Investigation to find out how temperature affects the rate of reaction

of HCl that I will use will be 25cm ³. This was founded bypart of my preliminary work. The reaction overall between HCl and Na2S203is exothermic which means that the reaction gives out heat energy. Interms of bond breaking and making if a bond is broken energy isreleased. If a bond is made energy is used up. An exothermic reactionmeans that there are more bonds broken than there are made, willresults in heat energy being given out. This exothermic reaction meansthat the temperature will affect the results, as heat is beinggenerated when the reactants are reacting. In the graph below you cansee that there is a bigger temperature difference at 5cm³ as thecoordinates are further apart, to overcome this problem of theexothermic reaction affecting results we need a concentration of HClthat generates the least amount of heat. This is 25cm³ so this is thevolume of hydrochloric acid that will be used in the investigation.The actual figures are also shown below and you can see that, at 5cm3of HCl the time taken for the reaction to take place and finish was 90seconds at 30°C, but at 32°C the time taken for the reaction to finishwas 80 seconds which means that there is a whole 10 second differencebetween the reaction times at only 2°C more. At 25cm3 of HCl the timetaken for the reaction to finish was 15 seconds at 30°C and 13 secondsat 32°C, the difference here is 2 seconds and by using this volume ofHCl I am using a volume of HCl that is more reliable due to giving outthe least amount of heat, as the coordinates are the closest at 25cm3.GRAPHMethodEquipmentThe equipment that I will need for this practical part of theinvestigation is;- Pipette- Burette- Conical flask – 250ml- Stopwatch- Goggles- Measuring cylinders- Clamp stand- Sodium Thiosulphate – 25cm3- Hydrochloric acid – 50cm3- Thermometer- Cross board- Kettle- Pen- PaperSafety IssuesI will have to consider safety issues whilst carrying out myinvestigation for obvious reasons. Goggles will have to be worn toprevent any substances that are being used going into my eyes. Safetyin the actual lab has to be considered, all stools have to be pushedunder the desk out of the way and coats and bags have to be out of theway in the corner of the classroom where the coat pegs are. Also Ihave to be alert to respond to any instructions that the teacher hasto put forward. The practical work has to be taken seriously and‘playing about’ cannot be an option, as this could endanger otherpeople as well as me. Fair Test I will have to keep this investigation a fair test so that my resultsare accurate as much as they can be. The way that I will keep it afair test is that I will only have one variable and I will accuratelymeasure my substances. I will also when investigating at differenttemperatures only proceed with my practical work when the temperatureis at its most accurate. Also the same person has to be judging whenthe cross disappears as people have different eye sight levels. Somemay be very short sighted and another very sharp and this differencecan alter the supposed rate of reaction.How will Temperature be changed?The temperature of HCl and Na2S2O3 will not be altered by heating thesubstances in a test tube over a Bunsen burner because, you areheating the substances directly above the flame although you may turnoff the Bunsen when it says 50°C the conduction of heat energy stillcarries on and the true actual temperature of it could really be 60°C.The most accurate way to get the temperatures equalling what you wantis to heat water and put the reactants into a test tube then put thetest tube in the beaker containing water and monitor the temperaturefrom there. This is a more reliable way, as the conduction that takesplace after you have heated them, is less than if you had have heatedthem directly which means it is more likely to be closer to its truetemperature.I plan to record my findings in a table like the one below;Temperature (°C)1530456075Time taken for cross to disappear (S)Time taken on second trail (S)The reason that I plan to repeat my experiment is because by havingtwo sets of results I wish to find an average to increase thereliability of the results.Average = Results Total number of entriesWhen I have worked out an average I will put my workings in a tablelike the one below and calculate rate;Temperature (°C)1530456075Average (S)RateIndependent variable: TemperatureDependant variable: Rate of reactionStep by Step Method1- Gather equipment needed and put on safety goggles.2- Get the cross board and put a beaker on top of it3- Get ready with the stopwatch4- Fill burette with 50cm3 of Na2S2O35- Pour 25cm3 of HCl into pipette and release that into a test tube6- Put water into kettle7- Turn on kettle and heat the water8- Pour the heated water into two beakers9- Put the test tube that you have filled with HCl into the water10- Pour the 50cm3 of Na2S2O3 into a test tube11- Put this into the water filled beaker also12- Insert a thermometer into both of the test tubes13- Wait for temperature to reach desired rate14- Obtain the test tubes15- Pour both substances into the beaker which is on the top of thecross board16- Start the stopwatch17- Observe wait till cross is not visible18- Stop the stopwatch19- Record results20- Reset the stopwatch21- Wash out all containers that had substances in them22- Repeat step 4 and onwards all over again but make sure youmonitor at different temperatures.ResultsTemperature(°C)1530456075Time taken for cross to disappear (S)34:5328:0611:6304:3502:46Second attempt (S)51:0331:1013:4205:2503:01Working out Averages;15°C = 34:53 + 51:03 = 85:56 2 = 42:7830°C = 28:06 + 31:10 = 59:16 2 = 29:5845°C = 11:63 + 13:42 = 25:05 = 25:05 2 = 12:525 which rounded to is 12:5360°C = 04:35 + 05:25 = 09:60 2 = 04:8075°C = 02:46 + 03:01 = 05:47 2 = 02:735 which rounded to is 02:74Calculating RateTo calculate the rate of a reaction I need to divide 1 by the timetaken for the reaction to be completed using the averages that I haveworked out. I then will plot the graph for my average time againsttemperature and rate against temperature, and analyse my results.Rate = 1 Time15°C = 1 42:78 = 0.02337540930°C = 1 29:58 = 0.03380662645°C = 1 12:53 = 0.07980845960°C = 1 04:80 = 0.20833333375°C = 1 02:74 = 0.364963503Rounded to 4 decimal places is;15°C = 0.023430°C = 0.033845°C = 0.079860°C = 0.208375°C = 0.3650GRAPHConclusionBy looking at my graph (Temperature VS Rate of Reaction) I can see atrend; there is a negative correlation. As the temperature increasesthe time taken for the reaction to occur decreases, as the highernumbers are at the top of the graph this causes a ‘left to righttrendÂ’. This graph is inversely proportional as when one variableincreases the other decreases. I have noticed that by increasing thetemperature by 15°C each time the average rate of reaction half’s eachtime, this is not that accurate but it is the general rule. I wascorrect in my prediction by saying that by increasing temperature Iwould decrease the rate of reaction. I knew that when a collision tookplace (an encounter between two or more particles that come togetheror close to each other) the exchange or transfer of energy would takeplace if sufficient Ea is present. If there was not enough Ea then theparticles would not react. The collision theory stated that the morecollisions that particles undergo the greater chance of reactingparticles colliding which results in a more rapid rate of reaction.This would mean that if particles had more kinetic energy which meantthat they would move faster, than they would undergo more collisions.So theoretically speaking if the reactants have more kinetic energythey react quicker. This led me to think how kinetic energy isconnected to temperature. By increasing temperature you aretransferring energy to the particles, this is kinetic energy and theparticles move more frequently, therefore by increasing temperatureyou increase the kinetic energy of particles so there are moreparticles with more Ea required for a reaction to successfully combineand more successful collisions- thus a quicker reaction as temperatureis increased. I predicted a graph of what my results would look likebased on the background science that I knew.As you can see the actual graph that I did produce looks similar asthe one above. The values are not close to the actual ones that I didget but, the general trend is there. EvaluationI have not really found out anything major in this coursework but ithas strengthened my understanding of rates in reaction overall and howyou can combine different ideas to come to some sort of predictionabout whatever it may be. I am interested to see how reliable myresults are I will plot the two attempts of the investigation next toeach other and analyse them.Series one indicates the first attempt and series two the secondattempt. As you can see the dots are quite far apart especially at thelower temperatures, this means that there is a difference in rate ofreaction and if the experiment was done more precisely then the dotswould have been closer together and a more accurate experiment.Possible causes could have been because the different set of resultswas done on different days and a different person judged the crossdisappearing and the differences in eye vision would have caused thesenot so precise results. Also the anomalous results could have beendown to the volume of reactants and the actual temperature gettingmeasured wrongly. I think to have increased the reliability even moreI should have at least had 5 attempts at each temperature range thenhave had found an average, as well as getting a second opinion onmeasuring the volume of HCl and reading the temperature and the timetaken e.t.c. Also if I was to do this experiment again I would use alight meter to help indicate when the cross has disappeared as this isa more precise way to carry out the investigation.