1. Thermodynamics Mr. Leavings

2. Objectives Use the laws of thermodynamics to solve problems, identify energy flow within a system, determine the classification of various reactions

3. Thermodynamic Laws 0-(zeroeth law): temperature makes sense! 1-(1 st Law): The energy in the universe is constant (cant be created or destroyed!) 2-(2 nd Law): The entropy (S) of the universe is increasing 3-(3 rd Law): Entropy (S) of a perfect crystal at 0K is Zero

4. Entropy (S) Molar entropy is the amount of randomness or disorder in 1 mol of a substance. Units – J/Kmol Entropy always increases with temperature Entropy of a gas > liquid > solid Atoms in a gas have the most freedom of movement

5. Enthalpy (H) Enthalpy is the total energy of a system, like heat or thermal energy Molar enthalpy depends on temperature Quantities needed –Molar Heat Capacity (C) –Change in Temperature (  T) ΔQ = mC(ΔT)

6. Heat exchange accompanies chemical reactions. Exothermic: Heat flows out of the system (to the surroundings). Endothermic: Heat flows into the system (from the surroundings). Exo and Endothermic

7. Exothermic

8. Endothermic

9. Calorimeters

10. Bond Energy Bond energy values can be used to calculate approximate energies for reactions. ΔH = ∑(Bonds Broken) - ∑(Bonds Created)

11. C + O 2  CO 2 Energy ReactantsProducts  C + O 2 C O 2 -1103kJ + 1103 kJ/mol

12. 12 CaCO 3  CaO + CO 2 Energy ReactantsProducts  CaCO 3 CaO + CO 2 176 kJ CaCO 3 + 176 kJ  CaO + CO 2

13. 13 In terms of bonds C O O C O O Breaking this bond will require energy. C O O O O C Making these bonds gives you energy. In this case making the bonds gives you more energy than breaking them.

14. Bond Energy

15. Gibb’s Free Energy (G)  G =  H – T  S Describes spontaneity Negative  G  Reaction will occur on its own Positive  G  Reaction will not occur on its own

16. Gibbs Free Energy There are three possible conditions: If  G < 0 then the forward reaction is spontaneous. If  G > 0 then the forward reaction is not spontaneous. (However, the reverse reaction is spontaneous.) If  G > 0, work must be supplied from the surroundings to drive the reaction. If  G = 0 then reaction is at equilibrium and no net reaction will occur.

17. Temperature and sign of  G - + -Spontaneous at all temperatures + - +Nonspontaneous at any temperature. - - - Spontaneous at low temp. but will + be nonspontaneous at high temp. + + +Nonspontaneous at low temp. but - will be spontaneous at high temp. Sign of  H  S  GTemperature effect  H - T  S =  G

18. Calculation of  G o We can calculate  G o values from  H o and S o values at a constant temperature of 25ºC and pressure of 1 atm.Example. Determine  G o for the following reaction at 25 o C Equation N 2 (g) + 3H 2 (g) 2NH 3 (g)  H f o, kJ/mol 0.00 0.00 -46.11 S o, J/K. mol 191.50 130.68 192.3

19. Calculation of  G o  H o =  n p  H f o products -  n r  H f o reactants = 2 mol NH 3 ( -46.11 kJ / mol NH 3 ) = -92.22 kJ  S o =  n p S o products -  n r S o reactants = 2 mol NH 3 ( 192.3 J/K. mol NH 3 ) - [1 mol N 2 (191.5 J/K. mol N 2 ) + 3 mol H 2 (130.68 J/K. mol H 2 ) ] = -198.9 J/K

20. Calculation of  G o  G =  H - T  S -92.22 kJ - (298.2K) ( -0.1989 kJ/K) = -32.91 kJ Note. It was necessary to convert  S from units of J/K to kJ/K.  G f o, kJ/mol, = -32.91 kJ / 2 mol NH 3 = - 16.46 kJ/mol NH 3 This reaction would occur spontaneously under standard conditions at 25 o C.