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Rates of Chemical Reactions

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Introduction

Introduction The rate at which a chemical reaction occurs depends on several factors: the nature of the reaction, the concentrations of the reactants, the temperature, and the presence of possible catalysts. In this experiment we will study the kinetics of the reaction between iodine and acetone in acid solution: O CH3 - C - CH3 + I2 ? CH3 - C - CH2I + H+ + I- In order to determine what affects the rate of this reaction, we will be making 4 different mixtures, keeping constant the concentration and temperature in order to determine the order of each chemical in this reaction. Objective The objective of this lab is to determine the rate of the iodination of acetone by differing the amounts of reactants, finding the rate constant (k) and the order of the reaction in terms of each of the reactants. Equipment * 4.0 M Acetone * 1.0 M HCL * 0.0050 M I2 * Distilled Water * Three 50mL Burette * Two Double Burette Clamps * Support Stand * Three small funnels * 125mL Erlenmeyer ...read more.

Middle

: I. The first mixture was 20mL of H2O and 10mL of the other three. II. Second mixture was 20mL of Acetone and 10mL of the other three. III. Third mixture was 20mL of HCL and 10mL of the other three. IV. Fourth mixture was 20mL of I2 and 10mL of the other three. 6. For each mixtures we performed two runs. Reaction Rate Data Mixture Volume of 4.0M Acetone (mL) Volume of 1.0M HCL (mL) Volume of 0.005M I2 (mL) Volume of H2O (mL) 1st Run time in seconds 2nd Run time in seconds Temp. in degrees Celsius I 10 10 10 20 218sec 209sec 26� C II 20 10 10 10 91sec 104sec 25� C III 10 20 10 10 89sec 91sec 24� C IV 10 10 20 10 385sec 365sec 25� C Determination of Reaction Orders with Respect to Acetone, H+, and I2 Mixture Acetone H+ (I2)0 Rate=(I2)0/avg.time I 0.80 M 0.20 M 0.0010 M 4.68 x 10-6 II 1.6 M 0.20 M 0.0010 M 1.03 x 10-5 III 0.80 M 0.40 M 0.0010 M 1.11 ...read more.

Conclusion

x 10-5 /(0.80M)1.14(0.0010M)0.18(0.40M)1.24 = 1.11 x 10-5 /.0674647592 = 1.65 x 10-4 * Rate IV: k= 5.33 x 10-6 /(0.80M)1.14(0.0020M)0.18(0.20M)1.24 = 5.33 x 10-6 /.0325607537 = 1.64 x 10-4 CONCLUSION In the iodination of acetone, we found the order of the reaction with respect to acetone and HCl and find a value for the rate constant, k. Since the concentrations of acetone and HCl were much higher than that of I2, the concentrations of acetone and HCl changed very little. Thus the rate was determined by the time needed for iodine to be used up. Since iodine has color we can easily follow changes in iodine concentration visually. The equation, rate = k(A)m(H+)n(I2)p, can be simplified to rate = k[I2]/t since the values for acetone and HCl essentially remain constant during the course of any run. In mixtures I & IV the constant rate were similar, even though the iodine doubled in mixture IV, it is still the limited reagent. Mixtures II &III were similar do to the doubling of acetone in mixture II and HCL in mixture III. ...read more.

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