2. Thermodynamics Thermochemistry Carol Brown Saint Mary’s Hall

3. The 0th Law Heat flows from hot to cold

4. The 1st Law  In all changes, energy is conserved  The energy of the universe is constant  ∆E=q + w (both q and w are from the system’s point of view)

5. P∆V Work  Using the MKS system, determine the units for P∆V.

6. Calorimetry-Basics  Heat: The total amount of thermal energy contained in a sample of matter –Measured in calories or joules  Temperature: The average kinetic energy of the molecules in a sample of mater. –Measured in Kelvins or degrees Celsius

7. More basics  Specific heat capacity: The amount of heat it takes to raise one gram of a substance one degree Celsius. –Units: J/g o C  Heat capacity: The amount of heat it takes to raise a system one degree Celsius. –Units--J/ o C

8. Coffee Cup Calorimeter

9. q = mc∆T  How much heat, in kJ, is necessary to take 20 g of ice at -5 o C to steam at 100 o C? The specific heat of ice is 2.1 J/g o C; of liquid water is 4.2 J/g o C. The latent heat of fusion is 334 J/g. The latent heat of vaporization is 2268 J/g.  60.7 kJ

10. Potential Energy Diagrams-- Exothermic Reactions

11. Potential Energy Diagrams-- Endothermic Reactions

12. Important Terms  Enthalpy (heat of reaction)--The amount of heat lost or absorbed during the course of a reaction when the only work done is expansion or contraction at a constant pressure (P∆V). Change in enthalpy is a state function and is symbolized by ∆H.

13. What drives a reaction?  Enthalpy ∆H  Entropy S

14. What is a State Function?

15. Examples of State Functions  Enthalpy  Entropy  Internal Energy  Temperature  Pressure

16. Thermodynamics  Defining a system  System  Surroundings  Universe

17. Stoichiometric Thermochemistry  When 2.50 g of methane burns in oxygen, 125 kJ of heat is produced. What is the molar heat of combustion of methane under these conditions?

18. Stoichiometric Thermochemistry: Answer  When 2.50 g of methane burns in oxygen, 125 kJ of heat is produced. What is the molar heat of combustion of methane under these conditions?  -802 kJ/mol

19. Stoichiometric Thermochemistry: Your Turn  The oxidation of glucose is described by the following equation: C 6 H 12 O 6 + 6O 2 --> 6CO 2 + 6H 2 O ∆H o = -2816 kJ How much heat in kJ is produced by the oxidation of 1.0 g of glucose?

20. Stoichiometric Thermochemistry:Your Turn Answer  The oxidation of glucose is described by the following equation: C 6 H 12 O 6 + 6O 2 --> 6CO 2 + 6H 2 O ∆H o = -2816 kJ How much heat in kJ is produced by the oxidation of 1.0 g of glucose? 16 kJ

21. Another term  Standard molar heat of combustion: The amount of heat released when one mole of a substance is burned in oxygen. The measurements must be taken at standard thermodynamic conditions. i.e. 298 K and 1.00 atm pressure. Symbolized by ∆H o comb.

22. Still another important term  Standard molar heat of formation: The amount of heat lost or absorbed when one mole of product is formed from its elements in their most stable state. Again, the measurements must be taken at thermodynamic standard conditions. Symbolized by ∆H o f.

23. Is this an example of ∆H o comb, ∆H o f, neither, or both?  C + 1/2 O 2 --> CO  CO + 1/2 O 2 --> CO 2  CH 3 OH + 3/2 O 2 --> CO 2 + 2H 2 O  8 C + 9H 2 --> C 8 H 18  CH 4 + 2O 2 --> CO 2 + 2H 2 O  FeCl 2 + 1/2Cl 2 --> FeCl 3  S + O 2 --> SO 2

24. Hess’s Law  Since enthalpy is a state function, the change in enthalpy in going from some initial state to some final state is independent of the pathway. This means that in going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in a single step or in a series of steps. This principle is known as Hess’s Law.

25. Hess’s Law

26. Hess’s Law Problem #1 Given the following data: H 2 + 1/2O 2 --> H 2 O(l)∆H o =-285.8 kJ N 2 O 5 + H 2 O --> 2HNO 3 ∆H o = -76.6 kJ 1/2N 2 + 3/2O 2 + 1/2H 2_ --> HNO 3 ∆H o =-174.1 kJ Calculate ∆H o for the reaction 2N 2 + 5O 2 --> 2N 2 O 5

27. Answer Given the following data: H 2 + 1/2O 2 --> H 2 O(l)∆H o =-285.8 kJ N 2 O 5 + H 2 O --> 2HNO 3 ∆H o = -76.6 kJ 1/2N 2 + 3/2O 2 + 1/2H 2 -->HNO 3 ∆H o =-174.1 kJ Calculate ∆H o for the reaction 2N 2 + 5O 2 --> 2N 2 O 5 28.4 kJ

28. Another Hess’s Law Problem  Calculate ∆H o for the process Sb(s) + 5/2 Cl 2 (g) --> SbCl 5 (g) from the following information. Sb(s) + 3/2 Cl 2 (g) --> SbCl 3 (g) ∆H o -314 kJ SbCl 3 (g) + Cl 2 (g) --> SbCl 5 (g) ∆H o -80 kJ

29. Another Hess’s Law Problem: Answer  Calculate ∆H o for the process Sb(s) + 5/2 Cl 2 (g) --> SbCl 5 (g) from the following information. Sb(s) + 3/2 Cl 2 (g) --> SbCl 3 (g) ∆H o -314 kJ SbCl 3 (g) + Cl 2 (g) --> SbCl 5 (g) ∆H o -80 kJ Answer: -394 kJ

30. Spontaneity  The driving force of reactions. –Enthalpy –Entropy

31. Free Energy ∆G ∆G = ∆H - T∆S