Answers+to+Ch5

0200moll00155moll 0371lmolmin

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Unformatted text preview: n 2.22/1.85 = (0.0300/0.0250)m ; 1.20 = 1.20m ; m = 1 3.06/2.22 = (0.0275/0.0200)n ; 1.38 = 1.38n ; n = 1 1st order in S2 O8‐2 ; 1st order in I− ; 2nd order overall (b) rate = k × [S2 O8‐2 ] ×[I− ] (c) k =1. 15×10−4 mol/L · min/(0. 0200 mol/L)(0. 0155 mol/L) = 0.371 L/mol· min (d) rate = (0.371 L/mol· min)(0.105 mol/L)(0.0875 mol/L) = 0.00341 mol/L· min 29. (a) rate = k× [I− ]m × [BrO − 3 ]n × [H+ ]p 1.78× 10−4/ 0.889 × 10−4 = (0.0040/ 0.0020)m ; 2.0 = 2.0m ; m = 1 2 1...
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