# 3.1 Collision theory, measurement and mechanism  (Page 3/4)

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## Experiment : measuring reaction rates

Aim:

To measure the effect of concentration on the rate of a reaction.

Apparatus:

• 300 cm ${}^{3}$ of sodium thiosulphate (Na ${}_{2}$ S ${}_{2}$ O ${}_{3}$ ) solution. Prepare a solution of sodium thiosulphate by adding 12 g of Na ${}_{2}$ S ${}_{2}$ O ${}_{3}$ to 300 cm ${}^{3}$ of water. This is solution 'A'.
• 300 cm ${}^{3}$ of water
• 100 cm ${}^{3}$ of 1:10 dilute hydrochloric acid. This is solution 'B'.
• Six 100 cm ${}^{3}$ glass beakers
• Measuring cylinders
• Paper and marking pen
• Stopwatch or timer

Method:

One way to measure the rate of this reaction is to place a piece of paper with a cross underneath the reaction beaker to see how quickly the cross is made invisible by the formation of the sulfur precipitate.

1. Set up six beakers on a flat surface and mark them from 1 to 6. Under each beaker you will need to place a piece of paper with a large black cross.
2. Pour 60 cm ${}^{3}$ solution A into the first beaker and add 20 cm ${}^{3}$ of water
3. Use the measuring cylinder to measure 10 cm ${}^{3}$ HCl. Now add this HCl to the solution that is already in the first beaker (NB: Make sure that you always clean out the measuring cylinder you have used before using it for another chemical).
4. Using a stopwatch with seconds, record the time it takes for the precipitate that forms to block out the cross.
5. Now measure 50 cm ${}^{3}$ of solution A into the second beaker and add 30 cm ${}^{3}$ of water. To this second beaker, add 10 cm ${}^{3}$ HCl, time the reaction and record the results as you did before.
6. Continue the experiment by diluting solution A as shown below.
 Beaker Solution A (cm ${}^{3}$ ) Water (cm ${}^{3}$ ) Solution B (cm ${}^{3}$ ) Time (s) 1 60 20 10 2 50 30 10 3 40 40 10 4 30 50 10 5 20 60 10 6 10 70 10

The equation for the reaction between sodium thiosulphate and hydrochloric acid is:

$N{a}_{2}{S}_{2}{O}_{3}\left(aq\right)+2HCl\left(aq\right)\to 2NaCl\left(aq\right)+S{O}_{2}\left(aq\right)+{H}_{2}O\left(l\right)+S\left(s\right)$

Results:

• Calculate the reaction rate in each beaker. This can be done using the following equation:
$Rate\phantom{\rule{4pt}{0ex}}of\phantom{\rule{4pt}{0ex}}reaction=\frac{1}{time}$
• Represent your results on a graph. Concentration will be on the x-axis and reaction rate on the y-axis. Note that the original volume of Na ${}_{2}$ S ${}_{2}$ O ${}_{3}$ can be used as a measure of concentration.
• Why was it important to keep the volume of HCl constant?
• Describe the relationship between concentration and reaction rate.

Conclusions:

The rate of the reaction is fastest when the concentration of the reactants was the highest.

## Mechanism of reaction and catalysis

Earlier it was mentioned that it is the collision of particles that causes reactions to occur and that only some of these collisions are 'successful'. This is because the reactant particles have a wide range of kinetic energy, and only a small fraction of the particles will have enough energy to actually break bonds so that a chemical reaction can take place. The minimum energy that is needed for a reaction to take place is called the activation energy . For more information on the energy of reactions, refer to Grade 11.

Activation energy

The energy that is needed to break the bonds in reactant molecules so that a chemical reaction can proceed.

Even at a fixed temperature, the energy of the particles varies, meaning that only some of them will have enough energy to be part of the chemical reaction, depending on the activation energy for that reaction. This is shown in [link] . Increasing the reaction temperature has the effect of increasing the number of particles with enough energy to take part in the reaction, and so the reaction rate increases.

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When ∆H<0
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H<0
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no the circles show the sound detected by the observer
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equation for dissociation
under which topic
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acids and bases
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Newton's law
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it the bases that dissociate
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hey
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Holy
it only speed up the reaction not to affect the reaction
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and therefore at the end of reaction it is released .. meaning that it hasn't caused any changes
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also understand the graphs
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