9_12 Mathematics Algebra Ii Benchmark 3 Algebra 2 Topic 9 Test Paper-Based
Saturday, November 9, 2019
Rate Law and Activation Energy Essay
Introduction In this experiment we are analyzing the relationship between reaction rates at different concentrations and temperatures to determine the true rate constant, activation energy, reaction orders, and half-life of a reaction. The reaction of interest is the addition of a hydroxyl group to the nucleus of Crystal Violet. Crystal Violet, or hexamethylparaosaniline chloride for short, is a strongly colored purple dye with the chemical formula C25H30N3Cl and disassociates completely in solution. The relevant structure for this compound can be seen in figure 1 Figure 1 The base that is being used for the reaction is the strong base Sodium Hydroxide, or NaOH. This molecule also completely disassociates in water. Because measuring the concentrations of reactants is difficult in a simple lab setting, the reaction between Crystal Violet and Sodium Hydroxide will be measured through light absorbance. As the reaction between the chemicals takes place and the Crystal Violet receives the hydroxide the overall intensity of the purple color will decrease thus affecting the absorbance. The absorbance of the solution will be measured with a colorimeter as the reaction takes place and will be interpreted as a direct representation of concentration of Crystal Violet. After the reaction has taken place, through analysis of graphs plotting absorption vs. time, the natural log of absorption vs. time, and the inverse of absorption vs. time the reaction will be determined to be either zeroth, first, or second order with respect to crystal violet. From here the a pseudo rate constant can be determined, and using comparisons of different constants at different concentrations of NaOH solution and different temperatures, the reaction order with respect to hydroxide, the true rate constant for the reaction, and the activation energy for the reaction can all be determined with the following equations respectively. equation 1 Where k2ââ¬â¢ is the pseudo rate constant of the reaction using twice the initial OH- concentration as is used in the k1ââ¬â¢ reaction and n is equal to the reaction order with respect to OH-. equation 2 Where kââ¬â¢ is a pseudo rate constant based off of absorption and n is the reaction order with respect to OH- determined by equation 1. equation 3 Where k1 is the reaction constant at temperature T1, a is a constant that can be ignored due to the way the equation will be utilized, R is that gas constant, and Ea is the activation energy. Procedure The following materials were needed for the experiment: 4 100mL beakers 250mL beaker 2.5Ãâ"10-5M Crystal Violet Stock solution 0.10M NaOH Stock solution Distilled Water 10 dry plastic cuvettes and caps Stirring rod Vernier Colorimeter 50mL volumetric pipet 100à µL syringe 2 10mL vials Logger Pro software Vernier computer interface Hot plate Vernier temperature probe 1. First, 100mL of 0.10M NaOH solution was obtained using a 50mL volumetric pipet, and 0.05M was prepared using a the pipet, the stock 0.10M NaOH solution, and distilled water. 2. The Logger Pro software was engaged and both the Vernier colorimeter and temperature probe were plugged into the appropriate channels. The temperature of the room was measured and the colorimeter was calibrated by setting the 0% light and 100% light conditions. 3. The colorimeter was set to 565nm and 1mL of 2.5Ãâ"10-5M Crystal Violet solution was mixed with 1mL of 0.05M NaOH solution and quickly added to the colorimeter. Data correlating time, temperature, transmittance, and absorbance was then recorded for seven minutes as the reaction between the two solutions took place, and this data was saved. 4. This previous step was repeated two additional times with the 0.05M NaOH solution, and then three times with the 0.10M NaOH solution. 5. Last, two 10mL-vials of 0.05M NaOH and 2.5Ãâ"10-5M Crystal Violet solution were prepared in a warm bath solution on the hot plate. Once the temperature reached 35ÃÅ¡C and was recorded, steps BLANK through BLANK were repeated again twice with the heated solutions of Crystal Violet and 0.05M NaOH. All of the data that was collected was saved and distributed between the two lab partners and all excess solutions were disposed of properly under the fume hood. Results The following are the graphs obtained from the absorption and time recordings of the third run for the reaction between 1mL of 0.05M NaOH and 1mL of and 2.5Ãâ"10-5M Crystal Violet carried out at 22.62ÃÅ¡C. figure 2 figure 3 figure 4 These plots show that the reaction order with respect to crystal violet is clearly 1st order due to the great r2 value of the linear trend line. Since our pseudo rate constant based off of absorption is equal to the negative slope of our linear plot, our kââ¬â¢ in for the reaction of 1mL of 0.05M NaOH and 1mL of and 2.5Ãâ"10-5M Crystal Violet carried out at 22.62ÃÅ¡C is 0.1894. These next three plots are the graphs obtained from the absorption and time recordings of the first run for the reaction between 1mL of 0.10M NaOH and 1mL of and 2.5Ãâ"10-5M Crystal Violet carried out at 22.50ÃÅ¡C. figure 5 figure 6 figure 7 As expected, these results still indicate a reaction order of 1 with respect to crystal violet as demonstrated by the linear plot on the figure 6. Our kââ¬â¢ in for the reaction of 1mL of 0.10M NaOH and 1mL of and 2.5Ãâ"10-5M Crystal Violet carried out at 22.50ÃÅ¡C is 0.2993. Now that we have two pseudo reaction constants in which the OH- concentration differs by a factor of 2, we can use equation 1 to obtain the reaction order with respect to OH-. Since the reaction order must be an integer we can see that the n must be 1. It is now know that for the reaction, the reaction orders with respect to both reactants are 1. At this point, the true rate constant can be determined using equation 2, where n is 1, the initial concentration of OH- is 0.05, and the pseudo rate constant kââ¬â¢ is 0.1894. These next three plots are the graphs obtained from the absorption and time recordings of the first run for the reaction between 1mL of 0.05M NaOH and 1mL of and 2.5Ãâ"10-5M Crystal Violet carried out at 36.09ÃÅ¡C. figure 8 figure 9 figure 10 Once again it is apparent from the three plots that the reaction is first order with respect to crystal violet. However, the reason we performed this last kinetic run was to obtain a value for k at a different temperature. This way we have two sets of values for equation 3 with two temperatures, and two rate constants. With this information we can cut out the pre-exponential factor a and solve for the activation energy. But first k must again be calculated for the reaction at the new temperature. Doing this the same way as done in calculation 2, we obtain a reaction constant of 4.964 ââ¬â a higher value, which is to be expected with the increase in temperature. Now, manipulating equation 4 we obtain that equation 4 While plugging the proper values provides which after some arithmetic leads to a calculated Ea of 15,254.67J, or 15.25467kJ. The calculation for half-lives for the different conditions is simple, and just requires the following equation. equation 5 When using the rate constant found in calculation 1, t1/2 for the kinetic run for the reaction between 1mL of 0.05M NaOH and 1mL of and 2.5Ãâ"10-5M Crystal Violet carried out at 22.62ÃÅ¡C is found to be 0.183 seconds. Error Analysis In this experiment there are several things calculated and several sources of error to take into account. Error needs to be calculated for the rate constants k, for the half-lives, and for activation energy. The errors for the pseudo-rate constants are obtained using the LLS method. Once these are obtained the next step is to calculate the error in the true rate constants. When calculating the error in true rate constant once must apply both the error in the pseudo rate constant and the error in the measurement of volume for the 100à µL syringe as it pertains to the concentration of hydroxide. The error in the syringe is 0.02mL, which for 0.05M NaOH solution leads to an error in concentration of approximately 1Ãâ"10-3M and 2Ãâ"10-3M for 0.10M NaOH. Equation 2 is manipulated to solve for the true rate constant. The following equation is used to solve for the error in the true rate constant. equation 6 And when the derivatives are solved is equal to equation 7 And when the numbers are plugged in for the first kinetic run looks like calculation =.08 In other words, the rate constant for the first kinetic run came out to be 3.79à ±.08. Now when calculating the error in the half-life the only thing that has to be taken into consideration is the error in the rate constant, which was just calculated above. Using the same method, equation 5 is solved for half-life, and the error is calculated like so. equation 8 Which after the derivatives are solved is equal to equation 9 And of course after the correct values for example the first kinetic run are plugged in provides calculation = .004 And last but nowhere near least, is the error analysis for the activation energy. With this the error for the true rate constant must again be taken into consideration, and the error for the temperature probe. The error for the true rate constant has already been calculated, while the error for the temperature probe is provided in the lab manual as being à ±0.03K. Taking these into consideration, a very complex process follows. The same process as above was used but involving much more complicated and lengthy derivatives. First equation 3 was manipulated to the following form. equation 10 The derivative of this equation with respect to each variable (T1, T2, K1, and K2) was then taken squared, and multiplied by the square of the respective variables uncertainty. These were added up and the square root was taken as in the above methods. The end result was a calculated error of 2 KJ for the calculated activation energy of 15kJ. Figure 11 Overall this lab was very successful in the use of absorption as a method of monitoring change in concentration. The calculated errors all seem to be about what one might expect. This lab was very analytical outside of one glaring hole. You can see in figure 9 a slight curve in the plot that isnââ¬â¢t found on either figure 3 or figure 6. To me this seems to be because the reactants are heated up to a temperature around 35-36ÃÅ¡C, but once the chemicals are mixed and placed in the cuvette the temperature is no longer controlled as the reaction takes place for the following seven minutes. Thus, as the temperature falls the rate of the reaction slows, and the pseudo rate constant is lower than it should be. This of course leads to a rate constant lower than it should be, and then the activation energy is affected as well. If I were going to change one thing about the lab, I would try and do something to control the temperature as the reaction persisted. Aside from that, there is little room for error outside of obvious blunders. Conclusion A reasonable value for activation energy was calculated from the data collected in this experiment. There were no major mistakes made in the laboratory, and the calculations all went smoothly. This experiment demonstrated that there are creative ways around difficult problems in the laboratory, such as measuring absorption in place of concentration to follow the progress of a reaction. References- Alberty, A. A.; Silbey, R. J. Physical Chemistry, 2nd ed.; Wiley: New York, 1997. Department of Chemistry. (2013, Spring). CHEMISTRY 441G Physical Chemistry Laboratory Manual. Lexington: University of Kentucky
Subscribe to:
Post Comments (Atom)
No comments:
Post a Comment
Note: Only a member of this blog may post a comment.