1. Slide 1 of 58 7-5 The First Law of Thermodynamics  Internal Energy, U.  Total energy (potential and kinetic) in a system. Translational kinetic energy. Molecular rotation. Bond vibration. Intermolecular attractions. Chemical bonds. Electrons.

2. Slide 2 of 58 First Law of Thermodynamics  A system contains only internal energy.  A system does not contain heat or work.  These only occur during a change in the system.  Law of Conservation of Energy  The energy of an isolated system is constant  U = q + w

3. Slide 3 of 58 First Law of Thermodynamics

4. Slide 4 of 58 Functions of State  Any property that has a unique value for a specified state of a system is said to be a State Function. ◦Water at 293.15 K and 1.00 atm is in a specified state. ◦d = 0.99820 g/mL ◦This density is a unique function of the state. ◦It does not matter how the state was established.

5. Slide 5 of 58 Functions of State  U is a function of state.  Not easily measured.  U has a unique value between two states.  Is easily measured.

6. Slide 6 of 58 Path Dependent Functions  Changes in heat and work are not functions of state.  Remember example 7-5, w = -1.24  10 2 J in a one step expansion of gas:  Consider 2.40 atm to 1.80 atm and finally to 1.20 atm.

7. Slide 7 of 58 Path Dependent Functions w = (-1.80 atm)(1.36-1.02)L – (1.30 atm)(2.04-1.36)L = -0.61 L atm – 0.82 L atm = -1.43 L atm = -1.44  10 2 J Compared -1.24  10 2 J for the two stage process

8. Slide 8 of 58 7-6 Heats of Reaction:  U and  H Reactants → Products U i U f  U = U f - U i  = q rxn + w In a system at constant volume:  U = q rxn + 0 = q rxn = q v But we live in a constant pressure world! How does q p relate to q v ?

9. Slide 9 of 58 Heats of Reaction

10. Slide 10 of 58 Heats of Reaction q V = q P + w We know that w = - P  V and  U = q P, therefore:  U = q P - P  V q P =  U + P  V These are all state functions, so define a new function. Let H = U + PV Then  H = H f – H i =  U +  PV If we work at constant pressure and temperature:  H =  U + P  V = q P

11. Slide 11 of 58 Comparing Heats of Reaction q P = -566 kJ/mol =  H P  V = P(V f – V i ) = RT(n f – n i ) = -2.5 kJ  U =  H - P  V = -563.5 kJ/mol = q V

12. Slide 12 of 58 Changes of State of Matter H 2 O (l) → H 2 O(g)  H = 44.0 kJ at 298 K Molar enthalpy of vaporization: Molar enthalpy of fusion: H 2 O (s) → H 2 O(l)  H = 6.01 kJ at 273.15 K

13. Slide 13 of 58 Break the problem into two steps: Raise the temperature of the liquid first then completely vaporize it. The total enthalpy change is the sum of the changes in each step. Enthalpy Changes Accompanying Changes in States of Matter. Calculate  H for the process in which 50.0 g of water is converted from liquid at 10.0°C to vapor at 25.0°C. = (50.0 g)(4.184 J/g °C)(25.0-10.0)°C + 50.0 g 18.0 g/mol 44.0 kJ/mol Set up the equation and calculate: q P = mc H 2 O  T + n  H vap = 3.14 kJ + 122 kJ = 125 kJ EXAMPLE 7-8

14. Slide 14 of 58 Standard States and Standard Enthalpy Changes  Define a particular state as a standard state.  Standard enthalpy of reaction,  H °  The enthalpy change of a reaction in which all reactants and products are in their standard states.  Standard State  The pure element or compound at a pressure of 1 bar and at the temperature of interest.

15. Slide 15 of 58 Enthalpy Diagrams

16. Slide 16 of 58 7-7 Indirect Determination of  H: Hess’s Law  H is an extensive property.  Enthalpy change is directly proportional to the amount of substance in a system. N 2 (g) + O 2 (g) → 2 NO(g)  H = +180.50 kJ ½N 2 (g) + ½O 2 (g) → NO(g)  H = +90.25 kJ  H changes sign when a process is reversed NO(g) → ½N 2 (g) + ½O 2 (g)  H = -90.25 kJ

17. Slide 17 of 58 Hess’s Law  Hess’s law of constant heat summation  If a process occurs in stages or steps (even hypothetically), the enthalpy change for the overall process is the sum of the enthalpy changes for the individual steps. ½N 2 (g) + O 2 (g) → NO 2 (g)  H = +33.18 kJ ½N 2 (g) + ½O 2 (g) → NO(g)  H = +90.25 kJ NO(g) + ½O 2 (g) → NO 2 (g)  H = -57.07 kJ