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Chapter 5 Thermochemistry
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Thermodynamics Study of the changes in energy and transfers of energy that accompany chemical and physical processes. address 3 fundamental questions Will two (or more) substances react when they are mixed under specified conditions? If they do react, what energy changes and transfers are associated with their reaction? If a reaction occurs, to what extent does it occur?
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The First Law of Thermodynamics Exothermic reactions combustion of propane combustion of n-butane
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The First Law of Thermodynamics Exothermic reactions release specific amounts of heat as products Potential energies of products are lower than potential energies of reactants.
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The First Law of Thermodynamics 2 basic ideas of importance systems tend toward a state of minimum potential energy
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The First Law of Thermodynamics 2 basic ideas of importance systems tend toward a state of maximum disorder
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The First Law of Thermodynamics Also known as Law of Conservation of Energy The total amount of energy in the universe is constant. Energy can be converted from one form to another but cannot be created.
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Some Thermodynamic Terms System - substances involved in the chemical and physical changes under investigation for us this is what is happening inside the beaker Surroundings - rest of the universe outside the beaker Universe - system plus surroundings
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Some Thermodynamic Terms Thermodynamic State of a System - set of conditions that describe and define the system number of moles of each substance physical states of each substance temperature pressure State Functions - properties of a system that depend only on the state of the system capital letters
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Some Thermodynamic Terms state functions are independent of pathway climbing a mountain, taking two different paths E 1 = energy at bottom of mountain E 1 = mgh 1 E 2 = energy at top of mountain E 2 = mgh 2 E 2 - E 1 = mgh 2 - mgh 1 = mg( h)
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Some Thermodynamic Terms Properties that depend only on values of state functions are also state functions examples: T P V
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Enthalpy Change, H Commonly, chemistry is done at constant pressure open beakers on a desk top are at atmospheric pressure H - enthalpy change change in heat content at constant pressure H = q p H rxn - heat of reaction H rxn = H products - H reactants H rxn = H substances produced - H substances consumed
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Thermochemical Equations Thermochemical equations are a balanced chemical reaction plus the H value for the reaction. for example: coefficients in thermochemical equations must be interpreted as numbers of moles 1 mol of C 5 H 12 reacts with 8 mol of O 2 to produce 5 mol of CO 2, 6 mol of H 2 O, and releasing 3523 kJ is referred to as one mole of reactions
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Thermochemical Equations Equivalent method of writing thermochemical equations H < 0 designates an exothermic reaction H > 0 designates an endothermic reaction
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Standard States & Standard Enthalpy Changes Thermochemical standard state conditions T = 298.15 K P = 1.0000 atm Thermochemical standard states pure substances in their liquid or solid phase - standard state is the pure liquid or solid gases - standard state is the gas at 1.00 atm of pressure gaseous mixtures - partial pressure must be 1.00 atm aqueous solutions - 1.00 M concentration
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Standard Molar Enthalpies of Formation, H f o Standard molar enthalpy of formation symbol is H f o defined as the enthalpy for the reaction in which one mole of a substance is formed from its constituent elements for example:
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Standard Molar Enthalpies of Formation, H f o Standard molar enthalpies of formation have been determined for many substances and are tabulated in Table 5.3 and Appendix C in the text. Standard molar enthalpies of elements in their most stable forms at 298.15 K and 1.000 atm are zero.
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Hess’s Law Hess’s Law in a more useful form any chemical reaction at standard conditions, the standard enthalpy change is the sum of the standard molar enthalpies of formation of the products (each multiplied by its coefficient in the balanced chemical equation) minus the corresponding sum for the reactants
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Hess’s Law Example: Calculate H o 298 for the following reaction from data in Appendix C C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(l)
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Hess’s Law Example: Calculate H o 298 for the following reaction from data in Appendix 1
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Hess’s Law Application of Hess’s Law and more algebra allows us to calculate the H f o for a substance participating in a reaction for which we know H rxn o, if we also know H f o for all other substances in the reaction.
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Hess’s Law Example: Given the following information, calculate H f o for H 2 S (g)
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Hess’s Law
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