Chapter 6: Energy Relationships in Chemical Reactions. Energy Relationships in Chemical Reactions

Transcription

1 Energy Relationships in Chemical Reactions Chapter 6 Chapter 6: Energy Relationships in Chemical Reactions I. Energy, work, heat II. Internal Energy, E Energy transfer ΙΙΙ. ΔH (q p ) vs ΔE (q v ) IV. Thermochemical Equations V. Heats of Formation; ΔH 0 f VI. Hess Law State Functions Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 6: Energy Relationships in Chemical Reactions VII. Calorimetry a. specific heat/ heat capacity b. Constant Pressure Calorimetry (coffee cup) c. Constant Volume Calorimetry (bomb) Energy is the capacity to do work or transfer heat Thermal energy is the energy associated with the random motion of atoms and molecules Chemical energy is the energy stored within the bonds of chemical substances Nuclear energy is the energy stored within the collection of neutrons and protons in the atom Electrical energy is the energy associated with the flow of electrons Potential energy is the energy available by virtue of an object s position Kinetic energy Kinetic energy is due to an object s mass and motion 1

2 Energy is the capacity to do work or transfer heat Heat : Process of energy transfer as a result of temperature difference Work: Energy transferred due to a change in position or directed energy transfer from a process Energy Changes in Chemical Reactions Heat is the transfer of thermal energy between two bodies that are at different temperatures. Temperature is a measure of the average thermal energy. Temperature = Thermal Energy or Heat When energy is transferred from one object to another it appears as heat and/or work! 90 0 C 40 0 C greater thermal energy Internal Energy, E: E Total of all possible types of energy within a substance 1.Kinetic Energy; Energy due to mass and motion a) Translational b) Vibrational c) Rotational E K = E T + E V + E R 2.Potential Energy; Stored energy due to an object s position a) Bond Chemical Energy b) Nucleus Nuclear Energy c) Atom E P = E B + E N + E A E TOTAL = E K + E P Units : Energy, work, and heat Calorie heat needed to increase the temperature of 1.00 gram of water by 1.00 C Joule or (Newton *meter) 1 calorie = J 2

3 Energy is the capacity to do work or transfer heat Heat : Process of energy transfer as a result of temperature difference Work: Energy transferred due to a change in position or directed energy transfer from a process When energy is transferred from one object to another it appears as heat and/or work! Exothermic process :any process that gives off heat transfers thermal energy from the system to the surroundings. 2H 2 (g) + O 2 (g) 2H 2 O (l) + energy; -q H 2 O (g) H 2 O (l) + energy; -q Endothermic process: any process in which heat has to be supplied to the system from the surroundings. energy + 2HgO (s) 2Hg (l) + O 2 (g); +q energy + H 2 O (s) H 2 O (l); +q q vs. ΔH 1. q is amount of heat released/absorbed for a specific amount of material. 2. ΔH is heat released/absorbed by 1 mole of material under constant pressure conditions. Change in Enthalpy (ΔH) is the Heat Lost or Gained at Constant Pressure: ΔH = q P q=nδh 3

4 Work: Most work done by a chemical reaction involves expansion/compression of gases. 2H 2 (g) + O 2 (g) 2H 2 O (l) + energy 3 moles gas o moles gas When energy is transferred from one object to another it appears as heat and/or work! ΔE = q + w work = -PΔV = -Δn g RT Expansion: +ΔV so w Gas pushes against atmosphere so system does work or work is lost by system Compression: -ΔV so + w Atmospheric pressure does work on system or work is added to system Si -28% of Earth s Crust but when in Contact With Air it Forms SiO 2. When energy is transferred from one object to another it appears as heat and/or work! Si (s) + O 2 (g) SiO 2 (s) ; ΔE = -908 kj/mole ΔE = q + w ΔE = = -908 Si (s) + O 2 (g) -> SiO 2 (s) ; ΔE = -908 kj/mole Internal Energy Si + O kj/mole SiO 2 Energy Reactants = Energy Products + KE Released 1. Heat of reaction (enthalpy) Si (s) + O 2 (g) -> SiO 2 (s) ; ΔΗ = -911 kj/mole 2. Work work = -PΔV = -Δn g RT = -(0-1mols)(8.314 X10-3 kj/molek)(298 K) -(-1)(2.5kJ) or about +3 KJ 4

5 Two ways that q can be measured 1. Constant Pressure q p = ΔE w q p =ΔE +PΔV 2. Constant Volume; ΔV =0 q v = ΔE +PΔV q v = ΔE+ P(0) q v = ΔE q v ΔE Thermochemical Equations The stoichiometric coefficients always refer to the number of moles of a substance q p ΔΗ H 2 O (s) H 2 O (l) ΔH = 6.01 kj If you reverse a reaction, the sign of ΔH changes H 2 O (l) H 2 O (s) ΔH = kj If you multiply both sides of the equation by a factor n, then ΔH must change by the same factor n. 2H 2 O (s) 2H 2 O (l) ΔH = 2 x 6.01 = 12.0 kj q = nδh (Heat of fusion) Thermochemical Equations The physical states of all reactants and products must be specified in thermochemical equations. H 2 O (s) H 2 O (l) ΔH = 6.01 kj H 2 O (l) H 2 O (g) ΔH = 44.0 kj How much heat is evolved when 266 g of white phosphorus (P 4 ) burn in air? (H = heat of combustion) P 4 (s) + 5O 2 (g) P 4 O 10 (s) ΔH = kj 1 mol P kj 266 g P 4 x x = 6470 kj g P 4 1 mol P 4 5

6 Stoichiometry and Heat of Reaction Thermite Animation 2 Al (s) + Fe 2 O 3 (s) -> Al 2 O 3 (s) + 2 Fe(s) Hess s s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. (Enthalpy is a state function. It doesn t matter how you get there, only where you start and end.) NET RXN = RXN 1 + RXN 2 + RXN ΔH RXN = ΔH RXN 1 + ΔH RXN 2 + ΔH RXN3 6

7 Thermodynamics State functions are properties that are determined by the state of the system, regardless of how that condition was achieved. energy, pressure, volume, temperature Potential energy of hiker 1 and hiker 2 is the same even though they took different paths. State Functions Calculate the standard enthalpy of formation of CS 2 (l) given that: C(graphite) + O 2 (g) CO 2 (g) ΔH 0 = kj rxn S(rhombic) + O 2 (g) SO 2 (g) ΔH 0 = kj rxn CS 2 (l) + 3O 2 (g) CO 2 (g) + 2SO 2 (g) ΔH 0 = kj rxn 1. Write the enthalpy of formation reaction for CS 2 C(graphite) + 2S(rhombic) CS 2 (l) 2. Add the given rxns so that the result is the desired rxn. + C(graphite) + O 2 (g) CO 2 (g) ΔH 0 rxn = kj 2S(rhombic) + 2O 2 (g) 2SO 2 (g) ΔH 0 rxn = x2 kj CO 2 (g) + 2SO 2 (g) CS 2 (l) + 3O 2 (g) ΔH 0 = kj rxn Two Ways Thermochemical Equations Data Can Be Given 1. Thermochemical Equations 2. ΔH f0 Values; Standard Molar Enthalpy of Formation Under Standard State Conditions C(graphite) + 2S(rhombic) CS 2 (l) ΔH 0 rxn= (2x-296.1) = 86.3 kj 7

8 Standard enthalpy of formation (ΔH f0 ) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm (standard state conditions). ΔH f0 (Li 2 O) = kj/mole stands for 2 Li (s) + ½ O 2 (g) -> Li 2 O (s); ΔH 0 = kj/mole The standard enthalpy of formation of any element in its most stable form is zero. ΔH 0 (O 2 ) = 0 f ΔH 0 (O 3 ) = 142 kj/mol f ΔH 0 (C, graphite) = 0 f ΔH 0 (C, diamond) = 1.90 kj/mol f Determining ΔH 0 rxn aa + bb from ΔH 0 f values cc + dd ΔH 0 rxn = [ cδh 0 f (C) + dδh 0 f (D)] - [ aδh 0 f (A) + bδh 0 f (B)] Benzene (C 6 H 6 ) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is kj/mol. 2C 6 H 6 (l) + 15O 2 (g) 12CO 2 (g) + 6H 2 O (l) ΔH 0 rxn = ΣnΔH 0 f (products) - ΣmΔHf 0 (reactants) ΔH 0 rxn = [ 12ΔH 0 f (CO 6ΔH 0 2 ) + f (H 2 O) ] - [ 2ΔH 0 f (C 6 H 6 )] ΔH 0 rxn = ΣnΔH 0 f (products) - ΣmΔHf 0 (reactants) ΔH 0 rxn = [ 12x x ] [ 2x49.04 ] = kj kj 2 mol = kj/mol C 6 H 6 8

9 Obtaining ΔH rxn Experimentally Determining ΔH from a Temperature Change 1. Hess s Law 2. Using ΔH 0 f Values 3. Experimentally Constant Pressure Calorimetry (Coffee Cup Calorimeter) Different quantities of the same substance absorb/release different amounts of heat for same temperature change. Different substances can absorb the same amount of heat yet have a different temperature increase The specific heat (s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius. The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius. C = ms How much heat is given off when an 869 g iron bar cools from 94 0 C to 5 0 C? s of Fe = J/g 0 C Δt = t final t initial = 5 0 C 94 0 C = C q = msδt = 869 g x J/g 0 C x 89 0 C = -34,000 J Heat (q) absorbed or released: q = msδt q = CΔt Δt = t final - t initial 9

10 Calorimetry; Measurement of Heat Changes Constant-Pressure Calorimetry System Surroundings System; Chemical Reaction Surroundings; Everything Else (including water reactants and products are dissolved in) q sys = q water + q cal + q rxn q sys = 0 q rxn = - (q water + q cal ) q water = msδt q cal = C cal Δt Reaction at Constant P ΔH = q rxn / n No heat enters or leaves! 6.4 Coffee-Cup Calorimeter -q System = +q Surroundings -(c solid x mass solid x ΔT solid ) = + (c water x mass water x ΔT water ) Heat Changes at Constant Volume ΔT = T final -T initial Follow-Up Problem 6.4, p

11 Heat Changes at Constant Volume = Energy Changes at Constant Volume For most reactions: ΔE ~= ΔH q V = ΔE Bomb Calorimeter: -q sample = +q calorimeter = heat capacity x ΔT heat capacity = (heat capacity bomb + heat capacity water x mass water ) Constant-Volume Calorimetry q sys = q water + q bomb + q rxn q sys = 0 q rxn = - (q water + q bomb ) q water = msδt q bomb = C bomb Δt Reaction at Constant V ΔH = q rxn No heat enters or leaves! ΔH ~ q rxn ΔE = q rxn 11