covalent bond strengths - Bond H (kJ/mole) Bond Length C=C...

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Covalent Bond Strengths - Bond Enthalpies In a chemical reaction we'll have bonds broken, bonds formed, and energy either absorbed or emitted by the  reaction. Let's look closer at the enthalpy change associated with each individual bond broken or formed. Process H Δ (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 found this way are. .. Bond H Δ (kJ/mole) Bond H Δ (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 H (kJ/mole) Bond Length C=C 614 1.37 C C 839 1.20 C-N 305 1.43 C=N 615 1.38 Bond H (kJ/mole) Bond Length C N 891 1.16 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 H = ( H of bonds broken) - ( H of bonds formed) we obtain H = [(1 mole) (432 kJ/mole) + (1 mole) (154 kJ/mole)] - [(2 moles) (565 kJ/mole) ] = -544 kJ...
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covalent bond strengths - Bond H (kJ/mole) Bond Length C=C...

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