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The Nature of Energy John W. Moore Conrad L. Stanitski Peter C. Jurs http://academic.cengage.com/chemistry/moore Chapter 6 Energy and Chemical Reactions Energy (E (E) = the capacity to do work. Work (w) occurs when an object moves against a resisting force: w = −(resisting force) x (distance traveled) w = −F d All energy is either Kinetic or Potential energy. Stephen C. Foster • Mississippi State University The Nature of Energy Kinetic energy (E (Ek) - Energy of motion Energy Units joule (J) - SI unit (1 J = 1 kg m2s-2) macroscale = mechanical energy random nanoscale = thermal energy 2.0 kg mass moving at 1.0 m/s (~2 mph): periodic nanoscale = acoustic energy Ek = ½mv2 (m = mass, v = velocity of object) Potential energy (E (Ep) – Energy of position. Stored E. It may arise from: gravity: Ep = m g h (mass x gravity x height). charges held apart. bond energy. Energy Units Ek = ½ mv2 = ½ (2.0 kg)(1.0 m/s)2 = 1.0 kg m2 s-2 = 1.0 J 1 J is a relatively small amount of energy. 1 kJ (1000 J) is more common in chemical problems. Conservation of Energy calorie (cal) Originally: “The energy needed to heat of 1g of water from 14.5 to 15.5 °C.” It can only change form. Now: Total E of the universe is constant. 1 cal = 4.184 J (exactly) Dietary Calorie (Cal) - the “big C” calorie Used on food products. 1 Cal = 1000 cal = 1 kcal “Energy can neither be created nor destroyed”. Also called the 1st Law of Thermodynamics. 1 Conservation of Energy Energy and Working If an object moves against a force, work is done. • Lift a book you do work against gravity. The book’s Ep increases. • Drop the book: Ep converts into Ek The book does work pushing the air aside. A diver: a) Has Ep due to macroscale position. b) Converts Ep to macroscale Ek. c) Converts Ek,macro to Ek,nano (motion of water, heat) • The book hits the floor Energy and Working Energy, Temperature, and Heating In a chemical process, work occurs whenever something expands or contracts. Expansion pushes back the surrounding air. On heating a balloon: The gas heats up The gas expands; the balloon swells. The gas does work pushing back the rubber and the air outside it. Energy, Temperature, and Heating no work is done on the floor (it does not move). Ek converts to a sound wave and T of the book and floor increase (Ek converts to heat). Temperature is a measure of the thermal energy of a sample. Thermal energy • E of motion of atoms, molecules, and ions. • Atoms of all materials are always in motion. • Higher T = faster motion. Energy, Temperature, and Heating Heat • Thermal E transfer caused by a T difference. • Heat flows from hotter to cooler objects until they reach thermal equilibrium (have equal T ). Consider a thermometer. As T increases: Atoms move faster; on average get farther apart. V of the material increases. Length of liquid column increases. 2 Systems, Surroundings, and Internal Energy Systems, Surroundings, and Internal Energy Internal energy = E within the system because of nanoscale position or motion System = the part of the universe under study chemicals in a flask. my textbook. Einternal= sum of all nanoscale Ek and Ep Surroundings = rest of the universe (or as much as needed…) the flask. • nanoscale Ek = thermal energy • nanoscale Ep ion/ion attraction or repulsion perhaps the flask and this classroom. nucleus/electron attraction perhaps the flask and all of the building, etc. proton/proton repulsion ….. Universe = System + Surroundings Systems, Surroundings, and Internal Energy Internal energy depends on • Temperature Calculating Thermodynamic Changes Energy change = final E – initial E ΔE = Efinal – Einitial higher T = larger Ek for the nanoscale particles. A system can gain or lose E • Type of material nanoscale Ek depends upon the particle mass. SURROUNDINGS SYSTEM nanoscale Ep depends upon the type(s) of particle. SURROUNDINGS SYSTEM Efinal ΔE > 0 • Amount of material E in Einitial number of particles. double sample size, double Einternal, etc. ΔE positive: internal energy increases Calculating Thermodynamic Changes Einitial ΔE < 0 E out Efinal ΔE negative: internal energy decreases Heat Capacity • No subscript? Refers to the system: E = Esystem • E is transferred by heat or by work. • Conservation of energy becomes: ΔE = q + w heat SURROUNDINGS SYSTEM Heat transfer out q<0 Heat transfer in q>0 Heat capacity = E required to raise the T of an object by 1°C. Varies from material to material. work Specific heat capacity (c (c ) • E needed to heat 1 g of substance by 1°C. ΔE = q + w Work transfer in w>0 Work transfer out w<0 Molar heat capacity (c (cm) • E needed to heat 1 mole of substance by 1°C. Note the same sign convention for q and w 3 Heat Capacity Heat Capacity c (J g-1 °C-1) cm (J mol-1 °C-1) Substance For other amounts or for other T changes: Heat required = mass x specific heat x ΔT q = m c ΔT or… Heat required = moles x molar heat capacity x ΔT q = n cm ΔT Heat Capacity How much energy will be used to heat 500.0 g of iron from 22°C to 55°C? cFe = 0.451 J g-1 °C-1. Heat required = q = m c ΔT q = 500.0 g (0.451 J g-1 °C-1)(55−22)°C q = 7442 J = +7.4 kJ Elements C (graphite) Al(s) Fe(s) Cu(s) Au(s) Compounds NH3(ℓ) H2O(ℓ) H2O(s) CCl4(ℓ) CCl2F2(ℓ) Common solids wood concrete glass granite 0.720 0.902 0.451 0.385 0.129 8.65 24.3 25.1 24.4 25.4 4.70 4.184 2.06 0.861 0.598 80.1 75.3 37.1 132. 72.3 1.76 0.88 0.84 0.79 Heat Capacity 24.1 kJ of energy is lost by a 250. g Al block. If the block is initially at 125.0°C what will be its final T? (cAl = 0.902 J g-1 °C-1) q = m c ΔT ΔT = q / (m c) Cooling, q is negative: negative ΔT = −24.1 x 103 J 250. g(0.902 J g-1 °C-1) ΔT = Tfinal – Tinital = −107 °C + sign, E added to the system (the iron) Heat Capacity A 215 g block of Cu at 505.0°C is plunged into 1.000 kg of water (T = 23.4 °C) in an insulated container. What will be the final equilibrium T of the water and the Cu? (cCu = 0.385 J g-1 °C-1) q = m c ΔT qCu = (215. g)(0.385 J g-1 °C-1)(Tfinal− 505.0) qH2O = (1000. g)(4.184 J g-1 °C-1)(Tfinal− 23.4) qCu + qH2O = 0 Thus Tfinal = ΔT + Tinital = −107 + 125°C = 18°C Heat Capacity 215 g Cu (505.0°C) + 1000. g H2O (23.4 °C). Final T ? qH2O = -qCu 4184(Tfinal – 23.4) = -82.78(Tfinal – 505.0) (4184 + 82.78)Tfinal = 41804 + 97906 Tfinal = 32.7°C (conservation of E) qH2O = -qCu (Note: Tfinal must be between Thot and Tcold) 4 Conservation of Energy and Changes of State Conservation of Energy and Changes of State When heat is: Added to a system A liquid cools from 45°C to 30°C, transferring 911 J to the surroundings. No work is done on or by the liquid. What is ΔEliquid? q is positive the change is endothermic Removed from a system q is negative ΔEliquid = qliquid + wliquid the change is exothermic exothermic. H2O(ℓ) H2O(g) Water Boils: Steam Condenses: H2O(g) H2O(ℓ) endothermic here wliquid = 0 Heat transfers from the liquid to the surroundings: qliquid = -911 J (qsurroundings = +911 J) exothermic Work occurs as the sample expands or contracts. Overall: ΔE = q + w ΔEliquid = -911J Enthalpy: Heat Transfer at Constant P Conservation of Energy and Changes of State A system does 50.2 J of work on its surroundings and there is a simultaneous 90.1 J heat transfer from the surroundings to the system. What is ΔEsystem? Work done on the surroundings by the system Heat transfers from the surroundings to the system wsystem = -50.2 J qsystem = +90.1 J Because ΔE = q + w: At Constant V: ΔE = qV • subscript V shows fixed V • work requires motion against an opposing force. • constant V = no motion, so w = 0. At Constant P: ΔE = qP + watm= ΔH + watm • Subscript P shows fixed P. • watm = work done to push back the atmosphere • H = enthalpy enthalpy. ΔH = qp ΔEsystem = qsystem + wsystem ΔEsystem = -50.2 J +90.1 J = +39.9 J Freezing and Melting (Fusion) During freezing (or melting) Vaporization and Condensation ΔHfusion = qP = heat to melt a solid. • Substance loses (or gains) E, but… Change -50 Example: Convert 1 g of ice at -50°C to water at +50°C Temperature (°C) -25 0 25 50 • T remains constant. Ice is melting. T remains at 0°C Name enthalpy of fusion liq → gas enthalpy of vaporization liq → solid enthalpy of freezing gas → liq enthalpy of condensation 333 2260 −333 −2260 Water warms from 0 to 50°C Note: ΔHfusion = − ΔHfreezing 100 200 300 400 500 Quantity of energy transferred (J) etc. qfusion = −qfreezing Ice warms from -50 to 0°C 0 value for H2O (J/g) solid → liq 600 5 State Functions and Path Independence Thermochemical Expressions State functions Always have the same value whenever the system is in the same state. ΔH = qP can be added to a balanced equation. Two equal mass samples of water produced by: 1. Heating one from 20°C to 50°C. State functions 2. Cooling the other from 100°C to 50°C. H E have identical final H (and V, P, E…). P V T etc. State function changes are path independent. independent ΔH = Hfinal – Hinitial is constant. Thermochemical Expressions The thermite reaction produces tremendous heat: 2 Al + Fe2O3 → Al2O3 + 2 Fe ΔH° = – 851.5 kJ How much heat is released when 10.0 g of Al reacts with excess Fe2O3 at constant P ? = 10.0 g / 26.982 g/mol = 0.3706 mol Al ΔH° = −890.36 kJ ΔH° is the standard enthalpy change P = 1 bar. T must be stated (if it isn’t, assume 25°C). ΔH° is a molar value. Burn 1 mol of CH4 with 2 mol O2 to form 2 mol of liquid water and release 890 kJ of heat Change a physical state, change ΔH° : H2O(ℓ) vs. H2O(g) Where Does the Energy Come From? Bond Enthalpy (bond energy) • Equals the strength of 1 mole of bonds • Always positive It takes E to break a bond Separated parts are less stable than the molecule. Less stable = higher E • E is always released when a bond forms 2 Al ≡ 1 ΔH° = – 851.5. kJ Product is more stable than the separated parts. – 851.5 kJ qp = 0.3706 mol Al = –158 kJ 2 mol Al More stable = lower E Bond Enthalpies During a chemical reaction: Old bonds break: requires E (endothermic) New bonds form: releases E (exothermic) Overall, heat may be absorbed or released: Exothermic reactions (ΔH < 0) E is released. New bonds are more stable than the old, or More bonds are formed than broken. Both typically occur: 2 H(g) + 2 Cl(g) reactants products 2 HCl(g) Endothermic reactions (ΔH > 0) E is absorbed. endothermic ΔH= +678 kJ/mol energy less stability Bond Enthalpies H2(g) + Cl2(g) CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(ℓ) exothermic ΔH= -862 kJ/mol New bonds are less stable than the old, or Fewer bonds are formed than broken energy less stability nAl CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) ΔH° = −803.05 kJ products reactants 6 Measuring Enthalpy Changes Heat transfers are measured with a calorimeter. calorimeter Common types: • Bomb calorimeter. rigid steel container. filled with O2(g) and a small sample to be burnt. constant V, so qV = ΔE • Flame calorimeter. samples burnt in an open flame. constant P, so qp = ΔH • Coffee-cup calorimeter in lab (constant P). Measuring Enthalpy Changes Octane (0.600 g) was burned in a bomb calorimeter containing 751 g of water. T increased from 22.15°C to 29.12°C. Calculate the heat evolved per mole of octane burned. Ccal = 895 J°C-1. 2 C8H18(ℓ) + 25 O2(g) 16 CO2(g) + 18 H2O(ℓ) qreaction + qbomb + qwater = 0 qbomb = CcalΔT = 895 J°C-1 (29.12 – 22.15)°C = +6238 J qwater = m c ΔT = 751 g (4.184 J g-1 °C-1)(29.12 – 22.15)°C = +2.190 x 104 J Measuring Enthalpy Changes Bomb Calorimeter Measure ΔT of the water. Constant V: qV = ΔE Conservation of E: qreaction + qbomb + qwater = 0 or −qreaction = qbomb + qwater with qbomb = mcalccalΔT = CcalΔT A constant for a calorimeter Bomb Calorimetry T = 22.15 → 29.12°C. Heat per mol octane (0.600 g) burned. 2 C8H18(ℓ) + 25 O2(g) →16 CO2(g) + 18 H2O(ℓ) −qreaction = qbomb + qwater −qreaction = +6238 J + 2.190 x 104 J = 2.81 x 104 J = 28.1 kJ qreaction = −28.1 kJ Measuring Enthalpy Changes Measuring Enthalpy Changes Octane (0.600 g)… Calculate the heat evolved per mole of octane burned CoffeeCoffee-cup calorimeter Nested styrofoam cups prevent heat transfer with the surroundings. Molar mass of C8H18 = 114.23 g/mol. nC8H18 = (0.600 g) / (114.23 g/mol) = 0.00525 mol C8H18 Heat evolved /mol octane = −28.1 kJ 0.00525 mol = −5.35 x 103 kJ/mol = −5.35 MJ/mol Constant P. ΔT measured. q = qp = ΔH Assume the cups do not absorb heat. (Ccal = 0 and qcal = 0) 7 Measuring Enthalpy Changes 1.02 g of Mg was reacted with excess 1 M HCl(aq) (255.0 g) in a coffee-cup calorimeter. Tsoln rose from 22.0 to 41.6°C. cHCl = 3.90 J g-1°C-1 . Complete: Mg(s) + 2 HCl(aq) H2(g) + MgCl2(aq) ΔH = ? qsoln Measuring Enthalpy Changes 1.02 g Mg + 255.0 g of acid. T: 22.0 → 42.5°C. Molar ΔH = ? qrxn = –19.57 kJ = ΔH for 1.00 g = msoln c ΔT (msoln = macid + mMg ) = 256.0 g (3.90 J g-1 °C-1)(41.6 − 22.0)°C = 1.957 x 104 J nMg = 1.02 g So Conservation of E: qrxn = –qsoln qsoln + qrxn = 0 1 mol = 0.04196 mol 24.31 g ΔH = –19.57 kJ 0.04196mol 1 mol Mg ΔH = – 466 kJ or exothermic qrxn = –19.57 kJ = ΔH Hess’s Law Hess’s Law “If the equation for a reaction is the sum of the equations for two or more other reactions, then ΔH° for the 1st reaction must be the sum of the ΔH° values of the other reactions.” Another version: “ΔH° for a reaction is the same whether it takes place in a single step or several steps.” H is a state function Multiply a reaction, multiply ΔH. Reverse a reaction, change the sign of ΔH. 2 CO(g) + O2(g) → 2 CO2 (g) ΔH = −566.0 kJ Then 2 CO2(g) → 2 CO(g) + O2(g) ΔH = –1(–566.0 kJ) = + 566.0 kJ 4 CO2(g) → 4 CO(g) + 2 O2(g) Hess’s Law ΔH = –2(–566.0 kJ) = +1132.0 kJ Hess’s Law Use Hess’s Law to find ΔH for unmeasured reactions. Example It is difficult to measure ΔH for: 2 C(graphite) + O2(g) 2 CO(g) Some CO2 always forms. Calculate ΔH given: C(graphite) + O2(g) 2 CO(g) + O2(g) CO2(g) 2 CO2(g) No phase confusion… drop phases ΔH = −393.5 kJ ΔH = −566.0 kJ Want: Have: A B 2 C + O2 → C + O2 → 2CO + O2 → 2 CO CO2 2 CO2 +2 x A 2C + 2O2 → 2CO2 −1 x B 2CO2 → 2CO + O2 ??? −393.5 kJ −566.0 kJ 2(−393.5) = −787.0 −1(−566.0) = +566.0 2C + 2O2 + 2CO2 → 2CO2 + 2CO + O2 2 C + O2 → 2 CO −221.0 ΔH° = −221.0 kJ 8 Hess’s Law Hess’s Law Want: Have: A B C Determine ΔH° for the production of coal gas: 2 C(s) + 2 H2O(g) CH4(g) + CO2(g) Using: C(s) + H2O(g) CO(g) + H2(g) ΔH° = 131.3 kJ A CO(g) + H2O(g) CO2(g) + H2(g) ΔH° = −41.2 kJ B CH4(g) + H2O(g) CO(g) + 3 H2(g) ΔH° = 206.1 kJ C No phase confusion… drop phases Hess’s law problems often use a combustion or … Formation reaction Make 1 mol of compound from its elements in their standard states. H2 combustion: CH4 + CO2 CO + H2 CO2 + H2 CO + 3 H2 ??? 131.3 kJ −41.2 kJ 206.1 kJ +2 x A 2 C + 2 H2O → 2 CO + 2 H2 +262.6 −1 x C CO + 3 H2 → CH4 + H2O −206.1 +1 x B CO + H2O → CO2 + H2 −41.2 2C + 2H2O → CH4 + CO2 15.3 kJ Standard Molar Enthalpy of Formation Standard state = most stable form of the pure element at P = 1 bar. e.g. C standard state = graphite (not diamond) ΔHf° for any element in its standard state is zero. (take 1 mol of the element and make… 1 mol of element) 2 H2O(ℓ) ΔH° = −571.66 kJ but the formation reaction is: H2(g) + ½ O2(g) → → → → (After cancelling and adding) Standard Molar Enthalpy of Formation 2 H2(g) + O2(g) 2 C + 2 H2O C + H2O CO + H2O CH4 + H2O ΔHf° = −285.83 kJ 1 H2O(ℓ) ΔHf° (Br2(ℓ) ) = 0 ΔHf° (Br2(g) ) ≠ 0 at 298 K at 298 K f = formation Standard Molar Enthalpy of Formation Appendix J (25°C) Notes • Most are negative (formation releases E), but can be positive. • If the physical state changes, ΔHf° changes. Compound ΔHf°, kJ/mol Al2O3(s) aluminum oxide −1675.7 CaO(s) calcium oxide −635.09 CH4(g) methane −74.81 C2H2(g) acetylene +226.73 C2H4(g) ethylene +52.26 C2H6(g) ethane −84.68 C2H5OH(l) ethanol −277.69 H2O(g) water vapor −241.818 H2O(l) liquid water −285.830 NaF(s) sodium fluoride −573.647 Standard Molar Enthalpy of Formation ΔH° ={(nproducts)(ΔHf° products)} – {(nreactants)(ΔHf° reactants)} Example Calculate ΔH° for: CH4(g) + NH3(g) HCN(g) + 3 H2(g) ΔH° = ΔHf°(HCN) + 3ΔHf°(H2) − ΔHf°(NH3) – ΔHf°(CH4) = +134 + 3(0) − (−46.11) – (−74.85) = 255 kJ 9