'Rate Law and Activation Energy Essay\r'

'Introduction\r\nIn this taste we be analyzing the relationship mingled with response targets at antithetic tightfistednesss and temperatures to determine the true up count qualifyless, activating energy, response revisions, and half-life of a reception. The response of interest is the addition of a hydroxyl group to the burden of vitreous silica Violet. watch watch glass Violet, or hexamethylparaosaniline chloride for short, is a potently colored purple dye with the chemical mandate C25H30N3Cl and disassociates completely in response. The relevant structure for this complicated put forward be seen in code 1\r\nFigure 1\r\nThe base that is being use for the reception is the strong base atomic number 11 Hydroxide, or NaOH. This molecule also completely disassociates in water. Because criterion the c oncentrations of reactants is difficult in a simple science science research laboratoryoratory setting, the reaction among vitreous silica Violet and So dium Hydroxide provide be cargonful through with(predicate) wakeful absorbance. As the reaction between the chemicals takes in referable and the Crystal Violet receives the hydroxide the overall forcefulness of the purple color will decrease therefrom affecting the absorbance. The absorbance of the resolve will be measured with a tintometer as the reaction takes rush along and will be interpreted as a direct representation of stringency of Crystal Violet.\r\n afterward the reaction has taken place, through analysis of graphs plotting assiduousness vs. time, the natural log of density vs. time, and the inverse of denseness vs. time the reaction will be firm to be either zeroth, first, or second order with respect to crystal violet. From here the a phony enumerate eternal cig atomic number 18tte be determined, and victimisation comparisons of dissimilar aeonians at different concentrations of NaOH solution and different temperatures, the reaction order with re spect to hydroxide, the true rate unremitting for the reaction, and the activation energy for the reaction quite a little all be determined with the sideline equivalences respective(prenominal)ly.\r\nequating 1\r\nWhere k2’ is the dissembler rate constant of the reaction employ twice the initial OH- concentration as is used in the k1’ reaction and n is competent to the reaction order with respect to OH-. equivalence 2\r\nWhere k’ is a pretender rate constant based off of density and n is the reaction order with respect to OH- determined by compare 1.\r\n compare 3\r\nWhere k1 is the reaction constant at temperature T1, a is a constant that fag end be ignored due to the way the comparability will be utilized, R is that gas constant, and Ea is the activation energy.\r\n physical answer\r\nThe following materials were needed for the try:\r\n4 100mL beakers\r\n250mL beaker\r\n2.5Ã10-5M Crystal Violet Stock solution\r\n0.10M NaOH Stock solution\r\nDi stilled Water\r\n10 dry p proceedic cuvettes and caps\r\n move rod\r\nVernier Colorimeter\r\n50mL volumetric pipette\r\n100µL syringe\r\n2 10mL vials\r\nfellow Pro software system\r\nVernier computer interface\r\nHot plate\r\nVernier temperature analyze\r\n1. First, 100mL of 0.10M NaOH solution was obtained utilise 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 board was measured and the colorimeter was calibrated by setting the 0% light and 100% light conditions.\r\n3. The colorimeter was set to 565nm and 1mL of 2.5Ã10-5M Crystal Violet solution was mixed with 1mL of 0.05M NaOH solution and speedily added to the colorimeter. Data correlating time, temperature, transmittance, and absorbance was hence recorded for seven minutes as the reactio n between the devil solutions took place, and this teaching was saved.\r\n4. This previous step was repeated twain spare times with the 0.05M NaOH solution, and hence three times with the 0.10M NaOH solution. 5. Last, two 10mL-vials of 0.05M NaOH and 2.5Ã10-5M Crystal Violet solution were prepared in a warm bath solution on the hot plate. Once the temperature reached 35ËšC and was recorded, step BLANK through BLANK were repeated once again twice with the het solutions of Crystal Violet and 0.05M NaOH. entirely of the data that was collected was saved and distributed between the two lab partners and all excess solutions were disposed of seemlyly under the fume hood.\r\nResults\r\nThe following are the graphs obtained from the assiduousness and time recordings of the third die hard for the reaction between 1mL of 0.05M NaOH and 1mL of and 2.5Ã10-5M Crystal Violet carried issue at 22.62ËšC. paradigm 2\r\nfigure 3\r\nfigure 4\r\nThese plots show that the reaction order with respect to crystal violet is clearly 1st order due to the great r2 value of the elongated trend line. Since our pseudo rate constant based off of absorption is equate to the negative slope of our linear plot, our k’ in for the reaction of 1mL of 0.05M NaOH and 1mL of and 2.5Ã10-5M Crystal Violet carried place at 22.62ËšC is 0.1894.\r\nThese 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 2.5Ã10-5M Crystal Violet carried out at 22.50ËšC. figure 5\r\nfigure 6\r\nfigure 7\r\nAs 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 k’ in for the reaction of 1mL of 0.10M NaOH and 1mL of and 2.5Ã10-5M Crystal Violet carried out at 22.50ËšC 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-.\r\nSince the reaction order must be an integer we can see that the n must be 1. It is instantaneously 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 k’ is 0.1894.\r\nThese 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 2.5Ã10-5M Crystal Violet carried out at 36.09ËšC.\r\nfigure 8\r\nfigure 9\r\nfigure 10\r\nOnce again it is apparent from the three plots that the reaction is first order with respect to crystal violet. However, the fence 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, a nd 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 proposed 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 †a higher value, which is to be expected with the increase in temperature. Now, manipulating equation 4 we obtain that\r\nequation 4\r\nWhile plugging the proper values provides\r\nwhich after some arithmetic leads to a cypher Ea of 15,254.67J, or 15.25467kJ. The calculation for half-lives for the different conditions is simple, and effective requires the following equation.\r\nequation 5\r\nWhen using the rate constant prove in calculation 1, t1/2 for the kinetic run for the reaction between 1mL of 0.05M NaOH and 1mL of and 2.5Ã10-5M Crystal Violet carried out at 22.62ËšC is found to be 0.183 seconds.\r\nError Analysis\r\nIn this experiment there are some(prenominal) things calculate d and several sources of misplay to take into account. Error needs to be calculated for the rate constants k, for the half-lives, and for activation energy. The faultings for the pseudo-rate constants are obtained using the LLS method. Once these are obtained the next step is to calculate the delusion in the true rate constants.\r\nWhen 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 100µL 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 1Ã10-3M and 2Ã10-3M 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\r\nAnd when the derivative instruments are resolved is equal to\r\nequation 7\r\nAnd when the numbers are plugged in for the first kinetic run looks like calculation =.08\r\nIn other words, the rate constant for the first kinetic run came out to be 3.79±.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 fairish calculated above. Using the same method, equation 5 is solved for half-life, and the error is calculated like so.\r\nequation 8\r\nWhich after the derivatives are solved is equal to\r\nequation 9\r\nAnd of feast after the ready values for example the first kinetic run are plugged in provides\r\ncalculation = .004\r\nAnd 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 ±0.03K. Taking the se into consideration, a really complex process follows. The same process as above was used but involving oft more complicated and lengthy derivatives. First equation 3 was manipulated to the following form.\r\nequation 10\r\nThe 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.\r\nFigure 11\r\nOverall 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 approximately 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 isn’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 35-36ËšC, 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.\r\nThus, 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.\r\nConclusion\r\nA 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 slipway around difficult problems in the laboratory, such as measuring absorption i n place of concentration to follow the progress of a reaction.\r\nReferences-\r\nAlberty, A. A.; Silbey, R. J. Physical chemical science, second ed.; Wiley: New York, 1997. Department of Chemistry. (2013, Spring). CHEMISTRY 441G Physical\r\nChemistry Laboratory Manual. Lexington: University of Kentucky\r\n'