Rank these reactions in order from the fastest to the
slowest.
Combustion is the fastest and weathering is the slowest. The
order of reactivity from fastest to slowest is:
combustion, photosynthesis, rusting, weathering.
How did you decide which reaction was the fastest and which
was the slowest?
Answers could include mention of the fact that coal and paper
burn fast, whereas rocks do not disappear overnight.
Also plants use photosynthesis to make food and this
process has to happen relatively fast but not too fast.
Think of some other examples of chemical reactions. How fast
or slow is each of these reactions, compared with those
listed earlier?
Decomposition of hydrogen peroxide,
relatively fast
Synthesis of water, very fast
Any other reasonable answers
This video is a simple demonstration of how a change in surface area can affect the
average rate of a reaction.
You can see how quickly the fuel burns when spread over the table. Think about how much more fuel
would be needed to cook a meal if you had it spread over a large surface area rather than kept
in a container with a small surface area.
What is a reaction rate? (ESCMY)
In a chemical reaction, the substances that are undergoing the reaction are called the
reactants, while the substances that form as a result of the reaction
are called the products. The reaction rate describes
how quickly or slowly the reaction takes place. So how do we know whether a reaction is
slow or fast? One way of knowing is to look either at how quickly the reactants are
used during the reaction or at how quickly the products form. For
example, iron and sulfur react according to the following equation:
In this reaction, we can observe the speed of the reaction by measuring how long it takes
before there is no iron or sulfur left in the reaction vessel. In other words, the
reactants have been used. Alternatively, one could see how quickly the iron sulfide (the
product) forms. Since iron sulfide looks very different from either of its reactants,
this is easy to do.
In this case, the reaction rate depends on the speed at which the reactants (oxygen gas and
solid magnesium) are used, or the speed at which the product (magnesium oxide) is
formed.
Reaction rate
The average rate of a reaction describes how quickly reactants are
used or how quickly products are formed during a
chemical reaction.
The average rate of a reaction is expressed as the number of moles of reactant used, divided
by the total reaction time, or as the number of moles of product formed, divided by the
total reaction time.
Average reaction rate for:
the use of a reactant = \(\dfrac{
\text{moles reactant used}}{\text{reaction time (s)}}\)
the formation of a product = \(\dfrac{
\text{moles product formed}}{\text{reaction time (s)}}\)
Using the magnesium reaction shown earlier:
Average reaction rate of \(\text{Mg}\) being used = \(\dfrac{
\text{moles Mg used}}
{\text{reaction time (s)}}\)
Average reaction rate of \(\text{O}_2\) being used = \(\dfrac{
\text{moles O}_2\text{ used}}{\text{reaction time (s)}}\)
Average reaction rate of \(\text{MgO}\) being formed = \(\dfrac{
\text{moles MgO formed}}{\text{reaction time (s)}}\)
The rate of the reaction is \(\text{0,005}\) \(\text{mol·s$^{-1}$}\)
Reaction rates
Textbook Exercise 7.2
A number of different reactions take place. The table below
shows the number of moles of reactant that are used in a
particular time for each reaction.
Reaction
Reactants used (mol)
Time (s)
Reaction rate
(\(\text{mol·s$^{-1}$}\))
\(\text{1}\)
\(\text{2}\)
\(\text{30}\)
\(\text{2}\)
\(\text{5}\)
\(\text{120}\)
\(\text{3}\)
\(\text{1}\)
\(\text{90}\)
\(\text{4}\)
\(\text{3,2}\)
\(\text{90}\)
\(\text{5}\)
\(\text{5,9}\)
\(\text{30}\)
Complete the table by calculating the average rate of
each reaction.
The reaction rate is the number of moles used up
divided by the time in seconds.
Reaction
Reactants used
(mol)
Time (s)
Reaction
rate
(\(\text{mol·s$^{-1}$}\))
\(\text{1}\)
\(\text{2}\)
\(\text{30}\)
\(\text{0,067}\)
\(\text{2}\)
\(\text{5}\)
\(\text{120}\)
\(\text{0,042}\)
\(\text{3}\)
\(\text{1}\)
\(\text{90}\)
\(\text{0,011}\)
\(\text{4}\)
\(\text{3,2}\)
\(\text{90}\)
\(\text{0,036}\)
\(\text{5}\)
\(\text{5,9}\)
\(\text{30}\)
\(\text{0,2}\)
Which is the fastest reaction?
The fastest reaction is reaction \(\text{5}\)
Which is the slowest reaction?
The slowest reaction is reaction \(\text{3}\)
Iron reacts with oxygen as shown in the balanced reaction:
\(\text{2}\) \(\text{g}\) of \(\text{Fe}\) and
\(\text{0,57}\) \(\text{g}\) of \(\text{O}_{2}\) are
used during the reaction. \(\text{2,6}\) \(\text{g}\) of
\(\text{FeO}\) is produced. The reaction takes
\(\text{30}\) \(\text{minutes}\) to go to completion.
Average rate of reaction for \(\text{Fe}\) =
\(\dfrac{\text{0,0358}\text{
mol}}{\text{1 800}\text{ s}}=\text{1,99}
\times \text{10}^{-\text{5}}\text{
mol·s$^{-1}$}\)
Rate of reaction is number of moles \(\text{O}_{2}\)
used up per second.
Average rate of reaction for \(\text{O}_{2}\) =
\(\dfrac{\text{0,0178}\text{
mol}}{\text{1 800}\text{ s}}=\text{9,89}
\times \text{10}^{-\text{6}}\text{
mol·s$^{-1}$}\)
Rate of reaction is number of moles \(\text{FeO}\)
produced per second.
Average rate of reaction for \(\text{FeO}\) =
\(\dfrac{\text{0,0362}\text{
mol}}{\text{1 800}\text{ s}}=\text{2,01}
\times \text{10}^{-\text{5}}\text{
mol·s$^{-1}$}\)
Note that the rates of the individual reactions
follow the stoichiometric rates ratios in the
balanced equation:
\(\text{1,99} \times
\text{10}^{-\text{5}}:\text{9,89} \times
\text{10}^{-\text{6}}:\text{2,01} \times
\text{10}^{-\text{5}}\) is \(2:1:2\)
Two reactions occur simultaneously in separate reaction
vessels. The reactions are as follows:
m = n \(\times\) M = \(\text{0,021}\)
\(\text{mol}\) \(\times\)
\(\text{23,0}\)
\(\text{g·mol$^{-1}$}\)
= \(\text{0,48}\) \(\text{g}\)
Reaction rates and collision theory (ESCMZ)
It should be clear now that the average rate of a reaction varies depending on a number of
factors. But how can we explain why reactions take place at different speeds under
different conditions? Collision theory is used to explain the rate of a
reaction.
For a reaction to occur, the particles that are reacting must collide with one another. Only
a fraction of all the collisions that take place actually cause a chemical change. These
are called successful or effective collisions.
Collision theory
Reactant particles must collide with the correct energy and orientation for
the reactants to change into products.
Collision theory explains how chemical reactions occur and why reaction rates differ for
different reactions. It states that for a reaction to occur the reactant particles must:
collide
have enough energy
have the right orientation at the moment of impact
These successful collisions are necessary to break the existing bonds (in the
reactants) and form new bonds (in the products).
Collision Theory
Aim
To determine the best way to approach your friend, in order to link your
right arm with their left arm.
Method
Try different ways of approaching your friend:
back to back
front to back
side to front
side to side
front to front
etc
Results
Determine how hard it is to link arms in each of these positions.
Discussion
If you approach your friend from behind (facing their back) it is hard to
link arms. Approaching from their left (sideways so that your right side
is on their left), it is easy to link up.
Conclusion
You should have found that each method had a different level of difficulty
for linking arms. This is similar to how molecules (compounds) approach
in a reaction. The different ways you approached your friend represent
the different orientations of the molecules. The correct orientation
makes successful collisions possible.
Factors affecting reaction rates (ESCN2)
Several factors affect the average rate of a reaction. It is important to know these factors
so that reaction rates can be controlled. This is particularly important when it comes
to industrial reactions, where greater productivity leads to greater profits for
companies. The following are some of the factors that affect the average rate of a
reaction.
Nature of reactants
Substances have different chemical properties and therefore react differently, and at
different rates (e.g. the rusting of iron vs. the tarnishing of silver).
Oxalic acid is abundant in many plants. The leaves of the tea plant
(Camellia sinensis) contain very high concentrations of oxalic
acid relative to other plants. Oxalic acid also occurs in small amounts
in foods such as parsley, chocolate, nuts and berries. Oxalic acid
irritates the lining of the gut when it is eaten, and can be fatal in
very large doses.
In the nature of reactants, surface area and concentration experiments
learners are required to work with concentrated, strong acids. These
acids can cause serious burns. Please remind the learners to be careful
and wear the appropriate safety equipment when handling all chemicals,
especially concentrated acids. The safety equipment includes gloves,
safety glasses and protective clothing.
The nature of reactants
Aim
To determine the effect of the nature of reactants on the average
rate of a reaction.
Apparatus
You will need the following items for this experiment:
a spatula, two test tubes, a medicine dropper, a
glass beaker and a glass rod.
Method
Concentrated \(\text{H}_{2}\text{SO}_{4}\) can cause serious
burns. We suggest using gloves and safety glasses
whenever you work with an acid. Remember to add the acid
to the water and to avoid sniffing the acid. Handle all
chemicals with care.
Label one test tube \(\text{1}\).
Prepare an iron(II) sulfate solution in test
tube \(\text{1}\) by dissolving two spatula tips
of iron(II) sulfate in \(\text{10}\)
\(\text{cm$^{3}$}\) of water.
Label the other test tube
\(\text{2}\). Prepare a
solution of oxalic acid in test tube
\(\text{2}\) in the same way.
Prepare a separate solution of sulfuric acid by
adding \(\text{2}\) \(\text{cm$^{3}$}\) of the
concentrated acid to \(\text{10}\)
\(\text{cm$^{3}$}\) of water. Remember always to
add the acid to the water, and
never the other way around.
Add \(\text{2}\) \(\text{cm$^{3}$}\) of the sulfuric
acid solution to the iron(II) sulfate and oxalic
acid solutions respectively.
Using the medicine dropper, add a few drops of
potassium permanganate to the two test tubes.
Observe how quickly the potassium permanganate
solution discolours in each solution.
Results
You should have seen that the the potassium
permanganate discolours in the oxalic acid
solution much more slowly than in the iron(II)
sulfate solution.
These reactions can be seen in the following videos:
It is the oxalate ions
\((\text{C}_{2}\text{O}_{4}^{2-})\) and the
\(\text{Fe}^{2+}\) ions that cause the
discolouration. It is clear that the
\(\text{Fe}^{2+}\)ions react much more quickly
with the permanganate than the
\((\text{C}_{2}\text{O}_{4}^{2-})\) ions. The
reason for this is that there are no covalent
bonds to be broken in the iron ions before the
reaction can take place. In the case of the
oxalate ions, covalent bonds between carbon and
oxygen atoms must be broken first.
Conclusions
Despite the fact that both these reactants (oxalic acid and iron(II)
sulfate) are in aqueous solutions, with similar concentrations
and at the same temperature, the reaction rates are very
different. This is because the nature of the reactants can
affect the average rate of a reaction.
The nature of the iron(II) sulfate in solution (iron ions, ready to
react) is very different to the nature of oxalic acid in
solution (oxalate ions with covalent bonds that must be broken).
This results in significantly different reaction rates.
The \(\text{KMnO}_{4}\) with oxalic acid and iron(II) sulfate pictures are
screenshots from videos by katalofuromai and Aaron Huggard on Youtube.
Surface area (of solid reactants)
Surface area and reaction rate
Marble \((\text{CaCO}_{3})\) reacts with hydrochloric acid \((\text{HCl})\)
to form calcium chloride, water and carbon dioxide gas according to the
following equation:
Concentrated \(\text{HCl}\) can cause serious burns. We
suggest using gloves and safety glasses whenever you
work with an acid. Remember to add the acid to the water
and handle with care.
Prepare a solution of hydrochloric acid in the beaker
by adding \(\text{2}\) \(\text{cm$^{3}$}\) of
the concentrated acid to \(\text{20}\)
\(\text{cm$^{3}$}\) of water.
Place the marble chips into one test tube and the
powdered marble into a separate test tube.
Add \(\text{10}\) \(\text{cm$^{3}$}\) of the dilute
hydrochloric acid to each of the test tubes and
observe the rate at which carbon dioxide gas
(\(\text{CO}_{2}\)) is produced (you should see
bubbles of \(\text{CO}_{2}\)).
Results
Note (write down) what you observe.
Questions and discussion
Which reaction proceeds faster?
Can you explain this?
Conclusion
The reaction with powdered marble is faster. The smaller the pieces
of marble are (in this case the powdered form is smallest), the
greater the surface area for the reaction to take place.
Only the molecules at the surface of the solid can react with the
hydrochloric acid. The next layer of molecules can only react
once the surface molecules have reacted. That is, the next layer
of molecules becomes the surface.
The chips of marble are relatively large, so only a small percentage
of the molecules are at the surface and can react initially. The
powdered marble has much smaller solid pieces, so there are many
more surface molecules exposed to the hydrochloric acid. The
more molecules exposed on the surface (the greater the surface
area) the faster the reaction will be.
For the same amount of mass, smaller pieces of solid react faster as
shown in Figure 7.2.
Figure 7.2: a) A large particle,
b) small particles with the
same volume as the large
particle, c) The surface area
of
large
particles (shown in blue) is much smaller than
that of small particles (shown in red).
If we react \(\text{1}\) \(\text{g}\) of \(\text{CaCO}_{3}\) we find that the
reaction is faster if the \(\text{CaCO}_{3}\) is powdered when compared with the
\(\text{CaCO}_{3}\) being large lumps.
Explanation:
The large lump of \(\text{CaCO}_{3}\) has a small surface area relative to the same
mass of powdered \(\text{CaCO}_{3}\). This means that more particles of
\(\text{CaCO}_{3}\) will be in contact with \(\text{HCl}\) in the powdered
\(\text{CaCO}_{3}\) than in the lumps. As a result, there can be more
successful collisions per unit time and the reaction of powdered
\(\text{CaCO}_{3}\) is faster.
\(\color{red}{\textbf{Increasing the surface area of the reactants increases the rate
of the reaction.}}\)
The following video shows the effect of surface area on the time an effervescent
tablet takes to fully dissolve. The tablet is fully dissolved once the bubbles
(\(\text{CO}_{2}\) gas) stop forming.
Surface area, concentration and pressure all have the same effect on reaction
rate (an increase leads to a faster reaction rate). This is because in
each case an increase in the property leads to an increase in the number
of collisions in that phase of matter.
As the concentration of the reactants increases, so does the reaction rate.
Concentration and reaction rate
Aim
To determine the effect of reactant concentration on reaction rate.
Two beakers, two test tubes and a measuring cylinder.
Method
Do not get hydrochloric acid (\(\text{HCl}\)) on your hands.
We suggest you use gloves and safety glasses whenever
handling acids and handle with care.
When diluting a solution remember that if you want a
1:10 solution (1 part original solution in 10
parts water) measure \(\text{10}\)
\(\text{cm$^{3}$}\) of water in a measuring
cylinder and pour it into a beaker, then add
\(\text{1}\) \(\text{cm$^{3}$}\) of the original
solution to the beaker as well. \(\text{2}\)
parts concentrated acid to \(\text{20}\) parts
water will also be a 1:10 solution. Remember to
always add the acid to the
water, and not the other way around.
Prepare a solution of 1 part acid to
10 parts water (1:10). Label a
test tube
A and
pour \(\text{10}\)
\(\text{cm$^{3}$}\) of this
solution into the test tube.
Prepare a solution of 1 part acid to
20 parts water (1:20). Label a
test tube
B and
pour \(\text{10}\)
\(\text{cm$^{3}$}\) of this
solution into the test tube.
Take two pieces of magnesium ribbon of the
same length. At the same time,
put one piece of magnesium ribbon into test tube
A and the other into test tube B, and pay close
attention to what happens.
Make sure that the magnesium ribbon is long
enough so that your hand is not close to
the \(\text{HCl}\).
Write down what happened (what did you observe?) in each test tube.
Questions and discussion
Which of the two solutions is more concentrated, the
\(\text{1}\):\(\text{10}\) or
\(\text{1}\):\(\text{20}\) hydrochloric acid
solution?
In which of the test tubes is the reaction faster?
Suggest a reason for this.
How can you measure the average rate of this
reaction?
Name the gas that is produced?
Why is it important that the same length of magnesium
ribbon is used for each reaction?
Conclusions
The \(\text{1}\):\(\text{10}\) solution is more concentrated and
therefore this reaction proceeds faster. The greater the
concentration of the reactants, the faster the average rate of
the reaction. The average rate of the reaction can be measured
by the rate at which the magnesium ribbon disappears.
The greater concentration of the reactant means that there are more particles of
reactant (\(\text{HCl}\)) per unit volume of solution. Therefore the chance that
\(\text{HCl}\) particles will collide with the \(\text{Mg}\) particles will be
higher for the solution with the greater concentration. The number of successful
collisions per unit time will be higher and so the rate of the reaction will be
faster.
In this project the learners should design their own experiment in the
following format:
Aim
Apparatus
Method
They can also perform the experiment and write up results and conclusions as
well.
This experiment should focus on the effect of concentration
on the rate. The easiest way to do this is to vary the concentration of
the vinegar and keep the mass of baking soda constant.
Concentration and rate
Design an experiment to determine the effect of concentration on rate using
vinegar and baking soda.
Hint: mix water and vinegar to change concentration but keep
the total volume constant.
Pressure (of gaseous reactants)
As the pressure of the reactants increase, so does the reaction rate.
The higher the pressure, the more particles of gas per unit volume. Therefore there
are more collisions per unit time. The number of successful collisions per unit
time will be higher and so the rate of the reaction will be faster.
If the temperature of the reaction increases, so does the average rate of the
reaction.
In the temperature and reaction rate experiment make sure the learners do not
shake the test tubes. Shaking gives energy to the reaction and affects
the rate. The test tubes should be left as still as possible once the
effervescent tablets have been added.
Plastic bottles, such as those shown in the picture, can be used instead of
test tubes.
Temperature and reaction rate
Aim
To determine the effect of temperature on reaction rate.
Apparatus
Two effervescent tablets (e.g. Cal-C-Vita)
An ice-bath, two test tubes
Two balloons, two rubber bands
Method
Half fill two large test tubes with water. Label them
A and B.
Break two effervescent tablets in two or three pieces
and place them in the two balloons.
Fit one of these balloons tightly to test tube A and
one to test tube B, being careful not to drop
the contents into the water. You can stand the
test tube in a beaker to help you do this.
Place only test tube A into an ice-bath and
leave to equilibrate (come to the same
temperature). Approximately 10 minutes should be
enough.
At the same time lift the balloons on test tubes A
and B so that the tablets go into the water.
Do not shake either test tube.
\(\text{CO}_{2}\)(g) is released during this
reaction.
Observe how quickly the balloons increase in size and
write down your observations (which increases in
size faster).
Results
Note (write down) your observations.
Questions and discussion
Which balloon expanded faster?
Suggest a reason for the difference in rates.
Conclusions
The balloon on test tube B will expand faster. This is because the
higher temperature (room temperature rather than an ice bath)
leads to an increase in the average rate of \(\text{CO}_{2}\)
gas production.
The video below shows how much pressure can build up when \(\text{CO}_{2}\)(g) is
released during the reaction of an effervescent tablet with water.
The higher the temperature, the greater the average kinetic energy of the particles,
which means that the particles are moving faster.
Therefore:
particles moving faster means more collisions per unit time
(collision theory)
particles with higher kinetic energy are also more likely to react on
colliding as they have enough energy for the reaction to occur
(see Section 7.4 on the mechanism of reaction).
Adding a catalyst increases the reaction rate by lowering the energy required for a
successful reaction to take place. A catalyst speeds up a reaction and is
released at the end of the reaction, completely unchanged.
In the first catalyst and reaction rate experiment (with manganese dioxide
and hydrogen peroxide) it is important to note that hydrogen peroxide
can cause burns. The learners should wear safety equipment, as always
when handling chemicals. If the concentration of hydrogen peroxide is
too high the liquid hydrogen peroxide may splash out of the container
along with the oxygen gas. As a result the learners should be
particularly careful around the mouth of the containers.
In the second experiment the learners are again working with a strong acid
and should follow all the usual safety procedures.
Catalysts and reaction rate
Aim
Hydrogen peroxide decomposes slowly over time into water and oxygen.
The aim of this experiment is to determine the effect a catalyst
has on the reaction rate.
Figure 7.3: \(\text{H}_{2}\text{O}_{2}\) before the
addition of \(\text{MnO}_{2}\) (left) and after
the addition of \(\text{MnO}_{2}\) (right).
Results
Which chemical compounds are acting as catalysts in
these reactions?
What causes the bubbles that form in the reaction?
Conclusions
The bubbles that form are oxygen gas formed through the decomposition
of hydrogen peroxide. This would happen over time without the
presence of the catalyst. The manganese dioxide speeds up the
reaction significantly. The yeast speeds up the reaction, but
not as much as the manganese dioxide.
Do not get hydrochloric acid (\(\text{HCl}\)) on your hands.
We suggest you use gloves and safety glasses whenever
handling acids. Be especially careful when removing the
copper pieces from the test tube.
Place a few of the zinc granules in the test tube,
using tongs.
Measure the mass of a few pieces of copper and, using
tongs, keep them separate from the rest of the
copper.
Add \(\text{20}\) \(\text{cm$^{3}$}\) of
\(\text{HCl}\) to the test tube. You will see
that a gas is released. Take note of how quickly
or slowly this gas is released (use a stopwatch
or your cellphone to time this). Write a
balanced chemical equation for the chemical
reaction that takes place.
Now add the copper pieces to the same test tube. What
happens to the rate at which the gas is
produced?
Carefully remove the copper pieces from the test tube
(use tongs), rinse them in water and alcohol and
then weigh them again. Has the mass of the
copper changed since the start of the
experiment?
Results
During the reaction, the gas that is released is hydrogen. The rate
at which the hydrogen is produced increases when the
copper pieces (the catalyst) are added. The mass of the copper
does not change during the reaction.
Conclusions
The copper acts as a catalyst during the reaction. It speeds
up the average rate of the reaction, but is not changed itself
in any way.
We will return to catalysts in more detail once we have explored the mechanism of
reactions later in this chapter.
In the iodine clock experiment it is important that the learners start timing the
experiment as soon as the sulfuric acid and hydrogen peroxide solution is added
to the potassium iodide solution. There should be a sudden colour change from
colourless to purple when the sodium thiosulfate is used up and free iodine is
available in the solution. The free iodine is what gives the reaction the purple
colour.
This experiment is best done in groups (\(\text{3}\) - \(\text{4}\) if often a good
size). You can divide your class into groups and assign each group a different
experiment. Afterwards the groups can present their results and conclusions to
the class. If you have time you can also vary the concentration of the hydrogen
peroxide.
As always, learners need to work carefully with acids, in particular with the
concentrated acids. Remind them to always add the acid to the water.
Temperature, concentration and reaction rate
Aim
To determine the effect of temperature and concentration on the average
reaction rate of the iodine clock experiment. This experiment is best
done in groups.
Five beakers, a measuring cylinder, a hotplate, an ice bath,
a glass stirring rod, a stop-watch
Method
Preheat the hotplate to \(\text{40}\) \(\text{℃}\)
Label a beaker solution 1. Measure \(\text{75}\)
\(\text{ml}\) \(\text{H}_{2}\text{SO}_{4}\) into the
beaker. Add \(\text{25}\) \(\text{ml}\) 3%
\(\text{H}_{2}\text{O}_{2}\). Remember to use dilute
(\(\text{0,2}\) \(\text{mol·dm$^{-3}$}\)) sulfuric
acid.
The equations for what is occuring in this reaction are given below:
It is good scientific practice to vary only one factor at a time during an
experiment. Therefore, this experiment has two parts. First we will vary
the concentration of \(\text{KI}\), then we will vary the temperature:
Varying the concentration
Weigh out \(\text{0,5}\) \(\text{g}\) of
\(\text{KI}\) into a beaker and label it
A.
Weigh out \(\text{1}\) \(\text{g}\) of
\(\text{KI}\) into a different beaker
and label it B.
Add \(\text{20}\) \(\text{ml}\)
\(\text{Na}_{2}\text{S}_{2}\text{O}_{3}\)
to both beaker A and beaker B.
Add a spatula of soluble starch to both
beaker A and beaker B and stir with a
glass rod.
Measure \(\text{15}\) \(\text{ml}\) of
solution 1 with the measuring
cylinder. Get your stopwatch ready. Pour
the \(\text{15}\) \(\text{ml}\) of
solution 1 into beaker A and start
timing.
Stop timing when the solution starts to
change colour. Write down your time in
the table below.
Repeat step \(\text{5}\) with beaker B.
Beaker
Concentration (M)
Temperature
(℃)
Time (s)
A
approx. 0.15
room temperature
B
approx. 0.3
room temperature
Varying the temperature
Weigh out \(\text{0,5}\) \(\text{g}\) of
\(\text{KI}\) into a new beaker and
label it C.
Add \(\text{20}\) \(\text{ml}\)
\(\text{Na}_{2}\text{S}_{2}\text{O}_{3}\)
to beaker C.
Add a spatula of soluble starch to beaker C
and stir with a glass rod.
Measure \(\text{15}\) \(\text{ml}\) of
solution 1 with the measuring
cylinder.
Place beaker C in the ice bath.
Get your stopwatch ready. Pour the
\(\text{15}\) \(\text{ml}\) of solution
1 into beaker C and start timing. Stop
timing when the solution starts to
change colour. Write down your time in
the table below.
Repeat steps 1 - 4 (label the beaker
D).
Place beaker D on the hotplate. Then
repeat step 6
Beaker
Concentration (M)
Temperature
(℃)
Time (s)
A
approx. 0.15
room temperature
C
approx. 0.15
0
D
approx. 0.15
40
Beaker A has been included here because it has the same
concentration as beakers C and D, but is at a different
temperature.
Results
Make a table with the information for all the beakers. Include columns for
concentration, temperature, time, and reaction rate.
Questions and discussion
Did beaker A or B have the faster reaction rate?
Why did it have a faster reaction rate?
Did beaker A, C or D have the fastest reaction rate? Why?
Did beaker A, C or D have the slowest reaction rate? Why?
Conclusions
You will notice that the faster reaction rate occurs in the beaker with the
higher concentration of \(\text{KI}\). You should also see that the
higher the temperature, the faster the reaction rate.
This video shows how this experiment can be used as a clock with the concentration chosen so
that the experiment changes colour at a specific time (or with a particular part of a
song). This is why this experiment is known as the iodine clock reaction.
Write a balanced equation for the exothermic reaction between
\(\text{Zn}(\text{s})\) and \(\text{HCl}(ℓ)\). Also name three
ways to increase the rate of this reaction.
Write the equation for zinc and hydrochloric acid
The products must be a salt and hydrogen gas. Zinc ions have a charge of 2+
while chloride ions have a charge of 1-. Therefore the salt must be
\(\text{ZnCl}_{2}\).
There are more chloride ions and hydrogen atoms on the right side of the
equation. Therefore there must be \(\text{2}\) \(\text{HCl}\) on the
left side of the equation.
The volume of carbon dioxide that is produced during the
reaction is measured at different times. The results are
shown in the table below.
Time (mins)
Total Volume of \(\text{CO}_{2}\)
produced
(\(\text{cm$^{3}$}\))
\(\text{1}\)
\(\text{14}\)
\(\text{2}\)
\(\text{26}\)
\(\text{3}\)
\(\text{36}\)
\(\text{4}\)
\(\text{44}\)
\(\text{5}\)
\(\text{50}\)
\(\text{6}\)
\(\text{58}\)
\(\text{7}\)
\(\text{65}\)
\(\text{8}\)
\(\text{70}\)
\(\text{9}\)
\(\text{74}\)
\(\text{10}\)
\(\text{77}\)
Note: On a graph of production against time, it is the
gradient of the tangent to the graph that shows
the rate of the reaction at that time.
e.g.
Use the data in the table to draw a graph showing the
volume of gas that is produced in the reaction,
over a period of 10 minutes.
(Remember to label the axes and plot the graph on
graphing paper)
At which of the following times is the reaction
fastest: \(\text{1}\)
\(\text{minute}\); \(\text{6}\)
\(\text{minutes}\) or \(\text{8}\)
\(\text{minutes}\). Explain.
Time = 1 minute. This is where the the gradient of a
tangent to the graph is the steepest (the red
line on the graph). The steeper the gradient the
faster the rate at that time.
Suggest a reason why the reaction slows down over
time.
As the reaction proceeds the reactants are used up
(form products). With a lower concentration of
reactants the rate of the reaction decreases.
Use the graph to estimate the volume of gas that will
have been produced after 11 minutes.
Approximately \(\text{79}\) \(\text{cm$^{3}$}\)
How long do you think the reaction will take to stop
(give a time in minutes)?
Any answer between 15 and 25 minutes is reasonable.
To see this extend the line and find approximate
the time that the gradient flattens out.
If the experiment was repeated using a more
concentrated hydrochloric acid solution:
would the average rate of the reaction
increase or decrease from the one shown
in the graph?
The rate would increase.
draw a line on the same set of axes to show
how you would expect the reaction to
proceed with a more concentrated
\(\text{HCl}\) solution.
The red line indicates roughly how the
reaction would proceed. Note that the
reaction does not produce more carbon
dioxide, it just reacts faster.
