Investigate the Energy Change during Neutralisation Reactions Essay Example

Investigate the Energy Change during Neutralisation Reactions Essay * Calorimeter This polystyrene device will hold the acids in the neutralisation reaction. This has been used, as it is a very good insulator of heat so as little as possible will be lost.* Burette This glass device can hold 50 cm3 of liquid and will be used to pour the acid in 1cm3 amounts into the alkali. This device has been used, as it is very accurate and easy to use.* Thermometer This will be used to measure the temperature of the reaction after each 1cm3 of acid has been added.* Clamp Stand and Burette Holder These will be used to keep the burette in place so it does not fall over and break or spill acid.* Strong and Weak Acid Hydrochloric acid or sulphuric acid will be used for the strong acid and ethanoic acid will be used for the weak acid. 40 cm3 will be used to neutralise the alkali.* Strong and Weak Alkali: Sodium hydroxide will be used for the strong alkali and ammonia will be used for the weak. We will use 15cm3 of alkali at 1 mole per decimetre cube concentration. Preliminary WorkPreviously I performed a thermotitration with a strong acid and a strong alkali and I found that the following quantities would be suitable for my experiment: 15cm3 of alkali would neutralise well within 15% accuracy compared to the theoretical energy output. We also decided to use a 1-mole/dm3 concentration as we found 0.5 mole/dm3 was too little for a good result, and 2 moles/dm3 was too violent. The preliminary work also showed that it was necessary to use a good insulating material to house the reaction in and we also needed to use very accurate equipment.SafetyAs with any experiment there are safety issues that need to be followed. Many of the acids or alkalis used in this experiment are corrosives or irritants therefore safety goggles must be worn at all times. Also any spilt acids or alkalis need to be cleaned up immediately and any got onto skin or clothing needs to be washed up immediately. All equipment must be out of the way of walkways to avoid spillage a nd accidents. Smashed glass must be swept up immediately.Fair TestTo ensure this experiment is a fair test it is imperative that all tests are done in the same way. Any equipment to be used again after a test has been completed must be cleaned thoroughly to prevent any contamination in the next experiment that could affect the accuracy of the results.PlanWe will set up the equipment as shown below:As shown in the preliminary write up above I will be using 15cm3 of 1 mole/dm3 alkali and 40cm3 of acid at 1 mole/dm3. I will then use the burette to add 1cm3 at a time of acid to neutralise the strong alkali. After adding the strong acid I will then stir the solutions together to let them react and quickly take the temperature. I will then repeat this 40 times until all the acid has gone. I will then repeat this experiment for a strong acid against a strong alkali; a strong acid against a weak alkali; a weak acid against a weak alkali. Therefore I will be doing four experiments. To analys e the results I will be using the equation (mass x heat capacity of water (4.2) x change in temperature). This will give me the answer for the energy change in joules for however many moles there were in the concentration on acid. I will then need to find out the energy change for 1 mole so I will then work out the number of moles of acid there was using:Number of moles = volume (cm3) x concentration (mol/dm3)1000I will then divide the answer I got from the energy equation and divide this by the result of the above equation. Thus giving me the energy change for 1 mole of acid neutralising the alkali.PredictionFrom preliminary work and acid theory I can determine that in this experiment the reaction with the greatest energy change will be the strong acid reacting with the strong alkali. When an acid reacts with a base it dissociates to form ions. In the case of hydrochloric acid it will form H+ ions and Cl- ions. For this example almost all of the HCl will dissociate into ions howev er with a weak acid such as ethanoic acid only about 0.3% of the acid is dissociated into ions. The action of acids disassociating creates energy and therefore the stronger the acid the more energy. This rule also applies for the alkali bases. After doing some further preliminary work I found the theoretical values for energy produced in a neutralisation reaction. These values were as follows:Strength of acid/alkali (respectively)Energy (Joules per mole)Strong and StrongWeak and StrongStrong and WeakWeak and Weak-57,900-56,100-53,400-50,400As you can see this table proves my prediction to be right. I can use this table in my evaluation to see how accurate my experiment has been by working out the percentage error of my results.Prediction GraphsResultsSodium Hydroxide and Hydrochloric Acid (Strong and Strong) Start Temperature 19à ¯Ã‚ ¿Ã‚ ½CVolume of Acid (cm3)Temperature (à ¯Ã‚ ¿Ã‚ ½C)Cumulative Temperature Change (à ¯Ã‚ ¿Ã‚ ½C)1.020.01.00.920.51.51.221.52.51.222.03.01.022.53.51. 123.04.01.023.54.51.024.05.01.124.55.50.925.06.01.025.06.01.025.56.51.126.07.00.826.07.01.025.56.51.025.56.51.025.06.01.025.06.01.025.06.01.024.55.51.024.55.51.024.55.51.124.05.01.124.05.00.924.05.00.924.05.01.024.05.01.024.05.01.023.54.51.023.54.51.123.54.51.023.54.50.923.54.51.023.04.01.023.04.0Ammonia and Ethanoic Acid (weak and weak) Start Temperature 19.5à ¯Ã‚ ¿Ã‚ ½CVolume of Acid (cm3)Temperature (à ¯Ã‚ ¿Ã‚ ½C)Cumulative Temperature Change (à ¯Ã‚ ¿Ã‚ ½C)1.020.51.01.121.01.50.921.52.01.022.02.51.022.53.01.023.03.51.023.03.51.023.54.01.224.04.51.024.04.51.024.55.01.024.55.01.025.05.51.025.05.51.025.56.01.125.56.00.925.56.01.025.56.01.025.05.51.025.05.51.225.05.51.025.05.51.025.05.51.024.55.01.024.55.01.024.55.01.024.55.01.024.55.01.024.55.00.924.55.01.124.04.51.024.04.51.124.04.51.024.04.51.024.04.51.024.04.51.024.04.51.024.04.51.023.54.0Ethanoic Acid and Sodium Hydroxide (weak and strong) Start temperature 19.5à ¯Ã‚ ¿Ã‚ ½CVolume of Acid (cm3)Temperature (à ¯Ã‚ ¿Ã‚ ½C)Cumul ative Temperature Change (à ¯Ã‚ ¿Ã‚ ½C)1.019.50.01.019.50.00.919.50.01.020.00.50.920.00.51.220.51.01.020.51.00.921.01.50.921.52.01.022.02.51.222.02.50.822.53.01.023.03.51.023.54.01.124.04.51.224.04.50.724.55.01.025.05.51.025.05.51.025.05.51.025.05.50.925.56.01.125.56.01.025.56.01.026.06.51.126.06.50.926.06.51.026.06.51.325.56.00.625.56.00.825.56.01.025.56.01.425.05.51.025.05.51.025.05.51.025.05.51.125.05.50.925.05.51.025.05.51.025.05.5Hydrochloric Acid and Sodium Hydroxide (strong and strong) Start Temperature 19.5à ¯Ã‚ ¿Ã‚ ½CVolume of Acid (cm3)Temperature (à ¯Ã‚ ¿Ã‚ ½C)Cumulative Temperature Change (à ¯Ã‚ ¿Ã‚ ½C)1.020.00.50.922.02.51.224.04.51.225.56.01.026.06.51.127.07.51.028.08.51.029.09.51.128.59.00.928.59.01.028.08.51.028.08.51.127.58.00.827.07.51.027.07.51.026.57.01.026.57.01.026.06.51.026.06.51.026.06.51.025.56.01.025.56.01.125.56.01.125.56.00.925.56.00.925.06.01.025.06.01.025.06.01.025.06.01.025.06.01.125.06.01.024.55.50.924.55.51.024.55.51.024.55.50.924.55.51.224.55.51 .024.55.51.024.05.0AnalysisFor the first test (sodium hydroxide and hydrochloric acid) I will analyse the results using the equation:- (mass x heat capacity of water (4.2) x change in temperature)Therefore the amount of energy produced is:- (15 + 11.2) x 4.2 x 6.7 = -737.3 joulesHowever the number of moles of alkali was only 0.015 moles worked out by the equation: Number of moles = volume (cm3) x concentration (mol/dm3)1000So the amount of energy produced for 1 mole is -737.30.015This equals -49153.3 joules of energy.For the second test (ammonia and ethanoic acid) the results are as follows:Mass (from graph) = 15.8Change in temperature = 5.9- (15 + 15.8) x 4.2 x 5.9 = -763.2 joulesThis answer, then divided by 0.015 to give the answer for 1 mole of alkali gives an answer of -50,880.0 joules of energyFor the Third test (ethanoic acid and sodium hydroxide) the results are as follows:Mass (from graph) = 22.8Change in temperature = 6.9- (15 + 22.8) x 4.2 x 6.9 = -1095.4 joulesThis answe r, then divided by 0.015 to give the answer for 1 mole of alkali gives an answer of -73,026.7 joules of energyThe fourth test is a repeat of the first test (hydrochloric acid and sodium hydroxide) the results are as follows:Mass (from graph) = 7.8Change in temperature = 9.2- (15 + 7.8) x 4.2 x 9.2 = -881.0 joulesThis answer, then divided by 0.015 to give the answer for 1 mole of alkali gives an answer of -58,733.0 joules of energyI would normally average the two results (for the first and last experiment) because they are repeats however the last result is very close to the theoretical value and if they were averaged this would not be the case so I will take the 1st result to be anomalous.For ease of reading and analysis I will put these results into a table and calculate the accuracy of my results from the theoretical results found in my preliminary workStrength of acid/alkali (respectively)Result from ExperimentTheoretical ResultPercentage Error:my answer true answertrue answerSt rong and StrongStrong and StrongWeak and StrongWeak and Weak-49153.3-58,733.0-73,026.7-50,880.0-57,900-57,900-56,100-50,40018%1%30%1%Average Error13%On the whole my results do support my prediction as for the reactions between both strong and both weak solutions the results only have a 1% error margin. Therefore my results show me that there is positive correlation between the strength of acid and the energy given off as heat in the neutralisation reaction.I believe this has happened because as stated in my prediction when an acid reacts with a base it dissociates to form ions. In the case of hydrochloric acid it will form H+ ions and Cl- ions. For this example almost all of the HCl will dissociate into ions however with a weak acid such as ethanoic acid only about 0.3% of the acid is dissociated into ions. The action of acids disassociating creates energy and therefore the stronger the acid the more energy.EvaluationIf it was possible for me to repeat this experiment again I would use a much more accurate measuring device. I would use a computer to measure the amount of acid added to the mixture so that the correct amount is used. I would also have the calorimeter on a machine that would constantly shake it so that when the acid is added it is thoroughly mixed with the alkali. For the temperature I would use a computer to continuously record the temperature so any change is noted and I would conduct the experiment in a temperature controlled environment so factors like wind or humidity could not affect the temperature. Because of the equipment available it was not possible to use a lid effectively so if the experiment was repeated I would also use one of these. To keep the insulation of the heat to a maximum I would use a more effective insulating material than the one the calorimeter was made of to make the results more accurate.The method I used did not include any repetition of experiments due to the time restrictions. To improve the accuracy I would do ea ch experiment 5 times then take an average leading to more accurate results.There were many problems encountered in this experiment and it is reflected in some of our results. One of our results had an error of 30%. This could be due to many reasons. On the graphs it shows the acid being added in 1cm3 amounts yet this was for ease of presentation. The amounts added were not always the specified amount and this resulted in the time for the acid to be added was increased or decreased. This could lead to the results being anomalous.Another reason for anomalous results occurring could be human error in the measurement of temperature or amounts of liquid. This could lead to the accuracy of results being affected. Due to the nature of the school laboratory it is possible for the solutions used to become contaminated and the pH to become changed. This could also have altered the results for my experiment. Apart from these possibilities I believe that the test I conducted was a fair one and complied with the statements I made in my plan.I believe that if I carried out the work stated above I would be more confidant of my conclusions. This is because the results ascertained from the above procedures would be a great deal more accurate than the method I used in my experiment.