1. CHEMISTRY 161 Chapter 6 www.chem.hawaii.edu/Bil301/welcome.html

2. REVISION Measurement of Heat Changes  H m = ΔQ m = c mp ΔT c mp (H 2 O) = 75.3 J mol -1 K -1

3. Calculation of Heat Changes  H m = H m,products – H m,reactants REFERENCE SYSTEM molar volume at standard temperature and pressure V m = 22.4 l oxidation numbers of elements are zero

4. 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

5. 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

6. 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

8. 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)

9.  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

10. 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)

11. 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

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

13.  H rxn for S(s) + 3/2 O 2 (g) SO 3 (g)  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 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 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

15. Enthalpy of Solution NaCl(s) → Na + (aq) + Cl - (aq)  H O solution = ? NaCl(s) → Na + (g) + Cl - (g) Na + (g) + Cl - (g) → Na + (aq) + Cl - (aq)  H 1 O =+788 kJ/mol  H 2 O =-784 kJ/mol  H O solution = 788 kJ/mol – 784 kJ/mol = + 4 kJ/mol ( solutebility tables)

16. ENTHALPY, H NaCl(s) Na + (g) + Cl - (g) Na + (aq) + Cl - (aq)  H 1 O =+788 kJ/mol  H 2 O =-784 kJ/mol

18. SUMMARY Standard Enthalpy of Formation HfOHfOHfOHfO HfOHfOHfOHfO  H f O (element) = 0 kJ/mol Standard Enthalpy of Reaction Hess’s Law

19. Homework Chapter 6, p. 217-222 problems