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Chapter 6 Thermochemistry: pp. 235-274 6.1 The Nature of Energy • Energy – Capacity to do work or produce heat. – 1st 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 6.1 The Nature of Energy Lower Potential Energy Add Energy Higher Potential Energy 6.1 The Nature of Energy • 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 • 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 • 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 • 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: E = q + w • Where q is heat and w is work. 6.1 The Nature of Energy • Internal Energy – Sign reflects the system’s point of view. q + + - w + + - What we observe Endothermic, surroundings do work Endothermic, system does work Exothermic, surroundings do work Exothermic, system does work 6.1 The Nature of Energy • 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 • Work – For an expanding gas, ΔV is a positive quantity because the volume is increasing. Thus ΔV and w must have opposite signs: w = - PV 6.1 The Nature of Energy • 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 Δ E = negative Δ E = positive • Works is done on a gas and the process is exothermic. – l work l > heat – l work l < heat Δ E = positive Δ E = negative 6.2 Enthalpy and Calorimetry • Change in Enthalpy – State Function – ΔH = q at constant pressure – ΔH = Hproducts - Hreactants 6.2 Enthalpy and Calorimetry • Reality Check Consider the combustion of propane: C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(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 6.2 Enthalpy and Calorimetry • 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 • 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 • Calorimetry Heat = s m T – Where, • s = specific heat capacity (J/oCg) • m = mass (g) • ΔT = change in temperature (oC) 6.2 Enthalpy and Calorimetry • Reality Check 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 6.2 Enthalpy and Calorimetry • Reality Check 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. (sH2O = 4.18 J/°C·g and sFe = 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. 18oC 6.3 Hess’ Law • 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 N2(g) + 2O2(g) → 2NO2(g) • ΔH1 = 68 kJ This reaction also can be carried out in two distinct steps, with enthalpy changes designated by ΔH2 and ΔH3. N2(g) + O2(g) → 2NO(g) 2NO(g) + O2(g) → 2NO2(g) N2(g) + 2O2(g) → 2NO2(g) ΔH1 = ΔH2 + ΔH3 = 68 kJ ΔH2 = 180 kJ ΔH3 = – 112 kJ ΔH2 + ΔH3 = 68 kJ 6.3 Hess’ Law • Principle of Hess’ Law 6.3 Hess’ Law Characteristics of Hess’ Law • • 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 • Reality Check – Calculate the ΔH for the reaction N2H4 (l) + O2 (g) N2 (g) + 2H2O (l) 2NH3 (g) + 3N2O(g) 4N2 (g) + 3H2O(l) ΔH = -1010. kJ N2O(g) + 3H2 (g) N2H4 (l) + H2O(l) ΔH = -317 kJ 2NH3 (g) + ½ O2 (g) N2H4 (l) + H2O(l) ΔH = -143 kJ H2 (g) + ½ O2 (g) H2O (l) ΔH = -286 kJ 6.4 Standard Enthalpy of Formation Standard Enthalpy of Formation (ΔHof) • 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 Conventional Definitions of Standard States • 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 [N2(g), K(s)] in which it exists at 1 atm and 25°C. Heat of formation is zero. 6.4 Standard Enthalpy of Formation • Calculating Enthalpy of formation H°rxn = npHf(products) - nrHf(reactants) • Remember, – Elements enthalpy of formation is zero. – Don’t forget to multiply by coefficients from the balanced equation. 6.4 Standard Enthalpy of Formation Reality Check Calculate H° for the following reaction: 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) Given the following information: Hf° (kJ/mol) Na(s) 0 H2O(l) –286 NaOH(aq) –470 H2(g) 0 H° = –368 kJ