CHEMICAL KINETICS EXPERIMENT EQUIPMENT AND CHEMICALS: Chemicals Equipment (2) 15
ID: 484214 • Letter: C
Question
CHEMICAL KINETICS EXPERIMENT EQUIPMENT AND CHEMICALS: Chemicals Equipment (2) 150 ml beaker (2) 600 mL cylinder 25 ml, graduated cylinder (3) 10ml, graduated tubes (2) 25 x 150 mm 0.0500 M KI 0.00500 M NazSzoy 0.100 M H2O2 2% Soluble starch solution 0.100 M HCI 0.050 M (NH4 Moo. 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 with time. rates can the of be affected by 4 main factors: the nature of the reactants, concentration the reactants or temperat a the presence of a The be determined by the rate at which one of the products appears or the rate at which one of the reactants is used up. Consider the following equation, aA bB cC dD The rate law expression for this reaction would be the following Rate- klAl' UBP [2] where k the specific rate constant [A] and [Bl 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 Brespectively. 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: 4 H20 2 H30 H2O2 Chemistry 132. EXPERIMENT 24Explanation / 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:
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