Introduction
In this experiment, we mixed different amounts of reactants and measured the volume of O2 they produced so that we could determine the rate and order of the reaction. Three trials were done, the first trial acted as a baseline, then the following two trials each varied the concentration of a single reactant, allowing us to calculate the order of the reaction. Additionally, the temperature of the solution was changed to see what effect that would have on the rate. Using this information the activation energy for this reaction could be calculated and compared to the activation energy of the uncatalyzed reaction.
Calculations Converting time readings to elapsed seconds
Trial 1: 0:42=42 sec 1:42=102 sec 2:41=161 sec 3:32=212 …show more content…
This means that the rate of H2O production is twice that of O2. For every two H2O2 that disappear, one O2 is produced. This means that the rate of H2O2 disappearance is twice that of O2 production. If you used 0.20 KI instead of 0.10 KI, how would this affect: The slope of the graphs would increase by a factor of 2.76 because the concentration of KI would be doubled, a lot like the difference between trial 1 and trial 3. The rate for each trial would be 2.76 times larger, because the concentration of the reactants would be different. This would not change k for the reaction, it just changes the concentration with respect to volume. rate = …show more content…
The concentration of the reactants was varied such that the order of the reaction could be determined, and the temperature was also changed to determine the activation energy of the reaction. The rate law was found to be rate = k[KI]1[H2O2]1, and the activation energy was calculated to be 53.9 kJ/mol. This activation energy was 21.1 kJ/mol less than that of the uncatalyzed reaction, showing that the catalyst was effective at decreasing the activation energy, and increasing the rate of the reaction. One possible cause of error is the inconsistency of reactions, the reactants must collide in the correct orientation to react, so there is always a possibility for some small variation. When calculating the activation energy, warmer water was used to increase the temperature of the reactants; however, because the surroundings were still room temperature, the temperature may not have been consistent throughout the reaction, leading to an error in the observed rate, and consequently an error in the activation energy. A final potential error could come from the decision not to account for the vapor pressure of the water, or for the change in vapor pressure when the temperature of the water was increased in trial 4, this could have indicated a higher rate of production of O2 than was actually
In this experiment, we mixed different amounts of reactants and measured the volume of O2 they produced so that we could determine the rate and order of the reaction. Three trials were done, the first trial acted as a baseline, then the following two trials each varied the concentration of a single reactant, allowing us to calculate the order of the reaction. Additionally, the temperature of the solution was changed to see what effect that would have on the rate. Using this information the activation energy for this reaction could be calculated and compared to the activation energy of the uncatalyzed reaction.
Calculations Converting time readings to elapsed seconds
Trial 1: 0:42=42 sec 1:42=102 sec 2:41=161 sec 3:32=212 …show more content…
This means that the rate of H2O production is twice that of O2. For every two H2O2 that disappear, one O2 is produced. This means that the rate of H2O2 disappearance is twice that of O2 production. If you used 0.20 KI instead of 0.10 KI, how would this affect: The slope of the graphs would increase by a factor of 2.76 because the concentration of KI would be doubled, a lot like the difference between trial 1 and trial 3. The rate for each trial would be 2.76 times larger, because the concentration of the reactants would be different. This would not change k for the reaction, it just changes the concentration with respect to volume. rate = …show more content…
The concentration of the reactants was varied such that the order of the reaction could be determined, and the temperature was also changed to determine the activation energy of the reaction. The rate law was found to be rate = k[KI]1[H2O2]1, and the activation energy was calculated to be 53.9 kJ/mol. This activation energy was 21.1 kJ/mol less than that of the uncatalyzed reaction, showing that the catalyst was effective at decreasing the activation energy, and increasing the rate of the reaction. One possible cause of error is the inconsistency of reactions, the reactants must collide in the correct orientation to react, so there is always a possibility for some small variation. When calculating the activation energy, warmer water was used to increase the temperature of the reactants; however, because the surroundings were still room temperature, the temperature may not have been consistent throughout the reaction, leading to an error in the observed rate, and consequently an error in the activation energy. A final potential error could come from the decision not to account for the vapor pressure of the water, or for the change in vapor pressure when the temperature of the water was increased in trial 4, this could have indicated a higher rate of production of O2 than was actually