An experiment to determine the enthalpy changes using Hesss law Essay Example

An experiment to determine the enthalpy changes using Hesss law Essay The main idea behind this experiment is to find out the temperature difference between the room temperature and the final temperature. Sodium carbonate, sodium hydrogencarbonate and hydrochloric acid were used in this experiment. Sodium carbonate, also known as soda ash is got from the reaction of carbonic acid and sodium hydroxide while sodium hydrogencarbonate (baking soda) is a salt formed by the partial replacement of hydrogen by sodium.Data Collection1) Temperature change by using 3.3g of sodium hydrogencarbonateMass of the container on which the sample was weighed = 11.48gMass of the container and the crystals = 14.98gMass of the container after the crystals were added = 11.70gMass of the crystals that did not react = 00.20gMass of the crystals that reacted = 03.30gTime (s)Temperature ( à ¯Ã‚ ¿Ã‚ ½ C)023.03023.06023.09018.012015.015015.018015.021015.024015.527016.030016.02) Temperature change by using 1.88g of sodium carbonateMass of the container on which the sample was weigh ed = 11.48gMass of the container and the crystals = 13.48gMass of the container after the crystals were added = 13.60gMass of the crystals that did not react = 00.12gMass of the crystals that reacted = 01.88gTime (s)Temperature ( à ¯Ã‚ ¿Ã‚ ½ C)023.03023.06023.09029.012030.015030.018030.021030.024029.527029.030029.03) Temperature change by using 5.66g of sodium hydrogencarbonateMass of the container on which the sample was weighed = 11.48gMass of the container and the crystals = 18.48gMass of the container after the crystals were added = 12.82gMass of the crystals that did not react = 01.34gMass of the crystals that reacted = 05.66gTime (s)Temperature ( à ¯Ã‚ ¿Ã‚ ½ C)023.03023.06023.09021.012021.015021.018021.021021.024022.527022.030022.04) Temperature change by using 3.85g of sodium carbonateMass of the container on which the sample was weighed = 11.48gMass of the container and the crystals = 15.48gMass of the container after the crystals were added = 11.63gMass of the crystals th at did not react = 00.15gMass of the crystals that reacted = 03.85gTime (s)Temperature ( à ¯Ã‚ ¿Ã‚ ½ C)023.03023.06023.09028.512028.515028.518028.521028.524028.027028.030028.0UNCERTAINITIES+/- 0.01g : Digital weighing scale+/- 0.01s : Stop watch+/- 0.05cm3 : Measuring cylinder+/- 0.05à ¯Ã‚ ¿Ã‚ ½C : ThermometerOBSERVATIONS- When hydrochloric acid is added to sodium carbonate, some effervescence (bubbles appear) is observed because of the liberation of carbon dioxide gas. It is completely soluble and in the process it starts getting warmer.- The same thing happens when sodium hydrogencarbonate is added to hydrochloric acid except for the fact it cools down instead of getting warmer.CHEMICALS (QUALITATIVE DATA)1) Hydrochloric acid- It is colorless and odorless. It is a monoprotic acid, that is, it produces 1 hydrogen ion when completely dissolved in water. The molarity of hydrochloric acid used in this experiment is 2M.2) Sodium carbonate- Sodium Carbonate is a white, crystalline com pound soluble in water (absorbing moisture from the air) but insoluble in alcohol. It forms a strongly alkaline water solution. It is also known as soda ash3) Sodium hydrogencarbonate- sodium bicarbonate or sodium hydrogen carbonate, chemical compound, NaHCO3, a white crystalline or granular powder, commonly known as bicarbonate of soda or baking soda. It is soluble in water and very slightly soluble in alcohol.DATA PROCESSING AND PRESENTATIONFinding the enthalpy of reaction for the following equation:2NaHCO3 Na2CO3 + H2O +CO2a) Using 1.88g of sodium carbonate and 3.3g of sodium hydrogencarbonateEnthalpy cycle for the reaction?H12NaHCO3(s) + 2 HCl 2NaCl(aq) +2 CO2(g)+ 2 H2O(l)?H2 ?H3Na2CO3(s) + CO2(g) + H2O + 2 HCL(aq)Heat released during the reaction between HCl and NaHCO3 (Q1)= Mc?TSo, Q1 = 25g * 4.18 * (15-23)Q1 = 25* 4.18 * -8Therefore, Q1 = -836 J= -0.836KJCalculating the number of moles present in 3.3g of NaHCO3Number of moles = mass(g)/ molar massMass (g) = 3.3gMolar mass = 2 3 + 1 + 12 +(16*3)= 23 + 1 + 12 + 48= 84 g/molnumber of moles present in 3.3g of NaHCO3 = 3.3g/ (84g/mol)= 0.039 molesCalculating the amount of energy given out by 1 moleIf 0.039 moles of NaHCO3 give -0.836KJ of energy then1 mole would give out (-0.836/0.039 = -21.44KJ) of energyTherefore, ?H1 = -21.44 KJ/molCalculating ?H3 by the above method, that is, the reaction between Na2CO3 and HClQ2 = Mc?TSo, Q2 = 25 * 4.18 * (30-23)Q2 = 25 * 4.18 * 7Therefore, Q2 = 731.5J= 0.7315KJCalculating the number of moles present in 1.8g of Na2CO3Number of moles = mass (g)/ molar massMass (g) = 1.8gMolar mass = (23 * 2) + 12 +(16*3)= 46 +12 + 48= 106 g/molnumber of moles present in 1.8g of Na2CO3 = 1.8g/ (106g/mol)= 0.018 molesCalculating the amount of energy given out by 1 moleIf 0.018 moles of Na2CO3 give -0.731KJ of energy then1 mole would give out (-0.7315/0.018 = 40.64KJ) of energyTherefore, ?H3 = 40.64 KJ/molIn order to find the enthalpy of reaction for:2NaHCO3 Na2CO3 + H2O+ CO2; we use the Hes ss law which states that 2 ?H1 = ?H2 + ?H3?H2 = 2 ?H1 ?H3so, 2 ?H1 = 2 * -21.44 KJ/mol= -42.88KJ/mol?H3 = 40.64 KJ/molTherefore, ?H2 = -42.88 40.64= -83.52KJ/molb) Calculating the enthalpy change of reaction using 5.66g of sodium hydrogencarbonate, 3.85g of sodium carbonate and 50cmà ¯Ã‚ ¿Ã‚ ½ of HCl.Enthalpy cycle for the reaction?H12NaHCO3(s) + 2 HCl 2NaCl(aq) +2 CO2(g)+ 2 H2O(l)?H2 ?H3Na2CO3(s) + CO2(g) + H2O + 2 HCL(aq)Heat released during the reaction between HCl and NaHCO3 (Q1)= Mc?TSo, Q1 = 50g * 4.18 * (21.5-23.0)Q1 = 50* 4.18 * 1.5Therefore, Q1 = -331.50 J= -0.3135KJ `Calculating the number of moles present in 5.66g of NaHCO3Number of moles = mass (g)/ molar massMass (g) = 5.66gMolar mass = 23 + 1 + 12 + (16*3)= 23 + 1 + 12 + 48= 84 g/molnumber of moles present in 5.66g of NaHCO3 = 5.66g/ (84g/mol)= 0.0674 molesCalculating the amount of energy given out by 1 moleIf 0.0674 moles of Na2CO3 give -0.3135KJ of energy then1 mole would give out (-0.3135/0.0674 = -4.65KJ) of en ergyTherefore, ?H1 = -4.65 KJ/molCalculating ?H3 by the above method, that is, the reaction between Na2CO3 and HClQ2 = Mc?TSo, Q2 = 50 * 4.18 * (28.5-23)Q2 = 50 * 4.18 * 5.5Therefore, Q2 = 1149.5J= 1.1495KJCalculating the number of moles present in 3.85g of Na2CO3Number of moles = mass (g)/ molar massMass (g) = 3.85gMolar mass = (23 * 2) + 12 + (16*3)= 46 +12 + 48= 106 g/molnumber of moles present in 1.8g of Na2CO3 = 3.85g/ (106g/mol)= 0.036 molesCalculating the amount of energy given out by 1 moleIf 0.036 moles of Na2CO3 give 1.1495KJ of energy then1 mole would give out (1.1495KJ/0.036 = 31.93KJ) of energyTherefore, ?H3 = 31.93KJ/molIn order to find the enthalpy of reaction for:2NaHCO3 Na2CO3 + H2O+ CO2; we use the Hesss law which states that 2 ?H1 = ?H2 + ?H3?H2 = 2 ?H1 ?H3so, 2 ?H1 = 2 * -4.65 KJ/mol= -9.3KJ/mol?H3 = 31.93KJ/molTherefore, ?H2 = -9.3KJ/mol 31.93KJ/mol= -41.23KJ/molc) Error analysisDigital weighing scale: 1) 0.01/3.3 * 100= à ¯Ã‚ ¿Ã‚ ½0.3%2) 0.01/5.66 * 100= à ¯Ã‚ ¿Ã‚ ½0.18%Stop watch : 0.01/300 * 100= à ¯Ã‚ ¿Ã‚ ½ 3.3*10^-3Measuring cylinder : 1) 0.05/25 * 100= à ¯Ã‚ ¿Ã‚ ½0.2%2) 0.05/50 * 100= à ¯Ã‚ ¿Ã‚ ½0.1%Thermometer : 0.05/23 * 100= à ¯Ã‚ ¿Ã‚ ½0.22%Total percentage error = 0.22%+ 0.1% + 3.3*10^-3 + 0.18%+ 0.3%= à ¯Ã‚ ¿Ã‚ ½0.8033%Accounting for the error ?H2 = à ¯Ã‚ ¿Ã‚ ½ -41.23KJ/mol= à ¯Ã‚ ¿Ã‚ ½ -83.52KJ/molConclusionThe reaction between sodium hydrogencarbonate and HCl is endothermic, that is, heat is being absorbed in the reaction and the reaction between sodium carbonate and HCl is exothermic because temperature is given out to the surroundings.Also, in the second part of the experiment when the volume of HCl is increased and also the masses of sodium hydrogencarbonate andsodium carbonate is increased, the temperature difference in the reaction is less than before when the mass were less. The enthalpy of reaction is also decreased in the second part.Evaluation:Reasons for shortcoming in the answers are as follows:1) While t ransferring the HCl from the measuring cylinder to beaker, there is a possibility of leaving out some amount of HCl in the beaker itself.2) An analogue thermometer was used so the temperature may not have been accurate.3) Some systematic errors in the equipment might have led to some slight changes in the readings.Solution to the above problems:1) The first problem stated above is a personal error; So it can only be overcome by practice and improving ones concentration while doing the experiment.2) The second problem could have been overcome by the use of digital thermometer which is more accurate than an analogue thermometer.

Choosing The Right Admissions Essay Sample

Choosing The Right Admissions Essay SampleUc admission essay samples will serve as a guide to help you write your own admission essay. Many applicants nowadays want to write their own admissions essays. The applications for admission are in such demand that it is only natural for schools to look for professional writers to help them out. The importance of a good application cannot be overemphasized, especially for universities with a quota for each year of applicants for admission.There are many schools in the country who base their admissions on the numbers of students applying for admission. As a rule, applicants must be good candidates for admission in order to ensure the success of the institution in the long run. Applicants also need to demonstrate some special qualities such as being energetic, responsible, and well-prepared. Admissions officers, on the other hand, also have to be sure that all prospective students are intelligent enough to understand the contents of the admiss ions requirements.An admissions essay is not enough to ensure success of any school. In fact, there are many ways to improve your chances of being accepted. To this end, you should take the time to search for the perfect admissions essay sample. You can also seek assistance from professional writers or even local experts.You will find many samples of admissions essays online but you need to understand that not all online materials are relevant. There are many materials that should not be included in your application. This is because many online articles are written by nonprofessionals who do not know much about admissions and therefore will probably do less than no good for your future at that school.College applications are an integral part of high school. These will serve as your first impression of the institution you will attend. An admissions essay is the part of your application that will have the biggest impact on whether you are accepted or rejected. You need to include your strongest points and perhaps highlight other aspects of your personality that are also positive.One of the best ways to prepare your admission essay is to use one of the samples available online. Make sure that you take some time to read through the essay and then look at a few examples so that you can incorporate what you see into your own essay. Remember, you are the one in charge of the content of your admission essay. Look for examples that sound relevant to the subject matter of your school and curriculum and then try to create similar sections for yourself.Remember, an admission essay is the very first thing the admissions officer will look at when considering your application. He or she may only want to see your strong points and be convinced that you are a good fit for the institution before he or she looks at your admission essay. Admissions officers want applicants to be concise is key. They will spend more time on applicants who are able to express themselves in a cohere nt manner.You may want to seek professional advice if you are unsure about what to write. There are several colleges and universities that offer free or low-cost counseling services. While these are a great way to learn how to compose a good admissions essay, you can expect a lot of assistance in writing a good essay from the admissions advisors. Do not be afraid to ask questions because they will only help you prepare yourself better for your admission.