# The 120cm3 of CO2. I have proved this

The mass will simply be kept constant by having 5g of CaCO3 each time. But the surface area is difficult to keep constant considering that the chips are roughly the same size but irregular shapes and so the surface area may differ. There is no way to make the surface area constant exactly in each case. However, I can reduce the amount of difference between each experiment by using a lot of the chips. To keep the surface area constant in size 10-12mm chips I used 10g. Therefore I deduce that if I use 5g of 2-4mm chips the surface area will still remain constant to a certain extent.

I believe this because I am using a lot smaller chips and therefore will need a lower mass to keep the surface area basically equal. Therefore to keep the mass and surface area equal I will use 5g of 2-4mm sized CaCO3 chips. In my preliminary work I established that the reaction was exothermic and so there is a temperature rise this means that I have to keep it constant. I will keep the temperature constant by leaving it alone. One of my preliminary experiments proved two things. Firstly that the temperature rise is so small it can be ignored, but this isn’t a very good way of keeping a constant as it still makes a small difference.

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The second and more important factor is that the temperature rises from 21i?? C i?? 23i?? C each time and so is already a constant without me having to bother about trying to keep it constant through manual means. Therefore if I leave the experiment alone the temperature will remain constant. The last variable which I will have to make constant is the volume of the HCl. This is very simple I will use 8cm3 each time. I chose 8cm3 because it will produce 96cm3 of CO2 gas while if I had used 10cm3 of HCl it would have produced 120cm3 of CO2. I have proved this below: With 10cm3 1 i?? 10 = 0. 01 = 0. 005 i?? 24000 = 120cm3 1000 2 With 8cm3

1 i?? 8 = 0. 008 = 0. 004 i?? 24000 = 96cm3 1000 2 The experiment will be set out as shown below. Prediction I predict that my graph will look like this: The graph above illustrates that if you have the concentration you will do two things. Firstly you will have the volume of CO2 produced and secondly you will the rate at which the CO2 is produced. The rate will be measured in the first 40 seconds so that we have the steepest and fastest rate to compare. Therefore we will be able to compare the different rates at their optimum before the acid concentration has started to deplete. The gradient can easily be measured by the below equation:

Change in the y axis Change in the x axis I think that all this will happen because if you half the amount of H+ ions then the collisions will be reduced by a half as well as the H+ ions will have to find the CaCO3, if you half the amount of collisions then the speed at which the chips are broken down is made half as well. I. e. the rate is halved. The volume is halved as well because you will have half as many H+ ions than before, while keeping the volume of HCl acid the same, and so only a certain amount of CO2 will be able to be given off before the H+ ions have been fully used. FAnalysis As you can see from the graph above there is one anomalous point. This is on the 1M experiment and is highlighted grey for ease of identification. The first experiment is the one which threw the average off. The point, for some reason, is lower than the point before it.

This suggests that we had lost gas and then gained more gas in the span of 20 seconds. I do not think that this is likely. Therefore I believe that the anomalous point’s existence has to be because of experimental error. I believe that I must have miss read what the volume of CO2 was after 90 seconds. Apart from that one anomalous point the result are very good. They prove my prediction to be correct. My prediction was: * If you half the concentration then you will half the rate at which the CO2 is given off. * If you half the concentration you will half the volume of CO2 produced. The rate was to be calculated by: – Change in the y axis

Change in the x axis Therefore, using the above equation and reading from 40 seconds each time (as stated in my prediction) I get these rates: 1M HCl = 63 = 1. 575 40 0. 75M HCl = 47 = 1. 175 40 0. 5M HCl = 31. 5 = 0. 7875 40 0. 25M HCl = 15. 5 = 0. 3875 40 You can clearly see the relation if you double the gradient of the 0. 5M HCl acid you will equal the gradient of the 0. 5M HCl acid: 2 i?? 0. 25M HCl = 0. 5M HCl 2 i?? 0. 3875 = 0. 775 It is close enough 2 i?? 0. 5M HCl = 1M HCl 2 i?? 0. 7875 = 1. 575 Perfect match 1. 3333333… i?? 0. 75M HCl = 1M HCl 1. 3333333… i?? 1. 175 = 1. 566666… Again it is close enough.

The above calculations clearly prove that if you double the concentration you will double the rate of the CO2 loss. This means that if you double the amount of H+ ions in 8cm3 of water you will double the chance of collision. This effectively means that you will have twice as many collisions which in turn means that the CaCO3 will lose its CO2 twice as fast. This is proved by my results and as I stated this in my prediction I have proved my prediction to be correct. In my prediction I also stated that I would be getting the following amounts of CO2 1M HCL – 96cm3 0. 75M HCl – 72cm3 0. 5M HCl – 48cm3 0. 25M HCl – 24cm3

My actual results were: 1M HCl – 95. 8cm3 0. 75M HCl – 72. cm3 0. 5M HCl – 48. cm3 0. 25M HCl – 24. cm3 As you can clearly see again my prediction was accurate on the last three and a mere 0. 2cm3 out on the first. One other thing which is shown by results is that the time it takes to reach constant volume is also being doubled. We can see that: For 1M HCl it took 100seconds to reach constant volume of CO2. For 0. 75M HCl it took 130 seconds to reach constant volume of CO2. For 0. 5M HCl it took 190 seconds to reach constant volume of CO2. For 0. 25M HCl it took 360 seconds to reach constant volume of CO2.

This is further proof that if you half the concentration of the acid you will be halving the number of H+ ions and therefore you will double the time taken for the volume of CO2 being produced to remain constant. The above results can be matched with results that are calculated to show my results’ accuracy and reliability. If we take the M HCl’s result to be accurate then we can deduce: In converting 1M HCl to 0. 75M HCl it would be – 100 i?? 1. 333… = 133. 333… cm3 If we half the 1M HCl to 0. 5M HCl it would be – 100 i?? 2 = 200cm3 Therefore if we half the 0. 5M HCl to 0. 25M HCl it would be – 200 i??

2 = 400cm3 To a certain extent this supports my prediction although out of all off the calculated results these were the most different. I believe that my results aren’t so close to the calculated ones because I didn’t get all the CO2 given off because there was a tiny leak in my equipment or that I didn’t wait long enough for the reaction to have finished. Evaluation I believe that my results are very good, apart from one anomalous point all the others fit comfortably on a line of best fit. Not only this, but they all follow the same trend. That one anomalous point I will disregard due to experimental error.

All my rates and initial recordings are accurate and prove my prediction correct. However, the final volume of all the CO2 results aren’t perfect. This is why I repeated the experiments twice each, this makes the data more reliable and accurate as you then average those results to get a final set. The final readings were still out even after I repeated the experiment twice. I believe that this is because the experiment relies too much on the equipment. What I by this is that if there is one leak then the final readings aren’t accurate. And also that if there isn’t a leak you may have to wait a very long time for the final readings to show up.

So to get around this problem you could repeat the experiment 3 times and then produce an average. Or you could wait a very long time until the reaction has finished. Unfortunately it is very difficult to tell whether or not the reaction has finished by looking at it. You could add a catalyst to the, this would speed up the reaction and so maybe the final volume of CO2 would come up sooner. But the problem with this is that then the initial rate becomes very difficult to measure and so by human error there would be mistakes in the readings. In my preliminary work I found that some gas escaped before you could place the bung into place.

However, I did disregard that that would have made a big difference to the final volume of the CO2, but I could be wrong and maybe my final volumes are wrong because of this very reason. Therefore another improvement in the experiment is to add a sample bottle into the apparatus, this would keep the HCl away from the CaCO3. Therefore you could firmly place the bung into position and so no gas would escape in the time it takes to put the bung into place. Therefore there is another experiment which can be performed alongside this one to prove my prediction correct again or on its own for some more accurate results.

In this experiment you weigh out a certain amount of CaCO3, let’s say 10g. Then you add a certain amount of HCl, about 50cm3 and put both of these onto a weighing scale. The constant is the mass of the CaCO3 and the volume of the HCl. You vary the molarity, just like in my original experiment. After having placed the CaCO3 and HCl onto the weighing scale you simply tip the acid into the conical flask containing the CaCO3 and note the weight loss after every 10 seconds or 30 seconds or however many you want.