Chem 218 Exp 6

# To a then the rate of the reaction will double as the

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to [A], then the rate of the reaction will double as the concentration of A doubles; if the reaction was second order with respect to [A], then the rate of the reaction will quadruple when the concentration of A doubles; and so on and so forth. If was found that Rate = K[]^2[KI]. Therefore, the actual rate was 9.7 × (mol/L)/s. 2 H 2 O 2 (l) + I‾(aq) →2 H 2 O (l) + O 2 (g) + I‾(aq)

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Data Table 6.1 Reactants 3% 0.1 M KI H 2 O 50mL 50mL 100mL 50mL 25mL 125mL 25 mL 50mL 125mL Temperature of water bath: 21.2 °C Temperature of hot water bath: 72 °C Time (min) Volume of Trial 1 (mL) Volume of Trial 2 (mL) Volume of Trial 3 (mL) Volume of Trial 4 (mL 1 0 0 0 11 2 18 23 4 26 3 25 27 6 62 4 32 32 9 98 5 42 36 12 134 6 50 40 15 159 7 - 44 17 181 Total Volume: Trial 1- 50mL Trial 2- 44mL Trial 3- 17mL Trial 4- 181mL Average Rate Change for: Trial 1: 8.17mL/minute Trial 2: 6.29mL/minute Trial 3: 2.43mL/minute Trial 4: 25.86mL/minute Reaction Order with Respect to : 2 nd Order Reaction Order with Respect to : 0 Order Rate Law: 9.7 ×
Activation Energy: 1510 J/mol Sample Calculations : Average Rate Change for Trial 1: (0 + 18 + 7 + 7 + 9 + 8)/6 = 8.17mL/minute Reaction Order with Respect to : 2 nd Order Using trials 1 & 3: 8.17/2.43 = 3.36 3.36 = 2 nd order with respect to Rate Law: Rate = K[]^2[KI] [KI] = 0.3 M [] = 0.42 M Rate = 9.7 × (mol/L)/s Activation Energy: ln(/) = /R (1/ - 1/) ln(21/39) = /8.314 [(1/345.15) - (1/294.35)

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