Reaction Of Acid And Limestone example essay topic

Aim In this investigation I am aiming to find out how the concentration of acid affects the rate of reaction with Calcium Carbonate (Limestone). Hypothesis I predict that the higher the concentration of acid then the faster the rate of reaction. Calcium Carbonate + Hydrochloric Acid = Carbon Dioxide + Calcium Chloride CaCO 3 + 2 HCl = CO 2 + H 2 O + CaCl 2 I predict this because when a chemical reaction takes place, the atoms are broken loose from their arrangements and re-arranged into different groupings, forming new substances. The reaction above can only happen if the particles of acid came into contact with particles of limestone. In gases and liquids or in solutions, the particles are moving around freely and can bump into each other more easily. In this reaction of acid and limestone, if the acid is very concentrated this means that there will be plenty of acid particles close together in the solution.

Acid particles bump into the calcium carbonate very frequently and if they do this, the surface of the limestone will dissolve quickly. If the acid is very dilute then the particles will be more spread out. They will bump into the limestone a lot less often and so the reaction will take longer. A general rule could be: If the solution is more concentrated, the reaction will be faster In this reaction, the limestone is a solid. Therefore the particles in a solid cannot move around freely and only the particles on the surface of the limestone can react with the acid. The diagrams on the next page show the same mass of limestone.

On the left hand side diagram, there is one lump of limestone. On the right there are several smaller pieces: Variables The variables I will be considering are: Pressure at which the experiment ensues. Surface area of the limestone chip (s). Temperature at which the reaction takes place.

Volume of acid. Molarity of acid. Each of the variables above affects the rate of reaction in their own ways. The pressure of the experiment causes the reaction to occur quicker as with more pressure there will be more gaseous particles and less space for the particles, so the particles will strike each other more frequently. With the Surface area of the limestone chip (s), there will be more for the acid particles to react with, as the total surface area will be more thus more particles of the limestone chip. The temperature at which the reaction happens does not really have an effect in this case, but it can be an important variable in another investigation.

The volume of acid will make a difference but only up to a certain point as if you keep adding acid then the particles will just keep increasing but if too many acid particles are added then the acid particles bump into each other as well as the limestone. Molarity of acid is a key aspect in this experiment as if I use an acid with higher molarity there will be more acid particles to react with the limestone surface and so there will be more chance of collision. Methods set up my apparatus as on the next page. Once everything had been set up I got 2.00 grams of limestone chip (s) and I put them into a conical flask.

I decided on 2.00 grams of limestone through a preliminary process which consisted of testing and making our own judgements to see if we could get a reasonable volume of carbon dioxide, if I was to use too much limestone or too much acid then the reaction will produce too much gas too quickly. If there was not enough limestone or acid then not enough gas will be produced. That is why I used 2.00 grams of limestone and 15 ml of acid, as they seemed to work well with each other. Once I added the 15 ml of Hydrochloric acid to the limestone in the conical flask I put a bung on top of it and started the stopwatch.

I measured the volume of gas collected in a measuring cylinder and timed the experiment for 2 minutes. I collected the gas in a measuring cylinder but this is not an entirely accurate technique of recording results, but they were the only instruments available to me. I repeated this process again for 0.5 molar acid, 0.75 and 1 molar acid. I kept the volume of acid and the mass of limestone the same for each experiment as to keep it a fair test. Then I repeated the entire process again 3 times so that I could get average results.

During my experiment I managed to control all the variables to the best of my ability. I recorded all my results in a table, which you can also see later on in the coursework, and using the results I will also draw a graph. Results Volume of gas collected after 2 minutes Attempts Concentration of Acid 1 2 3 Average 0 0 0 0 0 0.25 6 5 6 5.6 0.5 9 11 14 11.33 0.75 16 16 13 15 1 23 20 18 20.3 Analysis Below is the working out for my graph and its gradient, all the symbols used and key for the graph is explained below. 5.6 + 11.37 + 15 + 20.3 = 52.27 52.27 / 5 = 10.4460.

I have got these numbers from my table of results and as you can see on my graph the line of best fit must go through this. Also on my graph you can see a triangle drawn with a broken line, I have done this to find the gradient of the line. Once I find the gradient of the line I will have an equation, which will enable me to find results, which I have not experimented with. Dg is the height of my vertical dotted line and Dc is the length of my horizontal dotted line. Dg / Dc = Gradient 7.2 / 6 = 1.2 So now if I use the equation of g = mc + c, which is the equation to find the gradient, and substitute the figures that I have worked out, then re-arrange the equation I should end up with an equation to find results. g = mc + c 10.446 = 1.2 x 0.5 + c Now if I re-arrange it. -c = 1.2 x 0.5 / 10.446 Now with this equation I am able to find results, which I have not even experimented with. However, this equation will only let you find out how much gas has been produced after 2 minutes from different concentrations of acid as long as the limestone and Hydrochloric acid amount to what was used in this experiment, which was 15 ml of Hydrochloric acid and 2 grams of limestone.

Evaluation In my experiment I have kept to my variables as best to my ability however there were some variables, which I could not sustain. Surface area of the limestone chip (s) was the biggest factor, which I could not uphold this was because each limestone chip consists of a different size and shape and I went on the variable of weight of the limestone chip (s). Therefore each of my 2.00 grams per experiment I accomplished might have had a bigger or smaller surface area each time. Another variable, which I had no control over, was the variable of temperature.

I anticipated that the investigation would take place at room temperature, which is 24 degrees. However, I did not have the equipment or the facilities to control the temperature so if the temperature changed during the investigation, which I think it, did because we experimented over several days, and then the experiment might not be entirely fair. Another factor, which I could not be in command of, was pressure. The pressure created by the weather; I couldnt take control of it because, once again I did not have the facilities. If the weather changed during the experiment then the pressure would have changed also. Once again this is a factor, which I could not of had any effect upon and this could have caused the experiment not to be entirely fair.

I believe that I could have made the test better by using more accurate equipment, for example a more accurate measuring cylinder to record results better, or an even harder task would be to try and work out the total surface area for each of the tests..