Science Investigation (Chemistry) Rate of Reaction Essay Example

Science Investigation (Chemistry) Rate of Reaction Essay Example Science Investigation (Chemistry) Rate of Reaction Essay Science Investigation (Chemistry) Rate of Reaction Essay Outline: I will investigate the relationship between the rate of a reaction and the amount of catalyst added to the reaction. I will use hydrogen peroxide and speed pu the decomposition of it by adding manganese(IV), MnO2 as catalyst.Fair test:In order to maintain fair test conditions I will control variables that will affect the rate of reaction.The factors that are able to affect the rate of reaction include:* Temperature* Surface Area* Concentration* CatalystSince I am investigating the relationship between rate of reaction and catalysts, the other 3 factors will remain constant throughout the experiment.Temperature: At higher temperatures, the faster moving particles have more kinetic energy. This means they are moving quicker and will collide more often and with more energy, causing an increase in the number of effective collisions.Surface Area: If more surface area is available, the number of particles of the solid reactant available for collision will increase. However, I am d ealing with hydrogen peroxide (liquid) then there isnt a surface area for me to control.Concentration: More concentrated solutions contain more particles in the same space, making them more likely to collide. The increase in the number of collisions with sufficient energy increases the rate of reaction.These explain why I would need to control those factors because they would affect my results dramatically. I will keep them constant:Temperature: room temperature, temperature of hydrogen peroxide (17.C)Surface area: n/aConcentration: 5 volumes (halved the maximum which is 10 volumes, ratio 1:1)However, for room temperature, I will do all experiments in the same room hopefully the temperature remains constant in that one room.Safety:Whilst this experiment is relatively safe, there are still safety considerations that must be adhered to:* Due to the dangerous nature of many of the compounds involved in this experiment, it is important that safety goggles or safety glasses are worn duri ng all tests.* It is equally important to avoid skin contact with the compounds, as some are irritants. This can be done by ensuring that a suitable method is followed precisely so that there are no unexpected events that would lead to skin contact.Accuracy and Reliability:I will take 3 measurements of oxygen produced at each reading I will take. I will then use the arithmetic mean of these measurements in my analysis to calculate average rate of reaction. This should help to minimize unreliable results. Whilst I am doing my experiment, I will note the results down on a table, in order to spot any obvious anomalies.The measurements themselves will be taken as carefully as possible. Time will be measured using the stop-watch (precise to à ¯Ã‚ ¿Ã‚ ½0.1 second), volume of gas produced with the gas syringe (precise to à ¯Ã‚ ¿Ã‚ ½1 cmà ¯Ã‚ ¿Ã‚ ½), the mass of the powdered manganese oxide with measuring balance (precise to à ¯Ã‚ ¿Ã‚ ½0.01g) and the volume of hydrogen peroxide with a m easuring cylinder (precise to à ¯Ã‚ ¿Ã‚ ½0.5ml)Prediction:I predict that the grater the amount of manganese oxide, MnO2, is added to the solution, the greater the rate of reaction decomposition.The formula for this decomposition is:H2O2 (aq) =; 2H2O (l) + O2 (g)Hydrogen peroxide =; Water + OxygenThe manganese dioxide is present to be acted as a catalyst and as with all catalysts would lower the activation energy required in order for this reaction to take place. In doing so, the rate of reaction will be increased. Since more catalyst is present, it means there are more surface area of the catalyst so more molecules will be able to react faster.Method:1. Set up apparatus as shown in the diagram.2. Use a measuring cylinder to measure 20ml of hydrogen peroxide and use another measure cylinder to measure 20ml of distilled water.3. Measure 0.02g of manganese oxide using a measuring scale (remember to put a paper tower on the scale first so that manganese oxide wouldnt be able to direct ly touch the scale, which might cause malfunction to the scale if manganese oxide goes in the internal of the scale)4. Put the manganese oxide into a flask and put the 20ml of distilled water into the flask as well.5. Reset (push the gas syringe back to 0cmà ¯Ã‚ ¿Ã‚ ½ if needed) the gas syringe if needed, and then put the 20ml of hydrogen peroxide into the flask and start the stop watch.6. Observe the gas syringe and take note of the volume of gas (in this case oxygen) produced every 10 sec, 20 sec, 30 sec, 40 sec, 50 sec and 60 sec.7. Repeat step 2-6 for 0.04g, 0.06g, 0.08g and 0.1g of manganese oxide but must clean the flask before using it again.8. After obtaining all of the results, repeat each one 2 more times.Diagram:Pretest:I did a few pretests:0.01g manganese oxide and 10 volume concentration of hydrogen peroxide (20ml):Time (sec)Volume of gas produced (cmà ¯Ã‚ ¿Ã‚ ½)00101520203031404150506060By using 0.01g of manganese oxide, the maximum reached is already 60cmà ¯Ã‚ ¿Ã‚ ½, therefore I thought if I used 0.1g, the limit of the gas syringe (100cmà ¯Ã‚ ¿Ã‚ ½) will be reached with less than 60 sec, so I have decided to try and reduce the concentration and see whether it is a good choice.0.1g manganese oxide and 5 volume concentration of hydrogen peroxide (20ml):Time (sec)Volume of gas produced (cmà ¯Ã‚ ¿Ã‚ ½)00101520313040404950566061Later, I tried using only 10ml of hydrogen peroxide but it wasnt enough for my experiment and it caused a limiting factor very soon in the experiment (when I tested 0.04g of manganese oxide) due to the shortage of hydrogen peroxide molecules for decomposing.The second pretest gave me good range of results to use and it didnt reach the gas syringes limit (100cmà ¯Ã‚ ¿Ã‚ ½) so that wouldnt limit my results. Also, in the result taking, I realized that I would need a lot of time to refill my catalyst and liquids, therefore I have chosen to only do 0.02, 0.04, 0.06, 0.08 and 0.1. This will give me a good range of results to calculate the rate of reaction and also sufficient for supporting my prediction.Obtaining:Here is the tables of results to show the volume of gas produced using different amounts of catalysts. I will also show the average:0.02g catalyst usedAmount of gas produced (cmà ¯Ã‚ ¿Ã‚ ½)Time (sec)Trial 1Trial 2Trial 3AverageAverage Gain in gas produced from previous (cmà ¯Ã‚ ¿Ã‚ ½)00000.000.00104343.673.67207687.003.333010111211.004.004014151514.673.675020192019.675.006023232423.333.670.04g catalyst usedAmount of gas produced (cmà ¯Ã‚ ¿Ã‚ ½)Time (sec)Trial 1Trial 2Trial 3AverageAverage Gain in gas produced from previous (cmà ¯Ã‚ ¿Ã‚ ½)00000.000.00105876.676.672013141313.336.673020191919.336.004029302829.009.675035353635.336.336042414041.005.670.06g catalyst usedAmount of gas produced (cmà ¯Ã‚ ¿Ã‚ ½)Time (sec)Trial 1Trial 2Trial 3AverageAverage Gain in gas produced from previous (cmà ¯Ã‚ ¿Ã‚ ½)00000.000.001010111110.6710.672022212021.0010.333032303331.6710.674040394140.008.335046484646. 676.676051505050.333.670.08g catalyst usedAmount of gas produced (cmà ¯Ã‚ ¿Ã‚ ½)Time (sec)Trial 1Trial 2Trial 3AverageAverage Gain in gas produced from previous (cmà ¯Ã‚ ¿Ã‚ ½)00000.000.001018181617.3317.332031302930.0012.673040414140.6710.674048474546.676.005055545554.678.006062616261.677.000.10g catalyst usedAmount of gas produced (cmà ¯Ã‚ ¿Ã‚ ½)Time (sec)Trial 1Trial 2Trial 3AverageAverage Gain in gas produced from previous (cmà ¯Ã‚ ¿Ã‚ ½)00000.000.001031313331.6731.672045474545.6714.003053525453.007.334060615960.007.005068676767.337.336073727272.335.00I have taken the chance to add in another column (average gain in gas produced from previous reading) to show whether there is a certain pattern.Analysis:My graph clearly shows that as I increase the amount of manganese oxide, I will increase the rate of decomposition of hydrogen peroxide. This supports my prediction fully (prediction: the grater the amount of manganese oxide is added to the solution, the greater the rate of re action decomposition of hydrogen peroxide:H2O2 (aq) = 2H2O (l) + O2 (g))We can clearly see that with the line of best fit drawn the points are not that far away from this line, showing a quite strong relationshipThe rate of the decomposition (when the reaction is at its 20th second) for 0.02g of catalyst used is 0.35cmà ¯Ã‚ ¿Ã‚ ½/sec, for 0.04g it is 0.67cmà ¯Ã‚ ¿Ã‚ ½/sec, for 0.06g it is 1.05cmà ¯Ã‚ ¿Ã‚ ½/sec, for 0.08g it is 1.5cmà ¯Ã‚ ¿Ã‚ ½/sec and for 0.10g it is 2.28cmà ¯Ã‚ ¿Ã‚ ½/sec. This also supports my prediction because the rate of decomposition is increasing when the amount of catalyst increases. I chose to use 20th second into the reaction to calculate the rate because at the 20th second, all the reactions hasnt yet used up or near used up the hydrogen peroxides, therefore not reaching the point where there arent enough, which will cause a slow down in the rate.In theory, the graph should show a steep gradient at the beginning of the decomposition. By looking at th e graph, it does show a steeper gradient at the beginning but not for the line of best fit of 0.02g and 0.04g, this might be because the rate of decomposition is too slow to see any real effect. However, the other 3 lines show a steep gradient in the beginning. The end of the lines (0.06g, 0.08g and 0.10g) all tend to approach a point where it will stop the reaction (can be seen by the curve). The line for 0.10g has the greatest curve. This is because there are most catalyst compared to the others, therefore increases the rate of decomposition the fastest. The manganese oxide has therefore acted as a catalyst and as with all catalysts has lowered the activation energy required in order for this reaction to take place. In doing so the rate of reaction has increased. So the more manganese oxide means lowering the activation energy more effectively. A catalyst basically works like:If without catalyst, the energy for the reaction to take place is relatively high, but if a catalyst is ad ded, the activation energy lowers, therefore enabling the reactants to react with less energy. The activation energy of the reaction is about 75 kJ/mol in the absence of catalyst. The activation energy of the reaction is about 58 kJ/mol if manganese oxide is added to the reaction.However, towards the bottom end of the graph the lines (0.06g, 0.08g, 0.10g) are a lot shallower compared to the start. This shallower lines show that it is finding it hard to produce oxygen at a fast rate because there are not enough molecules left to decompose itself.Some of the results could be wrong, however I can still pull some conclusions from this and one is that we can definitely see that as the amount of manganese oxide increases, the faster oxygen is produced.Evaluation:Overall I think I carried out this investigation to my greatest capabilities.The data that I collected was fairly reliable because of the strong correlation and the accuracy of my experiment was reasonable. It also was enough to s upport my prediction. However, there are some anomalies which might be caused by several errors such as:1. The measuring balance wasnt exactly accurate because sometimes the more I add, the more the mass reading on the balance decreases.2. My reaction time, because even though I try my fastest to obtain the results, I will still be around 0.5 1.5 sec off from the correct time to take the results.There was also a big problem:As soon as the Manganese Dioxide touched the Hydrogen Peroxide, oxygen was produced straightaway. So I lost quite a lot of gas before we put the bung on. I tried to counter this problem by adding the hydrogen peroxide last because liquid is easy to transfer to the flask compared to transferring manganese oxide because the powder might possibly stick to the top of the flask and unable to go down to the hydrogen peroxide. Also, I started the countdown the second the bung was fitted on properly as soon as the oxygen was going up the flask.A major cause for concern on accuracy was the reading of the overall measurement of the oxygen produced but also the measurements of hydrogen peroxide and water used for the experiment. So if these miss-readings were carried out through the whole investigation then my results will be quite a long way out.Looking back on experiment I believe that there are many ways to improve it. One way is the accuracy of the measuring of the catalyst this can be done by requesting a new scale which should not have any chances of any faulty measurements. This would make every measurement of catalyst I take virtually no faults at all.My graph is very basic and one way to improve this will be to increase my range of tests, this will allow me to know where exactly the point where no more oxygen can be created is. Maybe doing until 2 minutes (120 sec) will allow me to locate where it is. I think the timing was as accurate and reliable as it could be, but only a second or third experiment would back that up fully. The equipment we used was reasonable but not the best, but for the conditions we were based in it was the best we were going to get.I could investigate on what type of catalysts is more effective for the speeding up of the decomposition of hydrogen peroxide. Possible catalyst selections include:* Manganese IV oxide (MnO2)* Zinc oxide (ZnO)* Lead IV oxide (PbO2)* Aluminium oxide (Al2O3)* Iron III oxide (Fe2O3)* Copper II oxide (CuO)* Copper III oxide (Cu2O)This might help me understand more about catalyst and what type of metals are best for catalysts, such as: transition metals.I could investigate on the concentration of hydrogen peroxide and keeping the catalyst constant, and also I can investigate on the effects of pH with rates of reaction.