Dee Saenz Chem 1CL M 1 00 4 50 04 04 24 Experiment 14 Chemical Kinetics The purpose of the week s lab is to determine the effects of a concentration and temperature at the rate of the catalyzed decomposition of hydrogen peroxide while also learning how to measure the reaction rates by determining the rate law The objective of this lab was completed by following the steps accordingly and carefully In part one we had to determine the reaction rate measurements and the effect it has on concentration for three solutions In part two we also had to find the reaction rate measurements but instead found the effect it has on temperature Lastly for part three we had to recover the catalyst using a titration A method I used for this experiment was to make sure when calibrating to make sure to hold the syringe at the same level as the water level as the burette to make sure it is calibrated correctly resulting in accurate results To determine the rate law for the decomposition of hydrogen peroxide we had to serves as the catalyst for the reaction of since it helps speed the chemical decomposition reaction for the hydrogen peroxide without changing the reaction completely Recording the initial readings of the volume at an interval of solutions for the following three 2 2 2 measure the reaction rate of the concentration 2 2 2 2 2 10 0 1 Solution 1 5 2 2 10 15 2 Solution 2 10 2 2 10 10 2 Solution 3 5 2 2 20 5 2 2 2 We were able to determine the values for the rate law equation In which k is the rate constant x is hydrogen peroxide s reaction order and y is iodide s reaction order To determine the order of the reaction we first had to plug in three solutions prior into the rate law equation like so 1 5 10 2 10 10 3 5 20 3 0 5 1 2 0 5 1 concentration of We then begin by selecting two sets of data and where the concentration of are the same with that being solutions one and two Next we use division for over like so 2 2 3 1 Using this equation we then try to isolate x in a result of an answer of approximately 1 47 Using the same concept we can do the same by selecting two sets of data where the 1 2 are the same with that be over By isolating the y we were able to get a value of approximately 0 7001 For the rate of reaction we plotted the the points we got from the time in seconds when we were measuring the rate reaction with the calibration and the points for the volume of in mL in which they go in intervals of 2 0 and graphing these points gives us the slope which is the rate of reaction For the solutions one two and three we got a rate of reaction of 0 0203 0 0565 and 0 033 We can then use the rate law equation to plug in our values for x and y but this time solving for k We 2 0 000375 1 1 then get the same rate constant for all three solutions being 0 000375 1 1 0 000375 1 1 2 2 1 47 0 700 of also With this we can then plug in our values for the rate law equation with an average Lastly for the initial temperature we got values of 22 70 C 23 0 C and 23 0 C For part two the reaction rate was also measured In this part the solution is exactly the same as part one solution but the only difference occurring is the initial temperature for part two being extremely warm Solution 1 5 2 2 10 15 2 0 000375 1 1 0 0203 a rate constant of For part one s solution there is an initial temperature of 22 70 C with a corresponding rate of For part two s solution there is an initial 0 001256 temperature of 39 80 C a corresponding rate of and a rate constant of Between both solutions there was temperature k percent increase of 5 15 rate percent increase of 234 98 and a rate constant percent increase of 235 18 Temperature is proposed to increase the rate constant of a reaction by the Arrhenius equation because in the actual equation 0 068 1 1 In which k is the rate constant A is the pre exponential factor is the activation energy of the reaction R is the gas constant and T being the absolute temperature With this in mind temperature is highly involved in this equation and with that being said as temperature increases we can see that the rate constant of the reaction also increases Iodide ion being our catalyst assists in speeding up a chemical reaction without any permanent chemical changes occurring The catalyst changes the rate of the reaction by actually increasing the rate of reaction by providing an alternative route that requires less energy for this reaction to occur A proposed mechanism for this reaction with the catalyst is the balanced equation 2 2 2 2 42 2 4 4 90 10 4 is recovered by first adding approximately 2 drops of potassium chromate The percentage of to the flask Then setting up the titration apparatus by filling the burette with silver nitrate solution The reaction occurred was Which resulted in a yellow colored solution But by adding in the catalyst we were able to find the point of equilibrium with the color changing to a red orangish solution The color change allows us to understand that all of the iodide in the current solution has completely left allowing us to further calculate the moles of iodide in the solution compared to our initial value of With this we calculated the iodide percent recovery to be 98 This resulted in finding our solution to have a 5 00 10 4 concentration of hydrogen peroxide to be hydrogen peroxide to be 0 448 0 0747 compared to the actual concentration of In this experiment there could be multiple instances which can result in a source of error For our data specifically a source of error could be not cleaning each solution completely which could result in calcium to appear since sink water usually has some in it A possible source of error in determining the rate of reaction could be not holding the syringe to the water level of the actual burette when doing the calibration This could have resulted in giving us inaccurate results or in our case a really high or low value compared to the actual value A possible source of error for the effect of the effect of temperature on the rate could be not keeping the initial temperature inbetween the 35 to 40 C range This could result in also giving us inaccurate calculations since we would be using a too high or too low of a temperature to accurately tell us if temperature has an …