Sodium Thiosulphate and Hydrochloric Acid Coursework Essay Example for Free

atomic number 11 Thiosulphate and Hydrochloric Acid Coursework searchIn this experiment I bequeath be seeing how the assiduousness of Sodium Thiosulphate, Na2S2O3, and Hydrochloric Acid, HCl, affects the crop of the reception. In this chemical answer the solution turns milky icteric as sulphate is displaced and forms a solid precipitate.Na2S2O3 (aq) + 2HCl (aq) - 2NaCl (aq) + H2O (l) + SO2 (aq) + S (s)The straddle of a reaction bay window be represented as ? assimilation and so is the term? duration fetchn for the reactants to be officed up. Therefore I entrust be investigating if eachRate ? Na2S2O3x or/and Rate ? HCLYwhere X and Y atomic number 18 the orders of the reaction.To calculate the order equation and orders of the reaction I am going to use the log functionAs Rate ? Na2S2O3x and Rate ? HCLYRate = k1 Na2S2O3x Rate = k2 HClY bring forth logs logRate = logk1 + X logNa2S2O3 logRate = logk2 + Y logHCleq. of line Y = C + M X Y = C + M XThis shows how the power, and order of the reaction, can be found by drawing a graph of logRate over against logconcentration and the gradient of the line provide give the order of the reaction. The two equations can then frame in together to give an over both let in equationRate = k Na2S2O3xHClYExperimentally I will be changing the concentrations of each the reactants independently and then calculating the rate for each concentration. From these rates I will be able to draw a log graph of rate against concentrations and from my calculation I can find the orders for each of the reactants.VariablesThe independent inconsistent in the experiment is the initial concentrations of Sodium Thiosulphate and Hydrochloric Acid. I will be changing this variable and seeing how it affect the time worryn for the reaction to take place and and then how the concentration affects the rate of the reaction. However as only cardinal variable can be changed I will keep one reactant constant whilst changing the o ther to get a fair set of results. This will show me the affect of each reactant on the rate independently.The dependent variable is the one that will be monitored in the reaction. The dependent variable is the time taken for the reaction to take place which can be used to calculate the rate of the reaction. The regularity I will be using is by seeing how long it takes for a black ill-tempered underneath the conelike flask to disappear which will be the point in all the reactions where the concentration of the sulphur precipitate is fitted and so will the rate can be fairly calculated. To reduce errors I will expect to use my preliminary results to find a volume that will give a long copious time to measure accurately, but not too long to affect the rate which is calculated by dividing initial concentration by time.These graphs show how if the time taken for the reaction to reach a item concentration is too long it will not be a true reflection on the actual initial rate of the reaction.The first variable that must be compriseled is the temperature of the reactions. This is because temperature will affect the rate of the reaction as a higher temperature will cause an make up in the rate. If the temperature varies between reactions it will digest an affect on the results. To try and insure the corresponding temperature is used I will hold back all reactants at room temperature and carry egress all the reactions in one lesson. some other control is the the volume of reactants used and the depth of the solution. If this is changed the amount of precipitate that will need to be observed by will change and therefore the concentration at which the cross disappears will be different. To control this the analogous volumes will be used and the same conic flask will be used and therefore the depth will be the same. Similarly the cross used will need to be the same as a thicker cross would mean it will disappear at a different concentration of sulphur an d the results will not be fair. Finally I will not shake or move the conical flask once the reactants take in been put together as this again exponent affect the rate of the reaction.Equipment1. 200cm3 of Na2S2O31. 200cm3 of HCl2. 10 x 50cm3 beakers3. 2 x Graduated Pipettes4. Conical Flask5. stopwatchMethod1. Start by making the concentrations, using a graduate pipette measure come to the fore the correct volumes of Sodium Thiosulphate and put into 5 small 50cm3 beakers. Then add the corresponding amounts of distilled water and judge with the correct concentration.2. Secondly draw a black cross onto a piece of paper which is no bigger than the base of the conical flask, place the conical flask on top of the flask.3. Using another(prenominal) graduated pipette, place 25cm3 of 2M hydrochloric acid into the bottom of the conical flask.4. Whilst standing above the conical flask, looking like a shot down, simultaneously pour the first concentration of Na2S2O3 into the conical flash and start the stopwatch.5. Keep observing and when the yellow precipitate causes the cross to completely disappear stop the stopwatch.6. Record the time taken, wash out the conical flask thoroughly and repeat from step 3 but using the next dilution until all the concentrations have been reacted and recorded.7. Repeat all results and average times.8. Secondly replace the Sodium Thiosulphate with the HCl and copy above steps by measuring out correct concentrations and place into 5 more 50cm3 beakers.9. This time add 25cm3 of Na2S2O3 into the bottom of the conical flask before recording the time taken for the cross to disappear again, for all the concentrations.10. Repeat all results again for HCl concentrations.Initial Rates can then be calculated.Dilution TablesNa2S2O3 density / moldm-3Na2S2O3 (1M) / cm3Distilled wet / cm31.02000.81640.61280.48120.2416HCl Concentration / moldm-3HCl (2M) / cm3Distilled Water / cm32.02001.61641.21280.88120.4416Safety PointsAs concentrated Hydrochlor ic acid will be used make sure center of attention and hand protection is worn. In case of contact, immediately flush skin with plenty of water for at least 15 minutes and with eye contact get medical attention immediately. Sulphur is too present so make sure the room is well ventilated to allow fumes to escape. Take care with glass wear, such as pipettes as they are fragile and may break easily. If furrowed clear up all glass immediately.Time interpreted for Cross to Disappear / sConcentration of Na2S2O3 / moldm-312AverageRate /moldm -3 s -1logRate /log moldm -3 s -1logconc. / log moldm-31.09.219.469.340.107-0.9710.0000.813.7114.0013.860.072-1.142-0.0970.617.4317.2117.320.058-1.237-0.2220.423.6622.7123.190.043-1.367-0.3980.249.7850.1249.950.020-1.699-0.699Time Taken for Cross to Disappear / sConcentration of HCl / moldm-312AverageRate /moldm -3 s -127.259.478.360.1201.69.039.099.060.1101.28.888.768.860.1130.89.098.428.760.1130.49.378.679.020.111AnalysisAfter drawing a graph of l ogRate against logConcentration and using the calculations in my plan, the gradient of the line gave me a value of 0.957 which taking into account experimental error shows the order of the reaction of Sodium Thiosulphate is 1. Also when carrying out the investigation for how the rate changes with the concentration of HCl there is no noticeable change in rate when the concentration varies. Therefore I didnt draw a graph, as the table shows HCl concentration doesnt affect the rate of the reaction and must be zero order. I can therefore deduce that the rate equation for this reaction isRate ? Na2S2O31 and Rate ? HCL0Rate = C x Na2S2O3where C is a constant which is proportional to the rate constant. The reason why the actual rate constant, k, can be calculated is because in the experiments the change in concentration cannot be calculated. However this always the same value as it is the concentration at which the cross disappeared and therefore the rate was proportional to 1/time. The in tercept of the graph is equal to logC, -1.02, therefore C = 10-1.02 = 0.095s -1.To come together the results show that the rate of the reaction between Sodium Thiosulphate and Hydrochloric Acid is directly proportional to the concentration of Sodium Thiosulphate. This gives evidence to the mechanism of the reaction. In a most reactions the mechanism is not made up of one step but a series of intermediate stages where molecules form and break down. Each of these steps will have a rate which is due to how many molecules are colliding. For example if 2 molecules must conflict to for one product the reaction will be 2nd order. However the overall order for the reaction cannot be calculated using the stoichiometry in the overall equation. This is because the rate is determined by the rate of the dilatory step of the mechanism, known as the rate determining step. In this reaction although 2HCl molecules are needed for the reaction to take place however from the kinetics that I found HC l has no effect on the rate and is zero order. Therefore the slowest step in this reaction is the breakdown of 1 Sodium Thiosulphate molecules as the reaction is 1st order.EvaluationI feel the results I obtained allowed me to draw a graph and come to valid conclusions and be confident about calculating the order and rate equation for this reaction. I didnt have any obvious anomalies from my graph however there are some variations that meant I couldnt calculate a rate of exactly 1. Also from my table of results for changing the concentration of HCl, where all the rates must be the same for difference concentrations, there is a anomaly with the 2.0moldm-3 concentration.A limitation of my procedure was the way in which the rate was calculated using the disappearing cross method. This involves human judgement rather than a qualitative value that can be used to calculated the rate. There will also be an trueness issue as there is a reaction time in realising the cross has disappeared and stopping the stopwatch. Another limitation of the procedure at the start of the timing and I had to add the chemicals together and start the stopwatch. This again causes an error in the actual time period taken for the cross to disappear. Another error is in the variation of temperature. As well as environmental variations in temperature the reaction was exothermic and heat is produced. When the temperature increases the k constant will increase, this is because the molecules have more kinetic energy and there is a higher chance that they will collide and react. This means its is hard to control this factor which is affecting the rate of the reaction.There are also many accuracy errors caused by percentage error in measuring. There is a time error in the time taken when the cross disappears and the delay in stopping the time. When using the graduated there was an error of 0.05/20 = 0.25% error, human reaction time is about 0.2/9.34 = 2.1%. To reduce this error I would choose anoth er method of measuring the rate. Instead of using the cross method I would use a light intensity machine that could measure the change in colour over a period of time and stop and start timing when it had reach a certain value. This quantitative method of measuring should make the results more accurate. To control the temperature I could use a large water bath which could control the temperature throughout the reaction. Another method that could be used is to measure the products formed. The easiest would be using a gas syringe to measure the amount of SO2 formed. This again would give a more quantitative results and increase the accuracy and reliability.