{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Covalent Bond Strengths - bonds Bond H Δ(kJ/mole Bond...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Covalent Bond Strengths 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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Bond H Δ (kJ/mole) Bond H Δ (kJ/mole) C-O 358 Notice how multiple bonds are shorter and require more energy to break than single 
Background image of page 2
Background image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: bonds. Bond H Δ (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 H = Δ Σ ( H of bonds broken) -Δ Σ ( H of bonds formed) Δ we obtain H = Δ [(1 mole) (432 kJ/mole) + (1 mole) (154 kJ/mole)] - [(2 moles) (565 kJ/mole) ] = -544 kJ...
View Full Document

{[ snackBarMessage ]}