1. CHEMISTRY 161 Revision Chapters 4-6 www.chem.hawaii.edu/Bil301/welcome.html

2. 4. CHEMICAL REACTIONS 4.1. properties of solutions 4.2. reactions in solutions a) precipitation reactions b) acid-base reactions (proton transfer) c) redox reactions (electron transfer)

3. non-electrolyte weak electrolyte strong electrolyte methanol sugar ethanol water darkbright ionic compounds NaOH HCl H 2 SO 4 CH 3 COOH HCOOH HF medium

4. DISSOCIATION ‘breaking apart’ NaCl (s) → Na + (aq) + Cl - (aq) NaOH (s) → Na + (aq) + OH - (aq) HCl (g) → H + (aq) + Cl - (aq) strong electrolytes are fully dissociated

5. weak electrolyte CH 3 COOH(aq) + H 2 O(l) H 3 O + (aq) + CH 3 COO - (aq) NH 3 (aq) + H 2 O(l) NH 4 + + OH - strong electrolyte HCl(aq) + H 2 O(l) → H 3 O + (aq) + Cl - (aq) HNO 3 (aq) + H 2 O(l) → H 3 O + (aq) + NO 3 - (aq) monoprotic acids

6. 4.2.1. insoluble compounds 1.M + compounds (M = H, Li, Na, K, Rb, Cs, NH 4 ) 2. A - compounds (A = NO 3, HCO 3, ClO 3, Cl, Br, I) (AgX, PbX 2 ) 3. SO 4 2- (Ag, Ca, Sr, Ba, Hg, Pb) 4. CO 3 2-, PO 4 3-, CrO 4 2-, S 2- (exception: M + )

7. 4.2.2. ACID-BASE REACTION acid + base → salt + water (neutralization reaction) HCl(aq) + NaOH(aq) → NaCl(aq) + H 2 O(l) molecular equation – ionic equation – spectator ions

8. 4.2.3. OXIDATION NUMBER ionic compounds ↔ molecular compounds NaClHF, H 2 Na + Cl - ? electrons are fully transferred covalent bond ‘oxidation number’ charges an atom would have if electrons are transferred completely

9. TYPES OF REDOX REACTIONS 1.combination reactions A + B → C 2. decomposition reactions C → A + B 3. displacement reactions A + BC → AC + B 4. disproportionation reactions

10. [p] = Nm -2 = kg m -1 s -2 = Pa SI units 5. GASES

11. p × V = n × R × T 1. ideal gas equation R = 8.314 J / mol / K V m = 22.4 l 2. molar volume

12. 3. Kinetic Molecular Theory of Gases 4. Distribution of Molecular Speeds E kin = ½ M u 2 = 3/2 R T 5. Real Gas Law (p + (a n 2 / V 2 ) ) (V – n b) = n R T

13. Enthalpy of Reaction heat released or absorbed by the system at a constant pressure  H = H products - H reactants H final > H initial :  H > 0 ENDOTHERMIC H final < H initial :  H < 0 EXOTHERMIC 6. THERMOCHEMISTRY

14. measurement of heat changes  H m = ΔQ m = c mp ΔT c mp (H 2 O) = 75.3 J mol -1 K -1  H = ΔQ = (c mp × n + c cup ) ΔT

15. 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

16. 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

17. 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

18.  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

19. 1 atm = 760 mm Hg = 760 torr = 103,325 Pa R = 8.314 J / mol / K c mp (H 2 O) = 75.3 J / mol / K p V = n R T  H O rxn = ΣnΔH f 0 (prod) – ΣmΔH f 0 (react) E kin = ½ M u 2 = 3/2 R T ΔQ = (c mp × n + c cup ) ΔT

20. Come in Time! Do not forget your calculator!