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John E. McMurry • Robert C. Fay C H E M I S T R Y Fifth Edition Chapter 8 Thermochemistry: Chemical Energy Lecture Notes Alan D. Earhart Southeast Community College • Lincoln, NE Copyright © 2008 Pearson Prentice Hall, Inc. Energy Energy: The capacity to supply heat or do work. Kinetic Energy (EK): The energy of motion. Potential Energy (EP): Stored energy. Units: 1 cal = 4.184 J (exactly) 1 Cal = 1000 cal = 1 kcal Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/2 Energy Changes and Energy Conservation Law of Conservation of Energy: Energy cannot be created or destroyed; it can only be converted from one form to another. Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/3 Energy Changes and Energy Conservation Total Energy = EK + EP Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/4 Energy Changes and Energy Conservation Thermal Energy: The kinetic energy of molecular motion and is 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 © 2008 Pearson Prentice Hall, Inc. Chapter 8/5 Internal Energy and State Functions First Law of Thermodynamics: The total internal energy of an isolated system is constant. DE = Efinal - Einitial Copyright © 2008 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. DE = Efinal - Einitial Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/7 Internal Energy and State Functions First Law of Thermodynamics: The total internal energy of an isolated system is constant. DE = Efinal - Einitial Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/8 Internal Energy and State Functions CH4(g) + 2O2(g) CO2(g) + 2H2O(g) + 802 kJ energy DE = 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 © 2008 Pearson Prentice Hall, Inc. Chapter 8/9 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. Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/10 Expansion Work Work = Force x Distance w=Fxd Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/11 Expansion Work C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g) 6 mol of gas 7 mol of gas Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/12 Expansion Work Expansion Work: Work done as the result of a volume change in the system. Also called pressure-volume or PV work. Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/13 Energy and Enthalpy DE = q + w q = heat transferred w = work = -PDV q = DE + PDV Constant Volume (DV = 0): qV = DE Constant Pressure: qP = DE + PDV Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/14 Energy and Enthalpy qP = DE + PDV = DH 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. DH = Hfinal - Hinitial = Hproducts - Hreactants Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/15 The Thermodynamic Standard State C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g) DH = -2043 kJ C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) DH = -2219 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) DH° = -2043 kJ Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/16 Enthalpies of Physical and Chemical Change Enthalpy of Fusion (DHfusion): The amount of heat necessary to melt a substance without changing its temperature. Enthalpy of Vaporization (DHvap): The amount of heat required to vaporize a substance without changing its temperature. Enthalpy of Sublimation (DHsubl): The amount of heat required to convert a substance from a solid to a gas without going through a liquid phase. Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/17 Enthalpies of Physical and Chemical Change Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/18 Enthalpies of Physical and Chemical Change 2Al(s) + Fe2O3(s) 2Fe(s) + Al2O3(s) DH° = -852 kJ exothermic 2Fe(s) + Al2O3(s) 2Al(s) + Fe2O3(s) DH° = +852 kJ endothermic Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/19 Calorimetry and Heat Capacity Measure the heat flow at constant pressure (DH). Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/20 Calorimetry and Heat Capacity Measure the heat flow at constant volume (DE). Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/21 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 DT 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 DT 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 DT Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/22 Calorimetry and Heat Capacity Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/23 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 DT J Specific heat = 4.18 g °C Mass = 350 g Temperature change = 25 °C - 3 °C = 22 °C Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/24 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 © 2008 Pearson Prentice Hall, Inc. Chapter 8/25 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) DH° = -92.2 kJ Multiple-Step Process 2H2(g) + N2(g) N2H4(g) DH°1 = ? N2H4(g) + H2(g) 2NH3(g) DH°2 = -187.6 kJ 3H2(g) + N2(g) 2NH3(g) DH°1+2 = -92.2 kJ Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/26 Hess’s Law DH°1 + DH°2 = DH°1+2 DH°1 = DH°1+2 - DH°2 = -92.2 kJ - (-187.6 kJ) = 95.4 kJ Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/27 Standard Heats of Formation Standard Heat of Formation (DH°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) DH°f = -74.8 kJ 1 mol of 1 substance Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/28 Standard Heats of Formation Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/29 Standard Heats of Formation DH° = DH°f (Products) - DH°f (Reactants) aA + bB cC + dD DH° = [c DH°f (C) + d DH°f (D)] - [a DH°f (A) + b DH°f (B)] Products Reactants Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/30 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) DH° = ? DH° = [DH°f (C6H12O6(s))] - [6 DH°f (CO2(g)) + 6 DH°f (H2O(l))] DH° = [(1 mol)(-1260 kJ/mol)] [(6 mol)(-393.5 kJ/mol) + (6 mol)(-285.8 kJ/mol)] = 2816 kJ Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/31 Standard Heats of Formation 6CO2(g) + 6H2O(l) C6H12O6(s) + 6O2(g) DH° = 2816 kJ Why does the calculation “work”? 1. Reverse the “reaction” and reverse the sign on the standard enthalpy change. C(g) + O2(g) CO2(g) DH° = -393.5 kJ C(g) + O2(g) DH° = 393.5 kJ Becomes CO2(g) Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/32 Standard Heats of Formation 6CO2(g) + 6H2O(l) C6H12O6(s) + 6O2(g) DH° = 2816 kJ Why does the calculation “work”? 2. Multiply the coefficients by a factor and multiply the standard enthalpy change by the same factor. CO2(g) C(g) + O2(g) DH° = -393.5 kJ Becomes 6CO2(g) 6C(g) + 6O2(g) DH° = 6(393.5 kJ) = 2361 kJ Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/33 Standard Heats of Formation 6CO2(g) + 6H2O(l) C6H12O6(s) + 6O2(g) DH° = 2816 kJ Why does the calculation “work”? C6H12O6(s) DH° = -1260 kJ 6CO2(g) 6C(g) + 6O2(g) DH° = 2361 kJ 6H2O(l) 6H2(g) + 3O2(g) DH° = 1715 kJ 6C(s) + 6H2(g) + 3O2(g) 6CO2(g) + 6H2O(l) C6H12O6(s) + 6O2(g) DH° = 2816 kJ Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/34 Bond Dissociation Energies Bond dissociation energies are standard enthalpy changes for the corresponding bond-breaking reactions. Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/35 Bond Dissociation Energies H2(g) + Cl2(g) 2HCl(g) DH° = D(Reactant bonds) - D(Product bonds) DH° = (DH-H + DCl-Cl) - (2DH-Cl) DH° = [(1 mol)(436 kJ/mol) + (1mol)(243 kJ/mol)] - (2mol)(432 kJ/mol) = -185 kJ Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/36 Fossil Fuels, Fuel Efficiency, and Heats of Combustion CH4(g) + 2O2(g) CO2(g) + 2H2O(l) DH°c = [DH°f (CO2(g)) + 2 DH°f (H2O(l))] - [DH°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 © 2008 Pearson Prentice Hall, Inc. Chapter 8/37 Fossil Fuels, Fuel Efficiency, and Heats of Combustion CH4(g) + 2O2(g) CO2(g) + 2H2O(l) Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/38 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 DH). • favored by an increase in S (positive DS). Nonspontaneous processes are • favored by an increase in H (positive DH). • favored by a decrease in S (negative DS). Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/39 An Introduction to Free Energy Gibbs Free Energy Change (DG) DG = DH - T DS Enthalpy of reaction Temperature (Kelvin) Copyright © 2008 Pearson Prentice Hall, Inc. Entropy change Chapter 8/40 An Introduction to Free Energy Gibbs Free Energy Change (DG) DG = DH - T DS DG < 0 Process is spontaneous DG = 0 Process is at equilibrium (neither spontaneous nor nonspontaneous) DG > 0 Process is nonspontaneous Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/41