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Chapter 6 Thermochemistry

Chapter 6 Thermochemistry. Energy Energy = the capacity to do work or produce heat Conservation of Energy = energy can’t be created or destroyed; it can only change forms Potential Energy = due to position or composition Water behind a dam has energy that can be converted to work

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Chapter 6 Thermochemistry

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  1. Chapter 6 Thermochemistry • Energy • Energy = the capacity to do work or produce heat • Conservation of Energy = energy can’t be created or destroyed; it can only change forms • Potential Energy = due to position or composition • Water behind a dam has energy that can be converted to work • Attractive and Repulsive forces (gravity) govern this type • Gasoline burns to make heat: forces holding atoms together • Kinetic Energy = due to motion of the object • KE = ½mv2 • m = mass and v = velocity of the object • Conversion of energy types: PE  KE  PE + frictional heating

  2. Heat and Work • Heat = q = transfer of energy between objects at different temperatures • Temperature = KE of particles in random motion • Heat is not a substance, but our language often treats it that way • Work = w = force acting through a distance = F x d • Pathway: energy transfer might be the same with different results • A rough surface: mostly heat and little work • Smooth, steep surface: little heat and mostly work • State Function = property that only depends on present state • Value doesn’t depend on the path of how it got that way • Energy is a state function: DE = q + w • Work and Heat are not • Chemical Energy • Mechanical Energy = energy of the movement of objects (see above) • Chemical Energy = energy of the change in chemical bonds • CH4(g) + 2O2(g) -------> CO2(g) + 2H2O(g) + heat • Heat energy is liberated by rearranging the chemical bonds

  3. Dividing up the Universe • System = the specific reactants and products we are investigating • Surroundings = everything else in the Universe • Heat flow in chemical reactions • Exothermic = energy flows out of the system as heat • Products have a lower PE than the reactants • Heat can be viewed as a product • Heat released results in an increase in KE of surrounding particles

  4. b. Endothermic = energy flows into the system • N2(g) + O2(g) + heat -------> 2NO(g) • Heat can be viewed as a reactant • Heat absorbed results in less KE of the surrounding particles • Products have more PE than the reactants • D. Doing thermodynamics problems • 1. Thermodynamics = study of energy and its interconversions • 2. 1st Law of Thermodynamics = The energy of the universe is constant • 3. Internal Energy = E = sum of KE and PE;

  5. Thermodynamic quantities • Consist of two parts: a number (magnitude) and a sign (direction) • The sign reflects the system’s point of view • Heat flowing into a system has +q (endothermic) • Heat flowing out of a system = -q (exothermic) • Work is done on the system = +w • The system does work on the surrounding = -w • DE = q + w sums up the changes in energy during a process • Example: Find DE of an endothermic reaction if 15.6 kJ is flowing and 1.4 kJ work is done on the system. DE = q + w = 15.6 + 1.4 = 17.0kJ • 6. PV Work • Gases do work (expand) or have work done on them (compress) • Expansion of gas in engine cylinder provides mechanical work • i. Expanding gas = -w = (+P)(+DV) • ii. Compressed gas = +w = (+P)(-DV)

  6. Example: Find w for expansion of a gas from 46L to 64L at 15 atm. • w = -PDV = -(15atm)(18L) = -270Latm • 8. Example: Find DE if 1.3 x 108 J of heat expands a balloon from • 4.00 x 106 L to 4.50 x 106 L at 1.0 atm. • DE = q + w = q + (-PDV) = (1.3 x 108 J) - (1atm)(0.50 x 106 L) • = (1.3 x 108 J) - (5 x 105 Latm)(101.3 J/Latm) = 8 x 107 J • Enthalpy and Calorimetry • Enthalpy = H = E + PV • E, P, and V are all state functions, so H is also a state function • Consider a constant P process where only PV work is allowed • DE = q + w = q – PDV → q = DE + PDV • DH = DE + D(PV) • Since P is constant: D(PV) = PDV →DH = DE + PDV • DH = q At constant P, DH of a system = energy flow as heat • Heat of reaction = change in enthalpy = DH = Hproducts – Hreactants • If DH = + heat will be absorbed by the system = Endothermic • If DH = - heat will be released by the system = Exothermic

  7. Example: a. Find DH if 1 mol CH4 burns at const. P and 890 kJ heat is released. b. Also find DH if 5.8 g of CH4 burns. • q = DH = -890 kJ • Calorimetry = science of measuring heat • Calorimeter = device for measuring heat changes in a chemical reaction • Each substance changes temperature at a different rate = Heat Capacity • Specific Heat capacity = J/g•oC = Cs • Molar Heat capacity = J/mol•oC • Water has a high heat capacity • Very good coolant • Much higher than metals • 3. Constant-Pressure Calorimeter (q = DH) • Experiment done under atm. Pressure • 50.0 ml 1.0M HCl + 50.0 ml 1.0M NaOH at 25 oC (H+ + OH---> H2O) • T = 31.9 oC after the reaction

  8. Example: Calculate DH/mol for the following reaction • Ba2+ + SO42- -------> BaSO4(s) • 1L 1.00M Ba(NO3)2; 1.00L 1.00M Na2SO4; 25 oC start; 28.1 oC end • Constant Volume Calorimetry • If DV = 0, then w = 0 • DE = q + w = q • Combustion of 0.5269g octane (C8H18) in a bomb calorimeter with a heat capacity = 11.3 kJ/oC with a DT = 2.25 oC. Find DE. Example:

  9. Hess’s Law • Recall that Enthalpy is a State Function • Path doesn’t matter • As long as we know reactants and products, steps don’t matter for DH • Example: • N2(g) + O2(g) -------> 2NO(g) DH1 = 180 kJ • 2NO(g) + O2(g) -------> 2NO2(g) DH2 = -112 kJ • N2(g) + 2O2(g) -------> 2NO2(g) DHtotal = 68 kJ • Hess’s Law • You may sum steps in order to find overall DH • The DH for the reverse reaction will simply change signs (+/-) • If you multiply a reaction, you must multiply DH the same

  10. Explanation • Sign of DH depends on direction of heat flow. Heat flow is reversed if the overall reaction is reversed. • Xe(g) + 2F2(g) -------> XeF4(s) + 251kJ (DH = -251kJ) • XeF4(s) + 251kJ -------> Xe(g) + 2F2(g) (DH = +251kJ) • DH is an extensive property = depends on the amount of substance (an intensive property = depends only on identity of the substance) • Xe(g) + 2F2(g) -------> XeF4(s) + 251kJ (DH = -251kJ) • 2Xe(g) + 4F2(g) -------> 2XeF4(s) + 502kJ (DH = -502kJ) • B. Examples • 1. Example: Calculate DH for the conversion of graphite to diamond using the known DH’s for the combustion of graphite and diamond. • Cg(s) + O2(g) -------> CO2(g) DH = -394kJ • Cd(s) + O2(g) -------> CO2(g) DH = -396 kJ • b. Cg(s) + O2(g) -------> CO2(g) DH = -394kJ • CO2(g) -------> Cd(s) + O2(g) (reverse) DH = +396 kJ • Cg(s) -------> Cd(s) DH = +2 kJ

  11. 2. Example: Find DH for synthesis of B2H6 from B and H2. • Known Reactions • 2B(s) + 3/2O2(g) -------> B2O3(s) DH = -1273 kJ • B2H6(g) + 3O2(g) -------> B2O3(s) + 3H2O(g) DH = -2035 kJ • H2(g) + 1/2O2(g) -------> H2O(l) DH = -286 kJ • H2O(l) -------> H2O(g) DH = +44 kJ • Combining Reactions • 2B(s) + 3/2O2(g) -------> B2O3(s) DH = -1273 kJ • B2O3(s) + 3H2O(g) -------> B2H6(g) + 3O2(g) DH = +2035 kJ • 2B(s) + 3H2O(g) -------> B2H6(g) + 3/2O2(g) DH = +762 kJ • 3[H2(g) + 1/2O2(g) -------> H2O(l)] DH = -858 kJ • 3[H2O(l) -------> H2O(g)] DH = +132 kJ • 2B(s) + 3H2(g) -------> B2H6(g) DH = +36 kJ • c. Let the final needed reaction guide how you combine other reactions + = + + =

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