Investigating How Concentration and Temperature Affect the Rate of Reaction Essay Sample

Investigating How Concentration and Temperature Affect the Rate of Reaction Pages
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I aim to investigate how the concentration of potassium bromide affect the rate of reaction when added to a solution made up of potassium bromate(V), sulphuric acid and phenol.

The general equation for the reaction between bromide and bromate ions in acidic aqueous solution is: (1)

BrO3-(aq) + 5Br -(aq) + 6H+(aq) 3Br2(aq) + 3H2O(l)

I am going to alter the concentrations of potassium bromide, potassium bromate and sulphuric acid to find the orders of reaction with respect to each reactant. This will allow me to prove that the rate equation for this reaction is: (1, pg 230)

Rate = k[BrO3-][Br -][H+]2

I will be working out a value for the rate constant, k, and will also be investigating the affect temperature has on rate of reaction, using my results and the Arrhenius equation to work out the activation enthalpy.


The colour change in this reaction is from an orange, brought about by the methyl orange indicator, to a colourless solution. As bromine is produced in the reaction it, at first, bonds to the phenol. However, when all binding sites on phenol have been used up, the presence of excess bromine will turn the solution colourless. This shows that I am measuring the time taken for all binding sites on phenol to be used up, which explains the change from orange to colourless in the solution. This can be shown in the following equation:

2C6H5OH(aq) + 3Br2(aq) 2C6H2Br3OH(aq) + 6H+(aq) + 6Br -(aq)

The term ‘rate of reaction’ refers to how fast reactants are converted into products in a given reaction. A way to measure rate of reaction directly has not yet been discovered, and therefore we must measure the change in amount of product/reactant in a certain time. This can be shown by the following equation: (1, pg 228)

Rate = Change in Property

Time Taken

However, orders of reaction must also be taken into account when writing a rate equation. Where A and B are reactants, m and n are orders of reaction (with respect to A and B) and k is the rate constant:

Rate = k[A]m[B]n

Reactions can be either zero, first or second order with respect to each molecule used in the reaction. A reaction with order zero will be a horizontal line on a graph of concentration against rate of reaction. First order will be a straight diagonal line and second order will show a curve. This is shown in Figure 4 below: (2)

For a reaction with order zero:

Rate = k[A]0

This means, due to the fact that anything to the power of zero equals one, the equation for a zero order reaction is:

Rate = k

This shows that, for order zero reactions, the rate is independent of the concentration of ‘A’. It has no effect on the rate of reaction. If this is the case, the reactant in question is excluded from the rate equation, as it has no bearing on it.

In some reactions changing the concentration of a particular reactant has no effect on the rate. When this is the case it means there are two steps involved in the reaction – this is known as the reaction mechanism. Where there are two or more steps it is the slowest step that determines the overall rate. This is called the rate determining step. I will suggest a mechanism later in my investigation for the reaction, based on the rate equation.

The rate equation for the reaction I am investigating is shown in my ‘Aim’ section. This shows that the reaction is expected to be first order with respect to both bromate ions and bromide ions, and second order with respect to hydrogen ions.

The rate constant in a rate equation, shown as k, is not affected by a change in concentration but is affected by a change in temperature. This means that the temperature of an equation must always be known when working out the rate of reaction.

Several different things contribute to a change in rate of reaction which concerned in my experiment:

* Concentration

* Temperature

* Collision Frequency

* Activation Enthalpy

The Collision Theory can account for changes in both concentration and temperature affecting rate of reaction. This theory suggests that reactions occur when the particles of reactants collide, provided that they collide with a certain minimum kinetic energy.

Concentration of Reactants:

If the concentration of the reactant molecules is increased there will be more molecules. This means that the likelihood of collisions occurring is increased, so a reaction will happen sooner (as shown in figure 3). This increased the ‘rate of reaction’ as a reaction occurs in a shorter space of time. The diagrams below show this effect: (1, pg 221)


When to molecules collide, they must do so with enough energy for the reaction to take place. If the temperature of the reactants is increased, the energy of the molecules will increase. This also means that the molecules will move at a faster speed, meaning that collisions are likely to occur sooner. For many reactions happening at room temperature (roughly 20-25�C), it is known that the rate of reaction doubles each time the temperature is increased by 10�C.(1, pg 224) However, this is not true for all equations and even equations where this is true the value may be slightly different, for example 9 or 11 rather than 10�C. Also, the temperature change needed to double the rate of reaction will vary gradually as temperature is increased.

Increasing the Collision Frequency:

The frequency of a collision between two molecules in a reaction is proportional to the square root of the temperature in Kelvin.(3) For example, if the temperature is increased from 293K to 303K, a change from 20 to 30�C, the factor increase can be calculated as follows:


310 = 1.016

This is an increase of 1.6% for a temperature change of +10�C. As I have already mentioned, the rate of reaction will have approximately doubled with this rise in temperature – this is around a 100% increase. This shows that the effect of increasing collision frequency on the rate of reaction is almost negligible.

Activation Enthalpy: (1, pg 224/225)

The Activation Enthalpy is described as the minimum amount of kinetic energy needed before a collision will occur between two molecules. This energy is needed for bonds in the reactants to be broken and new bonds in the products to be formed.

An important factor in the rate of reaction is how much energy the particles involved have. This energy is increased when the temperature of the reaction is increased. The Maxwell-Boltzmann distribution is a good way of showing this:

Figure 3 above shows that as temperature increases, more molecules move at higher speeds and have a higher kinetic energy. Figure 4 shows the change when the temperature is increased by 10K. It shows that there is still a wide range of energies but now there are more particles with a higher kinetic energy. At the higher temperature of 310 Kelvin there are many more molecules which have reached the activation enthalpy level than at 300K. The amount of molecules reaching the activation enthalpy is about twice as many at 310K than at 300K, so this means that the reaction will go around twice as fast at 310K in comparison to the reaction taking place at 300K.

Put simply, “reactions go faster at higher temperatures because a larger proportion of the colliding molecules have the minimum activation enthalpy needed to react.”

The Arrhenius Equation (4)


k = rate constant

A = Arrhenius constant (also known as frequency factor)

e = mathematical value similar to pi (has a value of 2.71828)

Ea = the activation enthalpy

R = the gas constant (has a value of 8.31 J K-1 mol-1

T = temperature (in Kelvin)

k = Ae-Ea/RT

This equation can be rearranged by taking the log (to the base e) form of the equation. Log to the base e is denoted by ln:

ln k = In A – Ea


This can be further rearranged to give:

ln k = -Ea x 1 + ln A


If a graph of In k against 1/T is drawn, it should form a straight line of best fit with gradient -Ea/R.

This gradient can be measured and the activation enthalpy can be worked out.

Making Molar Solutions (5)

Molar (M) solutions are based on the number of moles of chemical in 1 litre of solution. A mole consists of 6.02×1023 molecules or atoms. Relative Formula Mass (RFM) is the weight of one mole of a chemical. This can be determined using a periodic table by adding the atomic mass of each atom in the chemical formula. I have shown below how to work out the RFM of a compound, using ‘X’ and ‘Y’ as elements, with ‘n’ and ‘m’ as their respective molecular masses:


X2 = 2 x n = 2n

Y3 = 3 x m = 3m

= 2n3m g/mole

Therefore, a 1M solution of X2Y3 consists of 2n3m grams of X2Y3 dissolved in enough water to make one litre of solution.

Once the molecular weight of a chemical is known, the weight (in grams) of chemical to dissolve in water for a molar solution less than 1M is can be calculated using the following formula:

grams = (desired molarity of solution) x (RFM of chemical) x (desired amount of solution/ml)


For example, to make 100 ml of 0.1M X2Y3 solution, you would work out how many grams of X2Y3 to use as is shown below:

grams of X2Y3 = [0.1 x (2n3m) x 100]


This would give you the amount of grams of X2Y3 required, with the rest of the solution being made up with distilled water.


Firstly I have decided to use burettes rather than measuring cylinders for measuring out my solutions. This is due to the fact that they are more accurate apparatus. The error associated with a 50cm3 burette is +/- 0.1cm3. This is far more accurate than the value for a 25cm3 measuring cylinder, which is +/- 0.5cm3. When using a burette the measurement should be read from the bottom of the meniscus. It should also be read at eye level to avoid a parallax error.

Each measurement in my actual experiment will be carried out 5 times to confirm my result.

Repetitions increase reliability by proving that each reading is not a ‘one off’ and can be achieved the majority of the time. An average can then be taken from the 5 repetitions. This average time can be used to work out reaction rate.

Volumes of 5cm3 have been used to keep the percentage errors to a minimum, although at the same time making sure that I have enough solution that I can see any colour change easily. This means that my percentage error for the burettes is as follows:

Percentage error = error

measured value x 100

So for measuring, for example, 5cm3 from a burette the error would be:

Percentage error = 0.1

5 x 100

= 2%

I need to decide upon the range of concentrations of potassium bromide I will use in my experiment. To do this I will test concentrations of 0.001, 0.002, 0.003, 0.004, 0.005, 0.006, 0.008 and 0.01mol/dm-3 of potassium bromide. I will then see how long each test took and decide how many of these trials I can do in the time I have available.

Secondly I need to find the best way of stirring the mixture whilst the reaction occurs. Preferably, I need a way of stirring which will reduce the time taken for the solution to go colourless. This would mean that I can fit in more repeats and more concentrations of potassium bromide can be tested. However, I also need a method which does not stir the mixture so vigorously that the colour change cannot be seen easily.

Pre-test Results:

Concentration of Potassium Bromide (mol/dm-3)

Time Take For the Mixture to Turn Colourless (seconds)

















Although all of these concentrations are shown to have worked, I do not have enough time available to conduct results including each concentration. If I am going to repeat each concentration several times for more accurate results I need to investigate a smaller range of concentrations. I have chosen to exclude the two lowest concentrations (0.001 and 0.002mol/dm-3) when conducting my actual experiment. This is because they took the longest time for the mixture to turn colourless. Whilst conducting this pre-test I used a magnetic stirrer. However, I was not sure whether this was the most appropriate method to use. To solve this I conducted another pre-test, which is shown in the table below:

Concentration of Potassium Bromide (mol/dm-3)

Time Taken When Stirred by Hand (seconds)

Time Taken When Stirred by Magnetic Stirrer (seconds)

Time Taken When Left Without Stirring (seconds)

























From my results a reliable trend can be seen fro the results obtained when stirring by hand and when using the magnetic stirrer. However, when the mixture is left without stirring anomalies arise, as is shown in the table. There is a reliable trend until the concentration of 0.005mol/dm-3, where the time taken for the mixture to turn colourless decreases rather than increasing. To ensure that I could see the colour chance easily, I placed the conical flask on a white tile. This meant that the colour change would be more obvious. I have decided to use the magnetic stirrer as it allows me to concentrate more on when the colour change occurs, rather than stirring the mixture. When I was stirring the mixture by hand, I found that I had a tendency to lose concentration and lose track of when the colour change occurred. Having a magnetic stirrer meant that all I needed to do was focus on the mixture and note the timing of the colour change. I believe this will help me in obtaining accurate results.

Temperature of Reactants (�C)

Time Take For the Mixture to Turn Colourless (seconds)













I also conducted a pre-test to trial my method for investigating how temperature change affects the reaction rate. I came across a few problems that need to be adjusted in my final experiment. I originally decided to just monitor the temperature of the water bath, but soon realised that this would most likely not share the same value as the solutions in the boiling tubes. To overcome this problem I placed a thermometer in each boiling tube, one in the tube containing potassium bromide and one in the solution of the other reactants. Once both had reached the correct temperature I was able to mix the two solutions, knowing that both were at the required temperature. I also discovered that when investigating temperatures above 40�C the reaction became too fast to record accurate timings. I would have lowered the concentration of potassium bromide, although I wanted to keep this constant to my previous experiments investigation how concentration affects rate of reaction. Instead, I chose the following range of temperatures: 10, 15, 20, 25, 30, 35 and 40�C. This range also makes it easier for me to keep the temperature of the reactants stable, as they do not fluctuate to far above or below room temperature.

For the concentrations of the other reactants used in this reaction I decided to alter them by the same amount and in the same range as I have chosen to alter the concentration of potassium bromide. This ensures that the experiment is as fair as possible, as all reactants are being varied by the same amount. It could be harder to work out a rate equation if I vary the concentrations of the reactants by different amounts to each other.

Method for Investigating Concentration Change



Reason For Use



To store solutions before adding to mixture. This ensures that the solutions are kept under the same conditions as each other, and that accurate amounts of solution can be measured out.


Potassium bromate(V) solution

Used as part of the reaction.


Sulphuric acid solution

Used as part of the reaction.


Phenol solution

Used as part of the reaction.


Methyl orange indicator

Used to show colour change in the reaction as phenol binding sites are used up by bromine.


Potassium bromide

Used as part of the reaction.


Distilled water

Used to dilute solutions to chosen concentrations.



Used for reaction to take place in.


White tile

Allows easy observation of colour change against a white background.


Magnetic stirrer

Allows me to concentrate on observing the colour change alone, rather than having to conduct stirring and observe colour change at the same time.



Used to measure the time taken for the solution to turn colourless.


Volumetric flash

Used to make up solutions accurately.



1. Take 5 burettes and label them as “potassium bromate(V)”, “sulphuric acid”, “phenol”, “potassium bromide” and “distilled water”. Wash them out with water and then with the corresponding solution. Fill each burette with its corresponding solution.

2. Add 5cm3 of potassium bromate(V) solution, 5cm3 of sulphuric acid solution and 5cm3 of phenol solution to a beaker. Add 4 drops of methyl orange indicator to the beaker. These amounts should be kept constant throughout the experiment to ensure that the test is as fair as possible. The same solution should be used in each trial to keep the test fair, as different solutions may have very slight differences in concentration.

3. Add 5cm3 of potassium bromide solution of the required concentration to another beaker. For most trials the potassium bromide will need mixing with distilled water to reach the desired concentration.

4. Place the first beaker onto a white tile and pour into it the potassium bromide solution. Start the stopwatch and continuously stir the mixture until it goes completely colourless. Record the time taken for this to happen.

5. Wash the beakers out with distilled water and repeat the experiment using concentrations of 0.003, 0.004, 0.005, 0.006, 0.008 and 0.01mol/dm-3 of potassium bromide. Repeat each concentration 5 times. All other factors such as temperature should be kept constant and the same apparatus should be used each time. Sulphuric acid concentration should be kept at 0.1mol/dm-3 when not being investigated, so that it is not so weak that the reaction takes place too slowly to be measured. All other solutions should be kept at 0.01mol/dm-3.

6. Once all concentrations of potassium bromide solution has been investigated and repeated, conduct experiments with respect to potassium bromate(V) and sulphuric acid. To do this, simply alter the concentration of the chosen reactant as was done when using potassium bromide. Concentration can be changed simply by taking a sample of the original solution and adding distilled water. Where M1 is the molarity of the solution before dilution, M2 is molarity after dilution, V1 is the volume of the solution before dilution and V2 is volume after dilution:

V1 = (M2 x V2)


For example, to make a 250ml solution with concentration 0.008M of potassium bromide from a solution of 0.01M, you would do the following:

V1 = (0.008 x 250ml)


= 200ml

This means that you would need to use 200ml of the original 0.01M solution with 50ml of distilled water to make 250ml of 0.008M solution.

Remember to keep all other solutions at a constant concentration throughout. Investigate the same range of concentrations as was used for potassium bromide, to ensure that a rate equation can easily be found.

Making Required Solutions

0.01M potassium bromate(V):


K = 1 x 39.1 = 39.1

Br = 1 x 79.9 = 79.9

O3 = 3 x 16.0 = 48.0

RFM = 167

grams needed = (0.01 x 167 x 250)


= 0.4175g dissolved in enough distilled water to make 250ml of solution.

1M Sulphuric acid:


H2 = 2 x 1.0 = 2

S = 1 x 32.1 = 32.1

O4 = 4 x 16.0 = 64.0

= 98.1

grams needed = (1 x 98.1 x 250)


= 24.525g dissolved in enough distilled water to make 250ml of solution.

0.0001M phenol:


C = 6 x 12.0 = 72.0

H = 6 x 1.0 = 6.0

O = 1 x 16.0 = 16.0

= 94.0

grams needed = (0.0001 x 94 x 250)


= 0.00235g dissolved in enough distilled water to make 250ml of solution.

0.01M potassium bromide:


K = 1 x 39.1 = 39.1

Br = 1 x 79.9 = 79.9

= 119.0

grams needed = (0.01 x 119 x 250)


= 0.2975g dissolved in enough distilled water to make 250ml of solution.

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