Chemical Reactions: Comprehensive Notes

Chemical Reactions

Physical vs. Chemical Changes

Physical Change

Definition: Does not produce any new chemical substances.
Reversibility: Often easy to reverse.
Energy: Not a lot of energy involved in formation (e.g., mixing two substances).
Example: Evaporation.

Chemical Change

Definition: New chemical substances are formed that have very different properties from the reactants.
Signs of Formation: Color change, precipitate formation, bubbles of gas produced.
Reversibility: Very difficult to reverse.
Energy: Energy changes occur (exothermic or endothermic).
Example: Burning wood.

Factors Affecting Rates of Reaction

Surface Area: of solid reactants.
Concentration: of reactants in solution or pressure of reacting gases.
Temperature: at which the reaction is carried out.
Catalyst: The use of a catalyst.

Collision Theory

Statement: For a reaction to occur:
- Particles must collide with each other.
- The collision must have sufficient energy to cause a reaction (i.e., enough energy to break bonds).
Activation Energy: The minimum energy that colliding particles must have to react.
Successful Collisions: Collisions which result in a reaction.
- If particles have sufficient energy (i.e., energy greater than the activation energy), the collision will be successful.

Factors Affecting Successful Collisions

Number of Particles: More particles per unit volume produce more frequent successful collisions.
Frequency of Collisions: A greater number of collisions per second will give a greater number of successful collisions per second.
Kinetic Energy of Particles: Greater kinetic energy means a greater proportion of collisions will have an energy that exceeds the activation energy, and the more frequent the collisions will be as the particles are moving quicker; therefore, more collisions will be successful.
Activation Energy: Fewer collisions will have an energy that exceeds higher activation energy, and fewer collisions will be successful.

Surface Area and Reaction Rate

Increasing the surface area of a solid will increase the rate of reaction.
More surface area of the particles will be exposed to the other reactant, producing a higher number of collisions per second.
If you double the surface area, you will double the number of collisions per second.

Effect of Surface Area on Rate of Reaction (Graph)

Steeper gradient at the start.
Becomes horizontal sooner.
Shows that with increased surface area of the solid, the rate of reaction will increase.

Concentration and Reaction Rate

Increasing the concentration of a solution will increase the rate of reaction.
More reactant particles in a given volume, allowing more frequent and successful collisions per second, increasing the rate of reaction.
For a gaseous reaction, increasing the pressure has the same effect as the same number of particles will occupy a smaller space, increasing the concentration.
The number of collisions is proportional to the number of particles present.

Effect of Concentration/Pressure on Rate of Reaction (Graph)

Steeper gradient at the start.
Becomes horizontal sooner.
Shows that the rate of reaction will increase.

Temperature and Reaction Rate

Increasing the temperature will increase the rate of reaction.
Particles will have more kinetic energy than the required activation energy.
Therefore, there will be more frequent collisions, and a higher proportion of particles have energy greater than the activation energy.
This causes more successful collisions per second, increasing the rate of reaction.

Effect of Temperature on Rate of Reaction (Graph)

Steeper gradient at the start.
Becomes horizontal sooner.
Shows that an increase in temperature will increase the rate of reaction.

Catalysts

Catalysts are substances that speed up the rate of a reaction without themselves being altered or consumed.
The mass of a catalyst at the beginning and end of a reaction is the same, and they do not form part of the equation.
Different processes require different types of catalysts, but they all work on the same principle of providing a different pathway for the reaction to occur that has a lower activation energy.

Reversible Reactions

A reversible reaction is one that can be reversed by changing the reaction conditions.
$A + B \rightleftharpoons C + D$

Testing for Water (Copper (II) Sulphate)

$CuSO4 \cdot 5H2O(s) \rightleftharpoons CuSO4(s) + 5H2O(g)$
When blue hydrous copper (II) sulphate is heated, the water evaporates, and white anhydrous copper (II) sulphate is left behind.
Forward reaction: backwards. Backwards reaction: forward.

Dynamic Equilibrium

In a closed system, the rate of the forward reaction is equal to the rate of the reverse reaction.
$H2O(l) \rightleftharpoons H2O(g)$
The concentration of reactants and products are no longer changing.

Le Chatelier’s Principle

If a dynamic equilibrium is disturbed, the system will oppose the disturbance until equilibrium has been reached again.
Factors affecting equilibrium:
1. Temperature
2. Concentration
3. Pressure
4. A catalyst or changing the surface area of reactants will not affect dynamic equilibrium, only the time for reaching equilibrium.

Temperature

For a chemical reaction, if $\Delta H$ is negative, the forward reaction is exothermic, and the reverse reaction is endothermic. If the $\Delta H$ is positive, the forward reaction is endothermic, and the reverse reaction is exothermic.
If the temperature is increased, the endothermic reaction is favored (heat removed).
If the temperature is decreased, the exothermic reaction is favored (heat added).

Pressure

Pressure of a gaseous system is altered by changing the volume of the container. If the volume is decreased, the pressure is increased. If the volume is increased, the pressure is decreased.
If pressure increases, the reaction where the number of moles decreases is favored.
If pressure decreases, the reaction where the number of moles increases is favored.

Concentration

Concentration of an equilibrium system can be manipulated by either removing reactants from the system or by adding products to the system.
If reactant is added on the left-hand side of the equation, the forward reaction is favored.
If product is added on the right-hand side of the equation, the backward reaction is favored.

Catalyst

The presence of a catalyst does not affect the position of equilibrium, but it does increase the rate at which equilibrium is reached.
The catalyst increases the rate of both the forward and backward reactions by the same amount (by providing an alternative pathway requiring lower activation energy).
The concentration of reactants and products is the same at equilibrium as it would be without the catalyst.

The Haber Process

Nitrogen - from the air
Hydrogen - from natural gas (methane)
Reaction conditions:
- Temperature: 450 ºC
- Pressure: 200 atm
- Catalyst: Iron
Exam Tip: Remember: These conditions are a compromise among yield, rate, safety, and cost.

Explaining Conditions

Temperature: 450 ºC
- A higher temperature would favour the reverse reaction as it is endothermic (takes in heat) so a higher yield of reactants would be made.
- If a lower temperature is used, it favours the forward reaction as it is exothermic (releases heat) so a higher yield of products will be made.
- However, at a lower temperature, the rate of reaction is very slow.
- So 450 ºC is a compromise temperature between having a lower yield of products but being made more quickly.
Pressure: 200 atm
- A lower pressure would favour the reverse reaction as the system will try to increase the pressure by creating more molecules (4 molecules of gaseous reactants), so a higher yield of reactants will be made.
- A higher pressure would favour the forward reaction as it will try to decrease the pressure by creating fewer molecules (2 molecules of gaseous products), so a higher yield of products will be made.
- However, high pressures can be dangerous, and very expensive equipment is needed.
- So 200 atm is a compromise pressure between a lower yield of products being made safely and economically.
Catalyst: Iron
- The presence of a catalyst does not affect the position of equilibrium, but it does increase the rate at which equilibrium is reached.
- This is because the catalyst increases the rate of both the forward and backward reactions by the same amount (by providing an alternative pathway requiring lower activation energy).
- As a result, the concentration of reactants and products is nevertheless the same at equilibrium as it would be without the catalyst. So a catalyst is used as it helps the reaction reach equilibrium quicker.
- It allows for an acceptable yield to be achieved at a lower temperature by lowering the activation energy required.
Exam Tip: The reaction conditions chosen for the Haber process are not ideal in terms of the yield but do provide a balance between product yield, reaction rate, and production cost. These are called compromise conditions as they are chosen to give a good compromise between yield, rate, and cost.

The Contact Process

Sulfuric acid is synthesized by the Contact process.
Concentrated sulfuric acid is used in car batteries, making fertilizers, soaps, and detergents.
First Stage: The production of sulfur dioxide, either by burning sulfur to oxidize the sulfur (equation shown below) or roasting sulfide ores.
- $S + O2 \rightarrow SO2$
Main Stage: The main stage in the Contact process is the oxidation of sulfur dioxide to sulfur trioxide using a vanadium(V) oxide, $V2O5$ , catalyst:
- The oxygen used in this stage is obtained from air.
- The sulfur dioxide from the burning of sulfur or roasting of sulfide ores.
The conditions for this main stage of production are:
- A temperature of 450 ºC
- A pressure of 2 atm (200 kPa)
- Vanadium (V) oxide catalyst
Once sulfur trioxide is formed, it undergoes more processes to produce sulfuric acid.
Exam Tip: Remember: These conditions are a compromise among yield, rate, safety, and cost.

Explaining Conditions in the Contact Process

Temperature: 450ºC
- The forward reaction is exothermic, so increasing the temperature shifts the position of equilibrium to the left in the direction of the reactants.
- Therefore, the higher the temperature, the lower the yield of sulfur trioxide.
- The optimum temperature is a compromise between a higher rate of reaction at a higher temperature and a lower equilibrium yield at a higher temperature.
Pressure: 2 atm
- An increase in pressure shifts the position of equilibrium to the right in the direction of a smaller number of gaseous molecules.
- However, the position of equilibrium lies far to the right (the equilibrium mixture contains about 96% sulfur trioxide).
- So, the reaction is carried out at just above atmospheric pressure because:
  - High pressures can be dangerous, and very expensive equipment is needed.
  - A higher pressure causes the sulfur dioxide to liquefy.
Catalyst: Vanadium (V) oxide
- The presence of a catalyst does not affect the position of equilibrium, but it does increase the rate at which equilibrium is reached.
- This is because the catalyst increases the rate of both the forward and backward reactions by the same amount (by providing an alternative pathway requiring lower activation energy).
- As a result, the concentration of reactants and products is nevertheless the same at equilibrium as it would be without the catalyst. So a catalyst is used as it helps the reaction reach equilibrium quicker.
- It allows for an acceptable yield to be achieved at a lower temperature by lowering the activation energy required.
Exam Tip: The reaction conditions chosen for the Contact Process are not ideal in terms of the yield but do provide a balance between product yield, reaction rate, and production cost. These are called compromise conditions as they are chosen to give a good compromise between yield, rate, and cost.

Redox Reactions

Oxidation and reduction take place together at the same time in the same reaction.

Oxidation

The gain of oxygen.
The loss of electrons.
Increase in oxidation number.

Reduction

The loss of oxygen.
The gain of electrons.
Decrease in oxidation number.

Redox in Terms of Oxygen

$2CuO(s) + 2C(s) \rightarrow 2Cu(s) + CO_2(g)$
Oxidation: Oxygen is gained $C \rightarrow CO_2$ . Carbon is oxidized and is the reducing agent.
Reduction: Oxygen is lost $CuO \rightarrow Cu$ . Copper(II) oxide is reduced and is the oxidizing agent.

Redox in Terms of Electrons

$Mg(s) + 2HCl(aq) \rightarrow MgCl2(aq) + H2(g)$
Oxidation: Loss of electrons. $Mg(s) \rightarrow 2e^- + Mg^{2+}(aq)$ . Magnesium is oxidized and is the reducing agent.
Reduction: Gaining of electrons $2H^+(aq) + 2e^- \rightarrow H_2(g)$ . Hydrochloric acid or hydrogen ions are reduced and are the oxidizing agent.

Identifying Redox Reactions Using Oxidation Numbers

The oxidation number (also called oxidation state) is a number assigned to an atom or ion in a compound which indicates the degree of oxidation (or reduction).
It shows the number of electrons that an atom has lost, gained, or shared in forming a compound.
It is written as a +/- sign followed by a number (not to be confused with charge which is written by a number followed by a +/- sign).
E.g., aluminum in a compound usually has the oxidation state +3.

Rules for Assigning Oxidation Numbers

The oxidation number of any uncombined element is zero (e.g., $H_2$ , $Zn$ ).
Many atoms or ions have fixed oxidation numbers in compounds:
- Group 1 elements are always +1
- Group 2 elements are always +2
- Fluorine is always -1
- Hydrogen is +1 (except for in metal hydrides like NaH, where it is -1)
- Oxygen is -2 (except in peroxides, where it is -1 and $F_2O$ where it is +2)
The oxidation number of an element in a mono-atomic ion is always the same as the charge (e.g. $Zn^{2+}$ ox # = +2, $Fe^{3+}$ ox # = +3, $Cl^-$ ox # = -1).
The sum of the oxidation numbers in a compound is zero (e.g. $NaCl$ : Na = +1, Cl = -1, sum = 0).
The sum of oxidation numbers in an ion is equal to the charge on the ion. (e.g. $SO_4$ : S = +6, 4O atoms = 4(-2)= -8 + 6 = -2).

Identifying Redox Reactions by Colour Changes

The tests for redox reactions involve the observation of a color change in the solution being analyzed.
Two common examples are acidified potassium manganate(VII) and potassium iodide.
Potassium manganate(VII), $KMnO_4$ , is an oxidizing agent which is often used to test for the presence of reducing agents.
When acidified potassium manganate(VII) is added to a reducing agent, its color changes from purple to colorless.
Potassium iodide, KI, is a reducing agent which is often used to test for the presence of oxidizing agents

When added to an acidified solution of an oxidising agent such as aqueous chlorine or hydrogen peroxide (H2O2), the solution turns a red-brown colour due to the formation of iodine, $I2$ : $2KI (aq) + H2SO4 (aq) + H2O2 (aq) \rightarrow I2 (aq) + K2SO4 (aq) + 2H_20 (l)$

The potassium iodide is oxidised as it loses electrons and hydrogen peroxide is reduced, therefore potassium iodide is acting as a reducing agent as it will itself be oxidised: $2I^- \rightarrow I_2 + 2e^-$

Oxidizing Agent

A substance that oxidizes another substance and becomes reduced in the process.
An oxidizing agent gains electrons as another substance loses electrons.
Common examples include hydrogen peroxide, fluorine, and chlorine.

Reducing Agent

A substance that reduces another substance and becomes oxidized in the process.
A reducing agent loses electrons as another substance gains electrons.
Common examples include carbon and hydrogen.
The process of reduction is very important in the chemical industry as a means of extracting metals from their ores.

Example

$CuO + H2 \rightarrow Cu + H2O$
In the above reaction, hydrogen is reducing the CuO and is itself oxidized as it has lost electrons, so the reducing agent is therefore hydrogen: $H_2 \rightarrow 2H^+ + 2e^-$
The CuO is reduced to Cu by gaining electrons and has oxidized the hydrogen, so the oxidizing agent is therefore copper oxide $Cu^{2+} +2e^- \rightarrow Cu$

Examples

$Mg(s) + 2HCl(aq) \rightarrow MgCl2(aq) + H2(g)$
- Write down oxidation numbers of each element
- Find the changes
- Write down the half reactions.

Mg → $Mg^{2+}$ + 2e − Increase in oxidation number is oxidation oxidation half reaction

2H + 2e − → $H_2$ Decrease in oxidation number is reduction reduction half reaction Loss of electrons Mg = reducing agent HCl or H = oxidizing agent

NSSCO CHEMISTRY TOPIC 6 / CHEMICAL REACTIONS / A LOUW.

2CuO(s) + 2C(s) 2Cu(s) + CO2(g). Write down oxidation numbers of each element Find the changes Write down the half reactions. Cu 2+ Cu +2 0

C CO2 0 +4 + 4e- + 2e- x2 2Cu + 4e- 2Cu Example 2.

Investigating The Rate of a Reaction

To measure the rate of a reaction, we need to be able to measure either how
- quickly the reactants are used up or
- how quickly the products are formed.
Three commonly used techniques are:
1. measuring mass loss on a balance.
2. measuring the volume of a gas produced.
3. measuring a reaction where there is a color change at the end of the reaction.

Measure the loss in mass of reactant – effect of temperature on the rate of the reaction.

Method

Dilute hydrochloric acid is heated to a set temperature using a water bath
Add the dilute hydrochloric acid into a conical flask
Add a strip of magnesium and start the stopwatch
Stop the time when the magnesium fully reacts and disappears
Repeat at different temperatures and compare results

Result

With an increase in the temperature, the rate of reaction will increase
This is because the particles will have more kinetic energy than the required activation energy, therefore more frequent and successful collisions will occur, increasing the rate of reaction

Mass loss in general Monitoring changes in mass

Many reactions involve the production of a gas which will be released during the reaction
The gas can be collected and the volume of gas monitored as per some methods above
Alternatively, the reaction can be performed in an open flask on a balance to measure the loss in mass of reactant
Cotton wool is usually placed in the mouth of the flask which allows gas out but prevents any materials from being ejected from the flask (if the reaction is vigorous)
This method is not suitable for hydrogen and other gases with a small relative formula mass, Mr, as the loss in mass may be too small to measure
!Exam Tip! There are many different methods of investigating the rate of reaction. Another method of gas collection you may see uses a gas syringe. Students may be required to devise and evaluate methods of investigating rates of reaction.

Measure the volume of gas formed - changing the surface area of the reactant

Method:

Add dilute hydrochloric acid into a conical flask.
Use a delivery tube to connect this flask to a measuring cylinder upside down in a bucket of water (downwards displacement).
Add calcium carbonate chips into the conical flask and quickly put the bung back into the flask.
Measure the volume of gas produced in a fixed time using the measuring cylinder and stop clock.
Repeat with different sizes of calcium carbonate chips (lumps, crushed and powdered).

Result:

Smaller sizes of chips cause an increase in the surface area of the solid, so the rate of reaction will increase.
This is because more surface area of the particles will be exposed to the other reactant so there will be more frequent and successful collisions, increasing the rate of reaction.

3. Measure the amount of gas formed - using a catalyst.

Method:

Add hydrogen peroxide into a conical flask
Use a delivery tube to connect this flask to a measuring cylinder upside down in a tub of water (downwards displacement)
Add the catalyst manganese(IV) oxide into the conical flask and quickly place the bung into the flask
Measure the volume of gas produced in a fixed time using the measuring cylinder
Repeat experiment without the catalyst of manganese(IV) oxide and compare results

Result:

Using a catalyst will increase the rate of reaction
The catalyst will provide an alternative pathway requiring lower activation energy so more colliding particles will have the necessary activation energy to react
This will allow more frequent and successful collisions, increasing the rate of reaction

Measuring the color change - by changing the concentration of one reactant

Method:

Measure 50 cm3 of sodium thiosulfate solution into a flask
Measure 5 cm3 of dilute hydrochloric acid into a measuring cylinder.
Draw a cross on a piece of paper and put it underneath the flask.
Add the acid into the flask and immediately start the stopwatch.
Look down at the cross from above and stop the stopwatch when the cross can no longer be seen.
Repeat using different concentrations of sodium thiosulfate solution (mix different volumes of sodium thiosulfate solution with water to dilute it).

Result:

With an increase in the concentration of a solution, the rate of reaction will increase.
This is because there will be more reactant particles in a given volume, allowing more frequent and successful collisions, increasing the rate of reaction.

Planning an experiment

List apparatus
Write down the method in chronological order.
Repeat with more than one variable
Always have a control variable.
Repeat experiment for each variable 3 times and get an average of each.

Evaluating Investigations of Rates of Reactions

When investigating rates of reaction, there are a number of different methods that can be used to carry out the same investigation
Evaluating what is the best method to use is part of good experimental planning and design
This means appreciating some of the advantages and disadvantages of the methods available. Interpreting data
Data recorded in rate studies is used to plot graphs to calculate the rate of a reaction
Plotting a graph until the completion of the reaction shows how the rate changes with time Over time the rate of reaction slows as the reactants are being used up so the line becomes less steep and eventually become horizontal, indicating the reaction has finished.
You can plot more than one run of a variable on the same graph making it easier to see how the variable influences the rate
The steeper the curve, the faster the rate of the reaction.
The curve is steepest initially, so the rate is quickest at the beginning of the reaction
As the reaction progresses, the concentration of the reactants decreases, and the rate decreases shown by the curve becoming less steep
When one of the reactants is used up, the reaction stops, the rate becomes zero and the curve levels off to a horizontal line
The amount of product formed in a reaction is determined by the limiting reactant:
- If the amount of limiting reactant increases, the amount of product formed increases
- If the amount of the reactant in excess increases, the amount of product remains the same. Drawing a tangent to the slope allows you to show the gradient at any point on the curve.
The steeper the slope, the quicker the rate of reaction.The volume of a gaseous product would increase to a maximum overtime, so the time levels out indicating the reaction is over
Since the volume and mass would be proportional, this could also be a graph of the mass of product versus time