Chap 8C - Chem 210 / Wang / Chapter 8.8 Strength of...

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Unformatted text preview: Chem 210 / Wang / Chapter 8.8 Strength of Covalent Bonds Is Is there any relationship between bond enthalpy and bond length? Covalent Bond Enthalpies • Is the enthalpy change for the breaking of a covalent bond. • Heat is required to break up the pair of electrons in a bond. • This is an endothermic process. 436 kJ kJ kJ Cl••Cl + 242 kJ/mole Cl••Cl 242 Cl• + •Cl Cl• •Cl bond enthalpy of Cl2 = 242 kJ/mole. 242 348 614 839 293 615 891 163 418 941 201 607 Note – The reverse process (bond formation) is an exothermic process. exothermic 358 799 1072 299 kJ Linking Bond Enthalpies to Heat of Reactions Identify some trends in bond enthalpy. Support your answer to each question answer below with actual data provided in Table 8.4. How does bond enthalpy relate to: (a) Atomic size? (b) Bond order? (c) Bond polarity? The enthalpy change of a reaction can be accounted for by bond breaking and bond formation. ∆H of a reaction = H of the products – H of the reactants products reactants = net result of: – energy produced by bonds formed produced + energy consumed by bonds broken consumed Try and explain the trends you have discovered, if any. Using bond energies to calculate ∆H0rxn for 2CO + O2 2 C O2 CO Use bond enthalpies to estimate the ∆H for CH4 (g)+ 2O2(g) CO2(g) + 2H2O (g) (g)+ CO ∆ H0rxn = ∆ H0reactant bonds broken + ∆ H0product bonds formed 146 495 Estimate the ∆Hf o of HCl (g) using bond of bond enthalpy enthalpy data (Table 8.4). answer answer Need to find the ∆H for this reaction: H2 (g) + Cl2 (g) 2 HCl(g) ∆Hrxn = ? HCl(g) bonds broken (H-H + Cl-Cl) – bonds formed (2 H-Cl) = 436 + 242 – 2×431 = -184 kJ for 2 moles HCl ∆Hf° per mole HCl = -184 kJ ÷ 2 = –92.0 kJ/mole. This This answer is very close to the actual value given in the “heat of formation” Table 5.3 (page 192) = -92.30 kJ/mole Estimate the ∆Hrxn for this reaction with bond enthalpies (all chemicals are gases): C2H4 + 3O2 2CO 2CO2 + 2H2O = 4 (C-H) + 2(O=O) – 2(C=O) – 4(H-O) (C4(H= 4 (413) + 2(495) – 2(799) – 4(463) (413) = - 808 kJ Using Table 5.3 heat of formation table (page 192) = (-393.5) + 2×(-241.8) – (-74.80) (= - 802.3 kJ (< 1% difference) (< Fall 2009/ UM-SJTU JI Enthalpy, H ∆Hrxn = bonds broken - bonds formed BOND BREAKING ∆ H01 = 2 x C≡O + O=O = 2x1072 + 495 = 2639 BOND FORMATION ∆ H02 = - sum of BE = - 4x C=O = - 4 x 799 = - 3196 2 CO + O2 ∆H0rxn= 2639-3196 = -557 kJ 2 CO2 2639- 1 ...
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