1 / 29

Chapter 6

Chapter 6. Thermochemistry. Thermochemistry. Deals with the energy changes Chemical reactions Determines whether a reaction is exothermic or endothermic Physical Changes Energy changes. Vocabulary. Energy - The ability to do work (E) in J Potential Energy – Energy of position (PE)

wallace-barrett
Download Presentation

Chapter 6

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 6 Thermochemistry

  2. Thermochemistry • Deals with the energy changes • Chemical reactions • Determines whether a reaction is exothermic or endothermic • Physical Changes • Energy changes

  3. Vocabulary • Energy - The ability to do work (E) in J • Potential Energy – Energy of position (PE) • Kinetic Energy – Energy of motion (KE) • Temperature – Random motion of particles • Heat – Transfer of energy (q) in J • Work – Force x Distance (W) in J • Enthalpy – Energy Change for reactions • (ΔH) in J

  4. Vocabulary • Pathway – How energy is transferred • State Function – Properties that depend on current state. (Independent of pathway) • Energy is a state function • Temperature, Volume, Pressure • Heat and Work are not state functions

  5. Example of Energy Change

  6. Endothermic Reactions • Energy is put into a reaction • Product bonds are weaker than reactant bonds http://www.bbc.co.uk/scotland/education/bitesize/higher/img/chemistry/calculations_1/pe_diags/fig10.gif

  7. Exothermic Reactions • Energy is released from a reaction • Product bonds are stronger than reactant bonds http://www.bbc.co.uk/scotland/education/bitesize/higher/img/chemistry/calculations_1/pe_diags/fig03.gif

  8. Signs for Exo vs. Endo.

  9. Calorimetry • Measuring heat associated with a reaction • Uses a calorimeter • Coffee Cup Calorimeter • At constant pressure q = -ΔH q=m*s* ΔT • q = heat, m = mass, ΔT = Change in Temp • s=specific heat capacity

  10. Specific Heat Capacity • Energy required to heat one gram of a substance 1 degree Celcius • Units J/g*ºC • Molar Heat Capacity = Just per mole • Metals have low heat capacities

  11. Example How much energy is required to heat 55.3 grams of Aluminum metal from 22.0ºC to 100.0 ºC? Specific heat of water is 4.18 J/g*ºC the specific heat of aluminum is 0.89 J/g*ºC

  12. Example #46 p. 282

  13. Homework • P. 281 #’s 32, 37,40, 43, 48

  14. Hess’s Law • In going from a particular set of reactants to a particular set of products, the change in enthalpy (H) is the same whether the reaction takes place in one step or in a series of steps • Energy is a state function

  15. Comparison • One step: N2(g) + 2O2(g) 2NO2(g)H1 = 68kJ • Two step N2(g) + O2(g) 2NO(g)H2 = 180kJ 2NO(g) + O2(g) 2NO2(g) H3 = -112kJ N2(g) + 2O2(g) 2NO2(g)H2 + H3 = 68kJ

  16. Hess’s Law Rules • Start with the end reaction in mind and work backward rearrange your equations • Work to cancel like terms • When reversing the equation the sign of ΔH must be changed • When multiplying an equation by a coefficient ΔH must be multiplied by the same coefficient

  17. Example #54 p. 283 NH3 ½ N2 + 3/2 H2H = 46 kJ 2H2 + O2 2H2OH = -486 kJ 2N2 + 6H2O 3O2 + 4NH3H = ? Based on enthalpy is this a useful synthesis?

  18. Example #52 p. 283 C4H4 + 5O2 4CO2 + 2H2O H = -2341kJ C4H8 + 6O2 4CO2 + 4H2OH = -2755kJ H2 + ½ O2 H2O H = -286kJ C4H4 + 2H2 C4H8H = ?

  19. Homework • P. 283 #’s 51,53,55,57

  20. Calculate how much energy (kJ) is required to heat 50.0 L of tap water from 10.0 ºC to 45.0 ºC. Assume the density of water is 1.00 g/mL and specific heat of water is 4.18 J/g* ºC • 7320 kJ • A McDonalds hamburger contains 250. Calories (that’s 2.50x105 calories). How many burgers would it take to supply the same amount of energy? 1 calorie = 4.18 J • 7.00 burgers

  21. Enthalpy of Formation • The change in enthalpy that accompanies the formation of one mole of a compound from its elements with all elements in their standard state • Symbol = ΔHfº • f is formation • Degree symbol means standard states

  22. Standard States • Standard States • Temperature is 25 ºC • Pressure = 1.00 atm • Concentration = 1.00 M • The standard state is how something exists at these conditions • O2(g), Na(s), Br2(l)

  23. Reactions • 4C(s) + 2H2(g) CH4(g) ΔHfº = -75 kJ/mol • H2(g) + ½ O2(g)  H2O(l) ΔHfº = -286 kJ/mol • Write the enthalpy of formation reaction for gaseous carbon dioxide. Don’t worry about the value • C(s) + O2(g)  CO2(g) • Values are in Appendix 4 (A21) • ΔHfº = -393.5 kJ/mol

  24. Enthalpy of Formation Values • Most values are negative • Compounds are more stable than element • All elements are zero • Even diatomic elements! • States matter • Liquid water is -286 • Gaseous water is -242 • Use this information to calculate enthalpy of the reaction!

  25. Standard Enthalpy of Reaction • Standard Enthalpy of Reaction is the energy change at standard conditions • Symbol = ΔHº • ΔHº = Σ ΔHfº(Products) - Σ ΔHfº(Reactants) • Remember to include coefficients • Remember elements are zero!

  26. Calculate the standard enthalpy of reaction for the following Al(s) + Fe2O3(s)  Al2O3(s) + Fe(s) Al = 0, Fe = 0, Fe2O3 = -826, Al2O3 = -1676

  27. Homework • P. 284 #63

More Related