| On Calcium Carbonate Essay, Research Paper
AimIn this investigation I am aiming to find out how the
concentration of acid affects the rate of reaction with Calcium Carbonate
(Limestone).HypothesisI predict that the higher the concentration of acid then
the faster the rate of reaction.Calcium
Carbonate + Hydrochloric Acid => Carbon Dioxide + Calcium Chloride ????????????????
CaCO3??????? +???????? 2HCl?????????????????? =>??????????
CO2???????????? +????? H2O? +? CaCl2???? 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 fasterIn 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: VariablesThe 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. MethodI 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 15ml of acid, as they seemed to work well with each other. Once I
added the 15ml 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 = 10.4460.25 + 0.5 + 0.75 + 1 = 2.5 2.5 / 5 = 0.5(10.446,0.5) is the co-ordinate
for my mean average point (??? ). 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 = 1.2So 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 /
10.446Now 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 15ml of Hydrochloric acid and 2 grams of limestone. EvaluationIn 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 couldn?t 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.
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