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Thermochemistry: Chemical Energy

Thermochemistry: Chemical Energy. Thermodynamics 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 E K = 1 / 2 mv 2 (1 Joule = 1 kg m 2 /s 2 ) (1 calorie = 4.184 J)

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Thermochemistry: Chemical Energy

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  1. Thermochemistry: Chemical Energy

  2. Thermodynamics 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 = 1/2mv2(1 Joule = 1 kgm2/s2) • (1 calorie = 4.184 J) • Potential Energy:stored energy

  3. Thermodynamics 02 • Conservation of energy law: Energy cannot be created or destroyed; it can only be converted from one form into another.

  4. 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

  5. Thermodynamics 04 • First Law of Thermodynamics: energy of an isolated system must be kept constant

  6. 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 ispositiveenergy flows into the system

  7. Thermodynamics 06

  8. Work 07 w = –PDV

  9. Sign of w 08 negativepositive w = -PDV  expansion positivenegative w = -PDV  contraction

  10. Work Units 09 w = -PDV (J or kJ) w = L x atm = 1L x 1000mL x 1cm3 x 1m3 1L 1mL (100cm)3 m2 101 x 103 kg ms2 1000 = 101 kgm2 = 101J s2

  11. Energy and Heat 10 Energy = Work + Heat q DE= w + = q -PDV When a person does work, energy diminishes w = negative DE = negative q =DE +PDV

  12. Heat and Enthalpy 11 • The amount of heat exchanged between the system and the surroundings is given the symbolq. • q = DE + PDV • At constant volume (DV = 0):qv = DE • At constant pressure:qp = DE + PDV enthalpy = DH

  13. State Functions 12 State Function: • value depends only on the present state of the system • path independent • when returned to its original position, overall change is zero

  14. State Functions 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) • temperature, total energy, pressure, density, volume, and enthalpy (∆H)  state properties • nonstate properties include heat and work

  15. Enthalpy 14 • Enthalpy or heat of reaction: • DH = H(products) - H(reactants) • States of the reactants and products are important!  (g, l, s, aq) • Thermodynamic standard state: P = 1atm, [ ] = 1M, T = 298.15K (25ºC)

  16. Standard Enthalpy of Reaction 15 Thermodynamic standard state: P = 1atm, [ ] = 1M, T = 298.15K (25ºC) Standard enthalpy of reaction (Hº) N2(g) + 3H2(g)  2NH3(g) DHº = -92.2kJ

  17. Enthalpy Changes 16 • Most changes in a system involve a gain or loss in enthalpy • Physical (melting of ice in a cooler) • Chemical (burning of gas in your car)

  18. Physical Changes 17 • Enthalpies of Physical Change:

  19. Chemical Changes 18 • Enthalpies of Chemical Change:Often called heats of reaction (DHreaction). • Endothermic:Heat flows into the system from the surroundings  DH is positive • Exothermic:Heat flows out of the system into the surroundings DH is negative

  20. Enthalpy Changes 19 • Reversing a reaction changes the sign of DHfor a reaction. • C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l) DH = –2219 kJ • 3 CO2(g) + 4 H2O(l)  C3H8(g) + 5 O2(g) DH = +2219 kJ • Multiplying a reaction increases DHby the same factor. • 3 x [C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l) ] • DH =(-2219kJ x 3) = –6657 kJ

  21. Example 20 • How much work is done (in kilojoules), and in which direction, as a result of the following reaction? w = -0.25kJ Expansion, system loses -0.25kJ

  22. Example 21 • The following reaction has DE = –186 kJ/mol. • Is the sign of PDV positive or negative? • What is the sign and approximate magnitude of DH? Contraction, PDV is negative, w is positive DH = DE + PDV DH = (-186kJ) + (1atm) (-1mole) DH = negative (slightly more than DE)

  23. Example 22 The reaction between hydrogen and oxygen to yield water vapor has DH° = –484 kJ. How much PV work is done, and what is the value of DE (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? PDV = -0.57kJ Contraction, so w is positive DE = -120.43kJ

  24. 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) PDV = 45.2kJ Expansion, so w = -45.2kJ

  25. Example 24 How much heat (kJ) is evolved or absorbed in each of the following reactions? 1.) Burning of 15.5 g of propane: C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l) DHº = –2219 kJ 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)DHº = +80.3 kJ -780kJ (exothermic) +1.24kJ (endothermic)

  26. Hess’s Law 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. • 3 H2(g) + N2(g)  2 NH3(g) DH° = –92.2 kJ

  27. Hess’s Law 26 (a) 2 H2(g) + N2(g) N2H4(g) DH°1 = ? • (b) N2H4(g) + H2(g) 2 NH3(g) DH°2 = –187.6 kJ (c) 3 H2(g) + N2(g) 2 NH3(g) DH°3 = – 92.2 kJ • DH°1 = DH°3 – DH°2 = (–92.2 kJ) – (–187.6 kJ) = +95.4 kJ

  28. Standard Heats of Formation 27 Where do DH° values come from? • Standard Heats of Formation (DH°f): enthalpy change for the formation of 1 mole of substance in its standard state • DH°f = 0 for an element in its standard state!

  29. Standard Heats of Formation 28 H2(g) + 1/2 O2(g)  H2O(l) DH°f = –286 kJ/mol 3/2 H2(g) + 1/2 N2(g)  NH3(g) DH°f = –46 kJ/mol 2 C(s) + H2(g)  C2H2(g) DH°f = +227 kJ/mol 2 C(s) + 3 H2(g) + 1/2 O2(g)  C2H5OH(g) DH°f = –235 kJ/mol

  30. Standard Heats of Formation 29 • Calculating DH° for a reaction: • DH° = DH°f (products) – DH°f (reactants) Heat of formation must be multiplied by the coefficient of the reaction • C6H12O6 (s) 2C2H5OH (l) + 2CO2 (g) DH° = [2DH°f(ethanol) + 2DH°f(CO2)] - DH°f (glucose)

  31. 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 Standard Heats of Formation 30 Some Heats of Formation, Hf° (kJ/mol)

  32. Bond Dissociation Energy 31 • Bond Dissociation Energy (D): Amount of energy needed to break a chemical bond in gaseous state D = Approximate DHº • DH° = D(reactant bonds broken) – D(product bonds formed) H2 + Cl2 2HCl • DH° = (DCl-Cl + DH-H) - (2 D H-Cl) • = [(1 mol)(243 kJ/mol) + (1)(436 kJ/mol] - (2)(432 kJ/mol) • = -185 kJ

  33. Bond Dissociation Energy 32

  34. Calorimetry and Heat Capacity 33 • Calorimetry: measurement of heat changes (q) for chemical reactions • Constant Pressure Calorimetry: measures the heat change at constant pressure q = DH • Bomb Calorimetry: measures the heat change at constant volume such that q = DE

  35. Calorimetry and Heat Capacity 34 Constant Pressure Bomb

  36. q C = T D Calorimetry and Heat Capacity 35 • Heat capacity {C}: amount of heat required to raise the temperature of an object or substance a given amount • Specific Heat: amount of heat required to raise the temperature of 1.00 g of substance by 1.00°C • Molar Heat: amount of heat required to raise the temperature of 1.00 mole of substance by 1.00°C

  37. Calorimetry and Heat Capacity 36

  38. 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 DH° (kJ) • CH4(g) + Cl2(g) CH3Cl(g) + HCl(g) DH° = –98.3 kJ • CH3Cl(g) + Cl2(g) CH2Cl2(g) + HCl(g) DH° = –104 kJ DH° = -98.3 + -104 = -202kJ

  39. Example 38 Calculate DH° (kJ) for the reaction of ammonia with O2 to yield nitric oxide (NO) and H2O(g), a step in the Ostwald process for the commercial production of nitric acid. 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) = [(4)(90.2kJ/mol) + (6)(-241.8)] - [(4)(-46.1) + (5)(0)] = -905.6kJ

  40. Example 39 Calculate DH° (kJ) for the photosynthesis of glucose from CO2 and liquid water, a reaction carried out by all green plants. 6CO2(g) + 6H2O(l) C6H12O6(s) + 6O2(g) = [(1 mole)(-1260kJ/mol) + (6)(0)] - [(6)(-393.5) + (6)(-285.8)] = 2816kJ

  41. Example 40 • Calculate an approximate DH° (kJ) for the synthesis of ethyl alcohol from ethylene: • C2H4(g) + H2O(g)  C2H5OH(g) • Calculate an approximate DH° (kJ) for the synthesis of hydrazine from ammonia: • 2 NH3(g) + Cl2(g)  N2H4(g) + 2 HCl(g)

  42. Introduction to Entropy 42 • Second Law of Thermodynamics: Reactions proceed in the direction that increases the entropy of the system plus surroundings. • A spontaneous process is one that proceeds on its own without any continuous external influence. • A nonspontaneous processtakes place only in the presence of a continuous external influence.

  43. Introduction to Entropy 43 • The measure of molecular disorder in a system is called the system’s entropy; this is denoted S. • Entropy has units of J/K (Joules per Kelvin). • DS = Sfinal – Sinitial • Positive value of DS indicates increased disorder. • Negative value of DS indicates decreased disorder.

  44. Introduction to Entropy 44

  45. Introduction to Entropy 45 • To decide whether a process is spontaneous, both enthalpy and entropy changes must be considered: • Spontaneous process: Decrease in enthalpy (–DH). Increase in entropy (+DS). • Nonspontaneous process: Increase in enthalpy (+DH). Decrease in entropy (–DS).

  46. Introduction to Entropy 39 • Predict whether DS° is likely to be positive or negative for each of the following reactions. Using tabulated values, calculate DS° for each: • a. 2 CO(g) + O2(g)  2 CO2(g)b. 2 NaHCO3(s)  Na2CO3(s) + H2O(l) + CO2(g)c. C2H4(g) + Br2(g)  CH2BrCH2Br(l)d. 2 C2H6(g) + 7 O2(g)  4 CO2(g) + 6 H2O(g)

  47. Introduction to Free Energy 40 • Gibbs Free Energy Change (DG): Weighs the relative contributions of enthalpy and entropy to the overall spontaneity of a process. • DG = DH – TDS • DG < 0 Process is spontaneous • DG = 0 Process is at equilibrium • DG > 0 Process is nonspontaneous

  48. Introduction to Free Energy 41 • Situations leading to DG < 0: • DH is negative and TDS is positive • DH is very negative and TDS is slightly negative • DH is slightly positive and TDS is very positive • Situations leading to DG = 0: • DH and TDS are equally negative • DH and TDS are equally positive • Situations leading to DG > 0: • DH is positive and TDS is negative • DH is slightly negative and TDS is very negative • DH is very positive and TDS is slightly positive

  49. Introduction to Free Energy 42 • Which of the following reactions are spontaneous under standard conditions at 25°C? • a. AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq) DG° = –55.7 kJ • b. 2 C(s) + 2 H2(g)  C2H4(g)DG° = 68.1 kJ • c. N2(g) + 3 H2(g)  2 NH3(g) DH° = –92 kJ; DS° = –199 J/K

  50. Introduction to Free Energy 43 • Equilibrium (DG° = 0): Estimate the temperature at which the following reaction will be at equilibrium. Is the reaction spontaneous at room temperature? • N2(g) + 3 H2(g)  2 NH3(g) DH° = –92.0 kJ DS° = –199 J/K • Equilibrium is the point where DG° = DH° – TDS° = 0

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