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Energy Relationships in Chemical Reactions

This chapter explores the various forms of energy in chemical reactions, including thermal, chemical, nuclear, electrical, and potential energies. It also covers thermochemical definitions, determining energy changes in a system, and energy changes in chemical reactions. Includes worked problems and thermochemical equations.

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Energy Relationships in Chemical Reactions

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  1. Energy Relationships in Chemical Reactions Chapter 6

  2. Energyis the capacity to do work • Thermal energy is the energy associated with the random motion of atoms and molecules • Chemical energy is the energy stored within the bonds of chemical substances • Nuclear energy is the energy stored within the collection of neutrons and protons in the atom • Electrical energy is the energy associated with the flow of electrons • Potential energy is the energy available by virtue of an object’s position 6.1

  3. Thermochemical Definitions System : That part of the Universe whose change we are going to measure. Surroundings : Every thing else that is relevant to the change is defined as the “surroundings”. Internal Energy : The sum of the kinetic and potential energy of all the particles in a system. Heat (q) : Is the energy transferred between a system and it’s surroundings as result in the differences in their temperatures only! Work (w) : The energy transferred when an object is moved by a force. Therefore: E = q + w

  4. Thermodynamics State functions are properties that are determined by the state of the system, regardless of how that condition was achieved. energy , pressure, volume, temperature Potential energy of hiker 1 and hiker 2 is the same even though they took different paths. 6.7

  5. Thermodynamics DE = q + w DE is the change in internal energy of a system q is the heat exchange between the system and the surroundings w is the work done on (or by) the system w = -PDVwhen a gas expands against a constant external pressure 6.7

  6. Determining Energy Change in a System Problem: In the internal combustion engine, the heat produced by the combustion of the fuel causes the carbon dioxide and water that is produced during the combustion to expand, pushing the pistons. Excess heat is removed by the cooling system. Determine the change in energy ( E) in J, kJ, and kcal if the expanding gases do 515 J of work on the pistons, and and the system loses 407 J of heat to the cooling system. q = - 407 J w = -515 J E = q + w = - 407 J + ( - 515 J) = - 922 J

  7. E is a state function, q and w are not.

  8. First Law of Thermodynamics ( Law of Conservation of Energy ) “The Total Energy of the Universe is Constant” E of Universe = ESystem + ESurroundings = 0

  9. Temperature = Thermal Energy 900C 400C Energy Changes in Chemical Reactions Heat is the transfer of thermal energy between two bodies that are at different temperatures. Temperature is a measure of the thermal energy. greater thermal energy 6.2

  10. Problems from pages 185-188 1-4, 6-11, 15, 16, 19, 21, 22, 24, 27, 28, 31, 32, 35-37, 39, 43, 44, 48, 50, 54, 56, 66 This week – Exp #9

  11. SURROUNDINGS SYSTEM Thermochemistry is the study of heat change in chemical reactions. The system is the specific part of the universe that is of interest in the study. closed isolated open energy nothing Exchange: mass & energy 6.2

  12. 2H2(g) + O2(g) 2H2O (l) + energy H2O (g) H2O (l) + energy energy + 2HgO (s) 2Hg (l) + O2(g) energy + H2O (s) H2O (l) Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings. Endothermic process is any process in which heat has to be supplied to the system from the surroundings. 6.2

  13. Change in Enthalpy = H Enthalpy is defined as the system’s internal energy plus the product of its pressure and volume. H = E + PV For a change in enthalpy: H = E + PV Exothermic and Endothermic Reactions: H = H final - H initial = H products - H reactants Exothermic : H final H initial H 0 Endothermic : H final H initial H 0

  14. For processes occurring at constant pressure the enthalpy change equals the heat gained or lost.H = qpenthalpy of reaction or heat of reaction. (Energy change + small correction factor.)

  15. Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure. DH = H (products) – H (reactants) DH = heat given off or absorbed during a reaction at constant pressure Hproducts < Hreactants Hproducts > Hreactants DH < 0 DH > 0 6.3

  16. H2O (s) H2O (l) DH = 6.01 kJ Thermochemical Equations Is DH negative or positive? System absorbs heat Endothermic DH > 0 6.01 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm. 6.3

  17. DH = -890.4 kJ CH4(g) + 2O2(g) CO2(g) + 2H2O (l) Thermochemical Equations Is DH negative or positive? System gives off heat Exothermic DH < 0 890.4 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm. 6.3

  18. 2H2O (s) 2H2O (l) H2O (s) H2O (l) H2O (l) H2O (s) DH = -6.01 kJ DH = 6.01 kJ DH = 2 x 6.01= 12.0 kJ Thermochemical Equations • The stoichiometric coefficients always refer to the number of moles of a substance • If you reverse a reaction, the sign of DH changes • If you multiply both sides of the equation by a factor n, then DH must change by the same factor n. 6.3

  19. How much heat is evolved when 266 g of white phosphorus (P4) burn in air? P4(s) + 5O2(g) P4O10(s)DH = -3013 kJ x H2O (l) H2O (g) H2O (s) H2O (l) 3013 kJ 1 mol P4 x DH = 6.01 kJ DH = 44.0 kJ 1 mol P4 123.9 g P4 Thermochemical Equations • The physical states of all reactants and products must be specified in thermochemical equations. = 6470 kJ 266 g P4 6.3

  20. The specific heat (s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius. The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius. C = ms Heat (q) absorbed or released: q = msDt q = CDt Dt = tfinal - tinitial 6.4

  21. How much heat is given off when an 869 g iron bar cools from 940C to 50C? s of Fe = 0.444 J/g •0C Dt = tfinal – tinitial = 50C – 940C = -890C q = msDt = 869 g x 0.444 J/g •0C x –890C = -34,000 J 6.4

  22. DH = qrxn Constant-Volume Calorimetry qsys = qwater + qbomb + qrxn qsys = 0 qrxn = - (qwater + qbomb) qwater = msDt qbomb = CbombDt Reaction at Constant V DH~qrxn No heat enters or leaves! 6.4

  23. Constant-Pressure Calorimetry qsys = qwater + qcal + qrxn qsys = 0 qrxn = - (qwater + qcal) qwater = msDt qcal = CcalDt Reaction at Constant P DH = qrxn No heat enters or leaves! 6.4

  24. 6.4

  25. Establish an arbitrary scale with the standard enthalpy of formation (DHo) as a reference point for all enthalpy expressions. f Standard enthalpy of formation (DH0) is the heat change that results when one mole of a compound is formed from its elements (in most stable form) at a pressure of 1 atm. f DHo (O2) = 0 DHo (O3) = 142 kJ/mol DHo (C, graphite) = 0 DHo (C, diamond) = 1.90 kJ/mol f f f f Because there is no way to measure the absolute value of the enthalpy of a substance, must I measure the enthalpy change for every reaction of interest? The standard enthalpy of formation of any element in its most stable form is zero. 6.5

  26. 6.5

  27. The standard enthalpy of reaction (DH0 ) is the enthalpy of a reaction carried out at 1 atm. rxn aA + bB cC + dD - [ + ] [ + ] = - S S = DHo DHo rxn rxn mDHo (reactants) dDHo (D) nDHo (products) cDHo (C) aDHo (A) bDHo (B) f f f f f f Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. (Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.) 6.5

  28. Hess’s Law of Heat Summation The enthalpy change of an overall process is the sum of the enthalpy changes of its individual steps. Example: Problem: Calculate the energy involved in the oxidation of elemental sulfur to sulfur trioxide from reactions: 1) S (s) + O2 (g) SO2 (g) H1 = -296.0 kJ 2) 2 SO2 (g) + O2 (g) 2 SO3 (g) H2 = -198.2 kJ 3) S (s) + 3/2 O2 (g) SO3 (g) H3 = ?

  29. If you add two or more equations to get a new equation, you must add the H’sto get theH for the new equation.

  30. C(graphite) + O2(g) CO2(g)DH0 = -393.5 kJ rxn S(rhombic) + O2(g) SO2(g)DH0 = -296.1 kJ rxn CS2(l) + 3O2(g) CO2(g) + 2SO2(g)DH0 = -1072 kJ rxn 2S(rhombic) + 2O2(g) 2SO2(g)DH0 = -296.1x2 kJ C(graphite) + 2S(rhombic) CS2 (l) C(graphite) + 2S(rhombic) CS2 (l) rxn rxn C(graphite) + O2(g) CO2(g)DH0 = -393.5 kJ + CO2(g) + 2SO2(g) CS2(l) + 3O2(g)DH0 = +1072 kJ rxn DH0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJ rxn Calculate the standard enthalpy of formation of CS2 (l) given that: 1. Write the enthalpy of formation reaction for CS2 2. Add the given rxns so that the result is the desired rxn.

  31. 2C6H6(l) + 15O2(g) 12CO2(g) + 6H2O (l) - S S = DH0 DH0 DH0 - [ ] [ + ] = rxn rxn rxn [ 12x–393.5 + 6x–285.8 ] – [ 2x49.04 ] = -6534.88 kJ = 12DH0 (CO2) 2DH0 (C6H6) f f = - 3267 kJ/mol C6H6 6DH0 (H2O) -6535kJ f 2 mol mDH0 (reactants) nDH0 (products) f f Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol. 6.5

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