Blindsight Research Paper Example | Topics and Well Written Essays - 1750 words

Blindsight - Research Paper Example The outcome is that there is a decrease in the visual affectability that reacts to luminance differentiate; the case being progressively extreme in the event of high spatial and low fleeting frequencies (Barbur, Harlow, and Weiskrantz, 1994). It is additionally a marvel that this affectability isn't completely polished off if there should arise an occurrence of low spatial and high transient frequencies and this is the explanation that few reports have been made of leftover visual limits; in this is incorporated identifying and segregating any upgrades that might be available inside the field imperfection. Such is the thing that occurs in constrained decision tests (Cowey, 2010). The marvel of blindsight doesn't simply express that it is typical vision however without mindfulness. Other than the loss of essential visual cortex, there is another issue that must be thought of. Retrograde degeneration of hand-off neurons inside the ensuing territories of the parallel geniculate core and accompanying transneuronal degeneration of as much as 90% of the retinal ganglion cells (particularly the P? ganglion cells) (Cowey, Stoerig, and Perry, 1989) is liable for the incredibly low differentiation affectability for low worldly and high spatial frequencies, with resulting harms to the limit of separating structure, decreased movement, and frequency (Cowey, 2010) †these aptitudes are normally needy upon the parvocellular framework (Schiller, Logothetis, and Charles, 1990). Writing Review The issue of blindsight in individuals has a specific excellent property. This property expresses that there is a chance of distinguishing and segregating an upgrade even without there being any abstract mindfulness. That implies, an individual experiencing blindsight does in any case have certain visual capacities, and two of these limits incorporate location and segregation of development (Weiskrantz, 1986). The patients experiencing blindsight are really heedless to cognizant visua l discernment however they do have the capacity of performing visual manual arriving at works, which implies that they can react to visual data in spite of the fact that they don't have any visual recognition (Sanders, et al., 1974). As per certain examinations cortically dazzle patients have the capacity of segregating the heading of movement of single spots (King, et al., 1996) and bars (Azzopardi and Cowey, 2001); they are better ready to separate the quicker moving things and this proposes their affectability to high transient frequencies is expanded (Barbur, Harlow, and Weiskrantz, 1994). The instances of blindsight that have just been distributed had been caused because of injuries in the visual cortex. These patients despite everything stick to their component of utilitarian vision, for example the capacity to distinguish development, to point accurately at light flashes without cognizant visual discernment, and to have the option to figure if there is an improvement in the v isual visually impaired field. The cortically visually impaired patients don't hold the capacity of separating the bearing of the upgrades development that doesn't change its area all around, for instance, gratings and arbitrary dab kinematograms that delineate change, near development, and movement top to bottom (Azzopardi and Cowey, 2001). There is as yet a chance of there being a distinction toward the path separation and irregular kinematograms, gratings, etc, because of the way that such improvements fret about different movement handling techniques that are dependable according to the varieties that may be available between them with respect to their neighborhood and worldwide highlights. At the point when this contention is viewed as it will be workable for there to be separation of course in the cortically visually impaired visual field on the grounds of a strategy which straightforwardly sees development data of the upgrade.

Visa G4 para trabajar en organizaciones internacionales

Visa G4 para trabajar en organizaciones internacionales La visa G-4 permite an una persona extranjera vivir en Estados Unidos para trabajar en una organizaciã ³n internacional como la ONU, el FMI o la OEA. En comparaciã ³n con otras visas de trabajo, la G-4 ofrece notables ventajas en materia de lã ­mite de tiempo, impuestos y familiares que pueden acompaã ±ar a su main. Tiene cierto parecido con la visa Aâ para diplomticos pero es diferente. Quiã ©nes pueden solicitar la visa G-4 Los extranjeros con contrato de trabajo en una organizaciã ³n internacional ubicada en EE.UU. Adems, este mismo tipo de visa se otorga a las personas que conforman el hogar recognizable de la persona contratada: su cã ³nyuge, los hijos de cualquier edad, los hijos de la pareja y, en algunos casos muy limitados, los padres e incluso otros familiares cercanos. Validez de la visa G-4 y move La visa G-4 es vlida por el tiempo que seã ±ala la misma. Adems, se puede prorrogar sucesivamente sin lã ­mite de aã ±os. En los 60 dã ­as anteriores a que la visa lapse, se tramita su extensiã ³n sin necesidad de salir del paã ­s. Se puede transferir de una organizaciã ³n internacional an otra, por ejemplo, de trabajar en Naciones Unidas pasar a ser un empleado del Fondo Monetario Internacional. Para ello debe cancelarse con el Departamento de Estado la registraciã ³n con la instituciã ³n unique y registrarse con la nueva. Permiso de trabajo para familiares de titulares visa G-4 En algunos casos, los familiares que acompaã ±an an EEUU a la persona contratada por la organizaciã ³n internacional pueden trabajar. Para ello necesitan obtener previamente un permiso de trabajo, que se conoce por las siglas EAD. Debe solicitarse al USCIS, que sã ³lo los yield en los casos de nacionales de paã ­ses que tienen un tratado de reciprocidad con Estados Unidos. Los permisos de trabajo se aprueban para el cã ³nyuge o pareja, los hijos solteros menores de 21 aã ±os, o de 23 si estn estudiando a tiempo completo en la universidad. Si tuviesen una minusvalã ­a fã ­sica o mental no se aplica el lã ­mite de edad, siempre y cuando sean solteros. Por otra parte, los hijos de extranjeros con visa G-4 pueden estudiar la educaciã ³n primaria y secundaria. Para los reciã ©n llegados, estos child las reglas bsicas del sistema educativo de los Estados Unidos. Impuestos de titulares de la visa G-4 Adems, otra de las ventajas de la visa G-4 es que no se tributa en EEUU el impuesto sobre la renta (personal expense) por el salario pagado por la organizaciã ³n internacional. Tramitaciã ³n de la visa G-4 y de su extensiã ³n Una vez que se ha encontrado una organizaciã ³n internacional que ofrezca un trabajo al que se puede aplicar, debe conseguirse que estã © de acuerdo en contratar. Una vez que se tiene la oferta de contrato ya debe iniciarse el proceso risk una oficina consular de los Estados Unidos.. Si la persona contratada se encuentra en Estados Unidos deber tambiã ©n tramitar la visa en un consulado y salir del paã ­s para obtenerla, ya que no se puede hacer este trmite dentro del paã ­s salvo en casos limitadã ­simos como el de los estudiantes internacionales con una visa F-1 en prcticas en la organizaciã ³n internacional que ahora los contrata con una G-4. En estos supuestos podrã ­a ser posible cambiar la visa a travã ©s del USCIS. Se debe pedir una cita y los documentos a presentar child el DS-160, un pasaporte vlido por al menos medio aã ±o y una carta de la organizaciã ³n internacional ofreciendo la contrataciã ³n y fotos. Como regla general, no es necesario presentarse an una entrevista y las tasas de la visa no se agnostic. Las oficinas consulares pueden pedir informaciã ³n y documentaciã ³n adicional, dependiendo de cada caso.â Estas visas se caracterizan por ser tramitadas inmediatamente. En cuanto a su validez, si una persona contratada con una visa G-4 se retira o es despedida debe abandonar EEUU en un periodo de 60 dã ­as junto con los familiares inmediatos tambiã ©n titulares de ese tipo de visado. Y en estos casos, si se deal de EEUU no se puede volver an entrar, aunque la fecha de vigencia de la visa siga siendo vlida. En cuanto a la renovaciã ³n, si el main de una visa G-4 debe renovar la visa y est presente en EE.UU. puede contactar con la Diplomatic Liaison Division del Departamento de Estado o con la Misiã ³n de EE.UU. risk Naciones Unidas, ya que en muchos casos es posible renovar la visa a travã ©s de esas oficinas llenando el formulario online DS-1648. Finalmente, cabe destacar que, en general, los maestros de la escuela internacional de Naciones Unidas no se consideran plantel de dicha organizaciã ³n internacional, por lo tanto no pueden beneficiarse de esta visa. Pero puede ocurrir que la ONU solicite expresamente una G-4 an una persona en specific y el consulado correspondiente puede aprobarla pero no es la regla general. Tributaciã ³n por parte de ciudadanos que trabajar en organizaciã ³n internacional en EE.UU. Por supuesto que los ciudadanos estadounidenses, los residentes permanentes legales y las personas con permiso de trabajo en los EEUU no necesitan una visa G-4 para trabajar en una organizaciã ³n internacional. En cuanto al pago de impuestos por el salario recibido por la organizaciã ³n internacional, debe tributarse de una forma particular. La visa G-4 como camino a la green card La G-4 puede ser un puente hacia la residencia permanente tanto durante la vida laboral de una persona como tras su jubilaciã ³n. En este à ºltimo caso si la persona ha vivido al menos 15 aã ±os en EEUU y, de ellos, tres y medio los ha residido en los siete aã ±os anteriores a retirarse, puede solicitar la green card para sã ­ misma y para su cã ³nyuge en los seis meses siguientes a su à ºltimo dã ­a de trabajo. Otras opciones para trabajar en Estados Unidos Para trabajar legalmente es necesario tener una situaciã ³n migratoria que lo permita. Por ejemplo, porque se es ciudadano americano, residente permanente lawful, se tiene una visa de trabajo o se est en una situaciã ³n que permite solicitar y obtener un permiso de trabajo. Tener en cuenta que trabajar cuando no se est autorizado es una violaciã ³n migratoria y, si las autoridades se enteran o lo sospechan dar lugar a que se cancele la visa. Frecuentemente, la persona se entera de que su visa es revocada cuando intenta regresar an Estados Unidos despuã ©s de un viaje an otro paã ­s y se encuentra con la desagradable sorpresa de que no la dejan entrar a paã ­s y la envã ­an de regreso a su lugar de procedencia.â Puntos Clave: Visa G-4  ¿Para quiã ©n es?: extranjeros con contrato de trabajo en una organizaciã ³n internacional ubicada en EE.UU.El main de la visa G-4  ¿puede obtener visa para familiares?: sã ­, en general para el cã ³nyuge, compaã ±ero domã ©stico e hijos solteros menores de 21 aã ±os. En algunas circunstancias, ese cã ­rculo de familiares puede ampliarse. Pueden trabajar en EE.UU. y estudiar.El main de la visa G-4,  ¿paga impuestos en EE.UU.?: No. Este artã ­culo es meramente informativo. No es asesorã ­a lawful .

Understanding Abandonment Issues and BPD

Understanding Abandonment Issues and BPD BPD Living With BPD Print Understanding Abandonment Issues and BPD By Kristalyn Salters-Pedneault, PhD Kristalyn Salters-Pedneault, PhD, is a clinical psychologist and associate professor of psychology at Eastern Connecticut State University. Learn about our editorial policy Kristalyn Salters-Pedneault, PhD Updated on September 19, 2019 JGI/Tom Grill / Getty Images More in BPD Living With BPD Diagnosis Treatment Related Conditions I am a 22-year-old who was diagnosed with borderline personality disorder (BPD) when I was 19. I think my BPD is related to the fact that I had a difficult childhood. Without getting into details, my dad wasnt around, and my mom just wasnt a great mom. My biggest problem now is that I cant seem to maintain relationships. Everyone leaves me. I cant keep a boyfriend for more than a few months, and even my friends dump me after a while. Whenever one of my relationships ends, I feel horrible, empty, and desperate. I do my best to try to win them back, but it never works. Why cant people just be good to me and stick around? Borderline Personality Disorder and Relationship Struggles The struggle with relationships that you describe is very common for people with borderline personality disorder (BPD). A key symptom of BPD is fear of abandonment.?? This symptom may cause you to need frequent reassurance that abandonment is not imminent, to go to great lengths to try to avoid abandonment  and to feel devastated when someone ends a relationship with you. But you are also describing another phenomenon thats common in BPD. People with BPD tend to have more unstable, chaotic relationships than others, and these relationships often end prematurely due to conflict.   Conflict Can Lead to Abandonment In many ways, its a double-whammy. People with BPD both fear abandonment and have symptoms that create conflict with others and often lead to abandonment, which then reinforces the fear. In addition, people with BPD are likely particularly attuned to the experience of being abandoned. So, even though it is painful for everyone to experience the end of relationships, the end of a relationship can feel particularly devastating for people with BPD. Ways to Stop the Unhealthy Cycle of Conflict and Abandonment The good news is that there are things you can do to try to stop this cycle. For example, in dialectical behavior therapy (DBT) a set of skills called the “interpersonal effectiveness” skills are taught. These skills can help you learn to be more effective in relationships, which can make those relationships stronger and more likely to last. If you arent getting DBT now, this may be something to talk to your therapist about. Schema-focused therapy also may be helpful in identifying and actively changing problematic ways of thinking that cause issues in your life. It can help you pinpoint unmet needs you have that youve been trying to get others to meet in an unhealthy way and find healthy ways to get those needs met instead. In addition, it can help to explore the roots of the abandonment issues with your therapist. It sounds like you had some experiences in your early childhood that would understandably leave you afraid of people leaving you. Talking about how those early experiences influence your current ways of viewing and interacting with the world may be helpful. With treatment, hard work and time, it is possible to have more stable relationships and learn to view both yourself and others in a more healthy and realistic manner.

Current Events in Sociological Context

Current Events in Sociological Context

The Industrial Revolution Was A Period Of Significant...

The Industrial Revolution was a period of significant social and political change, constituted through the rise of science, increasing democracy, urbanisation, the growth of the state globalisation. This period of time changed the way people worked and went about their everyday lives. The shift from agrarian societies to a system where people were paid for their labour, resulted in urbanisation. This was also aided by the mass production of goods being manufactured in factories in order to make a higher profit for goods (Germov Poole, 2011, p. 22). Marx, Weber and Durkheim are just three theorists that tries to explain the processes of industrialisation; they all had strong views surrounding the consequences that this period had on society. However, although they aimed to give valid arguments, there are criticisms towards some of their theories. This essay will examine their explanations of the problems faced by early industrial society as well as how they would interpret gender se gregation in the workforce; which is one social issue that many Australians are faced with today. For example, men dominate women in leadership positions across all industries in Australia. women are also more likely to only hold a management position in female-dominated industries, such as healthcare and education. Nationally, on average (in full-time work), men earn 16.2 per cent more in wages than women (Australian Bureau of Statistics, 2016) . These issues will be discussed in relation toShow MoreRelatedIndustrial Revolutions During Europe During The Industrial Revolution1298 Words   |  6 PagesIndustrial Revolution in Europe Before the industrial revolution, Europe was mostly dominated by farmers but as the industrial revolution progressed this changed dramatically. Industrial revolution had a significant impact in the process by making new demands that shaped the way of life through increased competition and technological innovation. Generally, it was a historical period that sparked in a stroke a number numerous changes in the economic, social and political dimensions. It is consideredRead MoreFrench Revolution1740 Words   |  7 PagesTHE FRENCH REVOLUTION The French Revolution, which erupted in 1789 marked a turning point inthe history of human struggle for freedom and equality. It put an end tothe age of feudalism and ushered in a new order of society. An outline of  this revolution will explain to you the kind of turmoil that occurred inEurope. This revolution brought about far reaching changes in not onlyFrench society but in societies throughout Europe. Even countries in othercontinents such as, India, were influencedRead MoreIndustrial Revolution1160 Words   |  5 PagesAmerica has changed in many ways. Towards the end of the 19th century, a significant change took place in the fundamental structure of the economy. That change was industrialization. During this time period, the United States of America changed from a large, agricultural country, to an urban industrial society. The process of industrialization began to take place in America, and eventually took over the economy during this period. Entrepreneurs and invento rs put together various machines and businessesRead MoreThe Second American Revolution891 Words   |  4 PagesSecond American Revolution Thesis Statement â€Å"The Civil War may also be termed as the second American Revolution in terms of the political, social and economic changes that occurred during the war† Introduction American Civil War was fought between 1861 and 1865. The war began because President Abraham Lincoln, elected in 1860, was very persistent on preserving the Union, which was threatened by the issue of slavery. The North was growing rapidly in wealth and population, and it was clear to the SouthernRead MoreThe Second American Revolution901 Words   |  4 PagesSecond American Revolution Thesis Statement â€Å"The Civil War may also be termed as the second American Revolution in terms of the political, social and economic changes that occurred during the war† Introduction American Civil War was fought between 1861 and 1865. The war began because President Abraham Lincoln, elected in 1860, was very persistent on preserving the Union, which was threatened by the issue of slavery. The North was growing rapidly in wealth and population, and it was clear to theRead MoreImpact Of Industrialisation On Patterns Of Urban Development1498 Words   |  6 Pageseconomic growth. Roughly without industrial revolution, a few cities such as Italy and Netherlands had their urban dweller proportion raised. Switzerland, Germany, England and France had the similar phenomena. Precious long peace during the nineteenth century had conferred European countries chances to develop in all dimensions, such as political mechanism, technology, arts and lifestyle. It was the Age of Progress. The industrious revolution, together with political context, national identity andRead MoreThe Great Expectations Of The Industrial Revolution1528 Words   |  7 PagesFor thousands of years, families put their children to work in whatever labor was necessary for survival; only wealthy and powerful children were able to avoid this fate. In 19th Century England, children of lower class were to work long hours in factories, warehouses, and coal mines for low wages and little food. They also were considered by most societies to be property of their parents. Children had little protection from governments who viewed them as having little to no civil rights outsideRead MoreThe Industrial Revolution Was A Movement1335 Words   |  6 PagesThe Industrial Revolution was a movement wh ich contributed to significant progress in America which began in the later 1700’s throughout the later 1800’s. During the industrial revolution, dramatic economic and cultural shifts took place. The discovery of the mineral wealth, technological advancements, and the construction of a nationwide railway changed the transportation industry as well as its labor force. These changes ushered in an intense need for manufacturing factories, laborers, and resourcesRead More U.S. History 1877-1933 Essay1028 Words   |  5 Pages â€Å"The United States emerged from a virulent, intense, and inhumane civil war and evolved into a new nation during this period. This transition was the culmination of political, economic, social, and cultural movements which transformed the nation. E Pluribus Unum - out of many United States, one nation; the United States was forged in the cauldron of these revolutions.quot;nbsp;nbsp;nbsp;nbsp;nbsp;-Arnold Toynbee, A Study of History nbsp;nbsp;nbsp;nbsp;nbsp;The above statement is one thatRead MoreExploitation of Children during the Industrial Revolution in Great Britain1867 Words   |  7 Pagesnineteenth centuries, Great Britain experienced a period of radical change and transformation. This era is referred to as the Industrial Revolution. It brought a surge of technological innovations, an increase in production, more world trade, and a rise in urban population. One of the most controversial and widely debated issues until today among historians is the use of child labor. Despite that this era led to massive economic growth and social development, it violated women’s rights and exploited

Should Drinking Age Be Increased to 21 Free Essays

Should drinking age be increased to 21. It is no secret, that people drink alcohol before they turn 21. Stories about drinking on college campuses and high school parties are very easy to find. We will write a custom essay sample on Should Drinking Age Be Increased to 21 or any similar topic only for you Order Now That is why underage drinking can be stopped if the law were changed. The major reason that people drink at an early age is because they feel it is fun and exciting to do something they are not supposed to. However, if it would be lowered to 18, then the trill of doing something illegal would vanish. Besides that, 18 is the age of adulthood in the United States, and adults should have the right to make their own decisions about alcohol consumption. What is also very important, that other countries had demonstrated that young people, who are allowed to drink at the age of 18, don’t go wild. Although the legal alcohol purchase age is 21, a majority of college students under this age consume alcohol in an irresponsible manner. This is because drinking by these youth is seen as enticing â€Å"forbidden fruit†, a â€Å"badge of rebellion against authority† and a symbol of adulthood. According to National Institute on Alcohol Abuse and Alcoholism the major argument for lowering the drinking age is that prohibitions have always provoked over – indulgence. Those of us who have attended college over the last 25 years can certainly attest to the fact that the law has done nothing to diminish freshman and sophomore access to alcohol. It has only pushed underage consumption underground. The statistics show that many underage people drive home after a night of drinking in order to hide it from their parents. If the drinking age were lowered, young adults would feel less pressure to notify their parents that they have been drinking. Ultimately, the greatest threat to people’s safety comes, when young adults drive home under the influence. In addition, we say that 18 year olds are adult enough to kill and die in the armed forces, change the course of a nation in the voting both, judge other adults on a jury, be prosecuted as an adult, enter into binding contracts, operate automobiles and heavy machinery, smoke tobacco, own and operate a business, have a bank account and credit card, own a house, be married and have a family of their own. As a Froma Harrop, nationally syndicated columnist wrote in her Feb. 9, 2010 article â€Å"Age Discrimination for the Young† â€Å"Being adult who cannot have a beer is highly, absurdly inconsistent. â€Å" According o her, in our imperfect world, the law has to draw lines, however arbitrary. But laws that only appear to address a problem by burdening young people aren’t wise, and they aren’t fair. For the past 20 years, the U. S. has maintained a Minimum Legal Drinking Age of 21, while in most other countries is 18, and in U. K. is as low as 16 in restaurants. A ccording to John Cloud, writer for Time Magazine, all those countries with drinking age under 21 tend to have fewer alcohol – related problems, than we do in the U. S. In those countries, people learn how to drink from an early age and do so in the safe and supporting environment of the home. Alcohol statistics say that teens coming from homes that had no alcohol and were not taught how to drink responsibly had more issues with alcohol. The behavior in these teens was less risky if parents allowed them to drink at home. In today’s world alcohol is and always going to be a problem no matter what age we are. However, there is no evidence of massive brain impairment, alcohol dependency, or underage alcohol abuse, which the experts tell us, will be the inevitable result of lowering the drinking age in the United States. By lowering drinking age, young adults would be allowed to drink in controlled environments such as restaurants, taverns, pubs and official school and university functions. In these situations responsible drinking could be taught through role modeling and educational programs. Mature and sensible drinking behavior would be expected. If the drinking age was lowered, it will help kids realize how important alcohol is, and how mature we must become when we are under the influence of alcohol. How to cite Should Drinking Age Be Increased to 21, Essay examples

Tyler Larsen Essay Example For Students

Tyler Larsen Essay Period: 8Biology Paper: Carbon DatingFelschDuring the first part of class we talked about Isotopes andcarbon dating. This subject caught my attention unlike other lessons, so Idecided to do my report on this topic. It is not very controversial, theonly controversy being if it is accurate or not. Carbon dating iscontroversial in that is shares some of the fundamental assumptionsinherent to all Radiometric Dating techniques. In order for Carbon Datingto have any value, Carbon-14, produced in our outer atmosphere as Nitrogen-14 and changed into radioactive Carbon-14 by cosmic-ray bombardment, andmust be at equilibrium in our atmosphere. In other words, the productionrate must be equal to the decay rate. Therefore, the question I pose isthis; is carbon dating an effective way of telling the date of artifacts?The first thing I will discuss is how carbon dating works. Carbon-14is the radioactive version of Carbon. Radiation from the sun strikes theatmosphere of the earth all day long. This energy produces radioactiveCarbon-14. This radioactive Carbon-14 slowly decays into normal, stableCarbon-12. Laboratory testing has shown that about half of the Carbon-14molecules will decay in 5730 years. After another 5730 years half of theremaining Carbon-14 will decay, leaving only of the original Carbon-14. We will write a custom essay on Tyler Larsen specifically for you for only $16.38 $13.9/page Order now It goes from to to 1/8, ect. In theory it would never totallydisappear, but after about 5 half lives the difference is not measurablewith any degree of accuracy. This is why most people say that carbon datingis only good for objects less than 30,000 years old. Since sunlight causes the formation of Carbon-14 in the atmosphere,and normal radioactive decay takes it out, there must be a point where theformation rate and the decay rate equalize. This is called the point ofequilibrium. Let me illustrate; if you were trying to fill a barrel withwater but there were holes drilled up the side of the barrel, as you filledthe barrel it would began leaking out the holes. At some point you would beputting water in and water would be leaking out at the same rate. You willnot be able to fill the barrel pas this point. In the same way Carbon-14 isbeing formed and is decaying out simultaneously. A freshly created earthwould require about 30,000 years for the amount of Carbon-14 in theatmosphere to reach this point of equilibrium because it would leak out asit is being filled. Tests indicate that the earth has yet to reachequilibrium. This would mean that the earth is not yet 30,000 years old. This also means that plants and animals that lived in the past had lessCarbon-14 in them than they do today. This one fact totally upsets dataobtained by Carbon-14 dating. Yet another example is a candle you find burning in a room. You couldmeasure the present height of the candle (say, seven inches) and the rateof burn (say, an inch per hour). In order to find the length of time sincethe candle was lit we would be forced to make some assumptions. We wouldobviously have to assume that the candle has always burned at the samerate, and the initial height of the candle. The answer changes based on theassumptions. Similarly, scientists do not know that the Carbon-14 decayrate has been constant. They do not know that the amount of Carbon-14 inthe atmosphere is constant. Present testing shows the amount of Carbon-14in the atmosphere has been increasing ever since it was first measured inthe 1950s. This may be tied in to the declining strength of the magneticfield, but this has not yet been proven. .u5c18d649bd00217f7ce892df4202e32f , .u5c18d649bd00217f7ce892df4202e32f .postImageUrl , .u5c18d649bd00217f7ce892df4202e32f .centered-text-area { min-height: 80px; position: relative; } .u5c18d649bd00217f7ce892df4202e32f , .u5c18d649bd00217f7ce892df4202e32f:hover , .u5c18d649bd00217f7ce892df4202e32f:visited , .u5c18d649bd00217f7ce892df4202e32f:active { border:0!important; } .u5c18d649bd00217f7ce892df4202e32f .clearfix:after { content: ""; display: table; clear: both; } .u5c18d649bd00217f7ce892df4202e32f { display: block; transition: background-color 250ms; webkit-transition: background-color 250ms; width: 100%; opacity: 1; transition: opacity 250ms; webkit-transition: opacity 250ms; background-color: #95A5A6; } .u5c18d649bd00217f7ce892df4202e32f:active , .u5c18d649bd00217f7ce892df4202e32f:hover { opacity: 1; transition: opacity 250ms; webkit-transition: opacity 250ms; background-color: #2C3E50; } .u5c18d649bd00217f7ce892df4202e32f .centered-text-area { width: 100%; position: relative ; } .u5c18d649bd00217f7ce892df4202e32f .ctaText { border-bottom: 0 solid #fff; color: #2980B9; font-size: 16px; font-weight: bold; margin: 0; padding: 0; text-decoration: underline; } .u5c18d649bd00217f7ce892df4202e32f .postTitle { color: #FFFFFF; font-size: 16px; font-weight: 600; margin: 0; padding: 0; width: 100%; } .u5c18d649bd00217f7ce892df4202e32f .ctaButton { background-color: #7F8C8D!important; color: #2980B9; border: none; border-radius: 3px; box-shadow: none; font-size: 14px; font-weight: bold; line-height: 26px; moz-border-radius: 3px; text-align: center; text-decoration: none; text-shadow: none; width: 80px; min-height: 80px; background: url(https://artscolumbia.org/wp-content/plugins/intelly-related-posts/assets/images/simple-arrow.png)no-repeat; position: absolute; right: 0; top: 0; } .u5c18d649bd00217f7ce892df4202e32f:hover .ctaButton { background-color: #34495E!important; } .u5c18d649bd00217f7ce892df4202e32f .centered-text { display: table; height: 80px; padding-left : 18px; top: 0; } .u5c18d649bd00217f7ce892df4202e32f .u5c18d649bd00217f7ce892df4202e32f-content { display: table-cell; margin: 0; padding: 0; padding-right: 108px; position: relative; vertical-align: middle; width: 100%; } .u5c18d649bd00217f7ce892df4202e32f:after { content: ""; display: block; clear: both; } READ: Alexander the great 3 EssayThis dating technique assumes that Carbon-14 has reached equilibrium. There is more Carbon-14 in our atmosphere today then there was at any timein the past. Thus, Carbon Dating is controversial. If theres more Carbon-14 in the atmosphere today than there was 50 years ago, then an animal thatdied 100 years ago would test at an artificially higher age. Many experiments have been done in attempts to change radioactivedecay rates, but these experiments have failed to produce any significantchanges. We have found that decay constants are the same at a temperatureof 2000 degrees Celsius or at a temperature of 186 degrees Celsius and arethe same in a vacuum or under pressure of several thousand atmospheres. Measurements of decay rates under differing gravitational and magneticfields also

Individual Investigation Essays

Individual Investigation   Essays Individual Investigation   Essay Individual Investigation   Essay Aim: The aim of this experiment is to answer the following question: What is the effect of temperature on the equilibrium constant for the hydrolysis of an ester? The reaction I aim to investigate is a reversible reaction where an ester (an organic compound with RCOOR group) is produced and. Esters are generally insoluble in water but are soluble in other solvents. Esters are formed in an esterification reaction where and carboxylic acid (RCOOH group) and an alcohol (ROH), react to form an ester. Below is the esterification reaction/ester hydrolysis reaction. Alcohol + Carboxylic acid Ester + Water The Equipment and apparatus I will be using in my investigation are as follows: * Safety Goggles * Test Tube x 5 * Test tube stoppers x 5 * Water Bath with thermostatic control * Test tube holder (suitable for use in water baths) * Burette * Clamp stand with clamp fixings * Safety Mat * Funnel * 250ml Beaker * 250ml Conical Flask * 10ml Measuring cylinder * Thermometer * White tile Safety goggles will be worn so that the risk of chemicals coming into contact with the eyes is lessened (see risk assessment). Test tubes are being used as the environment for the reaction to take place, they are glass so can be monitored and can be easily stored in a test tube holder in the water bath. Test tube stoppers will be used so that none of the water or ethanol can evaporate out of the tube, thus giving inaccurate results. Water baths will be used so that the temperature, can be maintained at different temperatures. Test tube holders will be used so that all the test tubes can be held safely without risk of tipping over. A burette will be used in the titration because it will less small amounts of liquid at a time, and also has an accurate scale up the side, so volumes will be easier to obtain. A Clamp stand will be used so that the burette is held in place above the solution of unknown concentration securely. A safety mat will be used so that any spillages will be kept off of the tables. A funnel will be used so that any pouring of liquids will be safer from the risk of spillage. Beakers of 250ml are used to contain liquid yet will have lips on them to aid pouring. Conical flasks will be used in the titration to stop any of the liquid hitting the side. A 10ml measuring cylinder will be used so that the reactants can be measured accurately. A thermometer will be used so that the temperature of the water bath can be verified. Risk assessment The chemicals I will be using are: * Hydrochloric acid (known concentration of 0.1moldm-3-150ml approx. * Sodium Hydroxide Solution (unknown concentration)- 150ml approx. * Sulphuric acid (unknown concentration)- 150ml approx. * Ethanol (95% concentration with 5% water)-500ml approx. * Ethanoic Acid (Glacial Acetic Acid 100%)-500ml approx. * Methyl Orange * Phenolphthalein The Hydrochloric acid I am using is corrosive, so if it comes into contact with skin it will burn. Therefore when I am transferring the chemical from the sealed jar to the burette, I will ensure that safety goggles are on; so that any spillage will not come into contact with the eyes. I will also ensure that I am using a funnel on the burette. This is so that spillages will be less likely. Also when cleaning I will make sure I am using excess water to clean anything where the acid has contacted because it may react violently with water in small quantities. Similarly the sulphuric acid I am using is corrosive, therefore I must be careful when adding it to the test tube in the reaction, to prevent overfilling and spillages. The Sodium Hydroxide I am using is also corrosive, so when pouring into beakers and the burette, I will use a funnel and wear safety glasses. The ethanol I am using is highly flammable so I will keep it away from any naked flames. It is also very toxic if ingested, therefore I must make sure that it will be kept well away from the face at all times. As the ethanol is used in the reaction it I will make sure it is measured out accurately and that no spillages take place. Ethanoic acid is its concentrated form is also corrosive and will cause burns if it comes into contact with the skin or eyes. There I will be wearing safety goggles when using this chemical to add to the test tube or when it is used in the titration. Phenolphthalein is very harmful even in small quantities, it causes water imbalance and is a strong laxative. Therefore this substance must be kept off of tables, hands or any equipment that is likely to be handled. When doing a acid-base titration, the reaction is exothermic so therefore I must make sure nobody touches the beaker when the two chemicals are mixing. Also the water baths will be set so that there is an element heating the water. This must not be touched as it can cause burns. Also the water may too hot to touch so, the water must be touched as little as possible. Another reason for not touching the water too much is that if water gets onto the floor there is a risk of people falling over; therefore this risk should be minimized Variables Although these do not affect the value of the reaction constant I will still aim to control some variables. The variables I am aiming to control are that of concentration and that of total pressure. As all reactants and products are in the liquid phase, keeping the total pressure constant will be just the atmospheric pressure. To control the variable of concentration I will measure accurately different reactants each time so that the same concentration and amounts will be obtained each experiment. I will also be controlling the chain length of the acid and alcohol I use in the reaction. If I change the chain length of the reactants different products are going to be produced and may affect the equilibrium constant of the reaction. For this reason I will use ethanol and ethanoic acid, as they are easily obtainable and they produce an ester that is also a colourless liquid. I will be varying temperature as feel this is feasible to do, by using a water bath. Also because concentration does not actually affect the value for Kc, it will just shift the equilibrium position, away from the part of the reaction with the highest concentrations. The temperature ranges I will be using for this experiment will be in divisions of five degrees Kelvin from 293K, 298K, 303K , 308K , 313K , 318K , 323K. In terms of degrees Celsius this equal to 20oC, 25oC, 30oC, 35oC, 40oC, 45oC, 50oC. I am using this range because I feel I will be able to notice a marked difference in the equilibrium constant. Also I do not want to use any value of temperature above 60oC as above this temperature ethanol will boil, and will be in the gaseous state, separated from this rest of the reaction (condensed on the sides of the test tube) and therefore unable to take part in it. Also if I go above 100oC then the water would boil and the water bath would then not be as effective at keeping temperature constant, as well as damaging the heating element. Other ways I will ensure that I will make my experiment fair, accurate and able to produce repeatable data will now be explained. 🙂 I will make sure that the water bath I am using is set to that temperature for about 15minutes before the experiment, so that the water is an even temperature throughout. 🙂 I will also store the chemicals that I will be using in the water bath so that the reaction starts at the desired temperature, and does not have to be raised to that temperature. 🙂 I will use sterilised equipment throughout so that there is no chance of contamination, as this could affect the reaction. 🙂 I will also use the same equipment all the way through to ensure that no other contaminants can enter the mixtures. After each experiment I will clean out all equipment with both distilled water and the substance that is to be placed in the equipment after cleaning. 🙂 I will always use the same initial volumes and concentration of the alcohol and carboxylic acid. This is so that the 🙂 I will also use the same amount of concentrated sulphuric acid catalyst for each experiment. This amount will be four drops from a small pipette. This is so that the overall concentration of acid will be easier to work out if the drops are constant. 🙂 When titrating my sample I will make sure I use the same concentration of sodium Hydroxide solution. This amount will be 0.4molsdm-3. This will be done so that the same concentration will be used in calculations throughout, my analysis, and will make the spotting of patterns easier, if the sodium Hydroxide solution is of the same concentration all the way through. 🙂 I will use the same amount of phenolphthalein in the sample to measure the colour. This amount will be 6 drops from a small pipette. This will be done because if different amounts were added there may be a colour change at a different point to when I expect one, so making the amount of indicator constant should eliminate this. 🙂 When moving the sample I take from the reaction vessel to be titrated I will move it quickly as possible so that the temperature will not go down, if it does then that would change the amount of acid and make the experiment inaccurate. I will help to maintain accuracy by placing the titration apparatus close to the water bath. I will also have the titration equipment already set up, so the titration could be done immediately. 🙂 I will also graph my results as I get them. This way I can make sure that any patterns can be spotted early and any errors in the investigation will be made clear at an early stage. The thing I will be measuring is the amount of Sodium Hydroxide solution it takes to neutralise. The point at which I know that the solution is neutralised is the point when the phenolphthalein changes colour from colourless to pink. In order to find out this I will use a conical flask on top of a white tile. The colour will be deemed changed when I can no longer shake away the pink colour. This will indicate there are excess hydroxyl ions present in the solution. These ions are present in the Sodium Hydroxide solution and when they come into contact with the hydrogen ions in the sample solution they perform this reaction: H+(aq)+ OH- (aq) H2O(l) The indicator I am using is called phenolphthalein and is a organic indicator which I am using because it changes colour strongly at pH 8. I am using it in a 1% alcoholic solution. In acidic conditions the lactone ring in the phenolphthalein molecule is closed and the solution it is in is colourless. However when all the hydrogen ions are reacted with hydroxyl ions, the phenolphthalein is still colourless, as the hydroxyl ions are used up initially. However when there is one drop more of alkali in the solution the solution turns pink, as there is now an excess of hydroxyl ions. This causes the lactone ring to open, yielding the triphenylcarbinol structure. This structure is red/pink colour.1. 2. From the measurement of Sodium Hydroxide solution I obtain I can work out the final values of the acid, alcohol , water and ester present in the solution. Because the reaction below only has one mole of each reactant/product involved the values of final concentration are easily worked out. Alcohol + Carboxylic acid Ester + Water The final value of acid is equal to the titre obtained, as the titre is the amount that is needed to neutralise the acid in the solution. The amount of alcohol at the end, is the same as the amount of acid as one mole of acid is reacted with one mole of alcohol. If I use 20cm3 initially of acid and alkali, then as the amount of water and ester must be the same, as they have the same Stoichiometric values. The amount of ester and water is equal to the initial amount of alcohol minus the final amount of alcohol, as everything that has reacted must be ester. Therefore if I get a titre of 12 cm3 then the concentration is as follows: Acid: 12 cm3 From this I can work out the value of which I am investigating; the equilibrium constant. The equilibrium constant is worked out with this equation: Kc = Concentration products Concentration reactants In the case of this investigation the value of the equilibrium constant is found by this equation: Kc = [Ester] [Water] [Acid][Alcohol] In order to find the concentration I must first find the moles of each chemical in the reaction and to find that I need to know the moles of acid obtained. In order to find the moles I must multiply the volume of titre I obtain by the concentration of the sodium hydroxide I am using. As the reaction of hydrogen ions with hydroxyl ions is one to one the moles of alkali used is the same as the moles of acid used if the reaction mixture is now neutral. As the concentration of alkali in this case is 0.4moldm-3 to find the moles of acid used I multiply 0.012(this value is in decimetres cubed) by 0.4 this gives the answer 0.0048. As one mole of acid reacts with one mole of alcohol the moles of alcohol must also be the same. In order to find the moles of water and ester formed I must find the difference between the initial moles of acid and the final moles of acid. This is because any substance that is not acid or alcohol must be ester and water, so must account for the difference between i nitial acid value and final value. If my initial acid titre after time = 0 is 25 cm3 then the initial acid volume is: 0.025 x 0.4 = 0.01moles Now the difference between 0.01 and 0.0048 is equal to is 0.0052. Therefore the moles of ester and water formed is 0.0052moles. Now I can use these values to work out the equilibrium constant. As the equilibrium constant is found from concentrations and concentration is equal to moles/volume it can be written in this form: [moles/volume]1 [mole/volume]2 [moles/volume] 3[moles/volume]4 As the volume is the same for all of the chemicals it is common to all terms in the equation. I can then therefore cancel it, so that the equilibrium constant can be worked out using just the moles of each chemical. In the case of the example I provided the equilibrium constant would be with the moles of ester and water on top divided by the product of acid and alcohol. 0.0052 x 0.0052 0.0048 x 0.0048 = 1.17 3 S.F. As you can see the equilibrium constant in this case has no units. This is because the values of the units on the top of the equation cancel out those on the bottom to make no units. Investigation method 1. Obtain all equipment and chemicals needed for the experiment as listed previously. 2. In order for me to start the investigation I must make up a solution of 1 moldm-3 Sodium Hydroxide solution. This is for the titrations I will do later on in the investigation. I will do this by dissolving one mole of Caustic soda crystals into one litre of distilled water. One mole of Sodium Hydroxide: Na = 23 O = 16 H = 1 Total = 40 Therefore I will dissolve 40g of sodium Hydroxide in one litre of distilled water. I will weigh the crystals on a balance and stir the solution vigorously. As the enthalpy of solvation of NaOH is quite high, initially the solution will get quite hot, therefore I must exercise care when handling it. I will then put a bung on the one litre volumetric flask for use later in the experiment. 3. Fill the water bath up with water, and set water bath to the desired temperature. In the case of my preliminary work I will set it to 20oC and 60oC. I will place a thermometer in the bath to make sure the correct temperature is reached. 4. I will then decant 200cm3 of glacial acetic acid and ethanol into conical flasks, so that I do not contaminate the stock bottles. I will make sure that all equipment is thoroughly cleaned with distilled water and the substance that it will be containing. 5. I will use a filling pipette with a 20cm3 filling tube to put ethanol into a boiling tube. Then add four drops of concentrated sulphuric acid into the boiling tubes. 6. Then measure out using the pipette 20cm3 of ethanoic acid into the boiling tube. 7. At the moment the reaction mixes take a 5cm3 sample of the reaction mixture to be titrated immediately to see how much acid there is initially. Then place boiling tube bungs on all five of the boiling tubes containing the mixture and shake vigorously. Then place all in the test tube rack in the water bath. 8. The titration involves putting 50cm3 of NaOH solution into a burette initially. I then place the sample of reaction in a conical flask upon a white tile. I then add 6 drops of phenolphthalein into the sample. I then take a measurement of the starting value of the burette, and turn the tap so that the NaOH mixes with the sample. I will allow the NaOH to run slowly so I can get exactly when the colour changes. I will then measure the final titre and find the amount of solution used in neutralising the acid. 9. Take a sample again of the reaction mixture after two days and repeat the titration. 10. Take a third sample of the reaction mixture after three days and repeat the titration. If the titre of the third is the same as the one done after two days the reaction has reached equilibrium and the third titre can be used as data for that temperature. 11. Repeat these steps for all repeats and temperatures. Preliminary results The table below shows the results obtained for three repeats at three temperature intervals: 25oC, 45 oC, 60 oC Experiment (after 3 days) Temperature (Celsius) Start cm3 End cm3 Titre Repeat 1 25 0 9.3 9.3 Repeat 2 25 9.3 19.0 9.7 Repeat 3 25 19 28.8 9.8 Average 9.6 Repeat 1 45 0 10.8 10.8 Repeat 2 45 10.8 20.9 10.1 Repeat 3 45 20.9 32.0 11.1 Average 10.7 Repeat 1 60 0 12 12 Repeat 2 60 12 22 11 Repeat 3 60 22 34.3 12.3 Average 11.8 From these results I can see that there is a general upward trend in the titre as temperature. This does not seem to be highly correlated, but is still a relationship that can be useful in making my prediction. However I feel there is room for improvement which I will explain more below. Modifications to Method From my preliminary results and observations I noticed some flaws that could affect my investigation if they are done in the real experiments. They are listed below with ways to change them and make the experiment more accurate. * One of the main flaws with my experiment is that when using a bunged boiling tube, I had to shake the mixture to homogenise it. The problem this caused is that after a while the reaction mixture turned an orange tint. This was because the H+ in the ethanoic acid reacted with the rubber bung causing it to mix with the reaction. The problem this caused was that some of the acid was used up in reacting with the rubber therefore the correct amount was not available to react with the alcohol. The way I have chosen to overcome this is by changing the equipment holding the reaction. I will use glass 30cm3 sample tubes with screw-in plastic tops. This will make sure that all of the acid is available to react with the alcohol in the reaction. * The second problem I found is that the titres I found were rather small. This means that the alkali solution I was using in the titration was too strong. This is because it had too many hydroxyl ions in a given volume, and therefore a small volume neutralised the acid. A way to combat this is to change the concentration of NaOH in the alkali solution. I will use 0.4 moldm-3. The benefit that this has is too make the titres larger and therefore easier to spot trends and patterns. * Another problem I found when using the water bath is that at 60oC, after 3days much of the water in the water bath had evaporated. This was a bad thing because it meant the water bath could not maintain the correct temperature for the reaction and it can damage the water bath. A modification I am making is to take out 60oC from my temperature range. I will then therefore make the temperature range 25oC to 50oC in 5oC intervals. This will make sure I have enough water to maintain the correct water bath temperature. A second addition I can make is to use polystyrene foam and place it on top of the water bath to stop the water evaporating at the higher temperatures. This is because the foam will absorb some of the water and make a layer on top of the water bath to stop evaporation. * Another thing I found when looking at my chemicals was that the ethanol I used was 5% methanol. This could damage the experiment as the differing chain length could give rise to a different value for the Equilibrium constant than what I should actually get. A way to overcome this is too use an alcohol in a more pure form. Therefore I shall use propan-1-ol. This is bottled in near pure form so I believe it is more suitable for my investigation. Therefor the reaction will now be: Propan-1-ol + Ethanoic Acid Water + Propyl Ethanoate CH3CH2CH2OH + CH3COOH H2O + CH3CH2CH2COOCH3 * The last modification I will make to my investigation will be to change the amount of sample I take. I will take 1cm3 instead of 5cm3. This is because I will take 3 samples of each reaction, and I think a smaller sample will produce a titre that is under 50cm3. Final Method 1.Obtain all equipment and chemicals needed for the experiment as listed previously. 2.In order for me to start the investigation I must make up a solution of 1 moldm-3 Sodium Hydroxide solution. This is for the titrations I will do later on in the investigation. I will do this by dissolving one mole of Caustic soda crystals into one litre of distilled water. One mole of Sodium Hydroxide: Na = 23 O = 16 H = 1 Total = 40 x 0.4 moles = 16g Therefore I will dissolve 16g of sodium Hydroxide in one litre of distilled water. I will weigh the crystals on a balance and stir the solution vigorously. As the enthalpy of solvation of NaOH is quite high, initially the solution will get quite hot, therefore I must exercise care when handling it. I will then put a bung on the one litre volumetric flask for use later in the experiment. 3.Fill the water bath up with water and set water bath to the desired temperature. In this case I will set it to 25oC, 30oC, 35oC, 40oC, 45oC and 50oC. I will place a thermometer in the bath to make sure the correct temperature is reached. 4. I will then decant 200cm3 of glacial acetic acid and propan-1-ol into conical flasks, so that I do not contaminate the stock bottles. I will make sure that all equipment is thoroughly cleaned with distilled water and the substance that it will be containing. 5. I will use a filling pipette with a 20cm3 filling tube to put ethanol into a sample tube. Then add four drops of concentrated sulphuric acid into the sample tubes. Then measure out using the pipette 20cm3 of ethanoic acid into the sample tube 6. At the moment the reaction mixes take a 1cm3 sample of the reaction mixture to be titrated immediately to see how much acid there is initially. Then place sample tube lids on all five of the tubes containing the mixture and shake vigorously. Then place all directly into the water bath to make sure the sample tube is almost fully submerged. 7. The titration involves putting 50cm3 of NaOH solution into a burette initially. I then place the sample of reaction in a conical flask upon a white tile. I then add 6 drops of phenolphthalein into the sample. I then take a measurement of the starting value of the burette, and turn the tap so that the NaOH mixes with the sample. I will allow the NaOH to run slowly so I can discover exactly when the colour changes. I will then measure the final titre and find the amount of solution used in neutralising the acid. 8. Take a sample again of the reaction mixture after two days and repeat the titration. 9. Take a third sample of the reaction mixture after three days and repeat the titration. If the titre of the third is the same as the one done after two days the reaction has reached equilibrium and the third titre can be used as data for that temperature. 10. Repeat these steps for all repeats and temperatures The esterification reaction that I am performing is a rather slow reaction, therefore an acid catalyst is needed to make the reaction faster. Even with this catalyst the reaction still takes at least three days to reach equilibrium. Equilibrium is reached when the rate of the reaction going forwards is the same as the rate of reaction going backwards. All reactions are in a sense an equilibrium reaction, however some reactions have a small value for enthalpy change of reaction, so the reaction is possible to go in either direction. Another form of this reaction exists where the ester reacts with hydroxyl ions to form an alcohol and the carboxylic acids conjugate base. This reaction goes to completion however the ion produced is not easily measured so this reaction is preferred. Another way to get the reaction to reach equilibrium faster would be to reflux the mixture. However I can not do this as when the mixture cools down the equilibrium would shift again making the heating pointle ss. In the first step, the ethanoic acid takes a proton (a hydrogen ion) from the concentrated sulphuric acid. The proton becomes attached to one of the lone pairs on the oxygen that is double-bonded to the carbon. N.B. curly arrows in some diagrams represent the movement of one or a pair of electrons.4. It is quite misleading to show the positive charge just on the oxygen, because is actual fact the charge is delocalised over the entire double bond with the carbon. Therefore it is more accurate to show the molecules like this: These two structures are not complete representations of the ion as it can occurs in both ways. In this case they are resonant hybrids as both can be used to contribute to the actual structure. The positive charge of the ion attracts the lone pair of electrons on the ethanol and the ethanol gives up its lone pair of electrons to bond to the ion. From this ion a proton is removed from the bottom oxygen and placed onto the next oxygen up. This is not a direct transfer, first the proton is taken in by an unreacted ethanol on its lone pair of electrons, and then it acts as an intermediate to transfer the proton onto the middle oxygen. The overall proton transfer is shown below. Now a water molecule is lost from the ion From this ion the positive charge is now still spread over the carbon. The last hydrogen on the ion is removed by the hydrogensulphate ion that was created at the beginning from the sulphuric acid. Therefore the ester is made along with water and the sulphuric acid is not used up in the reaction, so is the catalyst. The hydrolysis of the ester is the reverse reaction, and will be sped up by the same amount with the use of sulphuric acid. Therefore the catalyst will not actually change the equilibrium constant, it will just help the reaction get to equilibrium quicker. The actual catalyst in this case is the hydroxonium ion (H3O+). The hydrolysis is started by the lone pair of electrons on the oxygen in the carbon-oxygen double bond accepts a proton from the hydroxonium ion. Once again the positive charge is delocalised throughout the double bond and can by draw in a number of resonance hybrids. The lone pair of electrons on the water molecule then attacks the positive charge on this carbon Now a proton on the bottom oxygen is transferred to another lone pair on a water molecule and is eventually placed on the middle oxygen. Now one of the products of the reaction, ethanol is removed from the ion. With the charge spread across the carbon a proton is removed from the ion and placed on a water molecule, restoring the hydroxonium ion, and producing ethanoic acid. In order to find out what way the equilibrium shifts when the temperature changes it is useful to consider Le Chatliers principle. This states that Whatever constraint is placed on a system, the system will move to oppose the change. This change can be adding more of one of the reactants, or changing the temperature or pressure of the system. Initially as there is no ester or water the system will move to produce more ester and water. In the case of temperature, if the temperature increases the system will move to remove the heat. This means that the system will make the two products that cause heat to be removed from the system. The way of the reaction that takes in heat is the endothermic root. To ascertain which way is endothermic I must draw an enthalpy cycle. An enthalpy cycle can be used to find out what the enthalpy change of reaction is by knowing what the enthalpy change of formation of all the reactants and products are. Hesss Law states that the enthalpy change of a react ion will be the same regardless of the intermediate steps taken. The enthalpy change of formation is the change in enthalpy when one mole of a molecule is formed from its component elements in their standard states and under standard conditions i.e. room temperature 298K and 1atm pressure. The enthalpy change of formation of each chemical is shown below: Water -285.8kJmol-1 Ethanoic Acid -484.5 kJmol-1 Propan-1-ol -302.7 kJmol-1 Propyl Ethanoate -502.7 kJmol-1 5. Notice that all these values are negative this means that energy is given out into the system as energy is given out when bonds are made This is the enthalpy cycle for my reaction: ?Hr? CH3COOH + CH3CH2CH2OH H2O + CH3COOCH2CH2CH3 (-484.5 +-302.7) (-285.8 + -502.7) 5C(s) + 3/2O2(g) + 6H2(g) In order to find the enthalpy change of reaction I must follow the arrows to get to the ester and water. Firstly as I start by going down the first arrow I must take the negative value of it. -484.5 302.7 = -787.2 This value becomes +787.2 as I am going against this arrow. I then add the sum of the esters and waters enthalpy changes of formation to get the enthalpy change of reaction. -285.8 + -502.7 = -788.5 787.2 788.5 = -1.3 kJmol-1 From this I can show that the reaction is exothermic going in the direction of making ester and water. As this value is small it also fits in that this reaction is an equilibrium reaction. From this I can now make a prediction. As an increase in temperature forces the equilibrium to shift to remove heat energy, and the reaction root that is endothermic (removing heat) is making acid and alcohol from ester and water, I predict that as I increase temperature the more acid and alcohol is produced. In terms of the value of Kc I predict that this value decreases as I increase temperature. This is because if more acid is produced it means less ester and water is produced; and as the value for Kc is found by the product concentration divided by reactant concentration, more reactant means lower Kc. 1. Vogels textbook of quantitative chemical analysis 2.chem.ox.ac.uk/vrchemistry/FilmStudio/phenolphthalein/Symbols/equilibrium.jpg 3.ilo.org/public/english/protection/safework/cis/products/icsc/dtasht/_icsc05/icsc0553.htm 4. chemguide.co.uk/physical/catalysis/hydrolyse.html 5. ucdsb.on.ca/tiss/stretton/chem2/data09f.html Below is a list of the main areas of chemistry in my investigation and where they occur in the course. Area Of Investigation Area of Chemistry AS/A2 Enthalpy Cycles Developing Fuels AS Hydrolysis of Esters Whats in a Medicine? AS Acid-Base titrations Whats in a Medicine? AS Equilibrium Constant Engineering Proteins A2 Results Tables The table below shows the readings at each temperature, proving that the reaction has reached equilibrium. The readings for each of the temperatures are taken after one day two three and four. If the third and forth reading are the same then the rate of reaction going forward is the same as the rate of reaction going backwards so there is no net change in the amounts of reactants/products. Temperature (Celsius) Start reading (cm3) End reading (cm3) Titre(cm3) Moles of acid Initial Reading 25 0 22.5 22.5 0.00900 Second Reading 25 0 15.7 15.7 0.00628 Third Reading 25 0 9.9 9.9 0.00396 Final Reading 25 0 9.9 9.9 0.00396 Temperature (OC) Start reading (cm3) End reading (cm3) Titre (cm3) Moles of acid Initial reading 30 0 22.5 22.5 0.00900 Second reading 30 22.5 37.7 15.2 0.00608 Third reading 30 37.7 47.8 10.1 0.00404 Final Reading 30 0 10.1 10.1 0.00404 Initial reading 35 0 22.5 22.5 0.00900 Second reading 35 22.6 37.9 15.3 0.00612 Third reading 35 0 12.3 12.3 0.00412 Forth Reading 35 0 10.3 10.3 0.00412 Final Reading 35 0 10.3 10.3 0.00412 Initial reading 40 0 22.4 22.5 0.00900 Second reading 40 22.4 37.5 15.1 0.00604 Third reading 40 37.5 48.0 10.5 0.00420 Final Reading 40 0 10.5 10.5 0.00420 Initial reading 45 0 22.5 22.5 0.00900 Second reading 45 22.3 37.2 14.9 0.00596 Third reading 45 37.2 47.9 10.7 0.00428 Final Reading 45 0 10.7 10.7 0.00428 Initial reading 50 0 22.5 22.5 0.00900 Second reading 50 22.5 37.5 15.0 0.00600 Third reading 50 37.5 48.4 10.9 0.00436 Final Reading 50 0 10.9 10.9 0.00436 N.B the Results highlighted in red shows that the reaction did not reach equilibrium as fast as all of the others. Although this did not affect the results obtained, it will still be accounted for later on in the evaluation section. Below shows graphs for all of the temperatures showing time from the beginning of reaction to the reaching of equilibrium. The table of results shows all of the repeats I did for all of the temperature ranges. It also shows that moles of acid in the sample. Temperature (degrees Celsius) Beginning measurement End measurement Titre Amount of Acid Moles of Acid 25 0 9.8 9.8 9.8 0.00392 25 9.8 19.7 9.9 9.9 0.00396 25 19.7 29.5 9.8 9.8 0.00392 25 29.5 39.4 9.9 9.9 0.00396 25 0 9.8 9.9 9.9 0.00396 Average titre 9.9 9.9 0.00396 30 0 10.1 10.1 10.1 0.00404 30 10.1 20.2 10.1 10.1 0.00404 30 20.2 30.3 10.1 10.1 0.00404 30 30.3 40.4 10.1 10.1 0.00404 30 0 10.1 10.1 10.1 0.00404 Average titre 10.1 10.1 0.00404 35 0 10.3 10.3 10.3 0.00412 35 10.3 20.6 10.3 10.3 0.00412 35 20.6 30.9 10.3 10.3 0.00412 35 30.9 41.2 10.3 10.3 0.00412 35 0 10.3 10.3 10.3 0.00412 Average Titre 10.3 10.3 0.00412 40 0 10.4 10.4 10.4 0.00416 40 10.4 10.9 10.5 10.5 0.0042 40 20.9 31.5 10.6 10.6 0.00424 40 31.5 42 10.5 10.5 0.0042 40 0 10.5 10.5 10.5 0.0042 Average Titre 10.5 10.5 0.0042 45 0 10.7 10.7 10.7 0.00428 45 10.7 21.4 10.7 10.7 0.00428 45 21.4 32.1 10.7 10.7 0.00428 45 32.1 42.8 10.7 10.7 0.00428 45 0 10.7 10.7 10.7 0.00428 Average titre 10.7 10.7 0.00428 50 0 7.6 7.6 7.6 0.00304 50 7.6 18.6 11 11 0.0044 50 18.6 29.5 10.9 10.9 0.00436 50 29.5 40.4 10.9 10.9 0.00436 50 0 10.9 10.9 10.9 0.00436 N.b.Anaomalous result ignored Average titre 10.9 10.9 0.00436 Analysis of results Now for each repeat for each temperature I will work out the moles of each in the reaction, and hence the equilibrium constant for each repeat. 25oC- Repeat 1 Titre initial = 22.5cm Initial moles = volume x concentration of alkali Initial Moles of acid = 0.00900 Volume of Ethanoic acid in 1cm3 at the end = 9.8 Final moles = volume of acid x concentration of alkali Final moles of acid = 0.00392 As one mole of acid reacts with one mole of alcohol then the amount of alcohol must be the same. Final moles of Propan-1-ol = 0.00392 Final moles of Propyl Ethanoate equals difference between initial acid moles and final acid moles. 0.00900 0.00392 = 0.00508 As one mole of Ester makes one mole of water then the moles of water must be exactly the same as that of ester. From these values the value of the equilibrium constant can be worked out: Kc = 0.00508 x 0.00508 0.00392 x 0.00392 = 1.68 3 S.F. Applying this same method to all my other data I will now produce a table showing each repeat and the equilibrium constant of that experiment. I have recorded this table because I feel it makes it easier for me to draw a graph of the equilibrium constant against the temperature. Temperature (degrees Celsius) Initial Moles of Acid Moles of Acid Equilibrium Constant 25 0.00900 0.00392 1.68 25 0.00900 0.00396 1.62 25 0.00900 0.00392 1.68 25 0.00900 0.00396 1.62 25 0.00900 0.00396 1.62 Average 0.00900 0.00396 1.62 30 0.00900 0.00404 1.51 30 0.00900 0.00404 1.51 30 0.00900 0.00404 1.51 30 0.00900 0.00404 1.51 30 0.00900 0.00404 1.51 Average 0.00900 0.00404 1.51 35 0.00900 0.00412 1.40 35 0.00900 0.00412 1.40 35 0.00900 0.00412 1.40 35 0.00900 0.00412 1.40 35 0.00900 0.00412 1.40 Average 0.00900 0.00412 1.40 40 0.00900 0.00416 1.35 40 0.00900 0.0042 1.31 40 0.00900 0.00424 1.26 40 0.00900 0.0042 1.31 40 0.00900 0.0042 1.31 Average 0.00900 0.0042 1.31 45 0.00900 0.00428 1.22 45 0.00900 0.00428 1.22 45 0.00900 0.00428 1.22 45 0.00900 0.00428 1.22 45 0.00900 0.00428 1.22 Average 0.00900 0.00428 1.22 50 0.00900 0.00304 3.84 50 0.00900 0.0044 1.09 50 0.00900 0.00436 1.13 50 0.00900 0.00436 1.13 50 0.00900 0.00436 1.13 Average 0.00900 0.00436 1.13 Anomalous Result is ignored From this graph I can deduce that as the Temperature increased the value of Kc decreases. As I said this in my prediction, it goes some way to proving that my prediction was at least to some extent correct. Another conclusion I can draw from my graphs is that as the temperature increased the equilibrium constant changed exponentially. This is because on the graph it shows a slight curve. The reason for this is that as the temperature rises the reaction favours the production of the reactants. Therefore as the reactants increase the products decrease so the equilibrium constant goes down. There is a constant fractional change in the value of Kc as the temperature increases. The reason that the reaction favours the production of the reactants, ethanoic acid and propan-1-ol is because this direction of reaction takes in energy, i.e. its energy level is higher than the products that it makes. As the temperature increases the system tries to remove the extra heat energy being put in. The way it does this is by making the reaction that removes heat energy go faster. Therefore the reaction constant goes down..

Individual Investigation Essays

Individual Investigation   Essays Individual Investigation   Essay Individual Investigation   Essay Aim: The aim of this experiment is to answer the following question: What is the effect of temperature on the equilibrium constant for the hydrolysis of an ester? The reaction I aim to investigate is a reversible reaction where an ester (an organic compound with RCOOR group) is produced and. Esters are generally insoluble in water but are soluble in other solvents. Esters are formed in an esterification reaction where and carboxylic acid (RCOOH group) and an alcohol (ROH), react to form an ester. Below is the esterification reaction/ester hydrolysis reaction. Alcohol + Carboxylic acid Ester + Water The Equipment and apparatus I will be using in my investigation are as follows: * Safety Goggles * Test Tube x 5 * Test tube stoppers x 5 * Water Bath with thermostatic control * Test tube holder (suitable for use in water baths) * Burette * Clamp stand with clamp fixings * Safety Mat * Funnel * 250ml Beaker * 250ml Conical Flask * 10ml Measuring cylinder * Thermometer * White tile Safety goggles will be worn so that the risk of chemicals coming into contact with the eyes is lessened (see risk assessment). Test tubes are being used as the environment for the reaction to take place, they are glass so can be monitored and can be easily stored in a test tube holder in the water bath. Test tube stoppers will be used so that none of the water or ethanol can evaporate out of the tube, thus giving inaccurate results. Water baths will be used so that the temperature, can be maintained at different temperatures. Test tube holders will be used so that all the test tubes can be held safely without risk of tipping over. A burette will be used in the titration because it will less small amounts of liquid at a time, and also has an accurate scale up the side, so volumes will be easier to obtain. A Clamp stand will be used so that the burette is held in place above the solution of unknown concentration securely. A safety mat will be used so that any spillages will be kept off of the tables. A funnel will be used so that any pouring of liquids will be safer from the risk of spillage. Beakers of 250ml are used to contain liquid yet will have lips on them to aid pouring. Conical flasks will be used in the titration to stop any of the liquid hitting the side. A 10ml measuring cylinder will be used so that the reactants can be measured accurately. A thermometer will be used so that the temperature of the water bath can be verified. Risk assessment The chemicals I will be using are: * Hydrochloric acid (known concentration of 0.1moldm-3-150ml approx. * Sodium Hydroxide Solution (unknown concentration)- 150ml approx. * Sulphuric acid (unknown concentration)- 150ml approx. * Ethanol (95% concentration with 5% water)-500ml approx. * Ethanoic Acid (Glacial Acetic Acid 100%)-500ml approx. * Methyl Orange * Phenolphthalein The Hydrochloric acid I am using is corrosive, so if it comes into contact with skin it will burn. Therefore when I am transferring the chemical from the sealed jar to the burette, I will ensure that safety goggles are on; so that any spillage will not come into contact with the eyes. I will also ensure that I am using a funnel on the burette. This is so that spillages will be less likely. Also when cleaning I will make sure I am using excess water to clean anything where the acid has contacted because it may react violently with water in small quantities. Similarly the sulphuric acid I am using is corrosive, therefore I must be careful when adding it to the test tube in the reaction, to prevent overfilling and spillages. The Sodium Hydroxide I am using is also corrosive, so when pouring into beakers and the burette, I will use a funnel and wear safety glasses. The ethanol I am using is highly flammable so I will keep it away from any naked flames. It is also very toxic if ingested, therefore I must make sure that it will be kept well away from the face at all times. As the ethanol is used in the reaction it I will make sure it is measured out accurately and that no spillages take place. Ethanoic acid is its concentrated form is also corrosive and will cause burns if it comes into contact with the skin or eyes. There I will be wearing safety goggles when using this chemical to add to the test tube or when it is used in the titration. Phenolphthalein is very harmful even in small quantities, it causes water imbalance and is a strong laxative. Therefore this substance must be kept off of tables, hands or any equipment that is likely to be handled. When doing a acid-base titration, the reaction is exothermic so therefore I must make sure nobody touches the beaker when the two chemicals are mixing. Also the water baths will be set so that there is an element heating the water. This must not be touched as it can cause burns. Also the water may too hot to touch so, the water must be touched as little as possible. Another reason for not touching the water too much is that if water gets onto the floor there is a risk of people falling over; therefore this risk should be minimized Variables Although these do not affect the value of the reaction constant I will still aim to control some variables. The variables I am aiming to control are that of concentration and that of total pressure. As all reactants and products are in the liquid phase, keeping the total pressure constant will be just the atmospheric pressure. To control the variable of concentration I will measure accurately different reactants each time so that the same concentration and amounts will be obtained each experiment. I will also be controlling the chain length of the acid and alcohol I use in the reaction. If I change the chain length of the reactants different products are going to be produced and may affect the equilibrium constant of the reaction. For this reason I will use ethanol and ethanoic acid, as they are easily obtainable and they produce an ester that is also a colourless liquid. I will be varying temperature as feel this is feasible to do, by using a water bath. Also because concentration does not actually affect the value for Kc, it will just shift the equilibrium position, away from the part of the reaction with the highest concentrations. The temperature ranges I will be using for this experiment will be in divisions of five degrees Kelvin from 293K, 298K, 303K , 308K , 313K , 318K , 323K. In terms of degrees Celsius this equal to 20oC, 25oC, 30oC, 35oC, 40oC, 45oC, 50oC. I am using this range because I feel I will be able to notice a marked difference in the equilibrium constant. Also I do not want to use any value of temperature above 60oC as above this temperature ethanol will boil, and will be in the gaseous state, separated from this rest of the reaction (condensed on the sides of the test tube) and therefore unable to take part in it. Also if I go above 100oC then the water would boil and the water bath would then not be as effective at keeping temperature constant, as well as damaging the heating element. Other ways I will ensure that I will make my experiment fair, accurate and able to produce repeatable data will now be explained. 🙂 I will make sure that the water bath I am using is set to that temperature for about 15minutes before the experiment, so that the water is an even temperature throughout. 🙂 I will also store the chemicals that I will be using in the water bath so that the reaction starts at the desired temperature, and does not have to be raised to that temperature. 🙂 I will use sterilised equipment throughout so that there is no chance of contamination, as this could affect the reaction. 🙂 I will also use the same equipment all the way through to ensure that no other contaminants can enter the mixtures. After each experiment I will clean out all equipment with both distilled water and the substance that is to be placed in the equipment after cleaning. 🙂 I will always use the same initial volumes and concentration of the alcohol and carboxylic acid. This is so that the 🙂 I will also use the same amount of concentrated sulphuric acid catalyst for each experiment. This amount will be four drops from a small pipette. This is so that the overall concentration of acid will be easier to work out if the drops are constant. 🙂 When titrating my sample I will make sure I use the same concentration of sodium Hydroxide solution. This amount will be 0.4molsdm-3. This will be done so that the same concentration will be used in calculations throughout, my analysis, and will make the spotting of patterns easier, if the sodium Hydroxide solution is of the same concentration all the way through. 🙂 I will use the same amount of phenolphthalein in the sample to measure the colour. This amount will be 6 drops from a small pipette. This will be done because if different amounts were added there may be a colour change at a different point to when I expect one, so making the amount of indicator constant should eliminate this. 🙂 When moving the sample I take from the reaction vessel to be titrated I will move it quickly as possible so that the temperature will not go down, if it does then that would change the amount of acid and make the experiment inaccurate. I will help to maintain accuracy by placing the titration apparatus close to the water bath. I will also have the titration equipment already set up, so the titration could be done immediately. 🙂 I will also graph my results as I get them. This way I can make sure that any patterns can be spotted early and any errors in the investigation will be made clear at an early stage. The thing I will be measuring is the amount of Sodium Hydroxide solution it takes to neutralise. The point at which I know that the solution is neutralised is the point when the phenolphthalein changes colour from colourless to pink. In order to find out this I will use a conical flask on top of a white tile. The colour will be deemed changed when I can no longer shake away the pink colour. This will indicate there are excess hydroxyl ions present in the solution. These ions are present in the Sodium Hydroxide solution and when they come into contact with the hydrogen ions in the sample solution they perform this reaction: H+(aq)+ OH- (aq) H2O(l) The indicator I am using is called phenolphthalein and is a organic indicator which I am using because it changes colour strongly at pH 8. I am using it in a 1% alcoholic solution. In acidic conditions the lactone ring in the phenolphthalein molecule is closed and the solution it is in is colourless. However when all the hydrogen ions are reacted with hydroxyl ions, the phenolphthalein is still colourless, as the hydroxyl ions are used up initially. However when there is one drop more of alkali in the solution the solution turns pink, as there is now an excess of hydroxyl ions. This causes the lactone ring to open, yielding the triphenylcarbinol structure. This structure is red/pink colour.1. 2. From the measurement of Sodium Hydroxide solution I obtain I can work out the final values of the acid, alcohol , water and ester present in the solution. Because the reaction below only has one mole of each reactant/product involved the values of final concentration are easily worked out. Alcohol + Carboxylic acid Ester + Water The final value of acid is equal to the titre obtained, as the titre is the amount that is needed to neutralise the acid in the solution. The amount of alcohol at the end, is the same as the amount of acid as one mole of acid is reacted with one mole of alcohol. If I use 20cm3 initially of acid and alkali, then as the amount of water and ester must be the same, as they have the same Stoichiometric values. The amount of ester and water is equal to the initial amount of alcohol minus the final amount of alcohol, as everything that has reacted must be ester. Therefore if I get a titre of 12 cm3 then the concentration is as follows: Acid: 12 cm3 From this I can work out the value of which I am investigating; the equilibrium constant. The equilibrium constant is worked out with this equation: Kc = Concentration products Concentration reactants In the case of this investigation the value of the equilibrium constant is found by this equation: Kc = [Ester] [Water] [Acid][Alcohol] In order to find the concentration I must first find the moles of each chemical in the reaction and to find that I need to know the moles of acid obtained. In order to find the moles I must multiply the volume of titre I obtain by the concentration of the sodium hydroxide I am using. As the reaction of hydrogen ions with hydroxyl ions is one to one the moles of alkali used is the same as the moles of acid used if the reaction mixture is now neutral. As the concentration of alkali in this case is 0.4moldm-3 to find the moles of acid used I multiply 0.012(this value is in decimetres cubed) by 0.4 this gives the answer 0.0048. As one mole of acid reacts with one mole of alcohol the moles of alcohol must also be the same. In order to find the moles of water and ester formed I must find the difference between the initial moles of acid and the final moles of acid. This is because any substance that is not acid or alcohol must be ester and water, so must account for the difference between i nitial acid value and final value. If my initial acid titre after time = 0 is 25 cm3 then the initial acid volume is: 0.025 x 0.4 = 0.01moles Now the difference between 0.01 and 0.0048 is equal to is 0.0052. Therefore the moles of ester and water formed is 0.0052moles. Now I can use these values to work out the equilibrium constant. As the equilibrium constant is found from concentrations and concentration is equal to moles/volume it can be written in this form: [moles/volume]1 [mole/volume]2 [moles/volume] 3[moles/volume]4 As the volume is the same for all of the chemicals it is common to all terms in the equation. I can then therefore cancel it, so that the equilibrium constant can be worked out using just the moles of each chemical. In the case of the example I provided the equilibrium constant would be with the moles of ester and water on top divided by the product of acid and alcohol. 0.0052 x 0.0052 0.0048 x 0.0048 = 1.17 3 S.F. As you can see the equilibrium constant in this case has no units. This is because the values of the units on the top of the equation cancel out those on the bottom to make no units. Investigation method 1. Obtain all equipment and chemicals needed for the experiment as listed previously. 2. In order for me to start the investigation I must make up a solution of 1 moldm-3 Sodium Hydroxide solution. This is for the titrations I will do later on in the investigation. I will do this by dissolving one mole of Caustic soda crystals into one litre of distilled water. One mole of Sodium Hydroxide: Na = 23 O = 16 H = 1 Total = 40 Therefore I will dissolve 40g of sodium Hydroxide in one litre of distilled water. I will weigh the crystals on a balance and stir the solution vigorously. As the enthalpy of solvation of NaOH is quite high, initially the solution will get quite hot, therefore I must exercise care when handling it. I will then put a bung on the one litre volumetric flask for use later in the experiment. 3. Fill the water bath up with water, and set water bath to the desired temperature. In the case of my preliminary work I will set it to 20oC and 60oC. I will place a thermometer in the bath to make sure the correct temperature is reached. 4. I will then decant 200cm3 of glacial acetic acid and ethanol into conical flasks, so that I do not contaminate the stock bottles. I will make sure that all equipment is thoroughly cleaned with distilled water and the substance that it will be containing. 5. I will use a filling pipette with a 20cm3 filling tube to put ethanol into a boiling tube. Then add four drops of concentrated sulphuric acid into the boiling tubes. 6. Then measure out using the pipette 20cm3 of ethanoic acid into the boiling tube. 7. At the moment the reaction mixes take a 5cm3 sample of the reaction mixture to be titrated immediately to see how much acid there is initially. Then place boiling tube bungs on all five of the boiling tubes containing the mixture and shake vigorously. Then place all in the test tube rack in the water bath. 8. The titration involves putting 50cm3 of NaOH solution into a burette initially. I then place the sample of reaction in a conical flask upon a white tile. I then add 6 drops of phenolphthalein into the sample. I then take a measurement of the starting value of the burette, and turn the tap so that the NaOH mixes with the sample. I will allow the NaOH to run slowly so I can get exactly when the colour changes. I will then measure the final titre and find the amount of solution used in neutralising the acid. 9. Take a sample again of the reaction mixture after two days and repeat the titration. 10. Take a third sample of the reaction mixture after three days and repeat the titration. If the titre of the third is the same as the one done after two days the reaction has reached equilibrium and the third titre can be used as data for that temperature. 11. Repeat these steps for all repeats and temperatures. Preliminary results The table below shows the results obtained for three repeats at three temperature intervals: 25oC, 45 oC, 60 oC Experiment (after 3 days) Temperature (Celsius) Start cm3 End cm3 Titre Repeat 1 25 0 9.3 9.3 Repeat 2 25 9.3 19.0 9.7 Repeat 3 25 19 28.8 9.8 Average 9.6 Repeat 1 45 0 10.8 10.8 Repeat 2 45 10.8 20.9 10.1 Repeat 3 45 20.9 32.0 11.1 Average 10.7 Repeat 1 60 0 12 12 Repeat 2 60 12 22 11 Repeat 3 60 22 34.3 12.3 Average 11.8 From these results I can see that there is a general upward trend in the titre as temperature. This does not seem to be highly correlated, but is still a relationship that can be useful in making my prediction. However I feel there is room for improvement which I will explain more below. Modifications to Method From my preliminary results and observations I noticed some flaws that could affect my investigation if they are done in the real experiments. They are listed below with ways to change them and make the experiment more accurate. * One of the main flaws with my experiment is that when using a bunged boiling tube, I had to shake the mixture to homogenise it. The problem this caused is that after a while the reaction mixture turned an orange tint. This was because the H+ in the ethanoic acid reacted with the rubber bung causing it to mix with the reaction. The problem this caused was that some of the acid was used up in reacting with the rubber therefore the correct amount was not available to react with the alcohol. The way I have chosen to overcome this is by changing the equipment holding the reaction. I will use glass 30cm3 sample tubes with screw-in plastic tops. This will make sure that all of the acid is available to react with the alcohol in the reaction. * The second problem I found is that the titres I found were rather small. This means that the alkali solution I was using in the titration was too strong. This is because it had too many hydroxyl ions in a given volume, and therefore a small volume neutralised the acid. A way to combat this is to change the concentration of NaOH in the alkali solution. I will use 0.4 moldm-3. The benefit that this has is too make the titres larger and therefore easier to spot trends and patterns. * Another problem I found when using the water bath is that at 60oC, after 3days much of the water in the water bath had evaporated. This was a bad thing because it meant the water bath could not maintain the correct temperature for the reaction and it can damage the water bath. A modification I am making is to take out 60oC from my temperature range. I will then therefore make the temperature range 25oC to 50oC in 5oC intervals. This will make sure I have enough water to maintain the correct water bath temperature. A second addition I can make is to use polystyrene foam and place it on top of the water bath to stop the water evaporating at the higher temperatures. This is because the foam will absorb some of the water and make a layer on top of the water bath to stop evaporation. * Another thing I found when looking at my chemicals was that the ethanol I used was 5% methanol. This could damage the experiment as the differing chain length could give rise to a different value for the Equilibrium constant than what I should actually get. A way to overcome this is too use an alcohol in a more pure form. Therefore I shall use propan-1-ol. This is bottled in near pure form so I believe it is more suitable for my investigation. Therefor the reaction will now be: Propan-1-ol + Ethanoic Acid Water + Propyl Ethanoate CH3CH2CH2OH + CH3COOH H2O + CH3CH2CH2COOCH3 * The last modification I will make to my investigation will be to change the amount of sample I take. I will take 1cm3 instead of 5cm3. This is because I will take 3 samples of each reaction, and I think a smaller sample will produce a titre that is under 50cm3. Final Method 1.Obtain all equipment and chemicals needed for the experiment as listed previously. 2.In order for me to start the investigation I must make up a solution of 1 moldm-3 Sodium Hydroxide solution. This is for the titrations I will do later on in the investigation. I will do this by dissolving one mole of Caustic soda crystals into one litre of distilled water. One mole of Sodium Hydroxide: Na = 23 O = 16 H = 1 Total = 40 x 0.4 moles = 16g Therefore I will dissolve 16g of sodium Hydroxide in one litre of distilled water. I will weigh the crystals on a balance and stir the solution vigorously. As the enthalpy of solvation of NaOH is quite high, initially the solution will get quite hot, therefore I must exercise care when handling it. I will then put a bung on the one litre volumetric flask for use later in the experiment. 3.Fill the water bath up with water and set water bath to the desired temperature. In this case I will set it to 25oC, 30oC, 35oC, 40oC, 45oC and 50oC. I will place a thermometer in the bath to make sure the correct temperature is reached. 4. I will then decant 200cm3 of glacial acetic acid and propan-1-ol into conical flasks, so that I do not contaminate the stock bottles. I will make sure that all equipment is thoroughly cleaned with distilled water and the substance that it will be containing. 5. I will use a filling pipette with a 20cm3 filling tube to put ethanol into a sample tube. Then add four drops of concentrated sulphuric acid into the sample tubes. Then measure out using the pipette 20cm3 of ethanoic acid into the sample tube 6. At the moment the reaction mixes take a 1cm3 sample of the reaction mixture to be titrated immediately to see how much acid there is initially. Then place sample tube lids on all five of the tubes containing the mixture and shake vigorously. Then place all directly into the water bath to make sure the sample tube is almost fully submerged. 7. The titration involves putting 50cm3 of NaOH solution into a burette initially. I then place the sample of reaction in a conical flask upon a white tile. I then add 6 drops of phenolphthalein into the sample. I then take a measurement of the starting value of the burette, and turn the tap so that the NaOH mixes with the sample. I will allow the NaOH to run slowly so I can discover exactly when the colour changes. I will then measure the final titre and find the amount of solution used in neutralising the acid. 8. Take a sample again of the reaction mixture after two days and repeat the titration. 9. Take a third sample of the reaction mixture after three days and repeat the titration. If the titre of the third is the same as the one done after two days the reaction has reached equilibrium and the third titre can be used as data for that temperature. 10. Repeat these steps for all repeats and temperatures The esterification reaction that I am performing is a rather slow reaction, therefore an acid catalyst is needed to make the reaction faster. Even with this catalyst the reaction still takes at least three days to reach equilibrium. Equilibrium is reached when the rate of the reaction going forwards is the same as the rate of reaction going backwards. All reactions are in a sense an equilibrium reaction, however some reactions have a small value for enthalpy change of reaction, so the reaction is possible to go in either direction. Another form of this reaction exists where the ester reacts with hydroxyl ions to form an alcohol and the carboxylic acids conjugate base. This reaction goes to completion however the ion produced is not easily measured so this reaction is preferred. Another way to get the reaction to reach equilibrium faster would be to reflux the mixture. However I can not do this as when the mixture cools down the equilibrium would shift again making the heating pointle ss. In the first step, the ethanoic acid takes a proton (a hydrogen ion) from the concentrated sulphuric acid. The proton becomes attached to one of the lone pairs on the oxygen that is double-bonded to the carbon. N.B. curly arrows in some diagrams represent the movement of one or a pair of electrons.4. It is quite misleading to show the positive charge just on the oxygen, because is actual fact the charge is delocalised over the entire double bond with the carbon. Therefore it is more accurate to show the molecules like this: These two structures are not complete representations of the ion as it can occurs in both ways. In this case they are resonant hybrids as both can be used to contribute to the actual structure. The positive charge of the ion attracts the lone pair of electrons on the ethanol and the ethanol gives up its lone pair of electrons to bond to the ion. From this ion a proton is removed from the bottom oxygen and placed onto the next oxygen up. This is not a direct transfer, first the proton is taken in by an unreacted ethanol on its lone pair of electrons, and then it acts as an intermediate to transfer the proton onto the middle oxygen. The overall proton transfer is shown below. Now a water molecule is lost from the ion From this ion the positive charge is now still spread over the carbon. The last hydrogen on the ion is removed by the hydrogensulphate ion that was created at the beginning from the sulphuric acid. Therefore the ester is made along with water and the sulphuric acid is not used up in the reaction, so is the catalyst. The hydrolysis of the ester is the reverse reaction, and will be sped up by the same amount with the use of sulphuric acid. Therefore the catalyst will not actually change the equilibrium constant, it will just help the reaction get to equilibrium quicker. The actual catalyst in this case is the hydroxonium ion (H3O+). The hydrolysis is started by the lone pair of electrons on the oxygen in the carbon-oxygen double bond accepts a proton from the hydroxonium ion. Once again the positive charge is delocalised throughout the double bond and can by draw in a number of resonance hybrids. The lone pair of electrons on the water molecule then attacks the positive charge on this carbon Now a proton on the bottom oxygen is transferred to another lone pair on a water molecule and is eventually placed on the middle oxygen. Now one of the products of the reaction, ethanol is removed from the ion. With the charge spread across the carbon a proton is removed from the ion and placed on a water molecule, restoring the hydroxonium ion, and producing ethanoic acid. In order to find out what way the equilibrium shifts when the temperature changes it is useful to consider Le Chatliers principle. This states that Whatever constraint is placed on a system, the system will move to oppose the change. This change can be adding more of one of the reactants, or changing the temperature or pressure of the system. Initially as there is no ester or water the system will move to produce more ester and water. In the case of temperature, if the temperature increases the system will move to remove the heat. This means that the system will make the two products that cause heat to be removed from the system. The way of the reaction that takes in heat is the endothermic root. To ascertain which way is endothermic I must draw an enthalpy cycle. An enthalpy cycle can be used to find out what the enthalpy change of reaction is by knowing what the enthalpy change of formation of all the reactants and products are. Hesss Law states that the enthalpy change of a react ion will be the same regardless of the intermediate steps taken. The enthalpy change of formation is the change in enthalpy when one mole of a molecule is formed from its component elements in their standard states and under standard conditions i.e. room temperature 298K and 1atm pressure. The enthalpy change of formation of each chemical is shown below: Water -285.8kJmol-1 Ethanoic Acid -484.5 kJmol-1 Propan-1-ol -302.7 kJmol-1 Propyl Ethanoate -502.7 kJmol-1 5. Notice that all these values are negative this means that energy is given out into the system as energy is given out when bonds are made This is the enthalpy cycle for my reaction: ?Hr? CH3COOH + CH3CH2CH2OH H2O + CH3COOCH2CH2CH3 (-484.5 +-302.7) (-285.8 + -502.7) 5C(s) + 3/2O2(g) + 6H2(g) In order to find the enthalpy change of reaction I must follow the arrows to get to the ester and water. Firstly as I start by going down the first arrow I must take the negative value of it. -484.5 302.7 = -787.2 This value becomes +787.2 as I am going against this arrow. I then add the sum of the esters and waters enthalpy changes of formation to get the enthalpy change of reaction. -285.8 + -502.7 = -788.5 787.2 788.5 = -1.3 kJmol-1 From this I can show that the reaction is exothermic going in the direction of making ester and water. As this value is small it also fits in that this reaction is an equilibrium reaction. From this I can now make a prediction. As an increase in temperature forces the equilibrium to shift to remove heat energy, and the reaction root that is endothermic (removing heat) is making acid and alcohol from ester and water, I predict that as I increase temperature the more acid and alcohol is produced. In terms of the value of Kc I predict that this value decreases as I increase temperature. This is because if more acid is produced it means less ester and water is produced; and as the value for Kc is found by the product concentration divided by reactant concentration, more reactant means lower Kc. 1. Vogels textbook of quantitative chemical analysis 2.chem.ox.ac.uk/vrchemistry/FilmStudio/phenolphthalein/Symbols/equilibrium.jpg 3.ilo.org/public/english/protection/safework/cis/products/icsc/dtasht/_icsc05/icsc0553.htm 4. chemguide.co.uk/physical/catalysis/hydrolyse.html 5. ucdsb.on.ca/tiss/stretton/chem2/data09f.html Below is a list of the main areas of chemistry in my investigation and where they occur in the course. Area Of Investigation Area of Chemistry AS/A2 Enthalpy Cycles Developing Fuels AS Hydrolysis of Esters Whats in a Medicine? AS Acid-Base titrations Whats in a Medicine? AS Equilibrium Constant Engineering Proteins A2 Results Tables The table below shows the readings at each temperature, proving that the reaction has reached equilibrium. The readings for each of the temperatures are taken after one day two three and four. If the third and forth reading are the same then the rate of reaction going forward is the same as the rate of reaction going backwards so there is no net change in the amounts of reactants/products. Temperature (Celsius) Start reading (cm3) End reading (cm3) Titre(cm3) Moles of acid Initial Reading 25 0 22.5 22.5 0.00900 Second Reading 25 0 15.7 15.7 0.00628 Third Reading 25 0 9.9 9.9 0.00396 Final Reading 25 0 9.9 9.9 0.00396 Temperature (OC) Start reading (cm3) End reading (cm3) Titre (cm3) Moles of acid Initial reading 30 0 22.5 22.5 0.00900 Second reading 30 22.5 37.7 15.2 0.00608 Third reading 30 37.7 47.8 10.1 0.00404 Final Reading 30 0 10.1 10.1 0.00404 Initial reading 35 0 22.5 22.5 0.00900 Second reading 35 22.6 37.9 15.3 0.00612 Third reading 35 0 12.3 12.3 0.00412 Forth Reading 35 0 10.3 10.3 0.00412 Final Reading 35 0 10.3 10.3 0.00412 Initial reading 40 0 22.4 22.5 0.00900 Second reading 40 22.4 37.5 15.1 0.00604 Third reading 40 37.5 48.0 10.5 0.00420 Final Reading 40 0 10.5 10.5 0.00420 Initial reading 45 0 22.5 22.5 0.00900 Second reading 45 22.3 37.2 14.9 0.00596 Third reading 45 37.2 47.9 10.7 0.00428 Final Reading 45 0 10.7 10.7 0.00428 Initial reading 50 0 22.5 22.5 0.00900 Second reading 50 22.5 37.5 15.0 0.00600 Third reading 50 37.5 48.4 10.9 0.00436 Final Reading 50 0 10.9 10.9 0.00436 N.B the Results highlighted in red shows that the reaction did not reach equilibrium as fast as all of the others. Although this did not affect the results obtained, it will still be accounted for later on in the evaluation section. Below shows graphs for all of the temperatures showing time from the beginning of reaction to the reaching of equilibrium. The table of results shows all of the repeats I did for all of the temperature ranges. It also shows that moles of acid in the sample. Temperature (degrees Celsius) Beginning measurement End measurement Titre Amount of Acid Moles of Acid 25 0 9.8 9.8 9.8 0.00392 25 9.8 19.7 9.9 9.9 0.00396 25 19.7 29.5 9.8 9.8 0.00392 25 29.5 39.4 9.9 9.9 0.00396 25 0 9.8 9.9 9.9 0.00396 Average titre 9.9 9.9 0.00396 30 0 10.1 10.1 10.1 0.00404 30 10.1 20.2 10.1 10.1 0.00404 30 20.2 30.3 10.1 10.1 0.00404 30 30.3 40.4 10.1 10.1 0.00404 30 0 10.1 10.1 10.1 0.00404 Average titre 10.1 10.1 0.00404 35 0 10.3 10.3 10.3 0.00412 35 10.3 20.6 10.3 10.3 0.00412 35 20.6 30.9 10.3 10.3 0.00412 35 30.9 41.2 10.3 10.3 0.00412 35 0 10.3 10.3 10.3 0.00412 Average Titre 10.3 10.3 0.00412 40 0 10.4 10.4 10.4 0.00416 40 10.4 10.9 10.5 10.5 0.0042 40 20.9 31.5 10.6 10.6 0.00424 40 31.5 42 10.5 10.5 0.0042 40 0 10.5 10.5 10.5 0.0042 Average Titre 10.5 10.5 0.0042 45 0 10.7 10.7 10.7 0.00428 45 10.7 21.4 10.7 10.7 0.00428 45 21.4 32.1 10.7 10.7 0.00428 45 32.1 42.8 10.7 10.7 0.00428 45 0 10.7 10.7 10.7 0.00428 Average titre 10.7 10.7 0.00428 50 0 7.6 7.6 7.6 0.00304 50 7.6 18.6 11 11 0.0044 50 18.6 29.5 10.9 10.9 0.00436 50 29.5 40.4 10.9 10.9 0.00436 50 0 10.9 10.9 10.9 0.00436 N.b.Anaomalous result ignored Average titre 10.9 10.9 0.00436 Analysis of results Now for each repeat for each temperature I will work out the moles of each in the reaction, and hence the equilibrium constant for each repeat. 25oC- Repeat 1 Titre initial = 22.5cm Initial moles = volume x concentration of alkali Initial Moles of acid = 0.00900 Volume of Ethanoic acid in 1cm3 at the end = 9.8 Final moles = volume of acid x concentration of alkali Final moles of acid = 0.00392 As one mole of acid reacts with one mole of alcohol then the amount of alcohol must be the same. Final moles of Propan-1-ol = 0.00392 Final moles of Propyl Ethanoate equals difference between initial acid moles and final acid moles. 0.00900 0.00392 = 0.00508 As one mole of Ester makes one mole of water then the moles of water must be exactly the same as that of ester. From these values the value of the equilibrium constant can be worked out: Kc = 0.00508 x 0.00508 0.00392 x 0.00392 = 1.68 3 S.F. Applying this same method to all my other data I will now produce a table showing each repeat and the equilibrium constant of that experiment. I have recorded this table because I feel it makes it easier for me to draw a graph of the equilibrium constant against the temperature. Temperature (degrees Celsius) Initial Moles of Acid Moles of Acid Equilibrium Constant 25 0.00900 0.00392 1.68 25 0.00900 0.00396 1.62 25 0.00900 0.00392 1.68 25 0.00900 0.00396 1.62 25 0.00900 0.00396 1.62 Average 0.00900 0.00396 1.62 30 0.00900 0.00404 1.51 30 0.00900 0.00404 1.51 30 0.00900 0.00404 1.51 30 0.00900 0.00404 1.51 30 0.00900 0.00404 1.51 Average 0.00900 0.00404 1.51 35 0.00900 0.00412 1.40 35 0.00900 0.00412 1.40 35 0.00900 0.00412 1.40 35 0.00900 0.00412 1.40 35 0.00900 0.00412 1.40 Average 0.00900 0.00412 1.40 40 0.00900 0.00416 1.35 40 0.00900 0.0042 1.31 40 0.00900 0.00424 1.26 40 0.00900 0.0042 1.31 40 0.00900 0.0042 1.31 Average 0.00900 0.0042 1.31 45 0.00900 0.00428 1.22 45 0.00900 0.00428 1.22 45 0.00900 0.00428 1.22 45 0.00900 0.00428 1.22 45 0.00900 0.00428 1.22 Average 0.00900 0.00428 1.22 50 0.00900 0.00304 3.84 50 0.00900 0.0044 1.09 50 0.00900 0.00436 1.13 50 0.00900 0.00436 1.13 50 0.00900 0.00436 1.13 Average 0.00900 0.00436 1.13 Anomalous Result is ignored From this graph I can deduce that as the Temperature increased the value of Kc decreases. As I said this in my prediction, it goes some way to proving that my prediction was at least to some extent correct. Another conclusion I can draw from my graphs is that as the temperature increased the equilibrium constant changed exponentially. This is because on the graph it shows a slight curve. The reason for this is that as the temperature rises the reaction favours the production of the reactants. Therefore as the reactants increase the products decrease so the equilibrium constant goes down. There is a constant fractional change in the value of Kc as the temperature increases. The reason that the reaction favours the production of the reactants, ethanoic acid and propan-1-ol is because this direction of reaction takes in energy, i.e. its energy level is higher than the products that it makes. As the temperature increases the system tries to remove the extra heat energy being put in. The way it does this is by making the reaction that removes heat energy go faster. Therefore the reaction constant goes down..