Kinetics Lab Report 2

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University of Cincinnati, Main Campus *

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2040L

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Chemistry

Date

Dec 6, 2023

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docx

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6

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11/06/22 Kinetics Lab Report Data and Results: Trial Infinity Volume (mL) K-value 1 10.8 0.0128 2 6.4 0.0157 3 4.7 0.0219 4 1.9 0.009 5 1.9 0.0099 6 1.8 0.0099 Standard average k-value = 0.0168 (Trials 1-3) Diluted average k-value = 0.0096 (Trials 4-6) *Tables and Graphs are shown below Discussion: The purpose of this experiment was to determine the rate constant of a first-order reaction by collecting data (time and amount of base added) and to further understand reaction kinetics. Basically, two different trials were run for the collection of data for this experiment. The first three trials were the standard concentration, and the second set of trials were run with the diluted concentration. The standard concentration trials (1-3) had an average k-value of 0.0168 and an average infinity volume of 7.3mL. The diluted concentration trials (4-6) had an average k- value of 0.0096 and an average infinity volume of 1.87mL. The k-values were calculated from the slopes given on the graphs of ln((V in -V t )/ V in ) vs. Time (sec) plot. After looking through the data, we observed that concentration does not have an effect on the rate of reaction due to the k-values being similar (within 0.01) in all of the trials. The reason for the average infinity volume being lower in the diluted concentration trials is due to less of the compound being present in the solution. This would require less base to be added to the reaction to neutralize it sooner, but not at a quicker rate. Many errors could have occurred during this experiment that could have thrown off the results. The time recorded could have been skewed due to human error between watching the color changing and time at the same time. Along with that, the shade of the indicator could have been different each time, meaning the time recorded could have been different. The adding of the base as soon as the color change appeared was also a place where error could have occurred. Some of this could explain the drastic difference in the amount of base added to the first 3 trials. Questions: 1. (1/2^10) x 100% = 0.98% 2.
3. ln [A] – ln [A o ] = -kt ln ([A o ]/2[A o ]) = -kt ln (1/2) = -kt 1/2 .696/k = t 1/2 4. The units used to measure the volume of the base does not affect the calculated values of k because the units for volume are cancelled out in the equation for the first-order reaction. 5. The slope would be positive instead of negative but the meaning of the value would be understood in a different way. The slope of ln (V in / (V in -V t )) vs. time would have a slope opposite in direction and equal in magnitude. Trial Tables and Corresponding Graphs: Trial 1- Amount Added (mL) Time (sec) 1 17 1 33 0.5 49 0.5 65 0.5 81 0.5 87 0.5 113 0.5 129 0.5 145 0.5 161 0.5 177 0.5 193 0.5 209 0.5 225 0.2 241 0.2 257 0.2 273 0.2 289
Trial 2- Amount Added (mL) Time (sec) 1 1 1 10 1 21 1 37 0.5 56 0.5 71 0.5 101 0.5 151 0.2 211 0.2 281 Trail 3-
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Amount Added (mL) Time (sec) 1 9 0.5 23 0.5 36 0.2 47 0.2 58 0.5 67 0.5 77 0.5 83 0.2 91 0.2 101 0.2 126 0.2 181 Trial 4- Amount Added (mL) Time (sec) 1 22 0.5 74 0.2 181 0.2 241
Trial 5- Amount Added (mL) Time (sec) 1 21 0.5 69 0.2 166 0.2 229 Trial 6- Amount Added (mL) Time (sec) 1 21 0.2 49 0.2 93 0.2 161 0.2 219
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