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Energy Changes in Chemical Rxns Chapter 8 z z Thermochemistry: Chemical Energy Most reactions give off or absorb energy Energy is the capacity to do work or supply heat. Heat: transfer of thermal (kinetic) energy between two systems at different temperatures (from hot to cold) Metal bar in water Metal bar drilled 1 Types of Energy z z Work (w): energy transfer when forces are applied to a system Heat (q): energy transferred from a hot object to a cold one Radiant energyÆ heat from the sun Thermal energy Æ associated with motion of particles Potential energy Æ energy associated with object’s position or substance’s chemical bonds Kinetic energy Æ energy associated with object’s motion Energy and Energy Conservation z z z z Heat versus Temperature z z Describe the difference between the two. 2 SI unit of energy: J 1 J = 1 kg ⋅ m s2 z z z Molecular heat transfer 1 watt = 1 J/s,, so a 100 Watt bulb uses 100 J each second We often use the unit of kJ to refer to chemical heat exchanges in a reaction. 1 kJ = 1000 J Energy is also reported in calories: Amount of energy needed to raise 1 gram of water by 1oC 1 cal = 4.184 J; 1 Cal = 4184 J Cal (or kcal) is used on food labels Conservation of Energy Heat: form of energy transferred from object at higher temperature to one at lower temperature (from hot object to cold object) Thermochemistry: study of heat changes in chemical reactions, in part to predict whether or not a reaction will occur Thermodynamics: study of heat and its transformations First Law of Thermodynamics: Energy can be converted from one form to another but cannot be created or destroyed 1 System and Surroundings Endothermic vs Exothermic z Endothermic reaction: q is positive (q > 0) Reaction (system) absorbs heat Surroundings feel cooler z Exothermic reaction: q is negative (q < 0) Reaction releases heat Surroundings feel warmer z System loses heat (negative); gains heat (positive) Applications of heat emission/absorption Specific Heat and Heat Capacity z z z Specific heat (sp. ht.): amount of heat required to raise 1 gram of substance by 1oC Use mass, specific heat, and ΔT to calculate the amount of heat gained or lost: q = msΔT Æ ms = C Æ q = CΔT Heat capacity (C): amount of heat required to raise the temperature of a given quantity of a substance by 1oC; C = q / ΔT = J / oC Molar heat capacity (Cm): amount of heat that can be absorbed by 1 mole of material when temperature increases 1oC; q = (Cm) x (moles of substance) x (ΔT) = J / mol • oC 9 Specific Heat and Heat Capacity Practice Problem z z z z z Calculate the amount of heat transferred when 250 g of H2O (with a specific heat of 4.184 J/g·oC) is heated from 22oC to 98oC. q = msΔT Is heat being put into the system or given off by the system? If a piece of hot metal is placed in cold water, what gains heat and what loses heat? Which one will have a positive q value and which will have a negative q value? Worked Ex. 8.4, 8.5; Problems 8.10 – 8.12 2 Practice Problem Calorimetry and Heat Capacity z 34.8 g of an unknown metal at 25.2oC is mixed with 60.1 g of H2O at 96.2 oC (sp. ht. = 4.184 J/g·oC). The final temperature of the system comes co es to o 88 88.4oC C. Identify de y the eu unknown o metal. ea z Specific heats of metals: z z Al Fe Cu Sn 0.897 J/g·oC 0.449 J/g·oC 0.386 J/g·oC 0.228J/g·oC z Heat changes in a reaction can be determined by measuring the heat flow at constant co sa p pressure essu e Apparatus to do this is called a calorimeter. Heat evolved by a reaction is absorbed by water; heat capacity of calorimeter is the heat capacity of water. 13 Example z z z z z Enthalpies of Physical/Chemical Changes A 28.2 gram sample of nickel is heated to 99.8oC and placed in a coffee cup calorimeter containing 150.0 grams of water at 23.5oC. After the metal cools, the final temperature of th metal the t l and d water t iis 25 25.0 0oC. C qabsorbed + qreleased = 0 Which substance absorbed heat? Which substance released heat? Calculate the heat absorbed by the substance you indicated above. z z z z z Why does T become constant during melting and evaporating? M lti Melting, vaporization, i ti and sublimation are endothermic We can calculate total heat needed to convert a 15 gram piece of ice at -20oC to steam at 120oC. Enthalpy (H) describes heat flow into and out of a system under constant pressure Enthalpy (a measure of energy) is heat transferred per mole of substance substance. At constant pressure, qp = ΔH = Hproducts – Hreactants ΔH > 0 Æ endothermic (net absorption of energy from environment; products have more internal energy) ΔH < 0 Æ exothermic (net loss of energy to environment; reactants have more internal energy) Heating a Pure Substance (Water) z 14 Enthalpies of Phase Changes 2.080 J/goC z 2250 J/g z 4.184 J/goC z 334 J/g 2.09 J/g J/goC z Heat of fusion (ΔHfus): Amount of heat required to melt (solid Æ liquid) Heat of vaporization (ΔHvap): Amount of heat required to evaporate (liquid Æ gas) Heat of sublimation (ΔHsub): Amount of heat required to sublime (solid Æ gas) Why are there no values for ΔHfreezing, ΔHcondendsation, or ΔHdeposition? 17 3 Enthalpies of Reaction z Thermochemical Equations Determine if the following processes are endothermic or exothermic… z z Combustion of methane Reacting Ba(OH)2 with NH4Cl z Neutralization of HCl Melting Shows both mass and enthalpy relationships 2Al (s) + Fe2O3 (s) Æ 2Fe (s) + Al2O3 (s) ΔHo = -852 kJ Amount of heat given off depends on amount of material: 852 kJ of heat are released for every 2 mol Al, 1 mol CaCO3 (s) Æ CaO (s) + CO2 (g) Fe2O3, 2 mol Fe, and 1 mol Al2O3 19 Thermochemical Equations z z z z z Hess’s Law 2Al (s) + Fe2O3 (s) Æ 2Fe (s) + Al2O3 (s) ΔHo = 852 kJ How much heat is released if 10.0 grams of Fe2O3 reacts with excess Al? Worked Ex. 8.3; Problems 8.8, 8.9 z z What if we reversed the reaction? Heat would have to be put in to make the reaction proceed: z 2Fe (s) + Al2O3 (s) Æ 2Al (s) + Fe2O3 (s) ΔHo = +852 kJ Hess’s Law z If a compound cannot be directly synthesized from its elements, we can add the enthalpies of multiple reactions to calculate the enthalpy of reaction eac o in ques question. o Hess’s Law: change in enthalpy is the same whether the reaction occurs in one step or in a series of steps Look at direction of reaction and amount of reactants/products Hess’s Law Value changes sign with direction z Values of enthalpy change For a reaction in the reverse direction, enthalpy is numerically equal but opposite in sign z Reverse direction, heat flow changes; endothermic becomes exothermic (and vice versa); sign of ΔH changes Proportional to the amount of reactant consumed z z Twice as many moles = twice as much heat; half as many moles = half as much heat ΔHT = ΔH1 + ΔH2 + ΔH3 + …. Figure 8.5 23 24 4 Enthalpy of Chemical Reaction z Thermochemical equation: H2(g) + I2(s) → 2HI(g) z z z ΔH = +53.00 kJ z Two possible changes: Reverse the equation: 2HI(g) → H2(g) + I2(s) z Hess’s Law z ΔH = -53.00 kJ z Calculate ΔHo for 2NO (g) + O2 (g) Æ N2O4 (g) ΔHo = ? N2O4 (g) ( ) Æ 2NO2 (g) ( ) NO (g) + ½ O2 (g) Æ NO2 (g) ΔH Ho = 57.2 2 kJ ΔHo = -57.0 kJ Double the amount of material: 2H2(g) + 2I2(s) → 4HI(g) ΔH = +106.00 kJ 25 Hess’s Law Hess’s Law z We can use known values of ΔHo to calculate unknown values for other reactions P4 (s) + 3 O2 (g) Æ P4O6 (s) ΔH = -1640.1 kJ P4 (s) + 5 O2 (g) Æ P4O10 (s) ΔH = -2940.1 2940 1 kJ z What is ΔHo for the following reaction? z z P4O6 (s) + 2 O2 (g) Æ P4O10 (s) z ΔH = ? Worked Ex. 8.6, 8.7; Problem 8.13, 8.15 27 Hess’s Law z z z z z z Given: 2NH3(g) Æ N2H4(l) + H2(g) 1 N (g) + 32 H2(g) Æ NH3(g) 2 2 CH4O(l) ( ) Æ CH2O(g) (g) + H2(g) Standard Heats of Formation z ΔH = 54 kJ ΔH = -69 kJ ΔH = -195 kJ Find the enthalpy for the following reaction: N2H4(l) + CH4O(l) Æ CH2O (g) + N2(g) + 3H2(g) ΔH = ? kJ z z Standard heat of formation (ΔHof): heat needed to make 1 mole of a substance from its stable elements in their standard states ΔHof = 0 for a stable (naturally occurring) element Which of these have ΔHof = 0? CO(g), Cu(s), Br2(l), Cl(g), O2(g), O3(g), O2(s), P4(s) z Do the following equations represent standard enthalpies of formation? Why or why not? 2Ag (l) + Cl2 (g) Æ 2AgCl (s) Ca (s) + F2 (g) Æ CaF2 (s) 30 5 Standard Enthalpies of Formation z z Standard Enthalpies of Formation Can use measured enthalpies of formation to determine the enthalpy of a reaction (use Appendix B in back of book) ΔHorxn = ΣnΔHof (products) – ΣnΔHof (reactants) z Table 8.2: Some Common Substances (25oC) Σ = sum; n = number of moles (coefficients) z Direct calculation of enthalpy of reaction if the reactants are all in elemental form Sr (s) + Cl2 (g) Æ SrCl2 (g) ΔHorxn = [ΔHof (SrCl2)] – [ΔHof (Sr) + ΔHof (Cl2)] = -828.4 kJ/mol Heats of Formation z z z z z z Bond Dissociation Energies ΔHorxn = Σ ΔHof,products - Σ ΔHof,reactants Calculate values of ΔHo for the following rxns: 1) CaCO3 (s) Æ CaO (s) + CO2 (g) 2)) 2C6H6 ((l)) + 15O2 (g) Æ 12CO2 (g) + 6H2O ((l)) z z ΔHof values: CaCO3: -1207.1 kJ/mol; CaO: -635.5 kJ/mol; CO2: 393.5 kJ/mol; C6H6: 49.0 kJ/mol; H2O(l): -285.8 kJ/mol Worked Ex. 8.8, 8.9; Problems 8.16, 8.17 Bond Dissociation Energy (or Bond Energy, BE): energy required to break a bond in 1 mole of a gaseous molecule Reactions generally proceed to form H Bond Energy compounds with more stable (stronger) bonds (greater bond energy) 34 33 Bond Dissociation Energies z z z z 2 Bond Dissociation Energies Bond energies vary somewhat from one molecule to another so we use average bond dissociation energy (D) H-OH 502 kJ/mol Avg O-H = 453 H-O 427 kJ/mol kJ/mol H-OOH 431 kJ/mol 35 36 6 Bond Dissociation Energies z ΔHorxn = ΣBE (reactants) + - ΣBE (products) endothermic energy input z z Heats of Reaction z exothermic energy released z ΣBE(react) > ΣBE(prod) Æ endothermic ΣBE(react) < ΣBE(prod) Æ exothermic z z z Use only when heats of formation are not available, since bond energies are average values for gaseous molecules z Use bond energies to calculate the enthalpy change for the following reaction: N2(g) + 3H2(g) → 2NH3(g) ΔHrxn = [BEN ≡ N + 3BEH-H] + [[-6BE 6BEN-H] ΔHrxn = [945 + 3(436)] – [6(390)] = -87 kJ measured value = -92.2 kJ Why are the calculated and measured values different? 37 Heats of Reaction z z z 38 Thermochemistry Calculation Summary Use bond energies to calculate the enthalpy change for the decomposition of nitrogen trichloride: NCl C 3 (g) Æ N2 (g) + C Cl2 (g) How many distinct bond types are there in each molecule? How many of each bond type do we need to calculate ΔHrxn? z z z z z BE(N-Cl) = 200 kJ/mol Use q = msΔT (s = J/g·oC) If given mass of reactant, convert to moles and multiply by enthalpy to find total heat transferred If given multiple equations with enthalpies, use Hess’s Law If given ΔHof values: products – reactants If given bond energy (BE) values: +reactants + -products BE(N≡N) = 945 kJ/mol BE(Cl-Cl) = 243 kJ/mol 39 Practice Problems z z Practice Problems Identify how to set up the following problems: Calculate the ΔHo of reaction for: z C3H8(g): -103.95 103 95 kJ/mol; 393 5 f CO2(g): -393.5 kJ/mol; ΔHof H2O(l): -285.8 kJ/mol 8750 J of heat are applied to a 170 g sample of metal, causing a 56oC increase in its temperature. What is the specific heat of the metal? Which metal is it? z ΔHo f ΔHo = ? H2 (g) + F2 (g) Æ 2HF (g) ΔHo = -537 kJ C (s) + 2F2 (g) Æ CF4 (g) ΔHo = -680 kJ C3H8 (g) + 5O2 (g) Æ 3CO2 (g) + 4H2O (l) ΔHo C2H4(g ) + 6F2(g) Æ 2CF4(g) + 4HF(g) 2C (s) + 2H2 (g) Æ C2H4 (g) ΔHo = 52.3 kJ z Use average bond energies to determine the enthalpy of the following reaction (from Table 7.1). CH4 (g) + Cl2 (g) Æ CH3Cl (g) + HCl (g) (BEC-Cl = 328 kJ/mol) 7