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6.1 Exercises

1. How fast a chemical reaction goes is measured by reaction rate. Define reaction rate.

Reaction rate is defined as the amount (in moles or mass) of product formed or reactant consumed per unit

time. The units are therefore: mol dm-3

s-1

(for moles), g s-1

(for mass).

2. Below is a list of some of the experimental techniques that may be monitored to determine the rate

of a reaction:

• volumetric analysis of reaction in solution

• measuring changes in mass

• measuring changes in density

• measuring changes in pressure for gaseous reactions

• measuring changes in electrical conductivity

• observing and quantifying changes in colour

• observing and quantifying the formation of precipitation (for example, change in transparency)

• absorption spectroscopy

• gas chromatography

Since the rate of reaction is the change in concentration of reactants or products over time, then these

techniques must measure this change either directly or indirectly. Select two methods from the above

list and briefly outline the principles involved.

Volumetric analysis, changes in mass and pressure, absorption spectroscopy and gas chromatography, density

and electrical conductivity all quantitatively measure the changing number of moles of reactants and products in

the reaction mixture. The methods may either directly measure the number of moles in solution (i.e.

concentration) or measure properties (e.g. electrical conductivity) that reflect how many moles are in solution.

Changes in colour and precipitation (for example from a transparent solution to an opaque one) indirectly and

qualitatively measure the concentration of product forming.

3. The following graph plots the mass loss of a reaction mixture when excess calcium carbonate

reacts with 500 cm3 of hydrochloric acid in the apparatus below.

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a) Write a balanced chemical equation for the reaction, giving state symbols.

CaCO3(s) + 2HCl(aq) � CaCl2(aq) + CO2(g) + H2O(l)

b) Why is there a loss of mass of the flask’s contents as the reaction proceeds?

The balance will show a decrease is mass as the reaction proceeds because the CO2 evolved in the reaction is

a gas and escapes from the open system (one in which both mass and energy can be transferred between the

system and its surroundings).

c) What is the purpose of the cotton wool plug?

The plug is to prevent loss of the reaction solution from the flask (during effervescence) which would lead to

incorrect mass loss measurement.

d) Explain why the rate of reaction represented at point A differs from that at the time represented

by B on the graph.

There are fewer reactant molecules present at B than at A (because some have reacted and been converted to

products) and therefore the extent of reaction per unit time is less. This is why there is a slower rate at B than at

A and explains why the rate of reaction changes over time.

e) What information can be determined about the reaction rate by looking at points C and D?

By points C and D the slope of the graph is constant and flat showing that no more mass is lost over time. This

indicates that the reaction is complete.

f) The initial mass of the reaction mixture and the reaction vessel was 85.24g. State the final

reading on the balance.

According to the graph, the mass lost = 14g. Therefore the final mass is equal to the initial mass minus the

mass lost: 85.24g – 14g = 71.24g.

g) Determine the number of moles of CO2(g) produced and lost to the surroundings.

n =m

M Therefore we must determine the mass (m) of CO2(g) that was produced and the molar mass (M) of

CO2.

The mass lost as indicated on the graph is equal to the mass of CO2(g) produced = 14g.

MCO2 = atomic weight of carbon + 2 times the atomic weight of oxygen (found on the periodic table).

= 12.01 g mol-1

+ (2 x 16.00 g mol-1

)

= 44.01 g mol-1

Therefore, nCO2 =

mCO2

MCO2

=14g

44.01gmol -1= 0.32 mol (2 s.f.)

h) Determine the mass of calcium carbonate which reacted.

Using the balanced reaction equation in part a), the mole ratio of CO2(g):CaCO3(s) = 1:1

Therefore nCO2 = nCaCO3 = 0.32 mol

MCaCO3 = 100.1 g mol-1

(determined using the periodic table as described in part i))

By rearranging the n = m/M equation to isolate the unknown, m = nxM = 0.32 mol x 100.1 g mol-1

= 32g (2sf)

i) Determine the initial concentration of the hydrochloric acid solution considering the CaCO3(s)

was in excess so all of the original HCl(aq) was used during the reaction.

Using the balanced reaction equation in part a), the mole ratio CaCO3(s) : HCl(aq) = 1:2

Therefore 2nCaCO3 = nHCl

Therefore nHCl = 2 x nCaCO3 = 2 x 0.32mol = 0.64mol

The questions states that 500 cm3 of the HCl(aq) solution was used.

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Therefore, rearranging the n = C x V equation to isolate the unknown, C = n/V = 0.64mol/0.5 dm

-3 = 1.28 mol

dm3. The concentration of the HCl(aq) reactant solution was 1.28 mol dm

-3.

4. The reaction rate for ethanoic acid and methanol forming methyl ethanoate was determined by

monitoring the number of moles of ethanoic acid and methyl ethanoate at regular time intervals. The

reaction is:

CH3COOH(aq) + CH3OH(aq) ���� CH3COOCH3(aq) + H2O(l)

The results are given in this table:

a) Plot graphs of n (mol) versus time (s) for both CH3COOH and CH3COOCH3, clearly labelled, on

the one set of axes.

Number of Mol (ethanoic acid and methyl ethanoate) With Changing Time

0.00

0.20

0.40

0.60

0.80

1.00

1.20

0 20 40 60 80 100 120 140 160Time (s)

Co

ncen

trati

on

(m

ol d

m-3

)

Change in enthanoicacid concentration

Change in methylenthanoateconcentration

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b) Plot a graph for rate (1/time) versus CH3COOH concentration (mol dm

-3).

Rate of consumption of ethanoic acid

0

0.05

0.1

0.15

0.2

0.25

0.00 0.20 0.40 0.60 0.80 1.00

Ethanoic acid concentration (mol dm-3)

1/t

ime

(1

/s)

Ethanoic acidconcentration

c) Calculate the average rate of reaction from (a) using the reactant species.

Over 150s the concentration of CH3COOH has changed from 1.0 mol dm-3

to 0.1 mol dm-3

.

Average reaction rate

=∆ [CH3COOH]

∆ time

= 0.9 mol dm-3

/150 s = 0.006 mol dm

-3 s

-1

5. The following graphs have been drawn by a student to represent a reaction that is occurring

in a closed system. They show the change in concentration (mol dm–3) of species A, B and C

with time as the reaction proceeds. These changing concentrations were measured using

absorption spectrophotometry which links the change in absorbance to change in

concentration using the Beer-Lambert Law (absorbance of the species is proportional to

the concentration of the species).

a) Describe and explain the concentration changes with time for each chemical species in the

mixture (i) between 0 and 30 minutes, and (ii) between 30 and 60 minutes. Make sure to

indicate which species are reactants and which are products.

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i) A and C are products. Their concentrations increase from zero mol dm

-3 between 0 and 30 minutes.

B is a reactant. Its concentration decreases from a maximum 0.6 mol dm-3

over this time frame.

ii) No further reaction is occurring after 30 minutes and the concentrations of all species remain

constant after this time. The reaction has gone to completion.

b) According to the graphs, the reaction occurs like this

B ���� A + C

Balance this equation by using the graphs to determine the mole ratios of the reactants and

products and explain your reasoning.

Initially there is 0.6 mol of reactant B (t = 0 min) which forms 0.3 mol of product A and 0.6 mol of product C

when there is no further reaction (t = 30 min onwards). The mole ratio of B:A:C is therefore 2:1:2. The equation

is balanced like this:

2B � A + 2C

6.2 Exercises

1. Reaction kinetics studies the rate at which chemical and biological reactions occur.

a) What is the kinetic theory of matter?

Kinetic theory is an explanation of the behaviour of matter in terms of random particle motion. It explains many

phenomena in terms of the average kinetic energy of a group of particles and the collisions they undergo.

Kinetic theory describes how particles are in continuous motion and that this motion is reflected by their

temperature. The higher the temperature, the more energetic the particles and the faster they move.

b) Use the kinetic theory to explain the relationship between gaseous particles and their

temperature in kelvin.

Particles are in constant motion and their temperature in is directly proportional to their average kinetic energy.

Reading the temperature on a thermometer is, in fact, taking a direct measurement of the average kinetic

energy of the medium it is in. Because the size of one degree in the Kelvin and Celsius scales is the same, the

temperature of a sample measured in both Kelvin and Celsius is directly proportional to the average kinetic

energy of the gaseous particles in that sample.

c) In a mixture at a given temperature, which move faster, the smaller particles or the larger

particles? Explain your choice.

Kinetic energy is a function of the mass and velocity of the particles –

Ek = 1/2mυ2

Therefore the kinetic energy of a particle (Ek) is directly proportional to both its mass (m) and velocity (υ).

At a given temperature the particles will have a set average kinetic energy and therefore the smaller particles

will move faster, as described by the equation. The particles with smaller mass must correspondingly have a

larger speed in order to keep Ek constant.

2. The collision theory is used to explain the effects of concentration, temperature, pressure, particle

size and catalysts on the rate of reaction.

a) What is the collision theory?

Collision theory is the explanation that a reaction between two or more particles can only occur if those particles

first collide. It then details what other conditions must be achieved at the time of collision in order for the reaction

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to proceed. Collision theory qualitatively explains how variables such as concentration, temperature and particle

size affect the rate of reaction and it allows us to predict the outcome of changing these variables.

b) According to this theory what factors determine where or not a reaction occurs?

For a reaction to occur the reactant particles must collide. On collision they must be in the correct orientation

and also have at least the minimum energy required for reaction (the activation energy).

c) What does the number of successful reactant collisions depend upon?

It depends upon the frequency of collision between particles and how many of these collisions involve particles

that have sufficient combined energy to overcome the activation energy barrier.

See next page for a diagram summary of the collision theory.

3. Activation energy is an important concept in reaction kinetics.

a) What is activation energy, Ea?

The activation energy, Ea, is the minimum amount of energy that particles must have in order for a reaction to

result when they collide. It is the least amount of energy the reactants must have in order for the particles to be

converted from reactants to products on collision.

b) Draw an energy profile diagram to show the activation energy for an endothermic reaction.

c) Explain, with the help of an energy profile graph, the effect of a catalyst in terms of activation

energy.

Energ

y

Reaction progress

Ea, activation energy withou

Ea, activation energy with ca

reactants

products

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A catalyst increases the number of successful collisions between molecules by providing an alternative reaction

pathway of lower energy. By lowering the activation energy of the reaction, more colliding particles have the

amount of energy required on collision and this results in more successful collisions.

Collision Theory

For molecules to react, they must first collide:

A2

B2

A2

B2

A2

B2

slower moving molecules or incorrect orientation, no bonds break, no reaction

In a reaction, reactant molecules must collide and chemical bonds are stretched, broken and

formed to produce product. Only a fraction of the total collisions are effective and lead to the

formation of products. Others involve molecules with not enough energy to react, or have incorrect

orientation on collision. The amount of energy that reactant molecules require in order to react and

form products is known as the activation energy. This minimum amount of energy required is the

energy barrier that the molecules must overcome in order for the reaction to proceed. Some

reactant molecules do not have enough energy to overcome this energy barrier. Many reactions

that do not occur, or only happen slowly at room temperature, will proceed at higher speed when

the reactants are heated. Sometimes heat is used to provide the reactant molecules with enough

energy to react.

A2

B2

A2

B2

AB

AB

faster moving molecules with correct orientation, bonds break a reaction occurs

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4. The factors affecting rate of reaction are concentration, temperature, particle size, pressure and

catalysts. Manipulating these variables is all about increasing the number of overall collisions and

increasing the number of successful collisions.

State and explain the effect of each of these variables on the rate of a reaction.

Increasing the reactant concentrations means there will be more particles present in a given volume. This

results in more collisions per unit time, which translates into more successful collisions per unit time.

Similarly, increasing the pressure also means there are more particles in a given volume. This translates that

the particles will experience more collisions per unit time, resulting in more successful collisions and a faster

rate.

Decreasing the particle size for solid reactants (for example by crushing a coarse solid sample) means there is

a greater surface area on which the reaction may take place. This means more collisions and therefore more

successful collisions.

Increasing the temperature works two-fold; it increases the average kinetic energy of the particles so that they

move faster and undergo more collisions and it provides them with more energy so that more collide with the

required activation energy.

Introducing a catalyst increases the rate by providing an alternative pathway of lower energy so that more

particles have the required energy for reaction on collision.

5. Explain, in terms of the collision theory, how refrigeration aids in minimizing spoilage of food. Use a

suitable Maxwell-Boltzmann distribution diagram.

Refrigeration decreases the temperature of the food. On the molecular level this involves lowering the

temperature of the molecules that make up the food and the air around it. This means that these molecules

have a lower average kinetic energy. Maxwell-Boltzmann distribution graphs (see below) show how decreasing

the temperature decreases the number of molecules that have the energy required for reaction. In this way,

keeping food refrigerated therefore slows the rate of reactions that decompose the food. If the temperature is

increased, the reactions that decompose food increase in rate and the food decomposes more quickly.

Decreasing the temperature by 10 K decreases the number of molecules with energy less than Ea. This is

indicated by the areas under each curve above the Ea boundary. This area is greater for the curve at the higher

temperature (T + 10 K) and therefore there are more molecules that have the required energy for the

decomposition reaction. For the curve at the lower temperature (representing refrigeration), the area under the

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curve above the Ea is less and therefore fewer molecules have this required energy for the decomposition

reaction.

6. A student carried out an investigation to determine how the rate of reaction of bromomethane and

sodium hydroxide varies with initial bromomethane concentration.

The reaction is as follows:

CH3Br + OH- ���� CH3OH + Br

-

a) Plot a graph of initial bromomethane concentration versus reaction time.

Initial bromomethane concentration (mol dm-3) versus

Reaction Time (s)

0

100

200

300

400

0 0.5 1 1.5 2 2.5

Initial bromomethane concentration (mol dm-3)

Re

ac

tio

n T

ime

(s

)

b) Explain, in terms of collision theory, why the rate varies this way.

Increasing the concentration of one of the reactants increases the number of collisions there will be between

reactants per unit time. Therefore there will consequently be more successful collisions per unit time, increasing

the rate. This is reflected by the decrease in reaction time as the initial bromomethane concentration is

increased.

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7. Draw and qualitatively explain a Maxwell-Boltzmann energy distribution curve.

The Maxwell-Boltzmann curve displays the distribution of the kinetic energies in a sample of gas particles at a

certain temperature. The area under the graph is directly proportional to the number of gas particles and the

shape indicates the distribution of energies.

8. Draw and explain qualitatively Maxwell-Boltzmann energy distribution curves for the same gas at a

lower, and then at a higher temperature.

At a lower temperature, more gas molecules have energies close to the average energy, which itself is a

relatively low number. As the temperature is increased, the shape of the graph changes. The spread of energies

increases and more molecules have energies further from the average value, which itself is higher. Therefore at

a higher temperature (T + 10 K), more molecules will have the energy required for a given reaction. More

molecules are able to jump the Ea barrier and convert to products.

9. Catalysts speed up the rates of chemical reactions.

a) How do catalysts work?

Catalysts provide an alternative reaction pathway of lower activation energy which means more of the species

present have the required activation energy on collision.

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b) Draw a reaction profile graph to show the effect of a catalyst on the activation energy of a

reaction.

En

er g

y

Reaction progress

Ea, activation energy without catalyst

Ea, activation energy with catalyst

reactants

products

10. Draw and explain Maxwell-Boltzmann curves for reactions with and without catalysts.

For a given reaction the activation energy is set. On a Maxwell-Boltzmann distribution curve a vertical line can

be drawn at this energy indicating that all the particles under the curve at or above this line have the energy

required for reaction.

When a catalyst is used, the activation energy is lowered (Ea* on the diagram below). The boundary line is

therefore moved to this lower energy and there is now a larger number of particles with the required energy –

the area under the curve in this section is now greater.

Nu

mb

er

of

part

icl e

s

Energy EaEa*

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11. The terms homogeneous catalyst, heterogeneous catalyst and enzyme are used in reaction kinetics.

Explain each, using an example.

a) homogeneous catalyst

A homogeneous catalyst is a catalyst that is in the same phase as the reactants (ie solid, liquid or gas). A good

example are the chlorine free radicals that break down ozone in the atmosphere. Both Cl• and O3 are gaseous

particles.

b) heterogeneous catalyst

A heterogeneous catalyst is a catalyst that is not in the same phase as the reactants. These catalysts often

provide a surface for the reaction to occur on. One example is the use of the metal Ni in the hydrogenation of

vegetable oils to produce margarine. It is present as a very high surface area solid on which the vegetable oils

have their double bonds converted to single bonds by the addition of hydrogen.

c) enzyme

Enzymes are biological catalysts – they are proteins that catalyse physiological reactions (reactions in living

organisms). Almost all processes in a living cell need enzymes in order to occur at sufficient rates. An example

is DNA polymerase, an enzyme we would not be able to live without as it aids in DNA replication.

The Action of Catalysts Until 1900, little was understood about catalysis except that catalysts worked and were very useful. Then in 1901 the German chemist, Wilheim Ostwald, proposed that catalysts only change the rate of a chemical reaction. The same chemical reaction can in fact proceed without a catalyst but much more slowly; although in some cases the reaction is so slow it can barely be detected! Later, it was proposed that in the case of solid catalysts, reactants bonded to the surface and then moved away when the reaction was complete, allowing the surface to be re-used by more reactant molecules. Despite the many advances in catalytic chemistry however, new catalysts are still being discovered either by chance or by testing many materials until suitable ones are found. Supplying heat is more expensive than using a catalyst. Energy is expensive and some forms are damaging to the environment. Catalysts however remain unchanged and can be recovered and reused. Even expensive catalysts such as platinum and silver, are far more economical than heating. Although heating a reaction increases the rate, it often lowers the yield of chemical reactions. Also heating can increase the rate of reactions other than those giving the desired product. This again lowers the yield and makes the process less economical. This points to a very important property of catalysts, they are selective. Chemists are able to select the catalyst which brings about the desired reaction. The theory to explain catalysts is complex, but it is thought that catalysts react in two main ways: Catalysts either 1. react by providing an alternative reaction pathway which has lower activation energy than the original reaction, or 2. line the reactants up in such a way as to bring about a faster reaction (correct orientation) In speeding up reactions, catalysts lower the hump which represents activation energy in the reaction.

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