recorded. Hydrogen peroxide decomposition at a slower rate on its own, but with the presence of a catalyst, which is used as a reactant, helps to increase the pressure which in turns increases the reaction rate. Calculation 1. Calculate the molarity of a 3% mass/volume H2O2 (Part I) 3g H2O2 x 1 mol H2O2 x 1000 mL = 0.882 M 100 mL H2O 34g H2O2 1 L 2. 3% H 2 O 2 has a concentration of 0.882 M; we used 4 mL + 1 mL of KI (0.882 M) (4 mL) = 3.53 mmol H 2 O 2 (3.53 mmol H 2 O 2 / (5 mL total volume)= .706 M H 2 O 2
Error Analysis: During this three-part lab experiment, we only conducted part I of the experiment which is the decomposition of 3% H2O2 with 0.5 M KI solution, which we believed to be somewhat free of error, with our rate constant values being relatively constant. this Conclusion: The purpose mentioned is reinforced using figure 1 to figure 5 as references for the rates of reaction. The reaction rates were achieved with the amounts of hydrogen peroxide, potassium iodide and temperature as was expected. As the amount of hydrogen peroxide decreased, the reaction rate increased and as the amount of potassium iodide decreased, the reaction rate decreased. Using 20 degrees Celsius water also increased the reaction rate as was predicted. When heat is added to a reaction, the molecules involved in the reaction are given more energy and therefore move about the solution quicker. This then increases the chances of two different molecules colliding with each other and reacting, therefore increasing the reaction rate.
- Fall '19