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CHEMISTRY 161 Chapter 6 chem.hawaii/Bil301/welcome.html

CHEMISTRY 161 Chapter 6 www.chem.hawaii.edu/Bil301/welcome.html. REVISION. Measurement of Heat Changes. D H m = Δ Q m = c mp Δ T. c mp (H 2 O) = 75.3 J mol -1 K -1. Calculation of Heat Changes. D H m = H m,products – H m,reactants. REFERENCE SYSTEM.

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CHEMISTRY 161 Chapter 6 chem.hawaii/Bil301/welcome.html

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

  2. REVISION Measurement of Heat Changes DHm = ΔQm = cmpΔT cmp(H2O) = 75.3 J mol-1 K-1

  3. Calculation of Heat Changes DHm = Hm,products – Hm,reactants REFERENCE SYSTEM oxidation numbers of elements are zero molar volume at standard temperature and pressure Vm = 22.4 l

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

  5. C(s, graphite) + O2(g) Hreactants 0 ENTHALPY, H DHf0 = - 393.51 kJ mol-1 Hproducts -393.51 CO2(g)

  6. Standard Enthalpy of Formation C(s, graphite) + O2(g) CO2(g) DHf0 = - 393.51 kJ mol-1 CH4(g) DHf0 = - 74.81 kJ mol-1 C(s, graphite) + 2H2(g) DHf0 = - 46.11 kJ mol-1 ½ N2(g) + 3/2 H2(g) NH3(g) NO(g) DHf0 = + 33.18 kJ mol-1 (1/2) N2(g) + (1/2) O2(g)

  7. Standard Enthalpy of Reaction a A + b B → c C + d D a A + b B a × DHfO (A) + b × ΔHfO(B) Hreactants ENTHALPY, H DHOrxn = ΣΔHf0(prod) – ΣΔHf0(react) Hproducts c × DHfO(C) + d × ΔHfO(D) c C + d D

  8. Standard Enthalpy of Reaction DHOrxn = ΣnΔHf0(prod) – ΣmΔHf0(react) CaO(s) + CO2(g) → CaCO3(s) [kJ/mol] -393.5 -1206.9 -635.6 DHOrxn = -177.8 kJ/mol

  9. Standard Enthalpy of Reaction CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) DHOrxn = ΣnΔHf0(prod) – ΣmΔHf0(react) CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) 2 H2O(g) → 2 H2O(l) CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)

  10. CH4(g) + 2O2(g) Reactants - 802 kJ - 890 kJ ENTHALPY, H CO2(g) + 2H2O(g) - 88 kJ CO2(g) + 2H2O(l) Products

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

  12. S solid Indirect Path +O 2 direct path DH1 = + 3/2 O2 -320.5 kJ DH3 = -395.7 kJ SO gas 2 SO3 gas + 1/2 O 2 DH2 = -75.2 kJ DHrxn for S(s) + 3/2 O2(g) SO3(g) S(s) + O2(g)  SO2(g) DH1 = -320.5 kJ SO2(g) + 1/2 O2(g) SO3(g) DH2 = -75.2 kJ

  13. Enthalpy of Solution DHOsolution = ? NaCl(s) → Na+(aq) + Cl-(aq) DH1O=+788 kJ/mol NaCl(s) → Na+(g) + Cl-(g) DH2O=-784 kJ/mol Na+(g) + Cl-(g) → Na+(aq) + Cl-(aq) DHOsolution = 788 kJ/mol – 784 kJ/mol = + 4 kJ/mol ( solutebility tables)

  14. Na+(g) + Cl-(g) DH1O=+788 kJ/mol DH2O=-784 kJ/mol ENTHALPY, H Na+(aq) + Cl-(aq) NaCl(s)

  15. SUMMARY Standard Enthalpy of Formation DHfO DHfO (element) = 0 kJ/mol Standard Enthalpy of Reaction Hess’s Law

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

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