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Bond enthalpy

Bond enthalpy. Cl – Cl . . 2 Cl. Δ H = D(Cl – Cl) = 242 kJ. Takes 242 kJ to break a Cl – Cl bond. H. H. C. H. . C. +. 4H. Δ H = 1660 kJ. H. How much energy does it take to break a C – H bond?. D(C – H) = 1660/4 = 415 kJ. Bond enthalpies.

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Bond enthalpy

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  1. Bond enthalpy Cl – Cl  2 Cl ΔH = D(Cl – Cl) = 242 kJ Takes 242 kJ to break a Cl – Cl bond H H C H  C + 4H ΔH = 1660 kJ H How much energy does it take to break a C – H bond? D(C – H) = 1660/4 = 415 kJ

  2. Bond enthalpies Not all C – H bonds are exactly the same. It depends on the molecule it is bonding to. Table shows average bond enthalpy It is an estimate….not an exact value Table value for C – H is 413 kJ/mol

  3. H H H - C - H + Cl – Cl  H - C - Cl + H – Cl H H Can look upon as happening in two steps: 1) CH4 + Cl2 C + 4H + 2Cl Break all bonds 2) C + 4H + 2Cl  CH3Cl + HCl Reform bonds

  4. C + 4H + 2Cl Break C – H and Cl – Cl bonds Make C – H, C - Cl and H – Cl bonds ΔH1 > 0 ΔH2 < 0 Enthalpy (H) CH4 + Cl2 ΔHrxn CH3Cl + HCl

  5. Calculating ΔH for each step • CH4 + Cl2 C + 4H + 2Cl Breaking 4 C – H bonds and 1 Cl – Cl bond ΔH1 = 4 * D(C – H) + D(Cl – Cl) ΔH1 = 4 * 413 kJ + 242 kJ = 1894 kJ

  6. Calculating ΔH for each step • C + 4H + 2Cl  CH3Cl + HCl Forming the bond should be the reverse of the process of breaking the bonds. ΔH should be equal in magnitude and opposite in sign from the bond enthalpy ΔH2 = - [3 * D(C – H) + D(C – Cl) + D(H – Cl)] ΔH2 = - [3 * 413 kJ + 328 kJ + 431 kJ] = -1998 kJ

  7. ΔH for the reaction Using Hess’s law, ΔH for the reaction should be the sum of the steps: ΔHrxn = ΔH1+ ΔH2 = 1894 kJ + -1998 kJ ΔHrxn = -104 kJ

  8. ΔH for the reaction Note three of the C – H bonds broken in the first step were recreated in the second step ΔH for these bonds should be 0 Can look only at the new bonds broken and formed ΔH = Σ(bond enthalpies of bonds broken) - Σ(bond enthalpies of bonds formed)

  9. Calculate ΔH for the reaction: H H H – C – C – H + 7/2 O = O  2 O = C = O + H H 3 H – O – H ΔH = Σ(bond enthalpies of bonds broken) - Σ(bond enthalpies of bonds formed) ΔH = [6D(C – H ) + D (C – C) + 7/2D(O = O)] – [4D(C = O) + 6D(O – H)] ΔH = -1416 kJ

  10. Vsepr theory Talk about electron domains Electron domains are electron pairs – inrespective of whether they are bonding or non-bonding pairs Last year we looked at 3 different numbers of electron domains – 2, 3 or 4

  11. Electron domain geometry 2 electron domains B A B Linear geometry Example: Bond angle = 180° O = C = O

  12. A B B B B B Electron domain geometry Trigonal Planar 3 electron domains F B F F Bent [ ]- Trigonal Planar N = A O O Bond angle = 120°

  13. CH4 A A A B B B B B B B B B 4 electron domains Trigonal Pyramidal NH3 Tetrahedral Bent Bond angle = 109.5° H2O

  14. Expanded octet 5 electron domains Trigonal Bipyramidal Axial bond 90° Equatorial bonds 120° Ex: PCl5

  15. A B B B B Where does the lone pair go? F – S – F SF4 F F Equatorial bonds have more space Seesaw 2 bond angles: 90° and 120°

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