Chem 161-2010 Lecture 19

Chem 161-2010 Lecture 19 - CHEMISTRY 161-2010 LECTURE 19...

Info iconThis preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
Chem 161-2010 Lecture 19 1 CHEMISTRY 161-2010 LECTURE 19 ANNOUNCEMENTS ATTENDANCE QUIZZES EXAMS/QUIZZES Chem 161 Exam III, Wednesday, Nov. 17 th , 9:40 – 11:00 PM, H&P Ch. 6.7-9.5 In class review for exam III Tuesday, Nov. 16 th LECTURE QUIZ?
Background image of page 1

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

View Full DocumentRight Arrow Icon
Chem 161-2010 Lecture 19 2 PLAN FOR TODAY : CHAPTER 9 ENTHALPY OF REACTION FROM BOND ENERGIES BOND ENERGY/LENGTH • ALKANES, ALKENES, ALKYNES AND POLYMERS
Background image of page 2
Chem 161-2010 Lecture 19 3 Δ H reaction Approaches: (1) Hill & Petrucci: Using enthalpy of formation: Δ H reaction = Σ∆ H formationproducts - Σ∆ H formationreactants (2) Tavss: Using enthalpy of bond breakage and bond formation: ET: Note that enthalpy of bond breakage, and enthalpy of bond formation are exactly the same, except for a reversal in signs. Δ H reaction = Σ∆ H bondformationproducts - Σ∆ H bondformationreactants (3) Hill & Petrucci: Using enthalpy of bond breakage and bond formation: Δ H reaction = Σ∆ H bondsbroken - Σ∆ H bondsformed ET: Note that all semester, when we say change, it is final minus initial, or products minus reactants. However, in this case change is reactants minus products. I like consistency. This is not consistent. That’s why I recommend against this approach.
Background image of page 3

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

View Full DocumentRight Arrow Icon
Chem 161-2010 Lecture 19 4 BOND ENERGIES AND LENGTHS Δ H reaction o Based on enthalpy of formation from elements in ref. form Note: Elements (often diatomic, e.g., H 2 , N 2 , Cl 2 , etc) are the references and are given values of zero. Δ H reaction = Σ∆ H formationproducts - Σ∆ H formationreactants 2 NO(g) + 5H 2 (g) 2NH 3 (g) + 2H 2 O(g) 2 x 90kJ 5 x 0kJ 2 x -46kJ 2 x -242kJ Δ H reaction = ((2 x -46) + (2 x -242)) – ((2 x 90) + (5 x 0)) = -756 kJ o Based on enthalpy of formation from elements in atomic form (1) Tavss’ approach Δ H reaction = Σ∆ H bondformationproducts - Σ∆ H bondformationreactants Given: Δ H bond energy NO = 628 kJ/mol Bond energy is the quantity of energy required to break one mole of covalent bonds between atoms in the gas phase. Given Δ H bond energy NO , what is the Δ H formationNO? N-O(g) N•(g) + •O(g) Δ H formation = - Δ H bond energy Therefore, Δ H bondformation NO = -628 kJ/mol) H&P 20. Given the bond energies (N-to-O bond) in NO, 628 kJ/mol; H-H, 435 kJ/mol; N-H, 389 kJ/mol; O-H, 464 kJ/mol, calculate H for the reaction: 2 NO(g) + 5H 2 (g) 2NH 3 (g) + 2H 2 O(g) Δ H reaction = Σ∆ H bondformationproducts - Σ∆ H bondformationreactants 2 NO(g) + 5H 2 (g) 2NH 3 (g) + 2H 2 O(g) 2 x -628kJ 5 x -435kJ 2 x 3 x -389kJ 2 x 2 x -464kJ ((6 x -389) + (4 x -464)) – ((2 x -628) + (5 x -435)) = -759 kJ = Δ H reaction (2) Hill & Petrucci approach: Δ H reaction = Σ H bondsbroken - Σ H bondsformed Bonds Broken (i.e., reactant bonds) Bonds Formed (i.e., product bonds) H be H be 2 NO = 2 x +628 6 NH = 6 x +389 5 H 2 = 5 x +435 4 OH = 4 x +464 Δ H reaction = ((2 x 628) + (5 x 435)) – ((6 x 389) + (4 x 464)) = -759 kJ
Background image of page 4
Chem 161-2010 Lecture 19 5 S S l l i i g g h h t t l l y y m m o o r r e
Background image of page 5

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

View Full DocumentRight Arrow Icon
Image of page 6
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 19

Chem 161-2010 Lecture 19 - CHEMISTRY 161-2010 LECTURE 19...

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online