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Determination of a Rate Law Lab Report Essay

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Determination of a Rate Law
Megan Gilleland
10.11.2012
Dr. Charles J. Horn
Abstract: This two part experiment is designed to determine the rate law of the following reaction, 2I-(aq) + H2O2(aq) + 2H+I2(aq) + 2H2O(L), and to then determine if a change in temperature has an effect on that rate of this reaction. It was found that the reaction rate=k[I-]^1[H2O2+]^1, and the experimental activation energy is 60.62 KJ/mol.

Introduction
The rate of a chemical reaction often depends on reactant concentrations, temperature, and if there’s presence of a catalyst. The rate of reaction for this experiment can be determined by analyzing the amount of iodine (I2) formed. Two chemical reactions are useful to determining …show more content…

Find the Ln of [I-]0
Ln(0.015)=-4.19970508
7.Find [H2O2]0
Take (0.10 M H2O2)*(6.00mL)/ ( final volume)=0.015 M
8. Ln of [H2O2]0
Ln(0.015)= -4.19970508
9. Find the Ln of rate:
Ln(2.13675x10-5)=-10.753638
10. The last step for week one calculations is to calculate the average value of k.
Rate= k [I-]1[H2O2]. (2.13675*10-5 ) = k [0.015] [0.015] then solve for k. For this trial, k=0.09497.
This is then done for all trials. Then, once all five values of k are found, the average is taken by adding all five values of k and dividing by 5. The experimental k average is 0.105894M/s.
Table 2: Calculations Week 1 | | | | | | | | | | | | solution# | mol s2O3-2 | mol I2 | I2 | (rate) changeI2/change in temp | [I-]o | ln[I-]o | [H2O2]0 | ln[H2O2]o | ln rate | k | | 1 | 0.001 | 0.0005 | 0.0125 | 2.13675E-05 | 0.015 | -4.19970 | 0.015 | -4.19971 | -10.753 | 0.0949 | | 2 | 0.001 | 0.0005 | 0.0125 | 4.3554E-05 | 0.030 | -3.50655 | 0.015 | -4.19971 | -10.041 | 0.0967 | | 3 | 0.001 | 0.0005 | 0.0125 | 9.54198E-05 | 0.045 | -3.10109 | 0.015 | -4.19971 | -9.2572 | 0.1413 | | 4 | 0.001 | 0.0005 | 0.0125 | 0.000109649 | 0.045 | -3.10109 | 0.025 | -3.68888 | -9.1182 | 0.0974 | | 5 | 0.001 | 0.0005 | 0.0125 | 0.00015625 | 0.045 | -3.09776 | 0.035 | -3.35241 | -8.7640 | 0.0988 | | | | | | | | | | | k avg | 0.1059 | | | | | | | | | | | | | |

Data Week 2
Table 3:

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