CHEMICAL KINETICS EXPERIMENT 24 EQUIPMENT AND CHEMICALS: Equipment beaker (2) Ch

Question

CHEMICAL KINETICS EXPERIMENT 24 EQUIPMENT AND CHEMICALS: Equipment beaker (2) Chemicals 600 mL beaker (2) 0.0500 M KI 0.00500 M NazS20, 25 mL graduated cylinder (2) 0.100 M H2O2 2% Soluble starch solution 0.100 M HCI 10 mL graduated cylinder (3) 0.050 M 0NHA Moou 25 x 150 mm test tubes (2) 1.0°C Thermometer 2) INTRODUCTION: The study of the rate of a chemical reaction is known as chemical kinetics. The rate of a reaction is defined by the change in concentration of a reactant or a product with time. Reaction rates can be affected by 4 main factors: the nature of the concentration of the or products, temperature and the presence of a catalyst. The rate determined by determining the rate at which one the products appears or the rate at which one of the reactants is used up. Consider the following equation, aA bB cC dD [1] The rate law expression for this reaction would be the following: Rate- klAl" [BP [2] k the specific rate constant where: Al and LB] are molar concentrations of A and B x and y are exponents which tell the dependence of the reaction rate on the concentration of A and B respectively. X and y are known as the orders, the sum of which is called the overall order of the reaction. The value of x and y must be determined experimentally. They are typically whole numbers. such as 0, 1, 2, or 3. In this experiment you will study the slow reaction of potassium iodide with hydrogen peroxide in an acidic solution: [3] 4 H2O 2 H3O H2O 21 Chemistry 132. EXPERIMENT 24

Explanation / Answer

A. Original [KI] = 0.0500 M

[H2O2] = 0.100 M

[HCl] = 0.100 M

Total volume in each run is 50 ml, so to calculate diluted molarity, M1V1= M2V2 is used

1) Order of I-:

Using formula given in 3rd point of data analysis and calculations:

Using formula given in 3rd point of data analysis and calculations:

ln(0.0120/0.0050) / ln(0.01/0.005) = 0.875/0.693 = 1.3 (calculated) , 1 (rounded)

Order of H2O2:

ln(0.0120/0.0050) / ln(0.04/0.02) = 0.875/0.693 = 1.3 (calculated) , 1 (rounded)

Order of H3O+:

ln(0.0120/0.0093) / ln(0.02/0.01) = 0.875/0.693 = 0.37 (calculated) , 0 (rounded)

3. rate constant = rate of run1 / [I-][H2O2] = 0.0120/(0.01*0.04) = 30

Run [KI] [H2O2] [H3O+] 1 0.01 0.04 0.02 2 0.005 0.04 0.02 3 0.01 0.02 0.02 4 0.01 0.04 0.01

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