1. Topic: Thermodynamics Do Now: packet p.1

2. Every physical or chemical change is accompanied by energy change  Energy released = _________________  Energy absorbed = _______________ exothermic endothermic GLSGLS

3. Thermodynamics is the study of entropy and enthalpy changes that accompany chemical reactions Thermodynamics Tells us if a reaction will occur

4. The total amount of energy a substance contains depends on many factors, some of which are not totally understood…it’s impossible to know the total heat content of a substance. So scientists measure ΔH describes chemical PE stored in matter H = enthalpy = the heat content of a system at constant pressure. It describes chemical PE stored in matter ΔH = enthalpy (heat) of reaction ΔH = H products – H reactants --the difference between the enthalpy(heat) of the substances that exist at the end of the reaction and the enthalpy(heat) of the substances that exist at the beginning of the reaction

5. Thermochemical Equations balanced chemical equation shows physical state of all reactants & products gives energy change –It can be written 2 ways energy term can be written as reactant or product OR  H is given right after equation

6. 4Fe(s) + 3O 2 (g)  2Fe 2 O 3 (s) + 1625 kJ OR 4Fe(s) + 3O 2 (g)  2Fe 2 O 3 (s)  H = -1625 kJ NH 4 NO 3 (s) + 27 kJ  NH 4 + (aq) + NO 3 - (aq) OR NH 4 NO 3 (s)  NH 4 + (aq) + NO 3 - (aq)  H = 27 kJ

7. If ΔH is negative… - ΔH = exothermic -H products < H reactants -PE of Products < PE of Reactant -Example: 4Fe (s) + 3O 2(g)  2Fe 2 O 3(s) + 1625 kJ -ΔH = -1625kJ

8. If ΔH is positive… + ΔH = endothermic -H products > H reactants -PE of Products > PE of Reactant Example: NH 4 NO 3 (s) + 27 kJ  NH 4 + (aq) + NO 3 - (aq) -ΔH = +27kJ

9. Universe Environment System A B Since energy is conversed…the system changes in one direction and the surrounding have to change in the opposite direction A. Reaction is Exothermic, environment gets _________________ B. Reaction is Endothermic, environment gets _________________ warmer colder

10. For any reaction occurring at constant pressure ΔH = Q Q = mC  TQ = mC  T –Q = Energy change –m = mass of water –C = specific heat of water –  T = temperature change = T f – T i

11. Since there are different types of reactions, you have various ΔH’s ΔH comb = enthalpy (heat) of combustion – the enthalpy change for the complete bunring of one mole of the substance ΔH formation = enthalpy (heat) of formation – the enthalpy change for the formation of a compound from its constituent elements ΔH solution = enthalpy (heat) of solution – the enthalpy change when 1 mole of an ionic substance is dissolved in water. Look at Table I: Heats of Reaction Lets label the various types from above

12. Rxns 1-6: combustion rxnsRxns 1-6: combustion rxns   H = heat of combustion Rxns 7-18: formation (synthesis) rxnsRxns 7-18: formation (synthesis) rxns –Substance is formed from its elements –  H = heat of formation Rxns 19-24: dissolving equationsRxns 19-24: dissolving equations –  H = heat of solution Table I

13. Many other processes other than chemical reactions absorb or release energy like, Changes of state  H vaporization = molar heat of vaporization = amount of heat required to vaporize one mole of a liquid  H fusion = molar heat of fusion = amount of heat required to melt one moles of a solid

14. Energy depends on amount Remember – it takes more energy to heat up water in the ocean than to make a cup of tea

15. CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O (l)  H = -890.4 kJ 1 mole of methane + 2 mole of oxygen → 1 mole of carbon dioxide gas & 2 moles of liquid water reaction is ____________ (negative sign for ΔH) 890.4 kJ energy released per mole of CH 4 (g) burned exothermic What would happen if we had 2 moles of methane? Twice as much energy would be released 2 x 890.4 kJ = 1780.8 kJ will be released

16. Reactions: Energy depends on direction too! N 2 (g) + 3H 2 (g)  2NH 3 (g)  H = -91.8 kJ 2NH 3 (g)  N 2 (g) + 3H 2 (g)  H = _______ If reverse equation, reverse sign of  H 91.8 kJ

17. If we can’t calc. H for one individual substance how are we able to calc. ΔH???!

18. FYI: Hess’s Law (not on regents) Can add 2 or more equations by adding the  H’s Enables you to calculate  H for # of rxns Say you’re interested in 2S(s) + 3O 2 (g)  2SO 3 (g)

19. Have  H’s for the following: a) S(s) + O 2 (g)  SO 2 (g)  H = -297 kJ b) 2SO 3 (g)  2SO 2 (g) + O 2 (g)  H = 198 kJ 222 2SO 2 (g) + O 2 (g)  2SO 3 (g)   H = -198 kJ  H = -594KJ + (-  kJ   H = -792 kJ x (2)