PS2solutions - PROBLEMS 5(a(c 9(a(b 11(a 2NOCl g 2NO g Cl 2...

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PROBLEMS 5. (a) 2 2NOCl( ) 2NO( ) Cl ( ) g g 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 3 2 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 L min L mol min = × × R = 4.9 mol/L (d) 2 mol 0.00624 0.0500 mol/L (0.0911 mol/L) L min k = × × 2 2 15.0 L /mol min k = 17. (a) 8 2 1.6 10 mol/L min (0.020 mol/L) (0.030 mol/L) k × = × Solving for k ; 3 2 2 1.3 10 L /mol min k = × (b) 7 3 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) 6 3 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 2 8 rate [S O ] [I ] m n 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 2 8 S O , 1st order in I , 2nd order overall (b) 2 2 8 rate [S O ] [I ] k = × × (c) 4 2 2 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
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