PS2solutions

PS2solutions - PROBLEMS 5. (a) (c) 9. (a) (b) 11. (a)...

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PROBLEMS 5. (a) 2 2N O C l () 2N O ()C l() gg g →+ (b) [ ] NOCl rate = 2 t −∆ (c) rate = min L mol 0214 . 0 min L 0 . 16 342 . 0 = 9. (a) 3rd order in A (b) 1st order in A, 1st order in B: 2nd order overall 11. (a) 2 32 mol/L min L (mol/L) mol min = (b) 2 mol/L min L mol min (mol/L) = 13. (b) 2 2 2 mol L mol 0.833 0.42 [R] 0.633 Lm in L mol min ⎛⎞ × ⎜⎟ ⎝⎠ R = 4.9 mol/L (d) 2 mol 0.00624 0.0500 mol/L (0.0911 mol/L) k × 22 15.0 L /mol min k =⋅ 17. (a) 82 1.6 10 mol/L min (0.020 mol/L) (0.030 mol/L) k ×⋅ = × Solving for k ; 2 1.3 10 L /mol min k (b) 73 2 2 2 3.5 10 mol/L min 1.3 10 L/mol min (0.043 mol/L) [Br ] −− = × × × Solving for 2 [Br ] : 2 [Br ] = 0.15 mol/L (c) 63 2 2 2 2.0 10 mol/L min 1.3 10 L /mol min [NO] [NO] 4 = × × × Solving: [NO] = 0.18 mol/L 21. (a) 2 28 rate [S O ] [I ] mn k × 2.22 1.85 (0.0300 0.0250) m = ; 1.20 1.20 m = ; m = 1 3.06 2.22 (0.0275 0.0200) n = ; 1.38 1.38 n = ; n = 1 1st order in 2 SO , 1st order in I , 2nd order overall (b) 2 rate [S O ] [I ] k × (c) 4 1.15 10 mol/L min 0.371 L/mol min (0.0200 0.0155) mol /L k == × (d) 3 rate 0.371 L/mol min 0.105 mol/L 0.0875 mol/L 3.41 10 mol/L min ××
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CHAPTER 11 25. (a) + 3 rate [I ] [BrO ] [H ] p mn k −− × × 44 1.78 10 0.889 10 (0.0040 0.0020) m
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PS2solutions - PROBLEMS 5. (a) (c) 9. (a) (b) 11. (a)...

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