Chem 161-2009 homework 12th week

Chem 161-2009 - Chem 161-2009 12th week Hill Petrucci Homework Problems Chapter 9 problems 20,53,55,57,63,67,74,98,99,100 Chapter 9 problems 75 77

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
1 Chem 161-2009 12 th Chapter 9 problems: 20,53,55,57,63,67,74,98,99,100 Chapter 9 problems: 75, 77, 83 Chapter 10 problems: 3,21,23,25,29 Self-Assessment Questions 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) Note: I do this problem in the same way that we have done H reaction = Σ H formationproducts - Σ H formationreactants but here we use “enthalpy of bond formation” rather than “enthalpy of formation”. Δ H reaction = Σ H bondformationproducts - Σ H bondformationreactants The enthalpy of bond formation is the reverse sign of enthalpy of bond energy. One is the energy to form a bond; the other is the energy to break a bond. Reactants Products H bf H bf 2 NO = 2 x -628 6 NH = 6 x -389 5 H 2 = 5 x -435 4 OH = 4 x -464 ((6 x -389) + (4 x -464)) – ((2 x -628) + (5 x -435)) = -759 kJ Hill & Petrucci: Δ H reaction = Σ H bondsbroken - Σ H bondsformed Reactants Products 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 Note: This answer is different than in the Solutions Manual. The Solutions Manual is using the formula backwards, and getting the reverse answer (+759 kJ). More Lewis Structures 53. Write the simplest Lewis structure for each of the following. Comment on any unusual features of the structures. (a) NO (b) ClF 3 (c) BCl 3 (d) SeF 4 (a) . . . . . N = O : Unusual feature: free radical (b) Unusual feature: Central atom has more than an octet of electrons (an expanded shell ).
Background image of page 1

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

View Full DocumentRight Arrow Icon
2 . . .. . . . . : F Cl F : . . | . . : F : . . (c) Unusual feature: Central atom has less than an octet of electrons. . . : Cl : . . | . . : C l B Cl : . . . . (d) Unusual feature: Central atom has more than an octet of electrons. SeF 4 . . : F : . . . . | . . : F Se F : see-saw; a little under 180 o & a . . | . . little under 90 o ; polar : F : . . 55. Write Lewis structures for the following. Where appropriate, use the concepts of formal charge, expanded valence shells, and resonance to choose the most likely structure(s). (a) SSF 2 (b) I 3 - (c) H 2 CO 3 (d) CN - (e) SF 5 -
Background image of page 2
3 (f) BrO 3 - (a) .. .. .. .. :F S S F: .. .. .. .. This is a different structure for SSF 2 than in the Solutions Manual. Mine is symmetrical, which is generally how one writes unknown structures. Also, S2F2 is shown as my structure in S2F2- [email protected] . I have no reason to believe that the Solutions Manual structure is correct. (b)
Background image of page 3

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

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This document was uploaded on 11/02/2011 for the course GEN CHEM 162 at Rutgers.

Page1 / 13

Chem 161-2009 - Chem 161-2009 12th week Hill Petrucci Homework Problems Chapter 9 problems 20,53,55,57,63,67,74,98,99,100 Chapter 9 problems 75 77

This preview shows document pages 1 - 4. Sign up to view the full document.

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