, Fuel Efficiency, and Heats of Combustion CH4(g) + 2O2(g) CO2(g) + 2H2O(l) An Introduction to Entropy Spontaneous Process: A process that, once started, proceeds on its own without a continuous external influence. Entropy (S): The amount of molecular randomness in a system. Spontaneous processes are • favored by a decrease in H (negative ∆H). • favored by an increase in S (positive ∆S). Nonspontaneous processes are • favored by an increase in H (positive ∆H). • favored by a decrease in S (negative ∆S). Copyright © 2010 Pearson Prentice Hall, Inc. Chapter 8/45 An Introduction to Free Energy Gibbs Free Energy Change (∆G) ∆G = ∆H - T ∆S Enthalpy of reaction Temperature (Kelvin) Copyright © 2010 Pearson Prentice Hall, Inc. Entropy change Chapter 8/47 An Introduction to Free Energy Gibbs Free Energy Change (∆G) ∆G = ∆H - T ∆S ∆G < 0 Process is spontaneous ∆G = 0 Process is at equilibrium (neither spontaneous nor nonspontaneous) ∆G > 0 Process is nonspontaneous Copyright © 2010 Pearson Prentice Hall, Inc. Chapter 8/48 An Introduction to Free Energy