1. Chapter 6 Thermochemistry: pp. 235-274

2. 6.1 The Nature of Energy Energy – Capacity to do work or produce heat. – 1 st Law of Thermodynamics: Energy can not be created or destroyed, but converted to different forms. – The energy of the universe is constant. – Potential Energy: Energy due to position or composition – Kinetic Energy: Energy due to motion; depends on mass and velocity

3. 6.1 The Nature of Energy Lower Potential Energy Higher Potential Energy Add Energy

4. Energy Heat involves the transfer of energy between two objects due to a temperature difference. Work: Force acting over a distance Energy is a state function: Work and Heat are not – State Function: Property that does not depend in any way on the system’s past or future (only depends on present state). 6.1 The Nature of Energy

5. Chemical Energy System – part of the universe where a reaction is happening. Surroundings – everything else in the universe – Endothermic Reaction Energy flows into the system, absorbs energy from surroundings. – Exothermic Reaction Energy flows out of the system, surroundings gain energy. – Energy gained by surroundings must be equal to the energy lost by the system (opposite is true also). 6.1 The Nature of Energy

6. Reality Check – Is the freezing of water an exothermic or endothermic process? Why? – Define the system, surroundings, and direction of energy transfer: Methane is burning in a Bunsen burner in a laboratory. 6.1 The Nature of Energy

7. Internal Energy – Internal energy, E, of a system is the sum of the kinetic and potential energies of all the “particles” in the system. – A change to the internal energy of a system is described as: Where q is heat and w is work. 6.1 The Nature of Energy

8. Internal Energy – Sign reflects the system’s point of view. 6.1 The Nature of Energy ++Endothermic, surroundings do work +-Endothermic, system does work -+Exothermic, surroundings do work --Exothermic, system does work qw What we observe

9. Work – Work = P × A × Δh = PΔV  P is pressure.  A is area.  Δh is the piston moving a distance.  ΔV is the change in volume. 6.1 The Nature of Energy

10. Work – For an expanding gas, ΔV is a positive quantity because the volume is increasing. Thus ΔV and w must have opposite signs: 6.1 The Nature of Energy

11. Reality Check – Determine the sign of  E for each of the following with listed conditions An endothermic process that performs work. – l work l > heat – l work l < heat Works is done on a gas and the process is exothermic. – l work l > heat – l work l < heat 6.1 The Nature of Energy Δ E = negative Δ E = positive Δ E = negative

12. Change in Enthalpy – State Function – ΔH = q at constant pressure – ΔH = H products - H reactants 6.2 Enthalpy and Calorimetry

13. Reality Check 6.2 Enthalpy and Calorimetry Consider the combustion of propane: C 3 H 8 (g) + 5O 2 (g) → 3CO 2 (g) + 4H 2 O(l) ΔH = –2221 kJ Assume that all of the heat comes from the combustion of propane. Calculate ΔH in which 5.00 g of propane is burned in excess oxygen at constant pressure. -252 kJ

14. Calorimetry – Science of measuring heat – Specific heat capacity: The energy required to raise the temperature of one gram of a substance by one degree Celsius. – Molar heat capacity: The energy required to raise the temperature of one mole of substance by one degree Celsius. 6.2 Enthalpy and Calorimetry

15. Simple Calorimeter If two reactants are mixed and the solution gets warmer an exothermic reaction is happening. An endothermic reaction cools the solution. 6.2 Enthalpy and Calorimetry

16. Calorimetry – Where, s = specific heat capacity (J/ o C  g) m = mass (g) ΔT = change in temperature ( o C) 6.2 Enthalpy and Calorimetry

17. Reality Check 6.2 Enthalpy and Calorimetry A 100.0 g sample of water at 90°C is added to a 100.0 g sample of water at 10°C. The final temperature of the water is: a) Between 50°C and 90°C b) 50°C c) Between 10°C and 50°C

18. Reality Check 6.2 Enthalpy and Calorimetry You have a Styrofoam cup with 50.0 g of water at 10.  C. You add a 50.0 g iron ball at 90.  C to the water. (s H2O = 4.18 J/°C·g and s Fe = 0.45 J/°C·g) The final temperature of the water is: a) Between 50°C and 90°C b) 50°C c) Between 10°C and 50°C Calculate the final temperature of the water. 18 o C

19. Hess’ Law – 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 one step or in a series of steps. 6.3 Hess’ Law

20. This reaction also can be carried out in two distinct steps, with enthalpy changes designated by ΔH 2 and ΔH 3. N 2 (g) + O 2 (g) → 2NO(g) ΔH 2 = 180 kJ 2NO(g) + O 2 (g) → 2NO 2 (g) ΔH 3 = – 112 kJ N 2 (g) + 2O 2 (g) → 2NO 2 (g) ΔH 2 + ΔH 3 = 68 kJ ΔH 1 = ΔH 2 + ΔH 3 = 68 kJ 6.3 Hess’ Law N 2 (g) + 2O 2 (g) → 2NO 2 (g) ΔH 1 = 68 kJ

21. Principle of Hess’ Law 6.3 Hess’ Law

22. If a reaction is reversed, the sign of ΔH is also reversed. The magnitude of ΔH is directly proportional to the quantities of reactants and products in a reaction. If the coefficients in a balanced reaction are multiplied by an integer, the value of ΔH is multiplied by the same integer. 6.3 Hess’ Law Characteristics of Hess’ Law

23. Reality Check – Calculate the ΔH for the reaction 6.3 Hess’ Law 2NH 3 (g) + 3N 2 O (g)  4N 2 (g) + 3H 2 O (l) ΔH = -1010. kJ N 2 O (g) + 3H 2 (g)  N 2 H 4 (l) + H 2 O (l) ΔH = -317 kJ 2NH 3 (g) + ½ O 2 (g)  N 2 H 4 (l) + H 2 O (l) ΔH = -143 kJ H 2 (g) + ½ O 2 (g)  H 2 O (l) ΔH = -286 kJ N 2 H 4 (l) + O 2 (g)  N 2 (g) + 2H 2 O (l)

24. Change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states. 6.4 Standard Enthalpy of Formation Standard Enthalpy of Formation (ΔH o f )

25. For a Compound  For a gas, pressure is exactly 1 atm.  For a solution, concentration is exactly 1 M.  Pure substance (liquid or solid) For an Element  The form [N 2 (g), K(s)] in which it exists at 1 atm and 25°C.  Heat of formation is zero. 6.4 Standard Enthalpy of Formation Conventional Definitions of Standard States

26. Calculating Enthalpy of formation Remember, – Elements enthalpy of formation is zero. – Don’t forget to multiply by coefficients from the balanced equation. 6.4 Standard Enthalpy of Formation  H° rxn =  n p  H f  (products) -  n r  H f  (reactants)

27. Calculate  H° for the following reaction: 2Na(s) + 2H 2 O(l) → 2NaOH(aq) + H 2 (g) Given the following information:  H f ° (kJ/mol) Na(s)0 H 2 O(l) –286 NaOH(aq) –470 H 2 (g)0 6.4 Standard Enthalpy of Formation Reality Check  H° = –368 kJ