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Characterization of the Rate Properties of Iodide Oxidation Reaction | CHEM 1, Papers of Chemistry

Material Type: Paper; Professor: Roland; Class: General Chemistry; Subject: Chemistry and Biochemistry; University: University of California-Santa Cruz; Term: Unknown 1995;

Typology: Papers

Pre 2010

Uploaded on 08/19/2009

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Download Characterization of the Rate Properties of Iodide Oxidation Reaction | CHEM 1 and more Papers Chemistry in PDF only on Docsity! 1 Characterization of the Rate Properties of an Iodide Oxidation Reaction Experiment 12 Experiment 12 Rate Properties of an I- Oxidation Reaction Goals: for the reaction of S2O8 2- with I-  Measure the reaction rate  Find the orders of reaction  Determine activation energy and pre-exponential factor Method:  Measure rates using different reactant concentrations  Measure rates at different temperatures 2 Reaction Rates Rate: Change in concentration of a reactant or product per unit time Kinetics: Study of reaction rates [ ] [ ] [ ] [ ] ∆t B∆ ∆t A∆ RateB:AFor ∆t B∆ ∆t A∆ RateB:AFor 2 1 2 +=−=→ +=−=→ Reaction Progress per Time 5 Arrhenius equation  k: rate constant  Ea: activation energy  T: absolute temperature  R: universal gas constant  A: orientation factor Energy & orientation requirements RT E a Ae k − = Energy Requirements 2. Activation energy, Ea  Required energy: Reactants → Transition state R R→P PP o te n ti a l E n er g y TS* Ea 6 Potential energy diagrams Exothermic Endothermic ∆H ∆H Temperature Increase T = Increase available energy  ↑ collision frequency and ↑ collision energy RT Ea Ae k − = 7 3. Pre-exponential factor, A  “Frequency factor”  Collision effectiveness / orientation requirements RT Ea Ae k − = Orientation effects © 1995-2002 by Prentice-Hall, Inc. Pearson Company 10 Rate ∆[S2O32-]used = [S2O32-]final – [S2O32-]init = – [S2O32-]init 1 mol I2 made / recycled = ½ mol S2O3 2- consumed ∆[I2]recycled = ½ ∆[S2O32-]used = ½ [S2O32-]init -2 64 -fast-2 322 OS I 2 OS 2 I +→+ SO 2I I 2 OS -2 42 slow--2 82 +→+ Rate Using: ∆[I2] = ½ [S2O32-]init in ∆t blue -2 32 blue -2 322 1 blue 2 t ]OS[ 2 1 t ]OS[ t ]I[ Rate ∆ ∆ ⋅= ∆ ∆ = ∆ ∆ = SO 2I I 2 OS -2 42 slow--2 82 +→+ [I2] amount recycled = ½ amount S2O3 2- used ∆t time until blue 11 Overview 1a) Reaction order m, S2O8 2-  Vary [S2O8 2-] at constant [I-] and T 1b) Reaction order n, I-  Vary [I-] at constant [S2O8 2-] and T 2) Contributions to k: Ea and A  Vary T at constant [S2O8 2-] and [I-] ∆t ][I ∆t ]O[S Rate - initial - ∆ == 2 2 32 Part 1a Reaction order with respect to [S2O8 2-] t ]O[S ][I]O k[SRate initial n-m- ∆ == − 2 2 322 82 Vary [S2O8 2-] Calculate Constant [I-] Run# 0.10 M K2S2O8 2%starch 0.010 M Na2S2O3 0.20 M KI 0.10 M K2SO4 1 1 1 2 5 4 2 2 1 2 5 3 3 3 1 2 5 2 4 4 1 2 5 1 5 5 1 2 5 0 12 m: First Order in S2O8 2-? )][I]O (k [S Rate n-m -2 82loglog = Take log each side Plot log Rate vs. log [S2O8 2-] n-m- ][I]O k[SRate 2 82= log ac=log a+ log c and log xb = b log x bxmy )] (k[I] O[S m Rate n-- +⋅= +⋅= logloglog 282 Reaction order , m n-m- ][I]O k[SRate 2 82= b x m y )] (k[I]O [S m Rate n-- +⋅= +⋅= logloglog 282 lo g R at e log [S2O8 2-] Slope = m 15 Slope ~1.0 so n = 1 Reaction order , n 1b I - I - Rate log[rate] log[I - ] Trial Time (s) M (mol/L) M/s --- --- 1 64 0.077 1.2E-05 -4.920 -1.114 2 86 0.062 8.9E-06 -5.048 -1.211 3 113 0.046 6.8E-06 -5.167 -1.336 4 180 0.031 4.3E-06 -5.369 -1.512 5 330 0.015 2.3E-06 -5.632 -1.813 I- dependence y = 1.0133x - 3.8118 R 2 = 0.9955 -5.7 -5.6 -5.5 -5.4 -5.3 -5.2 -5.1 -5.0 -4.9 -1.9 -1.7 -1.5 -1.3 -1.1 log[rate] lo g[ ra te ] Part 2 Temperature dependence (k) Find k at room temperature (each run) Calculate average k and σk n-m- ][I]O[S Rate k 2 82 = ∆t ]O[S Rate initial - 2 2 32= 16 Part 2 Determining Ea and A n-m-RT E ][I]O[S A eRate a 2 82 − =  Measure rate for constant concentrations (Part 1a, run#2)  T = 0°C, room T, 35°C, 45°C  Calculate k f(T) ][I]O[S Rate k n-m- == 2 82 A and Ea bxmy A TR E k a +⋅= +      ⋅−= ln 1 ln b a eA RmE = ⋅−= Take ln→RT E a Aek − = ln k 1/T Watch units! b = ln A R E m a−= Plot ln k vs. 1/T 17 Ea = -slope .R = 61 kJ/mol R = 8.314 J/K.mol A = ey-int =6×106 A and Ea T( o C) Time(s) Rate k 1/T ln[k ] 0 660 1.2E-06 2.0E-05 0.0037 -10.84 18 185 4.2E-06 7.0E-05 0.0034 -9.56 35 40 1.9E-05 3.3E-04 0.0032 -8.03 45 15 5.1E-05 8.7E-04 0.0031 -7.05 Temperature dependence y = -7274x + 15.65 R 2 = 0.9873 -11.80 -10.80 -9.80 -8.80 -7.80 -6.80 0.0031 0.0032 0.0033 0.0034 0.0035 0.0036 0.0037 1/T (1/K) ln [k ] Put it all together 1. Orders of reaction: m, n 2. Activation energy: Ea 3. Pre-exponential factor: A n-m-RT E ][I]O[S A eRate a 2 82 − =
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