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Lecture Note - Covalent Bond Strengths

# Lecture Note - Covalent Bond Strengths - Bond ΔH(kJ/mole...

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Covalent Bond Strengths For a given chemical reaction we'll have bonds being broken, bonds being formed, and energy either being absorbed or emitted by the reaction. Let's look a little closer at the change in enthalpy associated with each individual bond being broken or formed. Process ΔH (kJ/mole) CH 4(g) → CH 3(g) + H (g) 435 CH 3(g) → CH 2(g) + H (g) 453 CH 2(g) → CH (g) + H (g) 425 CH (g) → C (g) + H (g) 339 total = 1652 If we wanted to know what is the enthalpy change associated with breaking a C-H bond we find that it is slightly dependent on what molecule it is in. Thus, we take an average change in enthalpy when a C-H bond breaks as D C-H = 1652/4 kJ/mole = 413 kJ/mole. Other average bond enthalpy changes in kJ/mole are. .. Bond ΔH (kJ/mole) Bond ΔH (kJ/mole) H-H 432 C-H 413 H-F 565 C-C 347 H-Cl 427 C-N 305 C-O 358 Notice how multiple bonds are shorter and require more energy to break than single bonds.

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Unformatted text preview: Bond ΔH (kJ/mole) Bond Length C=C 614 1.37Å C≡C 839 1.20Å C-N 305 1.43Å C=N 615 1.38Å C≡N 891 1.16Å We can use these average bond enthalpy changes to calculate the approximate enthalpy change for reactions. For example, to calculate the change in enthalpy for the following reaction: H 2(g) + F 2(g) → 2 HF (g) we identify and count all the bonds that are broken (shown in red) and formed (shown in blue). H — H + F — F → 2 H — F Substituting the average bond enthalpies: D H-H = 432 kJ/mole, D F-F = 154 kJ/mole, D H-F = 565 kJ/mole in the expression ΔH = Σ (ΔH of bonds broken) - Σ (ΔH of bonds formed) we obtain ΔH = [ (1 mole) (432 kJ/mole) + (1 mole) (154 kJ/mole) ]- [ (2 moles) (565 kJ/mole) ] = -544 kJ...
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Lecture Note - Covalent Bond Strengths - Bond ΔH(kJ/mole...

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