Rates Of Reaction : Sodium Chloride + Sulphur + Sulphur Dioxide + Essay, Research Paper
Planning
This investigation is about rates of reaction and what affects them. In this
case I am going to look at hydrochloric acid and sodium thiosulphate which is a
precipitation reaction. They react as in the equations below: sodium thiosulphate +
hydrochloric acid -> sodium chloride +
sulphur + sulphur dioxide + water Na2S2O3(aq)
+ 2HCl(aq) -> 2NaCl(aq) + S(s) +
SO2(g) + H2O(l) A reaction will only occur where the particles of the
reactants meet and combine. This is called the collision theory. Therefore it
stands to reason that to increase the rate of reaction it is necessary to cause
more particles to collide harder and make it happen more often. There are
several ways to do this and these make up the variables for this experiment.
They are listed below along with predictions as to their affect on the
reaction. Increasing the pressure. By reducing the volume in
which the same amount of particles exist the pressure is increased. Once
the same number of particles are in a smaller area there is less space in
which to move and so the particles are more likely to hit each other. It
is therefore possible to predict that increasing the pressure will result
in an increase in the rate of reaction. I will not test this variable
because the school doesn’t have the facilities to test it. However
pressure is a continuous variable. Using a catalyst is another method I could use. A
catalyst is a separate substance which speeds up a reaction. After the
reaction has happened it gets left behind. This makes this variable
unsuitable for the type of experiment I am going to do. A catalyst is also
a discontinuous variable with only one likely useful catalyst emerging. Energy. By giving the particles extra energy they
will move faster. This means that they cover more ground and are therefore
more likely to hit each other which in turn makes the reaction faster. The
best way to give energy to a particle is as heat and so I can predict that
raising temperature will increase the rate of reaction. This is a
continuous, independent variable. I shall test this variable – see below. I predict that temperature is
proportional to rate of reaction. Concentration. Just as increasing the pressure will
increase the number of particles colliding, so will the concentration. By
putting more particles into the reaction, the chance of them colliding
increases and so the rate increases. This variable is continuous and
independent. I shall test this variable. I predict that by doubling the concentration of the acid, the rate
of reaction will double. Surface area. Particles can only collide when the
two sorts can meet. Therefore a reaction can only occur on the surface of
the material. Therefore by increasing the area of the material which is
available to collide the speed of the reaction will increase. I predict
that doubling the surface area will double the speed of the reaction. This
variable is continuous but I shall not test it because it is hard to
control the exact surface area of the two reactants as they both come in
an aqueous solution. I am going to test the two variables concentration and temperature.
Both of these are independent, continuous variables. I think that concentration
will have the biggest affect because the reaction is exothermic. Therefore even
while I am testing concentration, heat will be given out by the reaction which
will give more energy to the particles and so cause them to reach their
activation energy sooner. In addition to this, looking at the original
equation, it becomes clear that for every one mole of sodium thiosulphate, you
need two moles of hydrochloric acid. Therefore increasing the number of
hydrochloric acid particles will have a greater effect than if one were to
increase levels of sodium thiosulphate. I think that both concentration and energy are proportional
because: ·
doubling the number of particles doubles the
probability that they will collide
and ·
doubling the speed at which these particles travel will
double the distance they can travel in a set time and so double the probability
of them colliding. This proportionality can be expressed using algebra thus: X’
= XY’ / Y To carry out this experiment, I will need the following
equipment:
A3020 computer, light sensor, beaker, distilled water, sodium thiosulphate,
hydrochloric acid (stock bottle), electronic scales, thermometer, burette,
light, black paper, bunsen burner, tripod, mat. Firstly I shall test the variable "concentration of
HCl", testing five different strengths. I shall set up the equipment as in
the diagram below completely surrounding the light sensor with paper to ensure
that the only light which reaches it passes through the beaker containing the
reactants. As the reaction progresses, the sulphur will collect in the water
and form a cloudy solution. As more sulphur is formed, less light can get
through the solution and reach the sensor. I will put the hydrochloric acid
into the beaker and prepare the computer. I shall then put the sodium
thiosulphate into the beaker and start the computer reading. The computer
records light levels as a percentage of original levels against time and is
much more accurate than using a stop watch. I shall allow the reaction to take
place for 60 seconds. I shall then use the computer’s accurate analysis
facility to record how long it took for light levels to fall to 60% of the
original. Often one of the possible weaknesses in an experiment such
as this is that the different concentrations of acid are often made up
inaccurately. To solve this problem I shall use one large bottle of 0.5 molar
hydrochloric acid and use distilled water to dilute it to different
concentrations: 20, 40, 60, 80, and 100% acid. Because I need 20ml of acid and
20ml of sodium thiosulphate I shall use varying quantities of water. For
example, when making 20% concentration, I shall mix the water and acid
16ml/0.25ml respectively. After the experiment, I shall be able to draw a graph
comparing concentration and reaction time. If my prediction is correct, the
graph will be proportional. I shall back up my results for this section by using
results generated by another group using the optical method outlined in the
plan for the second variable below. I conducted the experiment as per my plan, although I had to
disregard the first few computer results as the system took a while to
configure. However I did several things to ensure the accuracy of my project.
These included: ·
Washing out the glassware with distilled water before
use and between measurements. This was designed to prevent any foreign ions
getting into the solution as this could damage the results. ·
Using an analogue thermometer when heating the
hydrochloric acid as this enables me to be more accurate than with a digital
thermometer. ·
Using a small measuring cylinder and funnel when
measuring out hydrochloric acid, water, and sodium thiosulphate rather than
using beakers. The results for the first variable are displayed in Table 1
below. There was only time to take measurements once for each concentration as
other groups needed to use the computer. However because the computer is very
accurate and because I also took results from another group, this will not pose
too great a problemConclusions
Before I can represent my data in graph form and then test my prediction, I
have to look at the way the data is laid out. I predicted that both variables
would be proportional. This implies that as temperature goes up, time taken
goes down. However because reaction time goes down, reaction rate is actually
increasing. The best way therefore to represent the results in graph form is to
draw a graph of concentration/temperature against the reciprocal of the time
taken. Graph 1 shows concentration against the reciprocal of the
time. However it is clear that it is not a straight line graph but rather a
curve, gradually getting steeper as molarity increases. It is clear that my
prediction was wrong and that the graph is not proportional. I can further test
this by running my results through the formula for proportionality. X’ = XY’ / Y so X’ = (0.056 x 60) /
20 = 0.168 If my prediction was correct the reciprocal of time taken
for 60% concentration should be 0.168. In fact it is 0.09. The slow growth of
the graph followed by a massive increase can be explained by looking at
activation energy. All of the reactions happened at room temperature (about
210C). Clearly this energy was only enough to push some of the particles beyond
their activation energy. However because the reaction is exothermic it gives
out energy and this energy pushes more particles to activation energy and these
in turn release more heat. More particles of HCI available to reaction with the
sodium thiosulphate means more heat given out and more particles being pushed
to activation energy. The investigation could have been improved by testing the
temperature variable on the computer as the stop watch I used could not cope
with the speed of the reaction. It would also have helped to test each
concentration more than once to ensure that the results were true. When using
the light sensor I should have covered the underside of the sensor with black
material rather than sticking on paper as this could have let in some light. In
addition I should have used an artificial source of light as the natural light
in the room was constantly changing as clouds pass in front of the sun. I could
also have used a burette to measure out the reactants although the measuring
cylinder was quite accurate. Squash Ball experiment"Squash Ball Experiment Input Variables: Pressure Of Air in Ball Type Of Surface Height Of Drop Temperature of Ball Material of Ball Acceleration Due To Gravity Mass Angle Of Surface Air Resistance Diameter of BallOutcome: Height Of BouncePrediction??????????????? The
squash ball will bounce higher as the temperature gets warmer. This is because
as it gets warmer the atoms in the ball vibrate more. This means that when it
hits the ground the atoms push each other way forcing the ball to bounce
higher. When the temperature is lowered the opposite occurs because the atoms
have less energy and therefor push each other further away. The graph would
look like this:The graph begins to level out because parts of the ball
begin to melt at certain temperatures as the atoms get more energy and break
their bonds turning the ball into a liquid. A theory, which links into this
experiment, is the kinetic theory. This is because the kinetic theory deals
with atoms vibrating as they receive more energy and they then break their
bonds. This is linked to this experiment as the squash ball’s atoms get more
energy and vibrate more before breaking their bonds to become a liquid when the
ball hits a critical temperature. I don’t think the graph will go through 0,0,
as even when the ball is at 0 degrees it will still bounce. I am using a large
range of results as well. DiagramMethod??????????????? We set up
the apparatus as shown in diagram and then heated the ball to a set
temperature. We then dropped it from 70 cm high and measured the bounce. We
then repeated that temperature another 4 times to gain an average. We had to be
careful with the Bunsen burner and so we wore goggles. To keep the experiment
fair the only thing, which we changed each time, was the temperature. We used
the same ball through out the experiment and checked the ball was at the same
temperature each time. We dropped it onto the same table from the same height
as well. The range of temperature we used was from 5 degrees Celsius to 70
degrees Celsius. Some of the results needed to be repeated to make sure that
they were accurate.ResultsTemperature (c)??? Measurements
(Cm) ??????????????? Result ???? 1???????? Result ??? 2????????? Result ??? 3 ???????? Result ???? 4???????? Result ???? 5???????? Average 5????????????? 10??????????? 11??????????? 13??????????? 12??????????? 11??????????? 11.9 10??????????? 15??????????? 21??????????? 20??????????? 19??????????? 13??????????? 17.6 20??????????? 20??????????? 23??????????? 21??????????? 26??????????? 24??????????? 22.8 30??????????? 25??????????? 29??????????? 26??????????? 26??????????? 23??????????? 25.8 40??????????? 21??????????? 21??????????? 22??????????? 26??????????? 28??????????? 23.6 50??????????? 30??????????? 30??????????? 29??????????? 28??????????? 25??????????? 28.4 60??????????? 31??????????? 33??????????? 32??????????? 35??????????? 36??????????? 33.4 70??????????? 37??????????? 31??????????? 33??????????? 35??????????? 37??????????? 34.6 ?Conclusion??????????????? From my
results I can conclude that as the temperature of the ball rises the height of
the bounce gets higher. This is in line with the kinetic theory, which defines
that as the ball gets hotter the atoms get more energy and vibrate more. When
the ball hits the surface then the atoms are pushed together and because they
are vibrating more they push each other further away causing the ball to bounce
higher. In this experiment the kinetic theory only lasts for a specific set of
temperatures. This is because when the ball hits a certain temperature it
starts to melt. At 0 degrees Celsius the ball will still bounce as the atoms
are still vibrating. The graph proves that the theory works for this
experiment, as it is a straight line to start with. However as the ball gets
nearer the critical temperature the extra height it bounces becomes less and
less. This is shown as the graph levels off. The sketch graph I drew in my
prediction matched the real graph showing that the science I used to explain my
prediction was correct.Evaluation??????????????? Looking
at my results I can say that they were quite reliable and accurate. I had one
anomalous result even after an average over five measurements. I can say that
looking at my results when I repeated results they were quite close together. I
think that I did the experiment quite well although I found it hard to spot
where the ball bounced too. This is why I did an average over 5 measurements.
To improve the experiment I would need to use specialist equipment like lasers
so I could be sure where the ball bounced too. Ways in which I could extend
this experiment are to use a different kind of rubber in the ball so that it
doesn’t melt at such a low temperature this way I could carry on to see whether
the kinetic theory is still right at higher temperatures. Also I would like to
see what happened when the ball was at 0 degrees Celsius. I would like to do
this to see whether the atoms still vibrated causing the ball to bounce. If it
did I would like to carry on getting lower and lower to see whether there was a
temperature where the atoms no longer vibrated (Absolute Zero)