rate of reaction Archives

What factors affect the rate of a reaction?

December 1, 2020December 1, 2020 by Veerendra

What factors affect the rate of a reaction?

Factors Affecting the Rate of Reaction:

The rate of a chemical reaction can be altered by changing the reacting conditions.
There are five factors that affect the rate of a reaction:

Total surface area of a solid reactant or particle size of a solid reactant
Concentration of a reactant in solution
Temperature of a reaction
Use of catalyst
Pressure for a reaction involving gaseous reactants

Table shows some examples of changing the reacting conditions which affect the rates of reactions in our daily occurrences.

Example	Explanation	Factor affecting the rate
(a) It is easier to start a fire using sticks rather than logs	This is because sticks have a bigger total exposed area than the logs. Hence, the rate of burning the sticks is higher than that of the logs.	Surface area (or particle size)
(b) Blowing air over glowing charcoal causes it to ignite brightly and quickly	This is due to increase in the concentration of oxygen/air. So, the rate of burning increases.	Concentration
(c) Milk turns sour faster on hot days than on cold days	On hot days, the temperature is higher. So, the rate of the milk turns sour is higher.	Temperature
(d) The presence of amylase in the saliva breaks starch to maltose	Amylase, as a biological catalyst/enzyme, speeds up the breakdown of starch to maltose.	Catalyst
(e) The manufacture of ammonia gas is carried out at high pressure N₂(g) + 3H₂(g) → 2NH₃(g)	At high pressure of the gaseous reactants, the rate of reaction is higher.	Pressure

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Application of factors that affect the rate of reaction

Application of the factors affecting the rate of reaction in daily life:

Knowledge of the factors affecting the rate of reaction can be applied to control the rates of many reactions that occur in our daily life.
These can be done by either speeding up or slowing down the reaction to suit our needs.

Table shows a few examples of applying the factors affecting the rate of reaction to control the rates of a few reactions in our daily life.

Example	Explaining the effect on the rate of reaction	Factor affecting the rate of reaction
(a) Burning of charcoal	Food can be cooked faster when smaller pieces of charcoal are used compared to bigger pieces of charcoal. This is because smaller pieces of charcoal have a larger total exposed surface area compared to bigger pieces. Hence, smaller pieces of charcoal can burn faster to produce more heat per second and the food is cooked faster.	Surface area of charcoal that controls its rate of burning
(b) Storing of food in a refrigerator	The decaying and decomposition of food by microorganisms are chemical reactions occurring in food that cause the food to turn bad. When the food is kept in a refrigerator, the food lasts longer. This is because the low temperature in the refrigerator slows down the decaying and decomposition of food by microorganisms.	Temperature that controls the rate of decaying of food by microorganisms
(c) Cooking of food in a pressure cooker	In a pressure cooker, the high pressure raises the boiling point of water to a temperature above 100°C. Thus, cooking is carried out at a temperature higher than 100°C in the pressure cooker. The higher temperature increases the rate of cooking. Hence, the food can be cooked faster.	Temperature that controls the rate of cooking
(d) Cooking of solid food with different sizes	Solid food in big chunks such as big chunks of potato have a smaller total surface area exposed to heat during cooking. Hence, the food will be cooked at a lower speed. For the food to be cooked faster, the big chunks of solid food must be cut into smaller pieces such as potato chips to increase its total surface area exposed to heat.	Surface area that controls the rate of cooking

Application of the factors affecting the rate of reaction in industrial processes

Knowledge of the factors affecting the rate of reaction is applied in choosing the optimum conditions to run an industrial process to achieve
(a) shorter time of production
(b) higher yield
(c) lower cost of production
It is very costly to run an industrial process at a very high temperature and pressure.
Catalysts are generally used to increase the rates of these processes.
The catalysts used enable the processes to be run in a shorter time at an optimum temperature and pressure to reduce the cost of production.

Three examples of industrial processes are:
(a) Haber Process

Haber process is an industrial process to manufacture ammonia on a large scale. In this process, nitrogen reacts with hydrogen to produce ammonia.
Table describes the effect of catalyst, temperature and pressure on the rate of production of ammonia and its yield.

Factors	Effect on the rate of reaction	Effect on the yield of ammonia	Conditions chosen
Catalyst	The use of catalyst increases the rate of production of ammonia	The quantity of ammonia (yield) produced remains the same with or without the catalyst	Iron is used as a catalyst to increase the rate of production of ammonia
Temperature	Increase in temperature will increase the rate of production of ammonia	Increase in temperature will decrease the quantity of ammonia (yield) produced	An optimum temperature of 450 to 550°C is chosen. A compromise is reached for a reasonable rate and yield.
Pressure	Increase in pressure will increase the rate of production of ammonia	Increase in pressure will increase the quantity of ammonia (yield) produced.	A high pressure of 200 to 300 atm is chosen to obtain a higher rate and yield.

So, Haber process is run in the industry as below:
In the Haber process, a mixture of nitrogen gas and hydrogen gas in the volume ratio of 1:3 is passed over iron powder (catalyst) at a temperature of 450 to 550°C and a pressure of 200 to 300 atm in the presence of molybdenum as a promoter.
These optimum conditions enable the Haber process to run at a higher rate to cut down the cost and time of production, and also to obtain a higher yield.

(b) Ostwald Process

Ostwald process is an industrial process to manufacture nitric acid on a large scale.
Ostwald process involves three stages, as shown in Table.

Stage	Reacting conditions	Chemical equation for the reaction
I	Ammonia gas is mixed with excess air and the mixture is passed over platinum (catalyst) at a temperature of 850°C and a pressure of 2 to 5 atm. Ammonia is oxidised to nitrogen monoxide. All these reacting conditions are chosen to ensure the reaction occurs at an optimum rate with an optimum yield.	4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(l)
II	Nitrogen monoxide gas from stage I is then reacted with excess air to produce nitrogen dioxide gas.	2NO(g) + 0₂(g) → 2N0₂(g)
III	Nitrogen dioxide gas from stage II is then mixed with excess air and dissolved in hot water at 80°C to produce nitric acid	4NO₂(g) + O₂(g) + 2H₂O(l) → 4HNO₃(aq)

(c) Contact process

Contact process is an industrial process to manufacture sulphuric acid on a large scale.
Contact process involves four stages, as shown in Table.

Stage	Reacting conditions	Chemical equation for the reaction
1	Sulphur is burnt in excess air to produce sulphur dioxide gas	S(s) + O₂(g) → SO₂(g)
II	Sulphur dioxide gas from stage 1 is mixed with excess air and the mixture is passed over vanadium(V) oxide, V₂O₅ (catalyst) at a temperature of 450°C and a pressure of 1 atmosphere. Sulphur dioxide is oxidised to sulphur trioxide. All these reacting conditions are chosen to ensure the reaction occurs at an optimum rate with an optimum yield.	2SO₂(g) + O₂(g) → SO₃(g)
III	Sulphur trioxide gas from stage II is then dissolved in concentrated sulphuric acid to form a liquid called oleum, H₂S₂O₇.	SO₃(g) + H₂SO₄(aq) → H₂S₂O₇(l)
IV	The oleum from stage III is then diluted with water to produce concentrated sulphuric acid of about 98%.	H₂S₂O₇(l) + H₂0(l) → 2H₂SO₄(aq)

Factors affecting the rate of reaction problems with solutions

1. Table shows the reacting conditions for three sets of experiments.

Set	Reacting condition
I	5 g of excess zinc powder + 100 cm³ of 0.4 moi dm^-3 sulphuric acid + 5 cm³ of 1 mol dm^–³ copper(ll) sulphate solution at room temperature
II	5 g of excess zinc powder + 50 cm³ of 0.4 mol dm^-3 sulphuric acid at room temperature
III	5 g of excess granulated zinc + 75 cm³ of 0.4 mol dm^–³ sulphuric acid at room temperature

Sketch the graphs of the volume of hydrogen gas liberated against time for sets I, II and III on the same axes. Explain how you obtain your answers.
Solution:
1. Compare the yield
What factors affect the rate of a reaction 2
From the table, number of moles of sulphuric acid reacted in set I > set III > set II.
Thus, the maximum volume of hydrogen gas liberated in set I > set III > set II.

2. Compare the rate

Initial rate of reaction for set I > set II because of the presence of catalyst in set I.
Initial rate of reaction for set II > set III because the total surface area of zinc used in set II is larger than that in set III.
Hence, the initial rate of reaction for set I > set II > set III
This means that the initial gradient of the curve for set I > set II > set III.

3. Sketch the graph
The graphs for the three sets of the experiments are:
What factors affect the rate of a reaction 4

2. 5 g of excess marble chips is added to 80.0 cm³ of 1.0 mol dm^–³ acid R in a conical flask. The conical flask is placed on an electronic balance. The mass of the conical flask and its contents is recorded at regular time intervals. Figure shows the graph of the mass of the conical flask and its contents against time.
What factors affect the rate of a reaction 5
(a) Write the chemical equation for the reaction if acid R is ethanoic acid.
(b) Why does the curve become horizontal after 180 s?
(c) (i) Describe the change in the rate of reaction during the first 180 s. Explain how you derive your answer from the graph.
(ii) Explain the change in the rate of reaction in (c)(i).
Solution:

(a) CaCO₃(s) + 2CH₃COOH(aq) → Ca(CH₃COO)₂(aq) + CO₂(g) + H₂O(l)
(b) The reaction has stopped after 180 s because all the acid R has reacted completely with excess marble.
(c) (i) Rate of reaction decreases with time during the first 180 s. This is because the magnitude of the gradient of the curve decreases with time.
(ii) The concentration of acid R and the total surface area of marble chips decrease with time, hence the rate of reaction decreases with time.

3. 1 g of excess magnesium powder is added to 50 cm³ of 0.5 mol dm^–³ hydrochloric acid at room temperature. Figure shows the curve obtained when the volume of hydrogen gas liberated against time is plotted.
What factors affect the rate of a reaction 6
The experiment is repeated using 25 cm³ of 0.5 mol dm^–³ hydrochloric acid at 80.0°C to replace 50 cm³ of 0.5 mol dm^–³ hydrochloric acid at room temperature. Copy the graph and sketch on the same axes the curve that you would expect to obtain for the second experiment. Explain how you obtain your answer.
Solution:
1. Compare the maximum volume of hydrogen gas liberated

From the above table, the number of moles of hydrochloric acid reacted in experiment I is twice the number of moles of hydrochloric acid reacted in experiment II.
Thus, the maximum volume of hydrogen gas liberated in experiment I is twice the maximum volume of hydrogen gas liberated in experiment II.
Hence, the maximum volume of hydrogen gas liberated in experiment II
= 1/2 x 600 cm³
= 300 cm³

2. Compare the initial rate

From the above table, the temperature of experiment II is higher than that of experiment I, whereas the other two factors remain constant.
Hence, the initial rate of reaction of experiment II is higher than that of experiment I.
In other words, the initial gradient of curve II is steeper than that of curve I.

3. Sketch the curve
Based on the above analysis, the curve for experiment II is shown below.
What factors affect the rate of a reaction 9

4. Experiment I is carried out to study the rate of reaction between 1.0 g of iron foil with 60 cm³ of 0.4 mol dm^–³ hydrochloric acid. The graph in Figure is obtained.
What factors affect the rate of a reaction 10
The experiment is repeated two more times using different volumes and concentrations of hydrochloric-acid as shown in Table.

(a) Write the chemical equation for the reaction.
(b) (i) Compare the initial rates of reaction for experiments I, II and III.
(ii) Give reasons for your answers.
(c) Determine the maximum volume of hydrogen gas liberated in experiments II and III.
Given that the relative atomic mass of Fe = 56 and the molar volume of any gas is 24 dm^–³ mol^–¹ at room conditions.
(d) By using the answers in (b) and (c), sketch the graphs of the volume of hydrogen gas liberated against time for experiments II and III on the same axes in Figure.
Solution:
(a) Fe(s) + 2HCl(aq) → FeCl₂(aq) + H₂(g)
(b) (i) Initial rate of reaction for experiment III > experiment I > experiment II
(ii) This is because the concentration of hydrochloric acid in experiment III > experiment I > experiment II.
What factors affect the rate of a reaction 12

What is the rate of the reaction?

December 1, 2020December 1, 2020 by Veerendra

What is the rate of the reaction?

Rate of reaction measures the speed at which the reactants are converted to the products in a chemical reaction.
For a reaction that occurs rapidly, the rate of reaction is high. Conversely, for a reaction that occurs slowly, the rate of reaction is low.
The time taken for a fast reaction is short, whereas the time taken for a slow reaction is long.
Hence, the rate of a particular reaction is inversely proportional to the time taken for the reaction.
Different chemical reactions occur at different rates. Some examples are illustrated in Table.

Type of reaction	Fast reaction	Slow reaction
Reaction involving liberation of a gas	Bubbles of carbon dioxide gas liberate rapidly when sodium carbonate powder reacts with dilute hydrochloric acid. Na₂CO₃(s) + 2HCl(aq) → 2NaCl(aq) + CO₂(g) + H₂O(l)	In photosynthesis, carbon dioxide reacts with water very slowly in the presence of sunlight and chlorophyll to produce glucose and oxygen gas. 6CO₂(g) + 6H₂O(l) → C₆H₁₂O₆(s) + 6O₂(g)
Precipitation reaction	When silver nitrate solution is added to sodium chloride solution, a white precipitate of silver chloride is formed immediately. AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)	When dilute hydrochloric acid is added to sodium thiosulphate solution, a yellow precipitate of sulphur appears only after a few seconds. Na₂S₂O₃(aq) + 2HCl(aq) → 2NaCl(aq) + S(s) + SO₂(g) + H₂O(l)
Heating a metal in air	When a small piece of potassium is heated in air, it burns rapidly to form a white solid of potassium oxide. 4K(s) + O₂(g) → 2K₂O(s)	When a small piece of copper is heated in air, it reacts slowly with oxygen in the air to form a black solid of copper(II) oxide. 2Cu(s) + O₂(g) → 2CuO(s)

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What does the rate of reaction measure?

Observable changes for measuring the rate of reaction:

1. When a reaction occurs, two obvious changes that occur are:

the quantity of a reactant decreases with time
the quantity of a product increases with time

2. The quantity of a reactant/product can be the

number of moles of a substance
mass of a solid
volume of a gas
concentration of a solution

3. If the changes of any of these quantities are visible and measurable during a reaction, then it can be used to measure the rate of that reaction.

4. Suitable measurable visible changes in a chemical reaction are:

volume of a gas liberated
formation of a precipitate
changes in the mass during a reaction
colour changes
changes in the electrical conductivity of the solution
temperature changes
pressure changes
changes in concentration of the solution of a reactant
pH changes

5. One of these measurable visible changes can be selected as a suitable quantity to determine the rate of a particular reaction.
The changes in this selected quantity can be measured by carrying out an experiment and the results are then analysed to determine the rate of that reaction.

6. Definition: Rate of reaction is defined as the change in a selected quantity during a reaction per unit time whereby the selected quantity can be any of the measurable visible changes in the reaction.
What is the rate of the reaction 3

7. Two examples to illustrate the meaning of rate of reaction.
Reaction between magnesium and dilute sulphuric acid
Mg(s) + H₂SO₄(aq) → MgSO₄(aq) + H₂(g)

In the reaction between dilute sulphuric acid and a magnesium ribbon, the following two changes are observed:
- The mass of magnesium (the reactant) decreases with time.
- The volume of hydrogen gas (the product) increases with time.
Hence, the rate of reaction between dilute sulphuric acid and magnesium can be determined by measuring the change in the mass of magnesium or the volume of hydrogen gas per unit time.
Quantitatively,
Reaction between ethanedioic (oxalic) acid, H₂C₂O₄and acidified potassium manganate(VII) solution.
- When excess aqueous ethanedioic acid is added to acidified potassium manganate(VII) solution, the purple colour of the solution slowly decolourises at room temperature.
- By measuring the time taken for the purple colour to decolourise, the rate of reaction can be determined.
- Rate of reaction is inversely proportional to the time taken for the purple colour to decolourise.
- Rate of reaction is reflected by the value of 1/time taken. The larger the value time taken of 1/time taken, the higher the rate of reaction.

Table shows the units for the rate of reaction measured in different ways.

Change in a selected quantity per unit time	Units for the rate of reaction
Change in mass per unit time	g s^-1 or g min^-1
Change in volume of a gas liberated per unit time	cm³ s^-1 or cm³ min^-1
Change in concentration of a reactant per unit time	mol dm^-3 s^-1 or mol dm^-3min^-1
Change in number of moles of a reactant per unit time	mol s^-1or mol min^-1

Rate of Reaction Experiment

Aim: To compare the rates of a few reactions.
Materials: Marble chips, 2 mol dm^-3 hydrochloric acid, 0.1 mol dm^-3 sodium thiosulphate solution, 1 mol dm^-3 lead(ll) nitrate solution.
Apparatus: 50 cm³ beakers, test tubes.

Procedure:

5 cm³ of 2 mol dm^-3 hydrochloric acid is poured into each of the three test tubes on a rack.
The test tubes are labelled I to III respectively.
One piece of marble chip is added into test tube I.
About 2 cm³ of 0.1 mol dm^-3 sodium thiosulphate solution is poured into test tube II and the mixture is shaken well.
About 2 cm³ of 1 mol dm^-3 lead(II) nitrate solution is poured into test tube III and the mixture is shaken well.
The changes are observed carefully. The rates of reactions in the three test tubes are compared.

Observations:

Test tube	Reactants	Observation
I	Marble chip and hydrochloric acid	Bubbles of a colourless gas are liberated rapidly, that is, effervescence occurs rapidly.
II	Sodium thiosulphate solution and hydrochloric acid	A yellow precipitate appears only after about 12 seconds.
III	Lead(II) nitrate solution and hydrochloric acid	A white precipitate is formed immediately.

Inferences:

The reaction between lead(ll) nitrate solution and hydrochloric acid is very fast.
The reaction between the marble chip and hydrochloric acid is moderately fast.
The reaction between sodium thiosulphate solution and hydrochloric acid is slow.

Discussion:

The chemical equation for the reaction in
The rate of reaction in ascending order is: rate in test tube II < rate in test tube I < rate in test tube III
The observable change that can be used to compare the rate of reaction in
(a) test tube I is the time taken for the effervescence to stop completely.
(b) test tube II is the time taken for the appearance of a yellow precipitate.
(c) test tube III is the time taken for the appearance of a white precipitate.

Conclusion:

The rate of reaction between sodium thiosulphate solution and hydrochloric acid is the lowest.
The rate of reaction between the marble chip and hydrochloric acid is moderately high.
The rate of reaction between lead(II) nitrate solution and hydrochloric acid is the highest.

How does the collision theory affect the rate of reaction?

December 1, 2020December 1, 2020 by Veerendra

How does the collision theory affect the rate of reaction?

Explaining the effect of size of a solid reactant/surface area on the rate of reaction using collision theory

When the size of a fixed mass of a solid reactant decreases, the rate of reaction increases.
This can be explained using the collision theory, as below:
(a) When the size of a fixed mass of a solid reactant becomes smaller, the total surface area exposed to collision with the particles of the other reactants increases, as shown in Figure.
(b) As a result, the frequency of collision among the reacting particles at the surface of the solid reactant increases.
(c) This causes the frequency of effective collision to increase.
(d) Hence, the rate of reaction increases.

Example: Two sets of experiments are carried out as shown below.
Set I: 1.0 g of granulated zinc is added to 20 cm³ of 0.5 mol dm^-3 sulphuric acid at 27.0°C.
Set II: 1.0 g of zinc powder is added to 20 cm³ of 0.5 mol dm^-3 sulphuric acid at 27.0°C.
The initial rate of liberation of hydrogen gas in set II is higher than that in set I. Explain the difference in the rate of reaction using the collision theory.
Solution:

Particle size of zinc powder in set II is smaller than that of granulated zinc in set I.
Thus, the total exposed surface area of 1.0 g of zinc powder in set II is larger than that of 1.0 g of granulated zinc in set I.
As a result, the frequency of collision between the hydrogen ions from sulphuric acid and the zinc atoms at the surface of zinc powder in set II is higher than that occurring at the surface of granulated zinc in set I.
This causes the frequency of effective collision between hydrogen ions and zinc atoms in set II to be higher than that in set I.
Hence, the initial rate of liberation of hydrogen gas in set II is higher than that in set I.

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Explaining the effect of concentration on the rate of reaction using collision theory

When the concentration of a reactant increases, the rate of reaction increases.
This can be explained using the collision theory, as below:
(a) When the concentration of the solution of a reactant increases, the number of particles per unit volume in this solution also increases.
(b) With more particles per unit volume, the frequency of collision among the reacting particles increases, as shown in Figure.
(c) Hence, the frequency of effective collision increases.
(d) As a result, the rate of reaction increases.

Example: Two experiments are carried out as shown below:

Experiment	Reacting conditions
I	50 cm³ of 0.20 mol dm^-3 sodium thiosulphate solution + 5 cm³ of 1.0 mol dm^-3 hydrochloric acid
II	50 cm³ of 0.12 mol dm³ sodium thiosulphate solution + 5 cm³ of 1.0 mol dm^-3 hydrochloric acid

(a) Which of the two experiments shows a higher rate of reaction?
(b) Explain the difference in (a) using collision theory.
Solution:

(a) Rate of reaction of Experiment I is higher than that of Experiment II

(b)

The concentration of sodium thiosulphate solution in Experiment I is higher than that in Experiment II.
So, the number of thiosulphate ions, S₂O₃^2- in one dm³of sodium thiosulphate solution in Experiment I is higher than that in Experiment II.
This causes the frequency of collision between the reacting particles (thiosulphate ions and hydrogen ions) in
Experiment I to be higher than that in Experiment II.
Hence, the frequency of effective collision in Experiment I is higher than that in Experiment II that results iii a higher rate of reaction in Experiment I.

Explaining the effect of pressure of gaseous reactants on the rate of reaction using collison theory

When the pressure of a reaction involving gaseous reactants increases, the rate of reaction increases.
This can be explained using the collision theory, as below:
(a) When particles of gaseous reactants are compressed to occupy a smaller volume, the pressure of the gaseous reactants increases, as shown in Figure.

(b) Thus, the number of gas particles per unit volume increases.
(c) With more gas particles per unit volume, the frequency of collision among the reacting particles increases.
(d) Hence, the frequency of effective collision increases, which results in a higher rate of reaction.

Explaining the effect of temperature on the rate of reaction using collision theory

When the temperature of a reaction increases, the rate of reaction increases.
This can be explained using the collision theory, as below.
(a) An increase in temperature will cause the kinetic energy of the reacting particles to increase.
(b) As a result, it leads to the following two changes:
(i) The reacting particles move faster and collide more often with one another. Hence, the frequency of collision increases, as shown in Figure.

(ii) The increase in kinetic energy causes more colliding particles to possess higher energy that is enough to overcome the activation energy.
(c) The above two changes result in a higher frequency of effective collision, and consequently a higher rate of reaction.

Example: Two sets of experiments are carried out as shown below.
Set X: 2 cm of magnesium ribbon is added to 50 cm³ of 1 mol dm^-3 sulphuric acid at room temperature.
Set Y: 2 cm of magnesium ribbon is added to 50 cm³ of 1 mol dm^-3 hot sulphuric acid at 80°C.
(a) Compare the time taken for the magnesium ribbon to disappear from sight for sets X and Y.
(b) Explain the difference in the time taken using the collision theory.
Solution:

The time taken in set Y is shorter than that in set X.
The temperature of reaction in set Y is higher than that in set X.
Thus, the kinetic energy of the hydrogen ions from the sulphuric acid that collide with the magnesium atoms at the surface of magnesium ribbon in set Y is higher than that in set X.
As a result, it leads to the following two changes:
(i) The hydrogen ions move faster and collide more often with the magnesium atoms in set Y compared to set X. The frequency of collision increases in set Y.
(ii) More collisions between hydrogen ions and magnesium atoms in set Y possess enough energy to overcome the activation energy compared to set X.
The above two changes result in a higher frequency of effective collision, and consequently a higher rate of reaction in set Y compared to set X.
Hence, the time taken for set Y is shorter than that in set X.

Explaining the effect of catalyst on the rate of reaction using collision theory

When a positive catalyst is used in a reaction, the rate of reaction increases.
This can be explained using the collision theory, as below:
(a) When a positive catalyst is used in a chemical reaction, it enables the reaction to occur through an alternative path which requires a lower activation energy, as illustrated in the energy profile diagrams in Figure.

(b) As a result, more colliding particles are able to overcome the lower activation energy.
(c) Hence, the frequency of effective collision increases, and consequently this results in a higher rate of reaction.

Example: Two sets of experiments are carried out as shown below.
Set Q: 2 g of granulated zinc is added to 50 cm³ of 0.2 mol dm^-3 hydrochloric acid at room conditions.
Set R: 2 g of granulated zinc is added to a mixture of 50 cm³ of 0.2 mol dm^-3 hydrochloric acid and 2 cm³ of 1 mol dm^-3 copper(II) sulphate solution at room conditions.
Explain why initial rate of set R is higher than that of set Q using the collision theory.
Solution:

Copper(II) sulphate solution in set R acts as a positive catalyst, whereas no catalyst is used in set Q.
The presence of copper(II) sulphate in set R enables the reaction between zinc and hydrogen ions from hydrochloric acid to occur through an alternative path which requires a lower activation energy.
As a result, more collisions between the hydrogen ions and zinc atoms at the surface of granulated zinc are able to overcome the lower activation energy in set R.
Hence, the frequency of effective collision increases. Consequently, this results in a higher rate of reaction for set R.

Appreciating scientist’s contributions

Scientists always practice problem-solving attitudes.
The patience, hard work and perseverance of scientists in doing their research and development had led to the discovery of the factors affecting the rate of reaction.
With this knowledge, we can control the rate of reaction in many activities in our daily life.
We must be thankful to those scientists for their contributions to improve the quality of life.

What is the collision theory in chemistry?

December 1, 2020December 1, 2020 by Veerendra

What is the collision theory in chemistry?

According to the kinetic theory of matter, particles of matter are in continuous motion and constantly in collision with each other.
For a reaction to occur, the particles of the reactants (atoms, molecules or ions) must touch each other through collision for bond breaking and bond formation to form the products.
However, not all collisions will result in a reaction to form the products.
According to the collision theory, only those collisions which
(a) achieve a minimum amount of energy, called activation energy, and
(b) with the correct orientation,
will result in reaction.
These types of collisions are known as effective collisions.
For particles that collide with energy less than the activation energy needed for reaction or with the wrong orientation, they simply bounce apart without reacting. These collisions are known as ineffective collisions.
An example to illustrate the collision theory
(a) (i) Hydrogen gas reacts with bromine gas to form hydrogen bromide gas
(ii) The chemical equation for the reaction is H₂(g) + Br₂(g) → 2HBr(g)
(b) Table illustrates three different situations that may happen during the collisions of hydrogen molecules with bromine molecules.

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Situations	Will it result in a reaction?
I. A hydrogen molecule and a bromine molecule collide with the correct orientation but the energy of the colliding particles is less than the activation energy, as shown in Figure.	This collision will not result in a reaction (Ineffective collision)
II. A hydrogen molecule and a bromine molecule collide with their total energy greater than or equal to the activation energy but with the wrong orientation, as shown in Figure.	This collision will not result in a reaction (Ineffective collision)
III. A hydrogen molecule and a bromine molecule collide with their total energy greater than or equal to the activation energy and with the correct orientation, as shown in Figure.	An effective collision occurs. This collision results in a reaction to form hydrogen bromide molecules.

What is Activation Energy?

Activation energy:

The meaning of activation energy can be visualised by sketching the energy profile diagrams, as shown in Figure.
The difference in energy between the energy of the reactants and the energy at the peak of the curve is the activation energy.
Activation energy is the energy barrier that must be overcome by the colliding particles of the reactants so that the reaction can occur.

What is the relationship between activation energy and collision theory?

Relationship between the frequency of collision and the activation energy with the rate of reaction:

According to the collision theory, the rate of a reaction is determined by the
(a) frequency of collision
(b) magnitude of activation energy
Frequency of collision
(a) If the frequency of collision for a reaction is high, then the frequency of effective collision that can result in a reaction is also high.
(b) Hence, the rate of reaction is high, and vice versa
(c) Figure illustrates this relationship.
Magnitude of activation energy
(a) The values of are different for different reactions.
(b) If the activation energy of a reaction is high, fewer number of colliding particles are able to achieve this high energy. Hence, the frequency of effective collision is low. As a result, the rate of reaction is low.
(c) Conversely, if the activation energy of a reaction is low, the rate of reaction is high.
The effect of the size of a solid reactant/surface area, concentration, pressure, temperature and catalyst on the rate of reaction can be explained using the collision theory.