1. CHEMISTRY 161 Chapter 6

2. Energy an Chemical Change 1.Forms of Energy 2.SI Unit of Energy 3.Energy in Atoms and Molecules 4.Thermodynamics 5.Calculation of Heat and Energy Changes 6.Measuring Heat and Energy Changes

3. 1. Forms of Energy 1. Kinetic energy energy of a moving microscopic or macroscopic object E = ½ m v 2 2. Radiant energy energy in form of photons (‘light’) (solar energy) E = h (h = Planck’s constant) (Chapter 7) 3. Potential energy energy by changing object’s position in height E = m g h (h = height)

4. 4. Thermal Energy energy associated with random motion of atoms and molecules E kin = ½ M v 2 = 3/2 R T (Chapter 5) M = m  N a 5. Chemical Energy energy stored in chemical bonds of substances LAW OF CONSERVATION OF ENERGY THE TOTAL ENERGY OF THE UNIVERSE IS CONSTANT EXP1

5. 1 Joule = 1 J 1 J = 1 Nm = 1 kg m 2 s -2 E kin = ½  m  v 2 1 cal = 4.184 J macroscopic versus microscopic 1 J vs. 1 kJ mol -1 2. SI Unit of Energy

6. Atoms – Kinetic and Thermal Energy E kin = ½ M v 2 = 3/2 R T 3. Energy in Atoms and Molecules gases are constantly in motion and hold a kinetic energy

7. Molecules – Kinetic, Thermal, & Potential Energy (N 2 ) molecules have different ‘internal’ (vibrational) energy when bond distances are changed EXP2 different bonds have different bond strength (stabilities) (H 2 vs. N 2 )

8. THERMODYNAMICS HEATCHANGE study of the energy associated with change 4. Thermodynamics reactants  products (different energies)

9. THERMOCHEMISTRY study of the energy associated with chemical change 2 H 2 (g) + O 2 (g) → 2 H 2 O(l)+ energy Hindenburg 1937Challenger 1986

10. Surrounding heat System ENDOTHERMICENDOTHERMIC heat Surrounding EXOTHERMIC 2 H 2 (g) + O 2 (g) → 2 H 2 O(l)2 HgO(s) → O 2 (g) + 2 Hg(l) EXP3

11. Energy 2 H 2 (g) + O 2 (g) 2 H 2 O(l) Exothermic (heat given off by system) NH 4 NO 3 (aq) Endothermic (heat absorbed by system) NH 4 NO 3 (s) + H 2 O (l)

12. QUANTIFICATION Enthalpy of Reaction Enthalpy is the heat release at a constant pressure (mostly atmospheric pressure)  H = H final - H initial  H = H products - H reactants H final > H initial :  H > 0 ENDOTHERMIC H final < H initial :  H < 0 EXOTHERMIC board

13. Energy 2 H 2 (g) + O 2 (g) 2 H 2 O(l) ExothermicEndothermic H final > H initial H final < H initial NH 4 NO 3 (s) + H 2 O (l) NH 4 NO 3 (aq)

14. Energy H 2 O(s) Endothermic H final > H initial H 2 O(l)  H = H final – H initial H 2 O(s) → H 2 O(l) ΔH = + 6.01 kJ mol -1

15. Energy H 2 O(s) Exothermic H final < H initial H 2 O(l)  H = H final – H initial H 2 O(l) → H 2 O(s) ΔH = - 6.01 kJ mol -1

16. THERMOCHEMICAL EQUATIONS CH 4 (g) + 2 O 2 (g) → 2 H 2 O(l) + CO 2 (g) ΔH=-890.4 kJ mol -1 Calculate the heat evolved when combusting 24.0 g of methane gas.

17. 5. Calculation of Heat and Enthalpy Changes  H m = H m,products – H m,reactants REFERENCE SYSTEM e.g. oxidation numbers of elements are zero molar

18. Standard Enthalpy of Formation HfOHfOHfOHfO HfOHfOHfOHfO heat change when 1 mole of a compound is formed from its elements at a pressure of 1 atm (T = 298 K)  H f O (element) = 0 kJ/mol  H f O (graphite) = 0 kJ/mol  H f O (diamond) = 1.9 kJ/mol

19. ENTHALPY, H H reactants H products C(s, graphite) + O 2 (g) CO 2 (g)  H f 0 = - 393.51 kJ mol -1 0 -393.51

20. C(s, graphite) + O 2 (g)CO 2 (g)  H f 0 = - 393.51 kJ mol -1 Standard Enthalpy of Formation C(s, graphite) + 2H 2 (g) CH 4 (g)  H f 0 = - 74.81 kJ mol -1 ½ N 2 (g) + 3/2 H 2 (g)NH 3 (g)  H f 0 = - 46.11 kJ mol -1 (1/2) N 2 (g) + (1/2) O 2 (g) NO(g)  H f 0 = + 33.18 kJ mol -1

22. ENTHALPY, H H reactants H products a A + b B → c C + d D Standard Enthalpy of Reaction a A + b B c C + d D  H O rxn = ΣΔH f 0 (prod) – ΣΔH f 0 (react) a ×  H f O (A) + b × ΔH f O (B) c ×  H f O (C) + d × ΔH f O (D)

23.  H O rxn = ΣnΔH f 0 (prod) – ΣmΔH f 0 (react) Standard Enthalpy of Reaction CaO(s) + CO 2 (g) → CaCO 3 (s) -635.6 -393.5-1206.9 [kJ/mol]  H O rxn = -177.8 kJ/mol

24. Standard Enthalpy of Reaction CH 4 (g) + 2 O 2 (g) → CO 2 (g) + 2 H 2 O(l)  H O rxn = ΣnΔH f 0 (prod) – ΣmΔH f 0 (react) CH 4 (g) + 2 O 2 (g) → CO 2 (g) + 2 H 2 O(g) 2 H 2 O(g) → 2 H 2 O(l) CH 4 (g) + 2 O 2 (g) → CO 2 (g) + 2 H 2 O(l)

25. Hess’s Law The overall reaction enthalpy is the sum of the reaction enthalpies of the steps in which the reaction can be divided

26. CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) ENTHALPY, H Reactants Products CO 2 (g) + 2H 2 O(g) - 802 kJ - 88 kJ - 890 kJ

27.  H rxn for S(s) + 3/2 O 2 (g)  SO 3 (g) S(s) + O 2 (g)  SO 2 (g)  H 1 = -320.5 kJ SO 2 (g) + 1/2 O 2 (g)  SO 3 (g)  H 2 = -75.2 kJ S solid SO 3 gas direct path + 3/2 O 2  H 3 = -395.7 kJ SO 2 gas +O 2 H1H1 = -320.5 kJ + 1/2 O 2 H2H2 = -75.2 kJ Indirect Path

28. THERMODYNAMICS quantitative study of heat and energy changes of a system the state (condition) of a system is defined by T, p, n, V, E CH 4 (g) + 2 O 2 (g) → CO 2 (g) + 2 H 2 O(l) 6. State Functions

29. the state (condition) of a system is defined by T, p, n, V, E STATE FUNCTIONS properties which depend only on the initial and final state, but not on the way how this condition was achieved Hess Law

30. ΔV = V final – V initial Δp = p final – p initial ΔT = T final – T initial ΔE = E final – E initial

31. Energy is a STATE FUNCTION IT DOES NOT MATTER WHICH PATH YOU TAKE ΔE = m g Δh

32. CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) ENTHALPY, H Reactants Products CO 2 (g) + 2H 2 O(g) - 802 kJ - 88 kJ - 890 kJ Hess Law

33. heat is spontaneous transfer of thermal energy two bodies at different temperatures T 1 > T 2 spontaneous T1T1 T2T2 Zeroth Law of Thermodynamics a system at thermodynamical equilibrium has a constant temperature Applications EXP4

34. First Law of Thermodynamics energy can be converted from one form to another, but cannot be created or destroyed CONSERVATION OF ENERGY

35. SYSTEM SURROUNDINGS THE TOTAL ENERGY OF THE UNIVERSE IS CONSTANT - +

36. ΔE system = ΔQ + ΔW First Law of Thermodynamics ΔQ heat change ΔW work done  Q > 0 ENDOTHERMIC  Q < 0 EXOTHERMIC ?

37. ΔW = - p ΔV mechanical work M the energy of gas goes up M the energy of gas goes down ΔV < 0 ΔV > 0

38. 6. Measurement of Heat Changes  H = ΔQ ∞ ΔT heat Surrounding System temperature increase (pressure is constant)

39. Where does the ‘heat’ go? EXP5

40.  H = ΔQ ∞ ΔT  H = ΔQ = const × ΔT  H =ΔQ = C  ΔT enthalpy change temperature change heat capacity  H = ΔQ = m s ΔT C = m  s s = specific heat capacity EXP6

41. specific heat capacity capability of substances to store heat and energy s = J g -1 K -1 the J necessary to increase the temperature of 1 g of a compound by 1 K

42.  H = ΔQ = m s ΔT 1.prepare two styrofoam cups 2. carry out chemical reaction in a compound with known s s (H 2 O) = 4.184 J g -1 K -1 3. measure temperature change 4. determine ΔH calorimeter

43. 100 ml of 0.5 M HCl is mixed with 100 ml 0.5 M NaOH in a constant pressure calorimeter (s cup = 335 J K -1 ). The initial temperature of the HCl and NaOH solutions are 22.5C, and the final temperature of the solution is 24.9C. Calculate the molar heat of neutralization assuming the specific heat of the solution is the same as for water. 1.Neutralization reactions 2.Redox reactions 3. Precipitation reactions  H = ΔQ = C ΔT  H = ΔQ = (c1 + c2) ΔT

44. Constant Volume Calorimeter ΔQ = (m s(H 2 O) + c bomb ) ΔT

45. Energy an Chemical Change 1.Forms of Energy 2.SI Unit of Energy 3.Energy in Atoms and Molecules 4.Thermodynamics 5.Calculation of Heat and Energy Changes 6.Measuring Heat and Energy Changes