Rate Law and Activation Energy Essay Example for Free

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 k2’ is the pseudo rate constant of the reaction using twice the initial OH- concentration as is used in the k1’ reaction and n is equal to the reaction order with respect to OH-. equation 2 Where k’ 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 100 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 35ËšC 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 22.62ËšC. 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 k’ in for the reaction of 1mL of 0.05M NaOH and 1mL of and 2.5Ãâ€"10-5M Crystal Violet carried out at 22.62ËšC 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 22.50ËšC. 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 k’ in for the reaction of 1mL of 0.10M NaOH and 1mL of and 2.5Ãâ€"10-5M 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-. 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 k’ 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 36.09ËšC. 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 – 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 22.62ËšC 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 100 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 3.79 ±.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 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  ±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 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. 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

The Wright brothers invent the airplane :: essays research papers

About one hundred years ago the planet earth was a much smaller place. On December 17, 1903 the Wright brothers, Wilbur and Orville, made history when they took off in flight and invented the first airplane. This is how the planet earth began shrinking geographically making it easier and quicker to travel over long distances. Wilbur was the older of the two brothers by four years. Wilbur was born in 1867 on a farm near Millville, Indiana and Orville was born in 1871 near Dayton, Indiana. As youngsters, Wilbur and Orville looked to their mother for mechanical expertise and their father for intellectual challenge. Milton, their father, brought them various souvenirs and trinkets he found during his travels for the church. One such trinket, a toy helicopter-like top, sparked the boys' interest in flying. In school, Wilbur excelled, and would have graduated from high school if his family had not moved during his senior year. A skating accident and his mother's illness and subsequent death kept him from attending college. Orville was an average student, known for his mischievous behavior. He quit school before his senior year to start a printing business. The two brothers were very intellectual and smart, but both did not ever get their high school diplomas. It just goes to show that even two of the best minds in our history didn’t have to go to college or even finish high school to become these great minds. The first time Wilbur and Orville referred to themselves as "The Wright Brothers" was when they started their own printing firm at the ages of 22 and 18. Using a damaged tombstone and buggy parts, they built a press and printed odd jobs as well as their own newspaper. In 1892, the brothers bought bicycles. They began repairing bicycles for friends, then started their own repair business. They opened up a bicycle shop in 1893, and three years later, made their own bicycles called Van Cleves and St. Clairs. While nursing Orville, who was sick with typhoid in 1896, Wilbur read about the death of a famous German glider pilot. The news led him to take an interest in flying. On May 30, 1899, he wrote to the Smithsonian Institution for information on aeronautical research. Within a few months after writing to the Smithsonian, Wilbur had read all that was written about flying. He then defined the elements of a flying machine: wings to provide lift, a power source for propulsio n, and a system of control.

Women Are Better Parents Than Men

Both mothers and fathers contribute to their full extent and provide us with utmost facilities. The try to give their children with best clothes to wear , provide good foods to eat etc , so that their children get proper development. All that parents earn is for their children so that they can have settled life . They provide their children with best possible education so that their children can stand on their own and compete in the world later on . Most of the fathers do jobs so that they best earn for their children and provide their families with a happy life. The fathers usually become the role models for their children . Meanwhile , the mother take full care of their children and teaches them manners of how to live . The both parents are the best teacher for any child . Both parties have their important role as parents , and an ideal child is one that gets the support of both . However , now the question arises that who is better between the two? Whom do the children love the most. The mothers are one those who give birth to a child and suffer the pain during the early nourishment of a child . Most of the working women , when turns into a mother , have to sacrifice jobs , earnings etc . The mothers are more nurturing by birth . Every wife has a dream to become a mother ,so that she can give her love and care that God has gifted to her . It is by-nature that mothers are more loving and close towards their children than fathers . It doesn ‘t mean that fathers are not close to their children , but the feeling that a mother is gifted , cannot be matched by the fathers . Women also stay with their children more than men because men tend to be busy working. Most women are usually at home with their children, This provide children someone to be close to and who understands you. Once a women becomes a mother , she leaves her everything and her main goal is to give the child best care . The best time for a mother is to spend with their children . The mothers are by-nature more emotionally to their children . The mothers do everything for their children without complaining to anyone . Such as she cleans off all the mess of their children and feeds them. Due to psychological perspective, is that children are more important to their mothers than to their fathers. So at the end if you reflect on the facts and how the society think of it always goes to the women side. The children are more into their mother than father because of the treatment, love and care they get from their mother.

Christianity vs Islam - 1485 Words

On the surface, Islam and Christianity appear to have very little in common, however, as you get deeper into areas such as rituals, beliefs, ethics, founders, and sacred objects, the two show strong mutual similarities, particularly in the fundamental areas. In this essay I will compare and contrast the doctrines that make up the worlds largest and most recognized religions, Christianity and Islam. The word Islam means surrender or submission, submission to the will of Allah, the one God. Muslims are those who have submitted themselves. The basic creed of Islam is brief: Both Muslims and Christians are monotheists, believing in the same god, referred to†¦show more content†¦The idea that people could have witnessed these events without having been amazed by them is, quite simply, ludicrous. Other cultures having witnessed this would certainly have offered their own explanations in keeping with their own cultural and religious beliefs. Surely a society existing at the time would have documented this miraculous event. Yet nowhere have such works been found. The Christian Bible is highly contradictory, not just to modern day Christian beliefs, but in and of itself. Todays society is of the belief that all people are created equal, and Christians submit that their god is of the same belief. Modern Christians believe that their god loves everyone, and that they are all equal. However, after Adam and Eve had eaten from the tree forbidden by god, this deity said to Eve I will intensify the pangs of your childbearing; in pain shall you bring forth children. Yet your urge shall be for your husband, and he shall be your master. (Genesis 3:16). This tells us that, according to the Christian religion, women shall naturally be dominated by men. This kind of behavior is not conducive to a being who believes in inherent equality. Women are repeatedly treated as objects and told to be submissive in the Bible. Had there been no sins of mankind, there would be no story of Christ. The nature of sin must then therefore be analyzed. It is accepted by Christians that god created everything. If this is true, then this same god createdShow MoreRelatedChristianity vs. Islam1148 Words   |  5 Pagesoccurred hundreds or even thousands of years earlier† (Hodges 48). So why is this? What makes it so necessary for peoples of a religion to wage war? No answer has been found to date (Hodges 14). God is God. There is no argument between Christianity and Islam as to the existence of a single, omnipotent, omnipresent, and all powerful being. Both religions accept that God is separate from humans and resides in another realm and plane of being called Heaven. All of the messengers from the faiths areRead MoreChristianity vs Islam1093 Words   |  5 PagesIslam vs. Christianity There are varieties of religions in the world. Islam and Christianity has over one billion followers and counting. 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