1 / 19

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.

steffi
Download Presentation

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

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related