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Thermochemistry Chapter 5
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Heat - the transfer of thermal energy between two bodies that are at different temperatures Energy Changes in Chemical Reactions Temperature - a measure of the thermal energy 90 °C 40 °C greater thermal energy Temperature = Thermal Energy greater temperature (intensive) (extensive)
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Exothermic process - gives off heat – transfers thermal energy from the system to the surroundings. 2H 2 (g) + O 2 (g) 2H 2 O (l) + energy H 2 O (g) H 2 O (l) + energy
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Endothermic process - heat has to be supplied to the system from the surroundings. energy + CaCO 3 (s) CaO (s) + CO 2 (g) energy + H 2 O (s) H 2 O (l)
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E system + E surroundings = 0 or E system = − E surroundings C 3 H 8 + 5O 2 3CO 2 + 4H 2 O Exothermic chemical reaction! Chemical energy lost by combustion = Energy gained by the surroundings system surroundings First law of thermodynamics – energy can be converted from one form to another, but cannot be created or destroyed. Internal energy of system
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Exothermic Energy relationships Fig 5.4 (a)
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Endothermic Energy relationships Fig 5.4 (b)
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Fig 5.5 Energy Diagram for the Interconversion of H 2 (g), O 2 (g), and H 2 O (l)
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Another form of the first law for E system E = q + w E 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 Table 5.1 pg 171
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Sign Conventions for Heat and Work Fig 5.6
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Thermodynamics State functions - properties that are determined by the state of the system, regardless of how that condition was achieved. energy, pressure, volume, temperature E = E final - E initial Fig 5.8
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Work Done By the System on the Surroundings V = V f – V i > 0 -P V < 0 w sys < 0 w = w final - w initial initialfinal Work is not a state function! Fig 5.12 Since the opposing pressure, P can vary: ViVi VfVf Expansion of a Gas:
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H = E + P V H = (q+w) − w H = q p So, at constant pressure, the change in enthalpy is the heat gained or lost. H = E + PV Enthalpy (H) - the heat flow into or out of a system in a process that occurs at constant pressure.
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H = heat given off or absorbed during a reaction at constant pressure H products < H reactants H products > H reactants
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Enthalpies of Reaction H 2 O (s) H 2 O (l) H = 6.01 kJ System absorbs heat Endothermic H > 0 6.01 kJ are absorbed for every 1 mole of ice that melts at 0 ° C and 1 atm. H = H (products) – H (reactants) = q p
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CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (l) H = -890.4 kJ System gives off heat Exothermic H < 0 890.4 kJ are released for every 1 mole of methane that is combusted at 25 ° C and 1 atm. Enthalpies of Reaction
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Enthalpy is an extensive physical property. H for a reaction in the forward direction is equal in magnitude, but opposite in sign, to H for the reverse reaction. 2H 2 O (s) 2H 2 O (l) H = 2 x 6.01 = 12.0 kJ H 2 O (s) H 2 O (l) H = 6.01 kJ H 2 O (l) H 2 O (s) H = - 6.01 kJ The Truth about Enthalpy
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H for a reaction depends on the state of the products and the state of the reactants. The Truth about Enthalpy H 2 O (s) H 2 O (l) H = 6.01 kJ H 2 O (l) H 2 O (g) H = 44.0 kJ
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How much heat is evolved when 155 g of iron undergoes complete oxidation in air? 4Fe (s) + 3O 2 (g) 2Fe 2 O 3 (s) H = -1118.4 kJ 155 g Fe 1 mol Fe 55.85 g Fe x = -776 kJ -1118.4 kJ 4 mol Fe x
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The specific heat (s) - the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius: Heat (q) absorbed or released: q = m·C s · T where T = T final - T initial Table 5.2
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How much heat is given off when an 869 g iron bar cools from 94.0 ° C to 5.0 ° C? C s of Fe = 0.45 J/(g ° C) T = T final – T initial = 5.0 ° C – 94.0 ° C = ‒ 89.0 ° C q = m·C s · T = (869 g) · (0.45 J/(g ° C)) · (– 89.0 ° C) = ‒ 34,800 J ≈ ‒ 3.5 x 10 4 J = ‒ 35 kJ
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Calorimetry: Measurement of Heat Changes H cannot be determined absolutely, however, ΔH can be measured experimentally Device used is called a calorimeter Typically, ΔT ∝ ΔH rxn
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Constant-Pressure Calorimetry No heat enters or leaves! q solution = m·C s · T = ‒ q rxn Because reaction at constant P: H = q rxn Figure 5.17
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(Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.) Standard enthalpy of reaction ( ) - the enthalpy of a reaction carried out at 1 atm. n H o (products) f = m H o (reactants) 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.
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Fig 5.20 An enthalpy diagram comparing a one-step and a two-step process for a reaction.
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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: H o (O 2 ) ≡ 0 f H o (O 3 ) = 142 kJ/mol f H o (C, graphite) ≡ 0 f H o (C, diamond) = 1.90 kJ/mol f Establish an arbitrary scale with the standard enthalpy of formation ( ) as a reference point for all enthalpy expressions. NO!! Standard enthalpy of formation ( ) - the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm.
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Table 5.3 Standard Enthalpies of Formation, at 298 K
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Calcium carbide (CaC 2 ) reacts with water to form acetylene (C 2 H 2 ) and Ca(OH) 2. How much heat is released per mole of calcium carbide reacted? The standard enthalpy of formation of CaC 2 is -62.76 kJ/mol. CaC 2 (l) + 2H 2 O (l) C 2 H 2 (g) + Ca(OH) 2 (s) HoHo rxn n H o (products) f = m H o (reactants) f ‒ f = HoHo rxn [ H o (C 2 H 2 ) + H o (Ca(OH) 2 ] –[ H o f (CaC 2 ) + H o f (H 2 O)] f HoHo rxn = [(226.77 kJ) + (–986.2 kJ) ] – [(-62.67 kJ) + 2(-285.83 ] = -125.1 kJ -125.1 kJ 1 mol CaC 2 = -125.1 kJ/mol CaC 2
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