Fall 07 HW9 solution

Fall 07 HW9 solution - Homework Assignment #9 Name_...

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Homework Assignment #9 Name____________________________ Laboratory Section (circle one) MW1.30 MW6.30 TTh7 1. Chapter 9, Problem 9, 13, 15, 23, 25, 27, 29, 31 [Show your work for Problem 9.30 (you will need to do this).], 40, 49, 58, 62, 68, 76 9.9 a) metallic b) covalent c) ionic 9.13 a) b) c) As Se Ga 9.15 a) 5A(15); n s 2 n p 3 b) 4A(14); n s 2 n p 2 9.23 a) 1A(1) b) 3A(13) c) 2A(2) 9.25 a) 7A(17) b) 6A(16) c) 3A(13) 9.27 a) CaO; O has a smaller radius than S. b) SrO; Sr has a smaller radius than Ba. 9.29 a) NaCl; Cl has a larger radius than F. b) K 2 S; S has a larger radius than O. 9.30 Lattice energy can be found from the fact that the sum of the energies from each step which equals the heat of formation for solid sodium chloride. o f H Δ + ) g ( Na ) s ( Na H Δ + ) g ( Cl ) g ( Cl H 2 2 2 1 Δ + - + + Δ e ) g ( Na ) g ( Na H + ) g ( Cl e ) g ( Cl H - - + Δ = lattice H Δ lattice H Δ = -41 kJ + 109 kJ + (1/2) (243 kJ) + 496 kJ + (-349 kJ) = 336.5 = 336 kJ The lattice energy for NaCl is less than that for LiF, which is expected since lithium and fluoride ions are smaller than sodium and chloride ions. 9.31 2 MgF H ° Δ = f H ° Δ - ( 1 H ° Δ + 2 H ° Δ + 3 H ° Δ + 4 H ° Δ + 2 5 H ° Δ ) 2 MgF H ° Δ = -1123 - (148 + 159 + 738 + 1450 + 2(-328)) kJ 2 MgF H ° Δ = -2962 kJ This value is larger due to greater ionic charge on Mg 2+ and greater numbers of ions, compared with those of NaCl. 9.40 a) H-F < H-Cl < H-I b) C=O < C-O < C-S c) N-H < N-O < N-S 9.49 CO 2 + 2 NH 3 (H 2 N) 2 CO + H 2 O Δ H rxn = Σ BE reactants - Σ BE products Δ H rxn = (2 BE C=O + 6 BE N-H ) - (4 BE N-H + BE C=O + 2 BE C-N + 2 BE O-H ) = [2(799) + 6(391)] - [4(391) + 745 + 2(305) + 2(467)] = 3944 - 3853 = 91 kJ
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9.58 a) I < Br < N b) Ca < H < F 9.62 a) Br Cl b) F Cl c) H O d) Se H e) As H f) S N 9.68 Increasing ionic character occurs with increasing Δ EN. a) Δ EN PCl = 0.9, Δ EN PBr = 0.7, Δ EN PF = 1.9 P-F > P-Cl > P-Br δ + δ - δ + δ - δ + δ - b) Δ EN BF = 2.0, Δ EN NF = 1.0, Δ EN CF = 1.9 B-F > C-F > N-F δ + δ - δ + δ - δ + δ - c) Δ EN SeF = 1.6, Δ EN TeF = 1.9, Δ EN BrF = 1.2 Te-F > Se-F > Br-F 9.76 a) 1) Mg( s ) Mg( g ) o H 1 Δ = 148 kJ 2) 1/2 Cl 2 ( g ) Cl( g ) o H 2 Δ = 1/2 (243 kJ) 3) Mg( g ) Mg + ( g ) + e - o H 3 Δ = 738 kJ 4) Cl( g ) + e - Cl - ( g ) o H 4 Δ = -349 kJ 5) Mg + ( g ) + Cl - ( g ) MgCl( s ) o H 5 Δ = -783.5 kJ (= o lattice H Δ (MgCl)) 6) Mg( s ) + 1/2 Cl 2 ( g ) MgCl( s ) o f H Δ (MgCl) = ? o f H Δ (MgCl) = o H 1 Δ + o H 2 Δ + o H 3 Δ + o H 4 Δ + o H 5 Δ = 148 kJ + 1/2 (243 kJ) + 738 kJ - 349 kJ - 783.5 kJ = -125 kJ b) Yes , since f H ° Δ for MgCl is negative, MgCl( s ) is stable relative to its elements. c) 2 MgCl( s ) MgCl 2 ( s ) + Mg( s ) ° Δ H = 1 mol ( o f H Δ , MgCl 2 ( s )) + 1 mol ( o f H Δ , Mg( s )) - 2 mol ( o f H Δ , MgCl( s )) = 1 mol (-641.6 kJ/mol) + 1 mol (0) - 2 mol (-125 kJ/mol) = -391.6 = -392 kJ d) No, o f H Δ for MgCl 2 is much more negative than that for MgCl. This makes the ° Δ H value for the above reaction very negative, and the formation of MgCl 2 would be favored. 2. Chapter 10, Problem 3, 5, 6, 9, 10, 14, 16, 18, 20, 27, 31, 30, 35, 37, 40, 41, 52, 53, 56, 73 10.3 For an element to obey the octet rule it must be surrounded by 8 electrons. To determine the number of electrons present (1) count the individual electrons actually shown adjacent to a particular atom, and (2) add two times the number of bonds to that atom. Using this method the structures shown give: (a) 0 + 2(4) = 8; (b) 2 + 2(3) = 8; (c) 0 + 2(5) = 10; (d) 2 + 2(3) = 8; (e) 0 + 2(4) = 8; (f) 2 + 2(3) = 8; (g) 0 + 2(3) = 6; (h) 8 + 2(0) = 8. All the structures obey the octet rule except: c and g.
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This note was uploaded on 02/12/2008 for the course CHEM 400 taught by Professor Gupta during the Fall '07 term at American River.

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Fall 07 HW9 solution - Homework Assignment #9 Name_...

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