Determining the of the Effect of the Concentration of Na2S2O3 on the Rate of Reaction Essay

In this experiment we reacted different concentrations of Na2S2O3 (aq) with a constant volume of HCl, and measured the time it took for the X drawn under the beaker in black marker to disappear. Uncertainty Details: 1. The uncertainty in the volume of Na2S2O3 (aq) and H2O (l) is given by manufacturer of the burettes. As we find the change in the volume in the burette, the uncertainties are added, and the uncertainty in the volume is  ±0.1cm ³ 2. The uncertainty in the HCl is given by the manufacturer of the measuring cylinder. 3. The uncertainty in the time is a rough estimate calculated by me trying to perfectly stop the stopwatch at 5 seconds three times in a row, and in all cases it was about 0.4 seconds reaction time. 4. The uncertainty in Total Volume of Na2S2O3 (aq) and H2O is found by adding the uncertainty in the volume of H2O and the uncertainty in the volume of Na2S2O3. Observations: 1. We stirred all solutions. 2. There is a small delay between when we started the stop watch and poured the HCl, as it is impossible to perfectly coordinate this. 3. Bad smell released. 4. The stirring speed was not the same for each reaction, though it was attempted to be replicated equally for each reaction. 5. The uncertainty given by the last figure on the stop watch was very inaccurate to use, therefore we calculated the reaction time instead to give a more true uncertainty. However this value has a range, so it is not necessarily accurate. Calculations To calculate the concentration of the Na2S2O3 in each trial, we use the equation: . As for both trials the volumes are all identical, we can simply calculate the concentrations for the first trial, and use them for the second. For the first solution, we apply the equation, and thus we do: (10.0cm ³/50.0cm ³)*0.2 à ¯ 0.04M. As for the uncertainty here, we must add the fractional uncertainty in the volume of sodium sulfate and total volume, and then multiply it by the concentration. The uncertainty in the initial concentration is unknown, so we do not use any value for it. So (0.1/10.0)+(0.2/50.0) = 0.014. 0.014*0.04 = 0.00056 à ¯ 0.0006. This can be repeated for all the other concentrations, and is shown in the following table: Concentration of Na2S2O3 (aq) (M) Uncertainty in Concentration (M) Time for Trial 1 ( ±0.4)(s) Time for Trial 2 ( ±0.4)(s) 0.0400 0.0006 125.2 133.2 0.0800 0.0007 61.4 65.1 0.1200 0.0009 40.0 36.7 0.160 0.001 29.1 29.8 0.2 Unavailable (0) 24.1 23.4 As in the last concentration no water is added, the whole solution has the same concentration as the initial concentration, so the uncertainty is unknown. Now as the volumes for both trials were identical, we can find an average of the times for both trials. To do this we add the 2 values and divide by 2. For the first one this would be (125.2+133.2)/2 = 129.2s. The uncertainty here would not be affected so it is still  ±0.4 for all times. Now that we have these results, we can find the order of the reaction with respect to Na2S2O3. Now as we know that in order for the x beneath the beaker to not be visible, a certain amount of the product must be produced, we assume the same amount of the products is produced in each solution. This then allows us to assume the same amount of the reactants is used up for the x to be formed in all experiments, so even though we do not know the change in concentration of each reaction, we know that it is about the same. Therefore if we plot 1/time against concentration, we should be able to see the relation between the concentration and the rate, even though we do not have the correct rate. Concentration of Na2S2O3 (aq) (M) Uncertainty in Concentration (M) 1/time (Rate) (mol dm-3 s-1) Uncertainty in Rate (mol dm-3 s-1) Now we can plot this: – As we can see in this graph, it is linear, and Rate is proportional to 1/time. This means that the order of the reaction with relation to Na2S2O3 is 1. Also as the gradient of the line is 0.2166, this tells us that in the rate equation K = 0.2166mol-1dm3s-1. So the rate equation is: Rate = 0.2166[Na2S2O3][HCl]y. However we do not know the order of HCl as we did not vary the volume of HCl. Conclusion To conclude, we have calculated the order of the reaction with respect to Na2S2O3 to be 1. This was efficiently experimentally calculated as shown by the graph above. The graph is very fitting, and there are no anomalous points on it. As the R ² value is so close to 1, we can see that our line fits very well, and that the results are quite precise. Also as we can see from the graph, while the y intercept is supposed to be 0, it is 0.0009. This is due to systematic error. While this is not 0 like would be ideally, this is not a problem as it is a very small number, and rather insignificant as it would be nearly impossible to have absolutely no systematic error. This error could have been caused by multiple things, though there were no factors that particularly affected the results significantly. The result is extremely accurate, as we were told by our teacher the expected order was 1. Evaluation Improvements Even though the x disappeared, this does not mean the same amount of precipitate was formed. As the x disappearing is a very unreliable method as the amount of precipitate formed could be more or less in each trial, even if the x disappears. This means we have to make the assumption that the same amount of precipitate was formed so that the same number of moles are used up, allowing us to find the rate and order. This added to our systematic error, thus less to slightly less accurate results as some points may have taken more or less time than needed. Also one of the most error causing points for sure in this experiment is deciding when the x had disappeared, as I recall countless times in which it had looked like it had disappeared, however it was not completely. However, I did attempt to stop the stop watch at the same point for each one to make it a fair test. As it was unclear at times whether or not the x had disappeared, this would have led to an increase in rate in some trials, and a decrease in rate in others, so the overall effect is unknown. The x drawn could have been drawn bigger and with thicker ink allowing it to stand out much more. This would have meant that as it was easier to see, once it had disappeared completely I would easily be able to tell that it had disappeared as it stands out more. Alternatively, a light meter could have been used, which detects the levels of light[1]. A light source can be place above the beaker, such as a simple lamp. Once enough precipitate has formed, the light meter should detect no light. The data can either be measured using a data logger, which would be started when the reaction was started, and automatically stopped by the light meter, or simply using a stop watch however starting and stopping the time according to the light meter. The temperature in this experiment was not maintained. Though the reactions all took place in the same room within a 1 hour range, the temperature may have varied in that time, so the rates could have gone up or down depending on the temperature of the room, which could have slightly affected our results. This would have also contributed to the systematic error in the experiment. Furthermore, the temperature during each trial may have also not remained constant, which could have led to slightly different calculated rates. The temperature could have been monitored during each trial so we can see when the rate could have been affected by a rise/fall in temperature. Also if the room was air-conditioned at a constant temperature, this would have meant the room temperature would stay the same (assuming no windows/doors are opened in the time). The uncertainty in the stop watch was much smaller than the actual uncertainty, so I attempted to find my reaction time, which was 0.4. However, when conducting the experiment it is impossible to tell if every time my reaction time was that, as it may have been more or less. This may have increased or decreased the uncertainty here. I could have taken a larger range of samples for my reaction time to get a more accurate value. As I poured the HCl and started the stop watch at the same time, this meant there was a small delay between when I poured the HCl in and when the stop watch was started. This means that the time was a little bit less than it had to be, once again adding to the slight systematic error. I could have gotten a fellow class mate to press the stop watch as soon as I poured the HCl in, so that there was a much small delay, and more precise results, as well as a smaller systematic error.