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Chapter 6 Thermochemistry: pp. 235-274
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6.1 The Nature of Energy Energy – Capacity to do work or produce heat. – 1 st Law of Thermodynamics: Energy can not be created or destroyed, but converted to different forms. – The energy of the universe is constant. – Potential Energy: Energy due to position or composition – Kinetic Energy: Energy due to motion; depends on mass and velocity
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6.1 The Nature of Energy Lower Potential Energy Higher Potential Energy Add Energy
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Energy Heat involves the transfer of energy between two objects due to a temperature difference. Work: Force acting over a distance Energy is a state function: Work and Heat are not – State Function: Property that does not depend in any way on the system’s past or future (only depends on present state). 6.1 The Nature of Energy
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Chemical Energy System – part of the universe where a reaction is happening. Surroundings – everything else in the universe – Endothermic Reaction Energy flows into the system, absorbs energy from surroundings. – Exothermic Reaction Energy flows out of the system, surroundings gain energy. – Energy gained by surroundings must be equal to the energy lost by the system (opposite is true also). 6.1 The Nature of Energy
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Reality Check – Is the freezing of water an exothermic or endothermic process? Why? – Define the system, surroundings, and direction of energy transfer: Methane is burning in a Bunsen burner in a laboratory. 6.1 The Nature of Energy
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Internal Energy – Internal energy, E, of a system is the sum of the kinetic and potential energies of all the “particles” in the system. – A change to the internal energy of a system is described as: Where q is heat and w is work. 6.1 The Nature of Energy
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Internal Energy – Sign reflects the system’s point of view. 6.1 The Nature of Energy ++Endothermic, surroundings do work +-Endothermic, system does work -+Exothermic, surroundings do work --Exothermic, system does work qw What we observe
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Work – Work = P × A × Δh = PΔV P is pressure. A is area. Δh is the piston moving a distance. ΔV is the change in volume. 6.1 The Nature of Energy
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Work – For an expanding gas, ΔV is a positive quantity because the volume is increasing. Thus ΔV and w must have opposite signs: 6.1 The Nature of Energy
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Reality Check – Determine the sign of E for each of the following with listed conditions An endothermic process that performs work. – l work l > heat – l work l < heat Works is done on a gas and the process is exothermic. – l work l > heat – l work l < heat 6.1 The Nature of Energy Δ E = negative Δ E = positive Δ E = negative
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Change in Enthalpy – State Function – ΔH = q at constant pressure – ΔH = H products - H reactants 6.2 Enthalpy and Calorimetry
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Reality Check 6.2 Enthalpy and Calorimetry Consider the combustion of propane: C 3 H 8 (g) + 5O 2 (g) → 3CO 2 (g) + 4H 2 O(l) ΔH = –2221 kJ Assume that all of the heat comes from the combustion of propane. Calculate ΔH in which 5.00 g of propane is burned in excess oxygen at constant pressure. -252 kJ
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Calorimetry – Science of measuring heat – Specific heat capacity: The energy required to raise the temperature of one gram of a substance by one degree Celsius. – Molar heat capacity: The energy required to raise the temperature of one mole of substance by one degree Celsius. 6.2 Enthalpy and Calorimetry
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Simple Calorimeter If two reactants are mixed and the solution gets warmer an exothermic reaction is happening. An endothermic reaction cools the solution. 6.2 Enthalpy and Calorimetry
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Calorimetry – Where, s = specific heat capacity (J/ o C g) m = mass (g) ΔT = change in temperature ( o C) 6.2 Enthalpy and Calorimetry
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Reality Check 6.2 Enthalpy and Calorimetry A 100.0 g sample of water at 90°C is added to a 100.0 g sample of water at 10°C. The final temperature of the water is: a) Between 50°C and 90°C b) 50°C c) Between 10°C and 50°C
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Reality Check 6.2 Enthalpy and Calorimetry You have a Styrofoam cup with 50.0 g of water at 10. C. You add a 50.0 g iron ball at 90. C to the water. (s H2O = 4.18 J/°C·g and s Fe = 0.45 J/°C·g) The final temperature of the water is: a) Between 50°C and 90°C b) 50°C c) Between 10°C and 50°C Calculate the final temperature of the water. 18 o C
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Hess’ Law – In going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. 6.3 Hess’ Law
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This reaction also can be carried out in two distinct steps, with enthalpy changes designated by ΔH 2 and ΔH 3. N 2 (g) + O 2 (g) → 2NO(g) ΔH 2 = 180 kJ 2NO(g) + O 2 (g) → 2NO 2 (g) ΔH 3 = – 112 kJ N 2 (g) + 2O 2 (g) → 2NO 2 (g) ΔH 2 + ΔH 3 = 68 kJ ΔH 1 = ΔH 2 + ΔH 3 = 68 kJ 6.3 Hess’ Law N 2 (g) + 2O 2 (g) → 2NO 2 (g) ΔH 1 = 68 kJ
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Principle of Hess’ Law 6.3 Hess’ Law
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If a reaction is reversed, the sign of ΔH is also reversed. The magnitude of ΔH is directly proportional to the quantities of reactants and products in a reaction. If the coefficients in a balanced reaction are multiplied by an integer, the value of ΔH is multiplied by the same integer. 6.3 Hess’ Law Characteristics of Hess’ Law
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Reality Check – Calculate the ΔH for the reaction 6.3 Hess’ Law 2NH 3 (g) + 3N 2 O (g) 4N 2 (g) + 3H 2 O (l) ΔH = -1010. kJ N 2 O (g) + 3H 2 (g) N 2 H 4 (l) + H 2 O (l) ΔH = -317 kJ 2NH 3 (g) + ½ O 2 (g) N 2 H 4 (l) + H 2 O (l) ΔH = -143 kJ H 2 (g) + ½ O 2 (g) H 2 O (l) ΔH = -286 kJ N 2 H 4 (l) + O 2 (g) N 2 (g) + 2H 2 O (l)
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Change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states. 6.4 Standard Enthalpy of Formation Standard Enthalpy of Formation (ΔH o f )
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For a Compound For a gas, pressure is exactly 1 atm. For a solution, concentration is exactly 1 M. Pure substance (liquid or solid) For an Element The form [N 2 (g), K(s)] in which it exists at 1 atm and 25°C. Heat of formation is zero. 6.4 Standard Enthalpy of Formation Conventional Definitions of Standard States
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Calculating Enthalpy of formation Remember, – Elements enthalpy of formation is zero. – Don’t forget to multiply by coefficients from the balanced equation. 6.4 Standard Enthalpy of Formation H° rxn = n p H f (products) - n r H f (reactants)
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Calculate H° for the following reaction: 2Na(s) + 2H 2 O(l) → 2NaOH(aq) + H 2 (g) Given the following information: H f ° (kJ/mol) Na(s)0 H 2 O(l) –286 NaOH(aq) –470 H 2 (g)0 6.4 Standard Enthalpy of Formation Reality Check H° = –368 kJ
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