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Transcript
Thermodynamics Chapter 8 Thermochemistry: Chemical Energy 01 • Thermodynamics: study of energy and it’s transformations • Energy: capacity to do work, or supply heat Energy = Work + Heat • Kinetic Energy: energy of motion EK = 1/2 mv2 (1 Joule = 1 kg⋅m2/s2) (1 calorie = 4.184 J) • Thermodynamics • 02 Conservation of energy law: Energy cannot be created or destroyed; it can only be converted from one form into another. Potential Energy: stored energy Thermodynamics 03 • Thermal Energy: kinetic energy of molecular motion (translational, rotational, and vibrational) • Heat: the amount of thermal energy transferred between two objects at different temperatures • Chemical Energy: potential energy stored in chemical bonds released in the form of heat or light 1 Thermodynamics • 04 First Law of Thermodynamics: energy of an isolated system must be kept constant Thermodynamics • • • 05 System ⇒ reactants + products Surroundings ⇒ everything else Energy changes are measured from the point of view of the system! ∆E is negative ⇒ energy flows out of the system • ∆E is positive ⇒ energy flows into the system • Thermodynamics 06 Work 07 w = –PΔV 2 Sign of w negative 08 Work Units 09 w = -PΔV (J or kJ) positive w = L x atm = w = -PΔV ⇒ expansion positive m2 1m3 1L x 1000mL x 1cm3 x 1L negative 1mL 101 x 103 kg (100cm)3 ms2 1000 w = -PΔV ⇒ contraction = 101 kgm2 = 101J s2 Energy and Heat Energy = Work + Heat ΔE = w + q = q - PΔV When a person does work, energy diminishes w = negative ΔE = negative 10 Heat and Enthalpy • 11 The amount of heat exchanged between the system and the surroundings is given the symbol q. q = ΔE + PΔV At constant volume (ΔV = 0): qv = ΔE At constant pressure: qp = ΔE + PΔV enthalpy = ΔH q = ΔE + PΔV 3 State Functions 12 State Function: • State Functions • value depends only on the present state of the system 13 State and Nonstate Properties: The two paths below give the same final state: N2H4(g) + H2(g) → 2 NH3(g) + heat (188 kJ) N2(g) + 3 H2(g) → 2 NH3(g) + heat (92 kJ) • path independent • temperature, total energy, pressure, density, volume, and enthalpy (∆H) ⇒ state properties • when returned to its original position, overall change is zero • nonstate properties include heat and work Enthalpy 14 • Enthalpy or heat of reaction: Standard Enthalpy of Reaction 15 Thermodynamic standard state: P = 1atm, [ ] = 1M, •ΔH = H(products) - H(reactants) T = 298.15K (25ºC) • States of the reactants and products are important! ⇒ (g, l, s, aq) Standard enthalpy of reaction (Hº) N2(g) + 3H2(g) → 2NH3(g) ΔHº = -92.2kJ • Thermodynamic standard state: P = 1atm, [ ] = 1M, T = 298.15K (25ºC) 4 Enthalpy Changes 16 Physical Changes Most changes in a system involve a gain or loss in enthalpy • • 17 Enthalpies of Physical Change: • Physical (melting of ice in a cooler) • Chemical (burning of gas in your car) Chemical Changes • Enthalpies of Chemical Change: Often called heats of reaction (ΔHreaction). 18 Enthalpy Changes • 19 Reversing a reaction changes the sign of ΔH for a reaction. C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) ΔH = –2219 kJ Endothermic: Heat flows into the system from the 3 CO2(g) + 4 H2O(l) → C3H8(g) + 5 O2(g) ΔH = +2219 kJ surroundings ⇒ ΔH is positive • Exothermic: Heat flows out of the system into the surroundings ⇒ ΔH is negative Multiplying a reaction increases ΔH by the same factor. 3 x [C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) ] ΔH =(-2219kJ x 3) = –6657 kJ 5 Example • 20 How much work is done (in kilojoules), and in which direction, as a result of the following reaction? Example • w = -0.25kJ 21 The following reaction has ΔE = –186 kJ/mol. • Is the sign of PΔV positive or negative? • What is the sign and approximate magnitude of ΔH? Expansion, system loses -0.25kJ Contraction, PΔV is negative, w is positive ΔH = ΔE + PΔV ΔH = (-186kJ) + (1atm) (-1mole) ΔH = negative (slightly more than ΔE) Example 22 The reaction between hydrogen and oxygen to yield water vapor has ΔH° = –484 kJ. How much PV work is done, and what is the value of ΔE (kJ) for the reaction of 0.50 mol of H2 with 0.25 mol of O2 at atmospheric pressure if the volume change is –5.6 L? PΔV = -0.57kJ Example 23 The explosion of 2.00 mol of solid TNT with a volume of approximately 0.274 L produces gases with a volume of 448 L at room temperature. How much PV (kJ) work is done during the explosion? Assume P = 1 atm, T = 25°C. 2 C7H5N3O6(s) → 12 CO(g) + 5 H2(g) + 3 N2(g) + 2 C(s) Contraction, so w is positive PΔV = 45.2kJ ΔE = -120.43kJ Expansion, so w = -45.2kJ 6 Example 24 How much heat (kJ) is evolved or absorbed in each of the following reactions? Hess’s Law • 1.) Burning of 15.5 g of propane: C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) 25 Hess’s Law: The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for the individual steps in the reaction. ΔHº = –2219 kJ 3 H2(g) + N2(g) → 2 NH3(g) ΔH° = –92.2 kJ -780kJ (exothermic) 2.) Reaction of 4.88 g of barium hydroxide octahydrate with ammonium chloride: Ba(OH)2·8 H2O(s) + 2 NH4Cl(s) → BaCl2(aq) + 2 NH3(aq) + 10 H2O(l) ΔHº = +80.3 kJ +1.24kJ (endothermic) Hess’s Law 26 (a) 2 H2(g) + N2(g) N2H4(g) ΔH°1 = ? (b) N2H4(g) + H2(g) 2 NH3(g) ΔH°2 = –187.6 kJ (c) 3 H2(g) + N2(g) 2 NH3(g) ΔH°3 = – 92.2 kJ Standard Heats of Formation 27 Where do ΔH° values come from? • Standard Heats of Formation (ΔH°f): enthalpy change for the formation of 1 mole of substance in its standard state • ΔH°f = 0 for an element in its standard state! ΔH°1 = ΔH°3 – ΔH°2 = (–92.2 kJ) – (–187.6 kJ) = +95.4 kJ 7 Standard Heats of Formation 28 ΔH°f = –286 kJ/mol H2(g) + 1/2 O2(g) → H2O(l) Standard Heats of Formation • 29 Calculating ΔH° for a reaction: ΔH° = ΔH°f (products) – ΔH°f (reactants) 3/ 2 H2(g) ΔH°f = –46 kJ/mol + 1/2 N2(g) → NH3(g) Heat of formation must be multiplied by the coefficient of the reaction ΔH°f = +227 kJ/mol 2 C(s) + H2(g) → C2H2(g) 2 C(s) + 3 H2(g) + 1/2 O2(g) → C2H5OH(g) ΔH°f = –235 kJ/mol Standard Heats of Formation C6H12O6 (s) ΔH° = [2ΔH°f(ethanol) + 2ΔH°f(CO2)] - ΔH°f (glucose) 30 Some Heats of Formation, ΔHf° (kJ/mol) Bond Dissociation Energy • CO(g) -111 C2H2(g) 227 Ag+(aq) 106 CO2(g) -394 C2H4(g) 52 Na+(aq) -240 H2O(l) -286 C2H6(g) -85 NO3-(aq) -207 NH3(g) -46 CH3OH(g) -201 Cl-(aq) -167 N2H4(g) 95.4 C2H5OH(g) -235 AgCl(s) -127 HCl(g) -92 C6H6(l) 49 Na2CO3(s) -1131 2C2H5OH (l) + 2CO2 (g) 31 Bond Dissociation Energy (D): Amount of energy needed to break a chemical bond in gaseous state D = Approximate ΔHº ΔH° = D(reactant bonds broken) – D(product bonds formed) H2 + Cl2 2HCl ΔH° = (DCl-Cl + DH-H) - (2 D H-Cl) = [(1 mol)(243 kJ/mol) + (1)(436 kJ/mol] - (2)(432 kJ/mol) = -185 kJ 8 Bond Dissociation Energy 32 Calorimetry and Heat Capacity • Calorimetry and Heat Capacity 34 Calorimetry: measurement of heat changes (q) for chemical reactions • Constant Pressure Calorimetry: measures the heat change at constant pressure q = ΔH • Bomb Calorimetry: measures the heat change at constant volume such that q = ΔE Calorimetry and Heat Capacity • 33 35 Heat capacity {C}: amount of heat required to raise the temperature of an object or substance a given amount C = q ΔT Specific Heat: amount of heat required to raise the temperature of 1.00 g of substance by 1.00°C Constant Pressure Bomb Molar Heat: amount of heat required to raise the temperature of 1.00 mole of substance by 1.00°C 9 Calorimetry and Heat Capacity 36 Example 37 The industrial degreasing solvent methylene chloride (CH2Cl2, dichloromethane) is prepared from methane by reaction with chlorine: CH4(g) + 2 Cl2(g) CH2Cl2(g) + 2 HCl(g) Calculate ΔH° (kJ) CH4(g) + Cl2(g) CH3Cl(g) + HCl(g) ΔH° = –98.3 kJ CH3Cl(g) + Cl2(g) CH2Cl2(g) + HCl(g) ΔH° = –104 kJ ΔH° = -98.3 + -104 = -202kJ Example 38 Example 39 Calculate ΔH° (kJ) for the reaction of ammonia with O2 to Calculate ΔH° (kJ) for the photosynthesis of glucose from yield nitric oxide (NO) and H2O(g), a step in the Ostwald process for the commercial production of nitric acid. CO2 and liquid water, a reaction carried out by all green plants. 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) 6CO2(g) + 6H2O(l) C6H12O6(s) + 6O2(g) = [(4)(90.2kJ/mol) + (6)(-241.8)] - [(4)(-46.1) + (5)(0)] = [(1 mole)(-1260kJ/mol) + (6)(0)] - [(6)(-393.5) + (6)(-285.8)] = -905.6kJ = 2816kJ 10 Example • 40 Calculate an approximate ΔH° (kJ) for the synthesis Example • What is the specific heat of lead if it takes 96 J to raise the temperature of a 75 g block by 10.0°C? • When 25.0 mL of 1.0 M H2SO4 is added to 50.0 mL of 1.0 M NaOH at 25.0°C in a calorimeter, the temperature of the solution increases to 33.9°C. Assume specific heat of solution is 4.184 J/(g–1·°C–1), and the density is 1.00 g/mL–1, calculate ΔH for the reaction. of ethyl alcohol from ethylene: C2H4(g) + H2O(g) → C2H5OH(g) • Calculate an approximate ΔH° (kJ) for the synthesis of hydrazine from ammonia: 2 NH3(g) + Cl2(g) → N2H4(g) + 2 HCl(g) 41 11