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Chapter 17 Energy and Chemical Change

Chapter 17 Energy and Chemical Change. Thermochemistry. The study of heat changes in chemical reactions. Law of Conservation of Energy. Defn – energy can be converted from one to another, but neither created nor destroyed Ex: potential to kinetic solar to chemical.

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Chapter 17 Energy and Chemical Change

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  1. Chapter 17 Energy and Chemical Change

  2. Thermochemistry • The study of heat changes in chemical reactions

  3. Law of Conservation of Energy • Defn – energy can be converted from one to another, but neither created nor destroyed • Ex: potential to kinetic solar to chemical

  4. Energy and Chemical Change • Energy (defn) – ability to do work or produce heat • Exists in two forms a) potential energy – stored energy b) kinetic energy – energy in motion

  5. Chemical Potential Energy • Defn – energy stored in a substance based on composition • Ex: methane (CH4) vs. propane (C3H8) H H H H C C C C H H H H H H H H Propane has more energy b/c it has more bonds

  6. Heat (q) • Defn – energy that goes from warmer object to colder object • Energy transfer moves from an area of high energy to an area of low energy

  7. Units of Heat • calorie (cal) – amount of heat required to raise the temp of 1 gram of water one degree Celsius • Calorie vs calorie 1 Calorie = 1000 calories = 1 kilocalorie (kcal) • Joule – SI unit of heat and energy 1 calorie = 4.18 J

  8. Comparing Specific Heat water (liquid) 4.18 water (ice) 2.03 ethanol 2.44 aluminum 0.897 iron 0.449

  9. Specific Heat • Defn – amount of heat required to raise temp of one gram of any substance by one degree Celsius q = m c ΔT change In temp heat mass specific heat J Unit: g ̊ C

  10. Specific Heat • If heat is released from the reaction: -q • If heat is absorbed into the reaction: +q • Change in temperature = tfinal - tinitial

  11. Sample problem (a) • The temperature of a 10.0 g sample of iron is changed from 50.4°C to 25.0°C with the release of 114 J of heat. What is the specific heat of iron? q = mcΔT 114 J = (10.0 g) c (25.4°C) c = 0.449 J/g·°C

  12. Sample problem (b) • If the temperature of 34.4 g ethanol increases from 25°C to 78.8°C, how much heat is absorbed by ethanol? (specific heat of ethanol = 2.44 J/g°C) q = mcΔT = (34.4 g) ( 2.44 J/g°C) (53.8°C) = 4515 J

  13. Heat In Chemical Reactions • Calorimeter – insulated device used to measure amount of heat absorbed or released during a chemical or physical process • Thermochemistry – study of heat changes in chemical reactions

  14. 3 parts we look at • 1) system – specific part of universe that contains the reaction • 2) surroundings – everything in universe other than the system • 3) universe – system + surroundings

  15. Enthalpy (H) • Defn – heat content of a system • Enthalpy (heat) of rxn (ΔHrxn) a) defn – change in enthalpy for a reaction b) formula ΔHrxn = Hproducts – Hreactants A + B  C + D

  16. Enthalpy (H) • c) endothermic vs exothermic rxns if +ΔHrxn = endothermic rxn if –ΔHrxn = exothermic rxn

  17. Reaction Energy Diagrams • This is an exothermic reaction – the reactants have more energy than the products, so energy has been released

  18. Reaction Energy Diagrams • This is an endothermic reaction – the reactants have less energy than the products, so energy has been absorbed

  19. Reaction Energy Diagrams • A: energy held by the activated complex • B: energy of the reactants • C: energy of the products • F: heat of reaction (ΔH) • I: activation energy

  20. EXOTHERMIC ENDOTHERMIC products reactants ΔH ΔH H H products reactants ΔH < 0 ΔH > 0

  21. Example reactions • 4 Fe + 3 O2 2 Fe2O3 + 1625 kJ i) exo- or endo-? ii) what is ΔHrxn? Exothermic (heat written on right side of equation) ΔHrxn = -1625 kJ Another way to write equation: 4 Fe + 3 O2 2 Fe2O3ΔHrxn = -1625 kJ

  22. 4 Fe + 3 O2 H ΔH = -1625 kJ 2 Fe2O3

  23. Example reactions • 27 kJ + NH4NO3 NH4+ + NO31- i) exo- or endo-? ii) what is ΔHrxn? Endothermic (heat written on left side of equation) ΔHrxn = +27 kJ Another way to write equation: NH4NO3 NH4+ + NO31- ΔHrxn = +27 kJ

  24. NH4+ + NO3- ΔH = +27 kJ H NH4NO3

  25. Standard Enthalpy of Formation (ΔHf°) • Defn – change in enthalpy when one mole of a compound is formed from its elements in their standard states • Standard State – normal physical state of substance at room conditions (25°C and 1 atm) ex:standard state of Hg is liquid N2 is gas

  26. Standard Enthalpy of Formation (ΔHf°) • Examples H2 (g) + S (s)  H2S (g) ΔHf° = -21 kJ S (s) + O2 (g)  SO2 (g) ΔHf° = -297 kJ

  27. Hess’s Law • Defn – overall enthalpy change of reaction is equal to the sum of the enthalpy changes of individual steps A  D ΔH = ? A + B  C ΔH = x C  D + B ΔH = y Overall: A  D ΔH = x + y

  28. Example problem #1 • Calculate the enthalpy of reaction, ΔH, for the reaction: 2 H2O2 2 H2O + O2 (a) H2 + O2  H2O2 ΔH = -188 kJ (b) 2 H2 + O2 2 H2O ΔH = -572 kJ

  29. Example problem #1 2 H2O2 2 H2 + 2 O2ΔH = +376 kJ 2 H2 + O2  2 H2O ΔH = -572 kJ 2 H2O2 + 2 H2 + O2  2 H2O + 2 H2 + 2 O2 2 H2O2 2 H2O + O2 ΔH = -196 kJ 1

  30. Example problem #2 • Calculate the enthalpy of reaction, ΔH, for the reaction: 2 S + 3 O2 2 SO3 (a) S + O2 SO2ΔH = -297 kJ (b) 2 SO3  2 SO2 + O2ΔH = 198 kJ

  31. Example problem #2 2 S + 2 O2 2 SO2ΔH = -594 kJ 2 SO2 + O2  2 SO3ΔH = -198 kJ 2 S + 3 O2 + 2 SO2 2 SO2 + 2 SO3 2 S + 3 O2 2 SO3 ΔH = -792 kJ

  32. Heat of Reaction (ΔHrxn) • Defn – amount of heat lost or gained in a reaction • Formula ΔHrxn = ΣΔHf (products) – ΣΔHf (reactants)

  33. Example problem • Find ΔHrxn for the following reaction. Is reaction endo- or exothermic? H2S + 4 F2  2 HF + SF6 ΔHrxn = = -1745 kJ -21 kJ 0 kJ -273 kJ -1220 kJ [(2)(-273) + (1)(-1220)] - [(1)(-21) + (0)(4)] exothermic

  34. Entropy (S) • Defn – measure of disorder or randomness in a system • Formula ΔS = Sproducts – Sreactants • Reaction Tendency • Nature favors a disordered state • The more entropy/disorder, the greater ΔS Unit: J/K

  35. Entropy (S) • Predicting ΔS - go to more disorder  + ΔS - go to less disorder  - ΔS • Ex problem: predict ΔSsystem for these changes a) H2O (s)  H2O (l) + ΔS; Solid to liquid is more disorder

  36. Entropy (S) b) 2 SO3 (g)  2 SO2 (g) + O2 (g) • Keep in mind: the reverse reactions have opposite signs + ΔS There are more product particles (3) than reactant particles (2)

  37. Entropy example problem • Calculate entropy change for this reaction: 2 PbO2 (s)  2 PbO (s) + O2 (g) ΔSrxn = = +209.2 J/K 66.6 68.7 205 - [(2)(66.6)] [(2)(68.7) + 205]

  38. Entropy, the Universe, and Free Energy • The universe entropy ΔSuniverse > 0 • Two natural universe processes a) things tend to go towards lower energy (-ΔH, exothermic) b) things tend to go towards higher disorder (+ΔS)

  39. Spontaneous Process • Defn – physical or chemical change that occurs with no outside intervention

  40. Free Energy (G) • Defn – energy available to do work • Gibbs Free Energy Equation ΔG = ΔH – TΔS T is in Kelvin • What does ΔG tell you? - ΔG  spontaneous rxn + ΔG  not spontaneous rxn

  41. Free Energy Problem • calculate the change in free energy, ΔG, for the reaction at 25°C. Is the reaction spontaneous or nonspontaneous? N2(g) + 3 H2 (g)  2 NH3 (g) ΔHrxn = -91800 J, ΔSrxn = -197 J/K

  42. Free Energy Problem ΔGrxn = ΔHrxn – TΔSrxn = -91800 J – (298 K)(-197 J/K) = -33100 J Spontaneous

  43. How does ΔH and ΔS affect spontaneity? -ΔH +ΔH Spontaneity depends on temp Always spontaneous +ΔS Spontaneity depends on temp never spontaneous -ΔS

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