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Chapter 6

Chapter 6. Thermochemistry. Energy and Work. In what terms d o chemists define energy? What are some of the many types of energy?. The ability to do work Work is a change in energy because of a process Kinetic , potential, thermal, chemical, nuclear, gravitational, etc.

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Chapter 6

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  1. Chapter 6 Thermochemistry

  2. Energy and Work • In what terms do chemists define energy? • What are some of the many types of energy? • The ability to do work • Work is a change in energy because of a process • Kinetic , potential, thermal, chemical, nuclear, gravitational, etc.

  3. Defining Energy Changes • What type of energy is the focus of thermochemistry? • How do we define these objects or bodies? • Heat • Exchange of thermal energy between two objects or bodies • Either as part of the: • System, or • Surroundings

  4. Defining Energy Changes • What are differences between the three types of systems? • Open – exchange of mass and energy • Closed – exchange of energy only • Isolated – no exchange of energy or mass

  5. Defining Energy Changes • What is the difference between exothermic and endothermic processes? • Exothermic • Releases heat to the surroundings • Endothermic • Absorbs heat from the surroundings

  6. Enthalpy • How can we quantify heat? • Impossible to measure the amount of heat in a substance • Must measure the change (Δ) in heat • Enthalpy (H) – measure of heat flow in a constant pressure system • Enthalpy can change during a physical or chemical change

  7. Enthalpy • What is the formula for enthalpy? • ΔH  Enthalpy • ΔH = Hfinal – Hintial • ΔH = Hproducts – Hreactants • ΔH > 0 = endothermic • ΔH < 0 = exothermic

  8. I never really thought about it that way before… • Determine whether the change in enthalpy is positive or negative? • Ice melting • Air around the melting ice • Water vapor condensing on a glass • Cold drink outside on a hot day • Combustion of methane • Neutralization of vinegar with baking soda • Ice absorbs heat, endothermic, + • Air loses heat, exothermic, - • Water vapor loses heat, exothermic, - • Cold drink absorbs heat, endothermic, + • Methane releases heat, exothermic, - • Reaction releases heat, exothermic, -

  9. Thermochemistry • What is the chemical equation for melting ice? • H2O(s) → H2O(l) • ΔH = 6.01 kJ • Positive • Endothermic

  10. Thermochemistry • What is the reaction for the combustion of methane? • CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) • ΔH = -890.4 kJ • Negative • Exothermic

  11. Thermochemistry Notes • What are some guidelines to thermochemical reactions? • Coefficients represent the number of moles • Reversing a reaction changes the sign • If you multiply the reaction by a constant, ΔH is also multiplied by the same constant • Always show the physical state of the substance (s, l or g)

  12. Practice Example 6.1: SO2(g) +½O2(g) → SO3(g) ΔH = -99.1 kJ Calculate the heat evolved when 74.6 g of SO2 is converted to SO3. • Answer = -115 kJ P4(s) + 5O2(g) → P4O10(s) ΔH = -3013 kJ Calculate the heat evolved when 266 g of white phosphorus burns (combusts) in air. • Answer = -6.48 x 103 kJ

  13. Enthalpy of Formation • What is the enthalpy of formation? • ∆Hºf • The heat change when one mole of a compound is formed from its elements at 1 atm • Table 6.3 (pg. 216) • Appendix 3 (pg. A-8)

  14. Enthalpy of Formation • Why is the enthalpy of formation so critical? • When we know the enthalpy of each piece we can determine the enthalpy of the entire reaction (∆Hºrxn)

  15. Enthalpy of Reaction • How do we directly calculate the enthalpy of reaction? • Consider this standard reaction: • aA + bBcC + dD • Lower case letters  molar coefficient (mol) • Upper case letters  enthalpy of formation of substances/chemicals (kJ/mol) • ∆Hºrxn = mƩ∆Hºf(products) - nƩ∆Hºf(reactants) • ∆Hºrxn = [c∆Hºf(C) + d∆Hºf(D)] - [a∆Hºf(A) + b∆Hºf(B)]

  16. Example • Example 6.6 (pg. 219) 2B5H9(l) + 12O2(g) 5B2O3(s) + 9H2O(l) Use Appendix 3 and the following equation to determine the ∆Hºrxn. The standard enthalpy formation of B5H9 is 73.2 kJ/mol. ∆Hºrxn = [c∆Hºf(C) + d∆Hºf(D)] - [a∆Hºf(A) + b∆Hºf(B)] ∆Hºrxn = [5(-1263.6) + 9(-285.8)] - [2(73.2) + 12(0)] • Answer = -9036.6 kJ

  17. Another Example • Example 6.6 – Practice (pg. 220) • 2C6H6(l) + 15O2(g)  12CO2(g) + 6H2O(l) • Calculate the heat released in kJ by the combustion of benzene. • ∆Hºf(C6H6) = 49.04 kJ/mol • ∆Hºrxn = [12(-393.5)+6(-285.8)]-[2(49.04)+15(0)] • ∆Hºrxn = [-6436.8]-[98.08] • ∆Hºrxn = -6534.9 kJ

  18. Practice Problem #1 Answer for 6.38 (a) = -1410.9 kJ Answer for 6.38 (b) = -1123.5 kJ

  19. Practice Problem #2 ∆Hºrxn = 177.8 kJ remember this is for one mole ∆Hº = 2.70 x 102 kJ

  20. Enthalpy of Combustion • What is the enthalpy of combustion (∆Hºc)? • Also called heat of combustion • Amount of heat released when a substance is combusted with oxygen • Always strongly negative • Table 12 in your data booklet

  21. Enthalpy of Solution • What is the enthalpy of solution (∆Hºsoln)? • Change in enthalpy when a solute is dissolved in a solvent • Can be exothermic or endothermic

  22. Bond Enthalpies • What are bond enthalpies? • Energy required to make and break chemical bonds • Also called bond dissociation energy • More details when we get to chapter nine

  23. Calorimetry • How do we measure heat changes? • Calorimetry is the measurement of heat changes • A calorimeter is the instrument of choice • 1 cal = 4.184 J

  24. Specific Heat and Heat Capacity • What information is needed to determine how much heat is exchanged? • q = mc∆t(IB use) • q = ms∆t(book use) • q = heat (J) • m = mass (g) • ∆t = change in temperature (˚C) • ∆t = tfinal– tinitial • c, s = specific heat (J/(g * ˚C))

  25. Specific Heat and Heat Capacity • What is specific heat (c, s)? • How is this related to heat capacity (C)? • The amount of heat needed (J) to raise the temperature of 1 gram of a substance by 1 ˚C • Water = 4.184 J/(g * ˚C) • The amount of heat needed to raise the temperature of a known mass of a substance by 1 ˚C • C = mc • q = C∆t

  26. Specific Heat and Heat Capacity • What is the difference between specific heat and heat capacity? • c = intensive property, not dependent on mass • C = extensive property, depends on the mass

  27. In other words…

  28. Specific Heat and Heat Capacity • Example 6.2 (pg. 210): A 466 g sample of water is heated from 8.50˚C to 74.60˚C. Calculate the amount of heat absorbed by the water in kJ. • q = mc∆t • q = (466 g)(4.184 J/(g *˚C))(74.60˚C – 8.50˚C) • q = 129000 J • q = 129 kJ

  29. Specific Heat and Heat Capacity Example 6.2 (pg. 210) • An iron bar of mass 869 g cools from 94˚C to 5˚C. Calculate the amount of heat release, in kJ, by the metal. • Specific heat capacity of iron • c = 0.444 J/(g * ˚C) • q = mc∆t • q = (869 g)(0.444/(g *˚C))(5˚C – 94˚C) • q = -34 kJ

  30. Getting Out of the Shower • Why do you get cold when you get out of the shower? • Water that remains on your body wants to evaporate • Water has a high specific heat , so it takes a lot of energy/heat to evaporate • The more energy/heat you lose the colder you feel… obviously

  31. Firewalking • How is it possible to walk across a bed of hot coals at temperatures of over 1000˚C without burning your feet? • Is it psychological? • Or is it simple physical chemistry?

  32. Firewalking World Record

  33. Firewalking • How is it possible to walk across a bed of hot coals at temperatures of over 1000˚C without burning your feet? • Water in feet, high heat capacity • Ash under feet, low heat capacity • The same amount of heat leaving the coals will lower their temperature much more than the same amount heat trying to raise the temperature in your foot

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