Matter and Its Properties

In this section, you will learn about how energy changes during chemical reactions (called enthalpy) and how to figure out these changes using experiments. We will also explore Hess’s Law, which helps you understand energy in a quantitative way. In addition, you will build important skills like critical thinking, problem solving, and scientific communication.

KEY IDEAS

• Atomisation energy: the energy required to break all the bonds in one mole of a substance to form individual atoms in the gas phase under standard conditions.

• Bond dissociation energy the energy required to break a specific chemical bond within a compound.

• Bond enthalpy is the average energy needed to split one mole of a specific type of covalent bond in a gaseous molecule.

• Enthalpy change is the enthalpy change within a reaction system.

• Enthalpy is a thermodynamic quantity which describes the energy of a system.

• Hess’s Law of constant heat summation states that the total enthalpy change of chemical reaction is equal to the sum of all the individual enthalpy changes.

• Lattice energy the energy required to separate 1 mole of an ionic solid to gaseous ions.

A system is the part of the Universe that we are focusing on in an experiment.

Everything else around it is called the surroundings. There are three types of systems:

1. Open system: This can exchange both heat and matter with the surroundings.

An example is heating water in an open beaker.

2. Closed system: This can exchange heat, but not matter, with the surroundings.

An example is heating water in a covered beaker.

3. Isolated system: This cannot exchange heat or matter with the surroundings.

An example is a well-insulated thermos (like a calorimeter). It should be noted that no system can be thought of as truly ideal, as some heat or matter will eventually escape to the surroundings. It is a term used to describe systems which will exchange a negligible fraction of heat/matter with their surroundings.

To measure how energy changes during a chemical reaction, scientists use a special quantity called enthalpy (H). Enthalpy is a combination of the system’s internal energy (U) and the product of its pressure (P) and volume (V). Enthalpy has the same units as energy; Joules.

Mathematically, we write this as, H = U + PV This means that enthalpy depends on things like the pressure and volume of the system. When the pressure stays the same, enthalpy is the amount of heat energy that is added or removed during the reaction. This is quite common in chemistry so is worth noting, an “isobaric” reaction.

The enthalpy change (ΔH) in a chemical reaction is found by comparing the enthalpy of the products (what we get after the reaction) with the enthalpy of the reactants (what we started with).

Enthalpy change (ΔH) is therefore defined as:

ΔH= H_(products)− Hreactants Endothermic and Exothermic Reactions When we study how energy changes during chemical reactions, we look at what is called the enthalpy of reaction (ΔHᵣₓₙ). This tells us whether a reaction gives off heat or takes in heat.

1. Exothermic reactions are reactions that release heat to the surroundings, making things around them warmer. When this happens, the enthalpy of the system decreases as the products have less energy/enthalpy than the reactants, and ΔHᵣₓₙis negative.

In exothermic reactions the system ends up with a lower heat content than the surrounding.

An example of an exothermic reaction is the combustion of methane CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ΔH = −890.4 kJ/mol

2. Endothermic reactions are reactions that absorb heat from the surroundings, making things cooler. In these reactions, the enthalpy of the system increases as the products have more energy/enthalpy than the reactants, and ΔHₓₙis positive.

In endothermic reactions, the system has more energy than the surroundings because it took heat from them. The difference in energy comes from the heat system absorbs!

An example of an endothermic reaction is the decomposition of calcium carbonate CaCO₃(s) → CaO(s) + CO₂(g) ΔH = +177.8 kJ/mol.

Exothermic and endothermic reactions can be represented by diagrams called Energy profile diagrams. See Figure 1.1.

Figure 1.1: Energy profile diagrams for exothermic and endothermic reactions

Activity 1.1 Demonstrating the concepts of chemical systems (Open, Closed, and Isolated) Materials needed: 3 identical plastic or glass bottles with lids, hot water, thermometers, towel (insulating material), stopwatch or timer, notebook pen and pencil Steps

1. Setup for the Demonstration

a. Bottle 1 (Open system): Pour the hot water into a bottle and leave the lid off.

b. Bottle 2 (Closed system): Pour the hot water into a bottle and tightly seal it with the lid.

c. Bottle 3 (Isolated system): Pour hot water into a bottle, seal it with the lid, and wrap it tightly with towel (insulating material).

2. Measure Initial Temperature

Measure and write down the initial temperature of the hot water in each bottle.

3. Prediction

a. What do you think will happen to the temperature of the water in each bottle over time?

b. Which bottles will lose heat fast, slow, or not at all?

c. Why do you think the bottle wrapped in towel stayed warm longer?

4. Observation

a. Leave the bottles for 10–15 minutes. Check and record the temperature at every 5 minutes.

b. Feel in your palm whether the bottle feels warm or cool and whether heat seems to be escaping.

5. Explain

a. what you observe about the temperature change of the hot water in Bottle 1 (Open System).

b. what you observe about the temperature change of the hot water in Bottle 2 (closed System).

c. what you observe about the temperature change of the hot water in Bottle 2 (isolated System).

6. Give examples of real-life examples of open, closed, and isolated systems at home.

Activity 1.2 Exploring Enthalpy and Enthalpy change Materials Needed: Chart paper and markers for group notes, access to a video on endothermic and exothermic reactions.

Steps

1. Your teacher will divide the class into small group (about 4–5 students per group).

2. Group discussion of some key terms

a. Activation Energy

b. Endothermic

c. Exothermic

d. Reactants

e. Products

f. Enthalpy

g. Energy

3. Use the link below to watch video on exothermic and endothermic reactions.

https://www.youtube.com/watch?v= 0MBVIXufFbM&t= 2s After the video:

a. Give real-life examples of an exothermic reaction.

b. Give real-life examples of an endothermic reaction.

c. Explain happens to the surroundings in an endothermic reaction.

d. Explain happens to the surroundings in an exothermic reaction.

e. Where does the energy come from in an exothermic reaction?

f. Where does the energy come from in an endothermic reaction?

g. Discuss why understanding of energy changes in chemical reactions is important.

Activity 1.3 Understanding how to represent enthalpy profile diagrams Materials Needed: Blank paper or graph paper, pencils, erasers, and rulers, markers or coloured pencils, reference charts or handouts explaining energy profile diagrams Steps

1. Introduction to enthalpy profile diagrams

a. Refer to Figure 1.1 for energy profile diagram for exothermic and endothermic reactions. Use these as a guide for constructing your own following the instructions below.

2. Draw an energy profile diagram for (Exothermic reaction) combustion of (CH₄), to produce carbon dioxide and water, releasing energy.

a. Draw and label vertical line representing enthalpy change (vertical ‘y’ axis).

b. Draw and label horizontal line representing reaction progress (horizontal ‘x’ axis).

c. Draw a horizontal line on the left for the reactants (CH₄+ O₂), representing a higher energy level.

d. Draw another horizontal line lower down on the right for the products (CO₂+ H₂O), showing that the products have less energy.

e. Show an upward curve, indicating the activation energy (Eₐ) (the energy needed to start the reaction).

f. Connect the product line with the peak of the activation energy curve.

g. Label the enthalpy change (ΔH) as negative, showing that energy was released.

3. Draw an energy profile diagram for (Endothermic reaction) like photosynthesis.

a. Draw and label vertical line representing enthalpy change (vertical ‘y’ axis).

b. Draw and label horizontal line representing reaction progress (horizontal ‘x’ axis).

c. Draw a horizontal line on the left for the reactants (CO₂+ H₂O) at a lower energy level.

d. Draw another horizontal line higher up on the right for the products (glucose + O₂), showing that the products have more energy.

e. Show an upward curve to represent the activation energy (Eₐ) required for the reaction to take place.

f. Connect the Eₐcurve with the product line.

g. Label the enthalpy change (ΔH) as positive, showing that energy was absorbed.

4. What are the differences between the exothermic and endothermic diagrams?

5. Where does the energy go/come from in each case?

6. Describe how you can see from a reaction profile that a reaction is exothermic or endothermic.

When a chemical reaction happens, energy is either taken in or given off, and we measure this energy change. But the amount of energy can change depending on the reaction conditions, like temperature and pressure. In order to compare processes, scientists use standard conditions to measure these energy changes.

Standard conditions are used to measure and compare chemical reactions easily.

The Standard conditions are:

1. Pressure: 1 atmosphere (1 atm or 101.3 kPa)

2. Temperature: 298 K (which is the same as 25°C or room temperature).

When we measure energy changes under these conditions, we call it standard enthalpy change. It shows how much energy (heat) is gained or lost during a reaction when everything is at the same pressure and starting temperature. We usually measure this energy in kilojoules per mole (kJ/mol), and we use the symbol ΔH° to show standard enthalpy change.

Standard State: This means the most stable form of an element or compound at room temperature and 1 atmosphere of pressure. For example, oxygen’s standard state is O₂gas because that’s how it exists in the air. For water, the standard state is liquid because that’s its most stable form at room temperature.

For reactions in solutions, scientists also use a standard concentration, which is usually:

1.0 mol/dm³ (This means 1 mole of a substance is dissolved in 1 dm³of water).

Understanding Enthalpy Changes

1. If the standard enthalpy change (ΔH°) is negative (ΔH° < 0), the reaction is exothermic, meaning it releases heat to the surroundings.

2. If the standard enthalpy change (ΔH°) is positive (ΔH° > 0), the reaction is endothermic, meaning it absorbs heat from the surroundings.

We can calculate the standard enthalpy change by looking at the energy of the products and reactants. This helps scientists compare how much energy is involved in different reactions and is very important in chemistry and industry!

Types of standard enthalpy changes

1. Standard enthalpy changes of reaction It is the quantity of heat involved when any valid stoichiometric molar quantities of reactants combine to form products under standard conditions. Any heat change measured under standard conditions is described as a standard enthalpy change of reaction. It is dependent on the stoichiometry of the balanced chemical equation.

Changes in the balanced coefficients of reactants and products in any chemical reaction will change the value of its enthalpy changes accordingly. It has a unit of J/mol (or kJ/mol).

For example;

N₂(g) + 3 H₂(g) → 2 NH₃(g) ∆Hᵣₓₙ °= − 92.38 k Jmol⁻¹1_ 2 N₂(g) + 3_ 2 H₂(g) → 2 NH₃(g) ∆Hᵣₓₙ °= − 46.19 k Jmol⁻¹3 N₂(g) + 9 H₂(g) → 6 NH₃(g) ∆Hᵣₓₙ °= − 277.14 k Jmol⁻¹Notice here this is the same reaction, the standard enthalpy of reaction changes based upon how much of the substances react. The value is not constrained in any way.

2. Standard enthalpy change of formation The standard enthalpy change of formation of a compound is the enthalpy change those results when one mole of the compound is formed from its elements under standard conditions (298K and 101.3 kPa), all reactants and products being in their standard states. It is given the symbol ∆H_(f) °E.g., C(s) + 2 H₂(g) → C H₄(g), ∆H_(f) °(C H₄) = − 74.8 kJ mol⁻¹. The standard enthalpy of formation of methane is -74.8 kJmol⁻¹.

The standard enthalpy of formation of all elements in their standard states is zero.

Using oxygen as an example, it exists in the most stable form as a diatomic molecule (O₂) in the gaseous form as compared to the other allotropic forms such as ozone (O₃). As such ∆H_(f) °(O₂) = 0 but ∆H_(f) °(O₃) = 142.2 kJ / mol while ∆H_(f) °(O) = 249.4 kJ / mol .

The most stable form of carbon under standard conditions is graphite, therefore ∆H_(f) °(C(graphite)) = 0

Table 1.1 describes some elements in their standard state; any other states will have associated enthalpies of formation.

Table 1.1: Elements in their standard state Substance ∆H_(f) °Substance ∆H_(f) °O₂(g) 0 C(graphite) 0 N₂(g) 0 S(rhombic) 0 H₂(g) 0 Ag(s) 0 F₂(g) 0 Ca(s) 0 I₂(g) 0 Cu(s) 0 Cl₂(g) 0 Hg(l) 0 Standard enthalpies of formation values help in calculating the standard enthalpy of reaction (Δ Hᵣₓₙ °), defined as the enthalpy of a reaction carried out under standard conditions.

For example, consider the hypothetical reaction aA + bB → cC + dD where a, b, c, and d are stoichiometric coefficients. For this reaction, Δ Hᵣₓₙ °is given by Δ Hᵣₓₙ °= [c ∆ H_(f) °(C) + d ∆ H_(f) °(D)] − [a ∆ H_(f) °(A) + b ∆ H_(f) °(B)] We can generalise Δ Hᵣₓₙ °= Σn ∆ H_(f) °(products) − Σm ∆ H_(f) °(reactants) where m and n are the stoichiometric coefficients for the reactants and products, respectively, and Σ (sigma) means “the sum of.”

Standard enthalpy of formation values also gives a measure of the stability of compounds upon their formation. Compounds with negative ∆H_(f) °values are more stable than compounds with positive ∆H_(f) °values.

3. Standard enthalpy change of combustion The standard enthalpy of combustion of a substance is defined as the enthalpy change that occurs when one mole of a substance is combusted completely in oxygen under standard conditions. (298K and 101.3 kPa), all reactants and products being in their standard states.

It is given the symbol ∆H_(c) °E.g., H₂(g) + 1/2 O₂(g) → H₂ O(l), ∆H_(c) °( H₂(g)) = − 285.8 kJ mol⁻¹.

The standard enthalpy of combustion of hydrogen is − 285.8 kJ mol⁻¹.

Notice here that fractional values for stoichiometric co-efficients can be used to balance the equation. This ensures only 1 mole of the substance is combusted.

Knowledge of ∆H_(c) °is applied in many fields including nutrition, industries, transportation and research.

Respiration and Combustion

Respiration is like a slow burning process in your body, where the food you eat gets “burned” to release energy that you need for daily activities. The amount of energy you get from food depends on what you eat. The main energy-giving foods are carbohydrates and fats.

In nutrition, the energy from these foods is measured in Calories. If you eat more carbohydrates than your body needs, the extra energy gets stored as fat. Eating too much can cause your body to store too much fat, which can lead to obesity.

In industries, the enthalpy of combustion is used to measure how efficiently machines like power plants or engines turn fuel into energy. Fuels with higher energy (higher ∆H_(c) °) are better for creating power. It also helps compare different fuels and even different forms of the same material, like allotropes.

4. Standard enthalpy change of neutralisation.

The standard enthalpy change of neutralisation is the enthalpy change when solutions of an acid and an alkali react together under standard conditions to produce 1 mole of water. It is assigned the symbol ∆Hₙ °.

When one mole of a monobasic acid reacts with one mole of a monobasic base under standard conditions the amount of energy released is almost the same as the enthalpy of formation of water from its ions.

HCl(aq) + NaOH(aq) → NaCl(aq) + H₂ O(l) ∆ H = − 57.5 kJ / mol HN O₃(aq) + KOH(aq) → KN O₃(aq) + H₂ O(l) ∆ H = − 57.3 kJ / mol Enthalpy of formation of water from its ions is represented by the equation;

H+(aq)+ OH−(aq) → H₂ O(l) ΔH = − 57.4 kJ / mol Strong acids and strong bases ionise completely in aqueous solution to give off hydrogen ions and hydroxyl ions therefore they tend to have higher enthalpies of neutralisation than the weak acid and weak bases (which do not fully ionise).

5. Standard enthalpy of solution and hydration.

The standard enthalpy of solution is the enthalpy change when one mole of an ionic substance is dissolved in excess water under standard conditions such that no further addition of water produces a further change in energy. It is assigned the symbol ∆Hₛₒₗₙ °.

e.g., NaCl (s)+ H₂ O(l) → Na+ (aq)+ Cl– (aq) ΔH = − 3.9 kJ / mol Standard enthalpy of hydration is the enthalpy change when one mole of gaseous ions is dissolved in water to give one mole of aqueous ions and a solution of infinite dilution. It can also be defined as the energy evolved when one mole of a gaseous ion is surrounded, coordinated and stabilised by a sheath of water molecules under standard conditions.

E.g., Na+(g) + H₂ O(l) → Na+ (aq) ΔH = − 406 kJ / mol

Activity 1.4 Understanding types of standard enthalpy changes Materials: Computer, pen, calculator and Worksheets with sample calculations, Step

1. Use the link below to watch the video on types of standard enthalpy changes.

https://www.nagwa.com/en/videos/298198084514/ Use the questions below to explore types of standard enthalpy changes.

2. Define and explain the following standard enthalpy changes:

a. Standard enthalpy change of reaction (∆Hᵣ °)

b. Standard enthalpy change of formation (∆H_(f) °)

c. Standard enthalpy change of combustion (∆H_(c) °)

d. Standard enthalpy change of neutralisation (∆Hₙ °)

e. Standard enthalpy change of solution (∆Hₛₒₗₙ °)

f. Standard enthalpy change of hydration (∆H_(hyd) °)

3. Give an example for each type of enthalpy change

a. Standard enthalpy change of reaction (∆Hᵣ °)

b. Standard enthalpy change of combustion (∆H_(c) °)

c. Standard enthalpy change of neutralisation (∆Hₙ °)

d. Standard enthalpy change of solution (∆Hₛₒₗₙ °)

e. Standard enthalpy change of hydration (∆H_(hyd) °)

f. Standard enthalpy change of formation (∆H_(f) °)

4. Calculating standard enthalpy change:

a. Standard enthalpy change of reaction (∆Hᵣ °) Formula: ∆Hᵣ °= Σ ∆ H_(f) °(products) − Σ ∆ H_(f) °(reactants) Let’s calculate the enthalpy change for the reaction: H₂(g) + Cl₂(g) → 2HCl(g) Given:

∆H_(f) °of HCl(g) = − 92 kJ / mol ∆H_(f) °of H₂(g) and Cl₂(g) = 0 kJ / mol (because they are elements in their standard states).

Calculation:

∆Hᵣ °= Σ ∆ H_(f) °(products) − Σ ∆ H_(f) °(reactants) = [2(− 92)] − [0 + 0] = − 184 kJ / mol The reaction releases 184 kJ of energy.

b. Standard enthalpy change of combustion (∆H_(c) °) Formula: ∆H_(c) °= Σ ∆ H_(f) °(products) − Σ ∆ H_(f) °(reactants) Calculate standard enthalpy of combustion of methane (CH₄):

Given:

C H₄(g) + 2 O₂(g) → C O₂(g) + 2 H₂ O(l) ∆H_(f) °of C H₄(g) = − 74.8 kJ mol⁻¹, ∆H_(f) °of C O₂(g) = − 393.5 kJ mol⁻¹, ∆H_(f) °of H₂ O(l) = − 285.8 kJ mol⁻¹Using the equation below, substitute the values into the equation.

∆H_(c) °= Σ ∆ H_(f) °(products) − Σ ∆ H_(f) °(reactants)

5. Discuss the significance of standard enthalpy changes in various fields (e.g., chemistry, physics, engineering) and explore real-life examples and applications (e.g., energy production, materials science).

Experimental determination of heat of combustion of alcohol Materials needed: Alcohol burner (containing ethanol), calorimeter (or a metal can), thermometer, Water (200 ml), balance, retort stand and clamp, heatproof mat.

Procedure

1. Measure initial mass of alcohol burner with the alcohol inside.

2. Add 200 ml of water to the calorimeter (or metal can).

3. Place the calorimeter on a stand above the alcohol burner, ensuring there is space for the flame to heat the water effectively.

4. Record initial temperature of water:

5. Light the alcohol burner and let the flame heat the water.

Stir the water gently to ensure even heating.

6. Allow the alcohol to burn until the temperature of the water rises by about 15–20 °C.

Extinguish the flame by capping the burner.

7. Record the final temperature of the water after extinguishing the flame.

8. Weigh the alcohol burner again to find the final mass of alcohol left in the burner.

Data

1. Initial mass of alcohol burner (mᵢₙᵢₜᵢₐₗ = x g).

2. Final mass of alcohol burner (m_(final) = y g).

3. Initial temperature of water (Tᵢₙᵢₜᵢₐₗ = Tᵢ).

4. Final temperature of water (T_(final) = T_(f)).

5. Mass of water = z g.

Calculations

1. Calculate Mass of Alcohol Burned

Mass of alcohol burned = (x − y) g

2. Calculate Temperature Change:

ΔT = T_(f) − Tᵢ

3. Calculate Energy Transferred to Water: Use the formula:

Q = m_(water) ⋅ c ⋅ ΔT Where, Q is the heat absorbed by the water (in joules), m_(water)is the mass of the water, c is the specific heat capacity of water (4.18 J/g°C), ΔT is the temperature change of the water.

4. Calculate Heat of combustion of the alcohol

a. Convert Q from joules to kilojoules (kJ) by dividing by 1,000.

b. Calculate the heat of combustion per gram of alcohol:

Heat of combustion per gram = Q___________________ mass of alcohol burned

c. If you want the heat of combustion per mole, multiply by the molar mass of the alcohol.

(for ethanol, C₂H₅OH, molar mass 46 g/mol)

Example Calculation

mᵢₙₜᵢₐₗ = x = 120 g m_(final) = y = 119 g Tᵢₙₜᵢₐₗ = Tᵢ = 22°C T_(final) = T_(f) = 42°C Mass of alcohol burned = (x − y) = 120 − 119 = 1.0 g ΔT = T_(f) − Tᵢ = 42 − 22 = 20°C Energy transferred to water:

Q = m_(water) ⋅ c ⋅ ΔT = 200 g × 4.18 J g⁻¹°C⁻¹× 20°C = 16,720 J = 16.72 kJ Heat of combustion per gram = Q___________________ mass of alcohol burned = 16.72_ 1 g = 16.72 k J g⁻¹Heat of combustion per mole (for ethanol, C₂H₅OH, molar mass 46 g/mol Heat of combustion per mole = 16.72 kJ g⁻¹× 46 g mol⁻¹= 769.12 kJ mol⁻¹Experimental Discussion Points

1. Why Stir the Water?

Stirring the water helps spread the heat evenly so that the thermometer shows a more accurate temperature. Without stirring, some parts might get hotter than others.

2. Comparing Your Results

Compare the energy value you calculated with the “official” value scientists have found in a lab. Do you notice any differences?

3. Why Are the Values Different?

The numbers might be different because some heat might have escaped to the surroundings instead of just heating the water.

Calculating % Difference To see how close you got, calculate the percentage difference between your result and the official value. Here’s how to do it:

% Difference = ( Official value − Your value_____________________ Official Value ) × 100 Improving the Experiment How could we improve our experiment? Maybe we could insulate the setup better to trap more heat or make sure the flame is steady.

Standard enthalpy of Neutralisation It is the heat released when one mole of an acid is neutralised by an alkali to produce one mole of water under standard conditions.

Experiment Title: Determination of the Standard Enthalpy of Neutralisation for a Strong Acid-Base Reaction.

In this experiment, we will find out how much heat is released when an acid reacts with a base to form water. This is called the enthalpy of neutralization.

Neutralisation reactions happen between an acid and a base and always release heat.

Materials need: Hydrochloric acid (HCl) solution, sodium hydroxide (NaOH)

solution, calorimeter (or a plastic cup in a beaker), thermometer, measuring cylinder, stirrer.

Procedure

1. Pour a specific mass (say, 50 g) of hydrochloric acid (HCl) into the calorimeter. Then, measure the same mass of sodium hydroxide (NaOH).

2. Measure and record the starting temperature of both the acid and base before mixing. They should both be at the same initial temperature.

3. Pour the sodium hydroxide into the hydrochloric acid. Stir the mixture gently record the highest temperature reached.

4. The temperature will rise because this reaction releases heat. The difference between the starting temperature and the highest temperature is the temperature change.

Calculating the Enthalpy of Neutralization

1. Find the heat released Use the formula:

Q = m × c × ΔT Where, Q is the heat released, m is the total mass of the solution, c is the specific heat capacity of water (4.18 J/g°C), here we will approximate the specific heat of the solution to that of water.

ΔT is the temperature change of the solution.

2. Calculate enthalpy of neutralisation Divide the heat released by the number of moles of acid or base that reacted to find the enthalpy of neutralisation in kJ/mol.

Example

If you get a temperature rise of 6°C with 100 g of solution, plug it into the formula:

Q = 100 × 4.18 × 6 = 2508 J Divide by the moles of acid or base used to get the enthalpy of neutralisation.

Calculations Use the calorimeter’s heat capacity to adjust for any heat lost or gained from the surroundings.

Find the moles of reactants using their mass, density and concentration (n = CV) (V = m/p).

Calculate the heat energy released per mole of water produced in the reaction.

Results and Analysis

Find the average enthalpy change for neutralisation from repeated trials.

Compare this value to the known value for strong acid-base reactions (-57 kJ/ mol). Calculate the % difference and think about why they differ.

Consider any sources of error and ways to improve the experiment.

Discussion Questions

1. Why is it important to stir the solution regularly?

2. How could we reduce the % difference between our results and the known value?

3. Would a colder room temperature affect the results? Why?

Conclusion: This experiment helps us understand the energy released when a strong acid reacts with a base. By comparing our results with the known value, we can check the accuracy of our experiment and improve it for future trials.

Standard enthalpy of solution It is the heat absorbed or released when one mole of an ionic substance is dissolved in excess water such that no further addition of water produces further heat change under standard conditions.

Experiment Title: Determination of the Standard Enthalpy of Dissolution for a Solute In this experiment, we will measure the energy change (enthalpy) that occurs when a solid dissolves in water. This process helps us understand how much heat is absorbed or released when a substance mixes with water.

Procedure

1. Pour a known mass of water into a beaker and measure its temperature.

2. Add a known mass of the solid (like salt) to the water.

3. Stir the mixture and watch for temperature changes. Record the final temperature once it stops changing.

4. Subtract the starting temperature from the final temperature to find the temperature change.

Data Needed

Mass of water in the beaker Mass of the solute (like salt) Temperature change (final - initial temperature) Specific heat capacity of water (4.18 J/g°C, a constant that tells us how much heat is needed to change water temperature). Here we will approximate the specific heat capacity of the solution to be that of water, for a more accurate result you could research the specific heat capacity of the concentration of solution you have made.

Calculation Using the data, we can calculate the heat energy (Q) absorbed or released:

Q = m × c × ΔT Where:

Q is the heat released, m is the total mass of the solution, c is the specific heat capacity of water (4.18 J/g°C), ΔT is the temperature change of the water.

Next, calculate the enthalpy change per mole by dividing Q by the moles of solute added (n = m_____ RAM).

Analysing results

1. Look at the official (theoretical) value for the solute and compare it to what we found our experiment. If they are not the same, that is okay! We will calculate the % difference to see how close they are.

2. Sometimes experimental results vary from official values due to small errors in setup or how we measured things.

3. Possible errors and improvements:

a. Stirring helps spread the heat evenly in the water, giving us a more temperature.

b. To improve accuracy, we could use a lid to reduce heat loss or ensure we measure everything carefully.

Conclusion This experiment shows us the energy change when a substance dissolves in water. Some substances absorb heat (endothermic), while others release heat (exothermic). This energy change, or “enthalpy of dissolution,” helps scientists understand how substances behave in solutions.

Activity 1.5 Experimental Determination of enthalpy changes In small groups, explore and measure different types of energy (enthalpy) changes by carrying out experiments.

Steps Your teacher will place you into groups.

Each group will do “circus experiments,” rotating through stations to explore different enthalpy changes.

Experiments

1. Enthalpy change of combustion

a. Measure the energy released when burning alcohol (e.g., ethanol) and food (e.g., maize or groundnut).

b. Procedure:

i. Burn the substance under a container of water.

ii. Measure the temperature increase of the water to determine the enthalpy change.

2. Enthalpy of Neutralisation

a. Measure energy changes when an acid (HCl) reacts with a base (NaOH).

b. Procedure:

i. Mix the acid and base in a cup and measure the temperature change.

ii. Record the temperature before and after the reaction to find the enthalpy change.

3. Enthalpy of Solution

a. Observe the energy absorbed or released when dissolving substances like NH₄Cl and CaCl₂in water.

b. Procedure:

i. Dissolve a measured amount of solute in water.

ii. Record the temperature change to determine the enthalpy change.

4. Calculations Use the formula ΔH = mcΔT:

m = mass of water/solution (in grams) c = specific heat capacity (for water, usually 4.18 J/g°C), for more accurate results remember to research the specific heat capacity of the solution you have created.

ΔT = change in temperature Calculate the energy change (ΔH) for each experiment using the recorded temperature changes.

5. Analysis and Interpretation

a. Interpret Results: Discuss if each reaction absorbed or released heat.

b. Presenting Results: Draw an energy profile diagram for each experiment.

c. Evaluate Experiments: What challenges did you encounter? How could be improved? How reliable are you results?

Activity 1.6 Digital learning

1. Go online and find a short video explaining how scientists the energy in food (like calories) and fuels (like gasoline or wood).

2. The video should cover

a. how burning food or fuel releases energy.

b. why measuring energy helps us understand which foods give us more energy activities and which fuels are more powerful.

Some compounds can’t be easily made by mixing their basic elements because the reaction is either too slow or creates the wrong products through unwanted side reactions. In these cases, scientists use an indirect method to figure out the energy change, called ∆H_(f)°. This method is based on Hess’s law.

Hess’s law states that if a reaction happens in several steps, the total energy change is the sum of the energy changes of each step.

Hess’s law helps to calculate the energy even if the reaction doesn’t happen in one step.

Calorimetry can be used to measure enthalpy changes (ΔH) for many chemical processes, but not all. Consider the hypothetical reaction below:

A + B → D Where A and B are the reactants, D is the product The overall energy change (ΔH) for a chemical reaction remains constant, regardless of the reaction pathway.

A + B → C ΔH₁ C + B → D ΔH₂ Then A + 2B → D ΔH₃ ΔH₃ = ΔH₁+ ΔH₂ Consider the example below, the oxidation of carbon to form carbon monoxide.

C(s) + 1__ 2O₂(g) → CO(g) When carbon reacts with oxygen, the primary product is CO₂even with insufficient oxygen is used. As soon as CO is formed, it reacts with O₂to form CO₂. Because the reaction cannot be carried out in a way that allows CO to be the sole product, it is not possible to measure the change in enthalpy for this reaction by calorimetry.

The enthalpy changes for the reaction between C(s) and O₂to form CO(g) can be determined indirectly, but, the enthalpy changes for other reactions for which values of ∆Hᵣₓₙcan be determined.

The enthalpy change of the reaction can be calculated using Hess’s law, which states that the total enthalpy change of chemical reaction is equal to the sum of all the enthalpy changes.

For example, the oxidation of carbon to carbon (IV) oxide.

This reaction occurs in two steps:

Step 1: the oxidation of C(s) to CO(g) (Equation 1) and

Step 2: the oxidation of CO(g) to CO₂(g) (Equation 2).

Adding the two equations to gives the overall equation for the oxidation of C(s) to CO₂(g) (Equation 3).

Equation 1: C(s) + 1 ⁄2 O₂(g) → CO(g) ° ∆Hᵣₓₙ₁ = ?

Equation 2: CO(g) + 1 ⁄2 O₂(g) → CO₂(g) ∆Hᵣₓₙ₂ = −283.0 kJ/mol Equation 3: C(s) + O₂(g) → CO₂(g) ∆Hᵣₓₙ₃ = −393.5 kJ/mol Hess’s law says that the enthalpy change for the overall reaction that is equation 3 is the sum of the enthalpy changes for reactions that equations 1 and 2. these values are then used to calculate the enthalpy change for reaction 1.

∆Hᵣₓn ° 3 = ∆Hᵣₓₙ ₁+ ∆Hᵣₓₙ ₂ −393.5 kJ/mol = ∆Hᵣₓₙ ₁+ (−283.0 kJ/mol) ∆_(Hrx)ₙ ₁ = −110.5 kJ/mol Remember that for these calculations it is imperative that the sign of the enthalpy change is preserved correctly during substitution and calculation of values.

Rules for Manipulating Thermochemical Equations

1. The opposite of an exothermic reaction is an endothermic reaction. As a result, the sign of an equation’s enthalpy must likewise be reversed or negated when it is stated in the other direction. To illustrate, the reverse of the equation: A + B → C ΔH = +x J/mol is the following equation: C → A + B ΔH = -x J/mol

2. Only when a substance is in the same physical state can it be eliminated from both sides of an equation.

3. If all the coefficients of an equation are multiplied by a factor, the value of its enthalpy must be multiplied by the same factor.

i. To illustrate, the factors of an equation: A + B → C ΔH = +x J/mol When an equation is multiplied by 2: 2A + 2B → 2C ΔH = 2 (+x J/mol)

4. When equations are added, the enthalpy changes are also added accordingly with proper signs A + B → C ΔH₁ C + B → D ΔH₂ Then A + 2B → D ΔH3 = ΔH₁+ ΔH₂

Example

Using Hess’s law to calculate the enthalpy change (∆Hᵣₓₙ _(°)) of the following reaction C (graphite) + 2 H₂(g) → CH₄(g) Data given from the table below:

Reactions ∆H

(i) C (graphite) + O₂(g) → CO₂(g) ∆H_(f) _(°)(i) = – 393.5 kJ mol⁻¹

(ii) H₂(g) + 1/2 O₂(g) → H₂O(l) ∆H_(f) _(°)(ii) = – 285.8 kJ mol⁻¹

(iii) CH₄(g) + 2 O₂(g) → CO₂(g) +2 H₂O(l) ∆Hᵣₓₙ _(°)(iii) = – 890.3 kJ mol⁻¹Solution C (graphite) + 2 H₂(g) → CH₄(g) ∆Hᵣₓₙ _(°)(A) = ?

∆Hᵣₓₙ _(°) = ∆H_(f) _(°)(products) – ∆H_(f) _(°)(reactants) (∆Hᵣₓₙ _(°)) = -890.3 kJ/mol ∆H_(f) _(°)product = ∆H_(f) _(°)(CO₂) + 2 x ∆H_(f) _(°)(H₂0) = -393.5 + 2(-285.8) = - 965.1 ∆H_(f)° reactants = x + 0

-890.3 = -965.1 – (x) X = -965.1 + 890.3 = - 74.8 kJ/mol Alternatively, using the rearrangement method, Maintain equation (i) Multiply equation (ii) by 2 Reverse equation (iii)

(iv) 2H₂(g) + O₂(g) → 2H₂O(l) ∆Hᵣₓₙ _(°)(iv) = – 2×285.8 kJ mol⁻¹= -571.6 kJ mol⁻¹(v) CO₂(g) + H₂O(l) → CH₄(g) + 2 O₂(g) ∆Hᵣₓₙ _(°)(v) = + 890.3 kJ mol⁻¹ equations (i) + (iv) + (v) with corresponding ∆Hᵣₓₙ _(°)values, to get equation (vi)

(vi) [C (graphite) + O₂(g)] + [2H₂(g) + O₂(g)] + [CO₂(g) +2 H₂O(l)] → CO₂(g) + 2 H₂O(l) + [CH₄(g) + 2O₂(g)] ∆Hᵣₓₙ _(°)(vi) = ∆Hᵣₓₙ _(°)(iv) + ∆Hᵣₓₙ _(°)(v) + ∆Hᵣₓₙ _(°)(i) 21 = (– 393.5 + (– 2×285.8 + 890.3) kJ mol–1 = –74.8 kJ mol⁻¹ cancel the common terms on the two sides of equation (vi), we get equation (vii)

(vii) C (graphite) + 2H₂(g) → CH₄(g) ∆Hᵣₓₙ _(°)(vii) = –74.8 kJ mol⁻¹BORN – HABER CYCLES Hess’ law is used to draw Born-Haber cycles, which are energy cycles for ionic molecules. Ionic compounds are created when constituent elements undergo a sequence of events for which, with the exception of the lattice energy, the majority of energy changes are directly measurable. A Born-Haber cycle is used to calculate lattice energy from the enthalpy data. Ionization energies, electron affinities, sublimation energies, bond dissociation energies, and their enthalpy of formation are all related to the lattice energies of ionic compounds.

Consider the reaction between lithium and fluorine:

Li(s) + 1/2 F₂(g) → LiF(s) For this reaction, the standard enthalpy change is -594.1 kJ/mol. Five distinct steps can be used to track the production of LiF from its constituent elements, and the sum of the enthalpy changes for the steps equals the enthalpy change for the entire reaction (-594.1 kJ/mol). Using Hess’s law, this pathway aids in the analysis of the energy shifts involved in the synthesis of ionic compounds.

1. Convert solid lithium to lithium vapor (the direct conversion of a solid to a gas is called sublimation):

Li(s) → Li(g) ΔH° = +155.2 kJ/mol The energy of sublimation for lithium is 155.2 kJ/mol.

2. Dissociate 1__ 2mole of F₂gas into separate gaseous F atoms:

1__ 2F₂(g) → F(g) ΔH° = + 75.3 kJ/mol The energy needed to break the bonds in 1 mole of F₂molecules is +150.6 kJ.

Here we are breaking the bonds in half a mole of F₂, so the enthalpy change is 150.6/2 , or 75.3 kJ.

3. Ionize 1 mole of gaseous Li atoms:

Li(g) → Li+(g) + e − ΔH° = + 520 kJ/mol This process corresponds to the first ionization of lithium.

4. Add 1 mole of electrons to 1 mole of gaseous F atoms.

F(g) + e − → F(g)− ΔH° = - 328 kJ/mol

5. Combine 1 mole of gaseous Li+ and 1 mole of F− to form 1 mole of solid LiF:

Li+(g) + F−(g) → LiF(s) ΔH° = ?

The reverse of step 5, energy + LiF(s) → Li+(g) + F −(g) According to Hess’ law ΔH °(ove rall)= ΔH₁ °+ ΔH₂ °+ ΔH₃ °+ ΔH₄ °+ ΔH₅ °−594.1 = 155.2 + 75.3 + 520 − 328 + x x = −594.70 − 422.5 x = −1016.6 kJ/mol

One or more pure compounds can be transformed into one or more other chemical substances through chemical reactions. In most cases, chemical bonds are produced in the products and broken in the reactants. These bond-breaking and bond-forming mechanisms are primarily responsible for the enthalpy change for any chemical reaction. A chemical bond must be broken with energy, and when a chemical bond is formed energy is liberated. Bond formation is an exothermic

activity that releases energy, whereas bond breaking is an endothermic process that absorbs energy. Consequently, estimating the total enthalpy change in a chemical reaction can be aided by knowledge of the energies of the individual bonds.

The average energy needed to break one mole of a specific type of covalent bond in a molecule is known as the bond enthalpy. The energy absorbed to break one mole of a specific kind of covalent bond in a gaseous molecule into free radicals under standard circumstances is known as the standard bond dissociation energy.

The enthalpy shift that occurs when a compound’s bonds are broken and its constituent atoms split apart into single atoms is known as the enthalpy of atomization. It can alternatively be described as the energy shift that takes place when a compound’s chemical bonds are broken, producing one mole of gaseous atoms as by-products.

Simple diatomic molecules like H₂, O₂, and Cl₂often have their bond enthalpies determined spectroscopically; however, complex compounds use Hess’s law to calculate bond enthalpies based on thermochemical data.

The experimentally determined bond enthalpy of the diatomic hydrogen molecule, for example, is H₂(g) → H(g) + H(g) ΔH° = + 436.4 kJ/mol According to this equation, breaking the covalent bonds in 1 mole of gaseous H₂ molecules requires 436.4 kJ of energy.

For the less stable chlorine molecule, Cl₂(g) → Cl(g) + Cl(g) ΔH° = + 242.7 kJ/ mol.

Bond enthalpies can also be directly measured for heteronuclear diatomic molecules, such as HCl, HCl(g) → H(g) + Cl(g) ΔH° = + 431.9 kJ/mol as well as for molecules containing multiple bonds:

O₂(g) → O(g) + O(g) ΔH° = + 498.7 kJ/mol The table below shows the bond enthalpies of some diatomic molecules Bond Bond Enthalpy (kJ/mol) H – H +436.4 O = O +498.7 Cl – Cl +242.7 H – O +460 ∆Hᵣₓₙ _(°) = ∑ΔH °bonds broke n − ∑ΔH °bonds forme d

Example

Calculate the enthalpy of reaction when one mole of water is formed from hydrogen gas and oxygen gas.

[Bond enthalpies in kJ/mol are H – H = +436, O = O = +498, H – O = +464]

Solution

H₂+ 1__ 2O₂→ H₂O Bonds broken: one mole of (H – H) and half of (O = O) = 436 + 1/2 (498) = +685 Bonds formed: H – O – H which is twice (H – O) = 2(464) = +928 ∆Hᵣₓₙ _(°) = ∑ΔH °bonds broke n − ∑ΔH °bonds forme d ∆Hᵣₓₙ _(°) = 685 − 928 ∆Hᵣₓₙ _(°) = − 243 kJ/mol

Activity 1.7 Hess’s Law and its applications Combustion reaction (combustion of methane)

1. In your own understanding, explain Hess’s law of constant heat summation and underlying principles.

2. In groups of five:

a. Write a balance equation for the reaction between carbon and hydrogen to form methane. ΔH for reaction is -74.8 kJ/mol

b. Break the methane into its constituent atoms and balanced the equation.

c. Write a balanced equation for the reaction between the carbon atom and oxygen to form carbon (iv) oxide. ΔH for the reaction is -393.5 kJ/mol

d. Write a balanced equation for the reaction between the hydrogen atom and oxygen to form water. ΔH for the reaction is -571.6 kJ/ mol

e. Combine equations ii, iii and iv and obtain the enthalpy of reaction for the reaction.

Activity 1.8 Calculating enthalpy change of reaction using Hess law

1. Use the information provided below to draw energy diagram for each reaction:

a. Reaction between carbon and hydrogen to form methane ΔH =

-74.8 kJ/mol

b. Reaction between carbon and oxygen to form carbon (iv) oxide ΔH = -393.5 kJ/mol

c. Reaction between hydrogen and oxygen to form water ΔH = -571.6 kJ/mol

2. a. Write a balanced reaction equation for the atomisation of nitrogen gas, the enthalpy change for this reaction is -45.9 kJmol⁻¹b. write a balanced equation for the atomisation of hydrogen gas, the enthalpy change this reaction is -436.4 kJmol⁻¹c. write a balanced reaction equation for the synthesis of ammonia from nitrogen gas and hydrogen gas, the enthalpy change for the reaction is -389.5 kJmol⁻¹

d. calculate the enthalpy change of formation of ammonia from the equations you have formed in iii above using Hess’s law.

Activity 1.9 A diagram of Born Haber cycle 1a.

i. Write the atomisation energy of sodium and fluorine

ii. Write the ionisation energy of sodium and electron affinity of fluorine

iii. Find the enthalpy of formation of sodium fluoride

iv. Determine the lattice energy of the sodium fluoride 1b.

i. Write the symbols of sodium and fluorine in their standard states

ii. Draw vertical lines on each of the sodium and fluorine.

iii. At the end of the lines, write the products of the sublimation of sodium and atomisation of fluorine.

iv. Extend the lines and indicate the products of the ionisation of sodium and electron affinity of fluorine

v. Draw lines to show the enthalpy of formation and the lattice energy of sodium fluoride.

2. a. In groups discuss bond energy and its importance in chemical reaction

b. Present your findings to the whole class.

Activity 1.10 Calculations of bond enthalpy of some molecules Study the table below Bond Bond enthalpy (kJ/mol) H-H +436.4 N-H +389 Cl-Cl +242.7 C-H +413 C= C +611 N≡N +945.4 C-C +345 C≡C +837

a. Write a balance equation for the formation of ammonia.

b. Identify the bonds formed and the bonds brokens in the process.

c. From the enthalpy values provided in the table above, calculate the enthalpy of reaction when one mole of ammonia is formed from hydrogen gas and nitrogen gas.

d. Write the equations for the formation of methane, ethene and ethyne.

e. Calculate the bond enthalpy of each of methane, ethene and ethyne.

f. Compare the bond enthalpy of methane, ethene, ethyne and relate it to their bond strength.

Review Questions 1.1

1. Identify the basic difference between exothermic and endothermic reactions in terms of energy transfer.

2. a. What is the standard enthalpy change of a reaction?

b. Name two types of standard enthalpy changes.

3. Describe a system in thermochemistry and differentiate between an open, closed, and isolated system in terms of energy exchange.

4. If burning 1 mole of methane (CH₄) produces -890 kJ/mol of energy, is this reaction exothermic or endothermic?

5. Draw an energy profile diagram for an exothermic reaction and label reactants, products, activation energy (Eₐ), and enthalpy change (ΔH).

6. The combustion of propane (C₃H₈) releases − 2,220 kJ / mol . How much energy is released when 2 moles of propane are burned?

7. Use the table to answer this question Reaction Starting temperature/ °C Final temperature/°C X 22 37 Y 25 18 Z 21 25

a. Decide whether each reaction is endothermic or exothermic, explain how you could tell.

b. Which reaction has the largest energy change?

8. Why do exothermic reactions feel warm while endothermic reactions feel cold?

9. Calculate the standard enthalpy change of a reaction if the enthalpy of the products is − 500 kJ and the enthalpy of the reactants is -200 kJ.

10. a. Design an experiment to determine if a given reaction is exothermic or endothermic and interpret the observed energy changes.

b. When hydrochloric acid reacts with ammonium hydroxide in a beaker, the temperature goes up.

HCl(aq) + N H₄ OH (aq) → N H₄ Cl (s) + H₂ O (l)ΔH = –53.4kJ / mol Draw the energy profile diagram and state whether the reaction is endothermic or exothermic, explain your answer.

11. Explain why fuels like methane (CH₄) are preferred for energy production over fuels with lower enthalpy of combustion.

12. You are given the following standard enthalpy changes:

a. Δ H_(f) °(CH₄) = − 74.8 kJ / mol

b. Δ H_(f) °(CO₂) = − 393.5 kJ / mol

c. Δ H_(f) °(H₂O) = − 241.8 kJ / mol The balanced combustion reaction for methane (CH₄) is:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

a. Using the given enthalpy of formation values, calculate the standard enthalpy change of combustion (Δ H_(c) °) for methane.

b. Explain how the result shows whether this reaction is exothermic or endothermic.

c. Discuss why knowing the enthalpy of combustion is important in real-life applications such as energy production.

Review Questions 1.2

1. What is the enthalpy change of combustion?

2. What is the enthalpy change of neutralisation?

3. What is the enthalpy change of solution?

4. Why is the enthalpy change of neutralisation usually the same for strong acids and bases?

5. Why is it important to know the enthalpy change of combustion for fuels?

6. Why does dissolving ammonium chloride in water cause the solution to cool?

7. How can you experimentally determine the enthalpy change of combustion of ethanol?

8. Describe how you would determine the enthalpy change of neutralisation between HCl and NaOH in a lab setting.

9. Describe how you would determine the enthalpy change of solution for a salt like KCl in the lab.

10. Compare your experimental enthalpy change of combustion for ethanol to the theoretical value. Discuss any differences and possible sources of error.

11. If the experimental enthalpy change of neutralisation differs from the theoretical value, what factors might explain this?

12. Compare your experimental enthalpy change of solution for KCl with the literature value and discuss any discrepancies. How could the experiment be improved?

13. In an experiment to determine the enthalpy change of combustion of ethanol, the initial mass of the ethanol burner (with ethanol in it) was measured as 150.0 g. Next, 100.0 g of water (equivalent to 100 ml) was added to a calorimeter, and the initial temperature of the water was recorded as 25.0 °C. The ethanol burner was placed under the calorimeter and lit.

As the ethanol burned, the heat produced caused the temperature of the water to increase. After the flame was extinguished, the final temperature of the water was recorded as 45.0 °C. The final mass of the ethanol burner, after the combustion process, was measured to be 148.50 g. The specific heat capacity of water used in the calculations is 4.18 J/g°C. Determine;

a. the amount of heat absorbed by the water.

b. the mass of ethanol burned.

c. the enthalpy changes of combustion per gram of ethanol.

d. the molar enthalpy changes of combustion of ethanol (molar mass of ethanol is 46.0 g/mol).

e. any two precautions you have to take in order to obtain correct result.

Review Questions 1.3

1. State Hess’s law of constant heat summation.

2. Given the following information:

P₄(s) + 3O₂(g) → P₄O₆(s) ΔH = −1640.1 kJ/mol P₄(s) + 5O₂(g) → P₄O₁₀(s) ΔH = −2940.1 kJ/mol what is the value of ΔH₍ᵣₓₙ₎for P₄O₆(s) + 2O₂(g) → P₄O₁₀(s)?

3. Consider the following data:

C(s) + O₂(g) → CO₂(g) ∆H = -394 kJmol⁻¹H2(g) + 1/2O₂(g) → H₂O(l) ∆H = -286 kJmol⁻¹C₄H₆(g) + 5/2O₂(g) → 4CO₂(g) + 3H₂O(l) ∆H = -254 kJmol⁻¹Determine the standard enthalpy of formation of C₄H₆.

4. a. Explain the term lattice energy

b. Use the following data to calculate for the lattice energy of sodium chloride:

∆H_(f)(NaCl) = - 411 kJmol⁻¹∆Hₐₜₒₘ(Na) = +107 kJmol⁻¹∆Hₐₜₒₘ(Cl) = +122 kJmol⁻¹IE(Na) = + 494 kJmol⁻¹E.A. (Cl) = - 349 kJmol⁻¹Calculate lattice energy of NaCl(s)

5. Define bond enthalpy

6. Distinguish between bond dissociation energy and bond energy.

7. Consider the following bond enthalpies Bond BDE (KJ/mol) H-C 436 C-C 348 C= C 612 N≡N 944 N-H 388

a. Write a balanced chemical equation for the formation one mole of NH₃.

b. Use the bond enthalpy data to calculate for the enthalpy of formation of NH₃.

In this section, you will explore Chemical Kinetics. You will learn how to measure reaction rates using experiments and understand factors that affect rates.

You will also study collision theory and how to use the concept of collisions to explain reaction rates and create rate equations from experiment results. Through problem-solving and analysing graphs, you will learn how to figure out reaction orders and solve rate problems.

KEY IDEAS

• Average rate of reaction: this is the average speed of a reaction over a specific time interval. Is calculated by dividing the change in concentration of a reactant or product by the time elapse

• Catalyst: a substance that increase the rate of reaction without being consume in the process.

• Initial rate of a reaction: the rate of reaction at a very beginning, at time t = 0

• Instantaneous rate of a reaction: this is the rate of reaction at a specific instant in time.

• Rate of reaction is a measure of how quickly a chemical reaction proceeds. This is normally measured by either tracking the reactant being consumed or product being formed

Chemical kinetics is the area of chemistry that studies how fast chemical reactions take place. This is called the reaction rate, and is important for things like creating medicines, reducing pollution, and processing food. Knowing reaction rates helps make these processes more efficient.

Rate of a chemical reaction is defined as the change in concentration (or moles/ mass) of a reactant or product per unit time.

Mathematically expressed as, Rate = ∆[A]_____ ∆ t , where A is the reactant or product being considered, the square brackets indicate concentration in mol/dm³.

the symbol ∆ indicates a change in a given quantity.

Consider the hypothetical reaction: aA + bB → cC + dD The rate can be expressed by the following equation rate = −1/a × ∆[A]_____ ∆ t = −1/b ∆[B]____ ∆ t = 1/c ∆[C]____ ∆ t = 1/d ∆[D]____ ∆ t Notice the negative symbols in front of the reactant stoichiometric coefficients too denote a consumption.

Example 1

Consider the thermal decomposition of calcium carbonate (CaCO₃) into calcium oxide (CaO) and carbon dioxide (CO₂) gas through the following equation:

CaCO₃(s) → CaO(s) + CO₂(g) Write expressions for the reaction rate in terms of the rate of change in concentration of the reactant (CaCO₃) and the product (CO₂)

Solution

From the balanced chemical equation above shows two moles of CaCO₃decompose to produce 1 mol of CO₂and 1 mole of CaO. From the equation the mole ratios of CO₂to CaCO₃and to CaO are 1:1 and 1:1, respectively. This means that the rate of change of [CaCO₃] and [CO₂] must be divided by its stoichiometric coefficient to obtain equivalent expressions for the reaction rate.

rate = ∆[C O₂]_______ ∆ t = ∆[CaC O₃]_________ ∆ t = ∆[CaO]_______ ∆ t

Example 2

Consider the thermochemical equation: 2N₂O₅(g) → 4NO₂(g) + O₂(g)

a. How is the rate at which N₂O₅disappears related to the rate at which N O₂ appears in the reaction?

b. If the rate at which N O₂ appears is 6.0 × 10⁻⁵moldm⁻³at a particular instant, at what rate is N₂ O₅ disappearing at this same time the rate at which N O₂ appears in the reaction.

Solution

2 N₂ O₅ (g) → 4N O₂(g) + O₂(g)

a. rate = −∆[N₂ O₅]_______ 2 × ∆ t = ∆[N O₂]_______ 4 × ∆ t ∆[ N₂ O₅]________ ∆ t = 2/4 × ∆[N O₂]_______ ∆ t ∆[ N₂ O₅]________ ∆ t = 1/2 × ∆[N O₂]_______ ∆ t So, the rate at which N₂O₅is consumed is equal to half the rate at which NO₂ is created.

b. But the rate of appearance of N O₂ ( ∆[N O₂]_______ ∆ t ) is 6.0 × 10⁻⁵moldm⁻³= the rate which N₂ O₅ disappears (∆[N₂ O₅]_ ∆ t ) = 2_ 4 × 6.0 × 10⁻⁵= 3.0 × 10⁻⁵mol dm⁻³Exercise Consider the reaction: 2H₂(g) + O₂(g) → 2H₂O(g) Write expressions for the reaction rate in terms of the rate of change of the concentration of each species. If the rate of consumption of H₂in this process is 3 × 10⁻⁴mol dm⁻³calculate the rate of consumption of O₂and the rate of appearance of H₂O.

Initial rate of a reaction When the reagents are first combined, that is the starting rate of reaction. The initial rate can be determined graphically or empirically. It is necessary to combine the chemicals and evaluate the rate of reaction as soon as possible in order to experimentally ascertain the beginning rate.

Measuring the time, it takes for a specific event to occur early in a reaction is a common practice in initial rate experiments. This could be the amount of time needed for a colour shift, the formation of a tiny, detectable amount of precipitate, or the production of a measurable volume of gas.

Average rate of reaction The change in reactant or product concentration over time period is known as the average rate of reaction for that time period. The average rate of reaction can be determined choosing two points in time say t₁ and t₂and measuring the concentration of the reactants (C₁ and C₂) at those points. The average rate of reaction for the reaction (a.v.g rate) = ∆[reactants]___________ ∆ t

a. v. g rate = C₂ − C₁_______ t₂ − t₁ The average rate of reaction can also be expressed from a balanced chemical equation taking into consideration the stoichiometric coefficients. Consider chemical reaction below:

aA+ bB → cC+ dD a.v.g rate = −1_ a ∆[A]_____ ∆ t = −1_ b ∆[B]____ ∆ t = 1_ c ∆[C]____ ∆ t = 1_ d ∆[D]_____ ∆ t Instantaneous rate of a reaction The instantaneous rate of a reaction is the reaction rate at any given point in time.

As the period of time used to calculate an average rate of a reaction becomes shorter and shorter, the average rate approaches the instantaneous rate.

Suppose we have a reaction that forms a product A. A graph of the concentration of A [A] against time t is usually not a straight line but a curve. The instantaneous rate of reaction is the slope of the line (the tangent to the curve) at any time (t).

Instantaneous rate = ∆[P]____ ∆ t (where ∆t is small) Measurement of rate of reaction by colour change Sodium thiosulfate solution reacts with dilute hydrochloric acid according to the reaction below:

Na₂S₂O₃(s) + 2HCl(aq) → 2NaCl(aq) + H₂O(l) + SO₂(g) + S(s) During the reaction, sulphur produces a cloudy yellow-white precipitate. The time it takes for this precipitate to reach a specific level of cloudiness can be used to measure the reaction time. From this time the rate can be determined.

Procedure SAFETY NOTE: Take care not to inhale any fumes given off in this reaction as SO₂(g) can cause respiratory irritation.

1. Using a measuring cylinder, add 50 cm³of dilute sodium thiosulfate solution to a conical flask.

2. Place the conical flask on a piece of white paper with a black cross drawn on it.

3. Using a different measuring cylinder, add 10 cm3 of bench dilute hydrochloric acid to the conical flask. Immediately swirl the flask to mix its contents, and start a stop clock.

4. Look down through the reaction mixture, ensure you observe from a height of more than 20cm to reduce exposure to fumes. When the cross can no longer be seen, record the time on the stop clock.

5. Measure and record the temperature of the reaction mixture, and clean the apparatus.

6. Ensure the still reacting mixture is poured into the alkaline bath to stop the reaction continuing.

7. Repeat steps 1 to 6 with different starting temperatures of sodium thiosulfate

solution.

8. Plot a graph of reaction rate on the vertical axis and temperature (°C) on the horizontal axis and draw a curve of best fit.

Measurement of rate of reaction by changes in volume of gas evolved Calcium carbonate reacts with dilute hydrochloric acid:

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

The volume of carbon dioxide gas produced can be measured using a gas syringe.

Procedure

1. Support a gas syringe with a stand, boss and clamp.

2. Using a measuring cylinder, add 50 cm3 of dilute hydrochloric acid to a conical flask.

3. Add 0.4 g of calcium carbonate to the flask. Immediately connect the gas syringe and start a stop clock.

4. Every 10 seconds, record the volume of gas produced.

5. When the reaction is complete, clean the apparatus.

6. Repeat steps 1 to 5 with different concentrations of hydrochloric acid.

7. Plot a graph of volume of gas (cm3) on the vertical axis and time (s) on the horizontal axis for each concentration of acid. You may wish to plot this data simultaneously on the same axes.

8. Draw a curve of best fit for each concentration.

Activity 2.1 Defining reaction rate and explaining its units

1. Define the term rate of reactions in your own words and explain its units.

2. Share with a friend your definition and idea the rate of reaction.

3. Which definition should be accepted and why?

4. Explain with examples the various ways initial rate, average rate and instantaneous rate can be written.

5. Tell your group the difference between instantaneous and average rates of reactions.

6. What can slow down or hasten the rate of reaction.

7. Share your definitions and ideas with the class and agree on definitions and to express the types of rates of reaction.

8. The examples will guide you on how to express both types of rates.

E → D. the concentration and time for the completion of the reaction are shown the table.

Time (s) [E]/moldm⁻³0 1.00 10 0.95 20 0.90 30 0.85 The instantaneous rate at t= 20s is calculated as:

∆[A] = ∆[A]₁ − ∆[A]₂ = 0.90 − 0.95 = − 0.05M ∆ t = t₂ − t₁ = 20 − 10 = 10s Instantaneous rate = (-0.05)/10s = 0.005 M/s Average rate from concentration-time graph Data Time/min [A]/moldm⁻³0 2.0 5 1.8 10 1.6 15 1.4 Average of reaction between 5-10 min.

From graph, ∆[A] = ∆[A]₂-∆[A]₁= 1.6-1.8 = - 0.2 M ∆t = t₂-t₁ = 10-5 = 5 min Average rate = ∆[A]____ ∆ t = − 0.2 M/5 min = 0.04 M / min

Activity 2.2 Measuring reactions rates using specific methods

1. a. With the method assigned to your group, plan and perform experiments to calculate the rate of reaction.

i. The procedure for measuring rate of reaction.

ii. The equipment and materials needed

• Potential sources of error associated with your method and how to them.

• Discuss the advantages and limitations of your assigned method

• How do your results compare with results in the literature?

b. Watch the video on CD-ROM encyclopaedia of science on rate of reaction and discuss,

i. the procedures

ii. limitations

iii. Potential sources of error.

iv. Each group present your procedure and finding to the class.

v. Compare and discuss your method and the methods that your friends used.

Reaction rates can vary greatly over a large range of time. While certain reactions can happen slowly over many years, others can happen at incredibly quick rates.

1. Temperature All chemical reactions proceed more quickly as the temperature increases.

This is because at higher temperature, the average kinetic energy of the particles increases, so more reacting particles have energy equal to or greater than the activation energy of the reaction. Additionally, because the reacting particles move faster at higher temperatures, there are more collisions between them, which raises the frequency of effective collisions.

As a result, more products are created per unit of time, increasing the pace of reaction.

Number of

molecules with a given energy At a higher temperature, the peak moves to a higher energy with a lower height.

At a higher temperature, a greater propostion of molecules exceeds E_(A).

Rate of reaction increases.

EA T1 T2

Figure 2.1: Effect of temperature on the kinetic energy distribution of molecules in a sample This explains why food can spoil quickly when left on the kitchen counter.

But that process is slowed down by the refrigerator’s lower internal temperature, keeping the same food fresh for days. In the chemistry laboratory, gas burners, hot plates, and ovens are frequently employed to speed up processes that move slowly at room temperature.

2. Concentration The rate of reaction increases when the concentration of one or more reactants rises. There is a greater likelihood of particle collisions when the reactant concentration rises; more collisions translate into a higher reaction rate.

3. Available surface area and nature of the reactants The rate of the reaction increases with the amount of surface contact between reactants. The reaction rate decreases with decreasing surface contact.

Note

It is also important to note that not all reactions depend on surface area. If both reactants are gases or liquids that mix together, then surface area is not a factor. Surface area is most prevalent when one of the reactants is in the solid phase.

4. Catalyst A catalyst is a material that increases a chemical reaction’s rate without being consumed by the reaction itself. A catalyst functions by offering a different path for the reaction to proceed. The activation energy of this other path is lower. This increases the proportion of reacting particles with energy greater than the activation energy.

As a result, the frequency of effective collision increases. More products are formed per unit time and the rate of reaction is higher.

Figure 2.2: Effect of catalyst on the rate of a reaction

5. Pressure The rate of chemical reactions with reactants and products in their gaseous form increases with increasing pressure. Increasing pressure results in either more particles in a given volume or the same number of gaseous particles being forced into a smaller volume. The particles are now closer together in both situations, which increases the number of collisions per unit time and the reaction rate.

Figure 2.3: Effect of pressure on the rate of a reaction Experiments to investigate the factors that affect rate of reaction

1. Experiment to investigate the effect of temperature on the rate of reaction Procedure

a. A water bath is used to heat diluted hydrochloric acid to a set temperature.

b. In a conical flask, add the warmed dilute hydrochloric acid.

c. Put a magnesium strip of known mass and surface area in the conical flask and start the stopwatch

d. When the magnesium has completely reacted and disappears, stop the time.

e. Compare the results after repeating at various temperatures, keep the mass of magnesium and available surface area of magnesium constant.

Result The rate of reaction will increase when the temperature increases. This is because the particles will collide more frequently and effectively due to their greater kinetic energy than the required activation energy, which will increase the rate of reaction.

Why was it important to keep the mass of surface area of the magnesium the same across the different reactions?

2. Experiment to investigate the effect of concentration on rate of reaction SAFETY NOTE: Take care not to inhale any fumes given off in this reaction as SO₂(g) can cause respiratory irritation.

Procedure

a. Fill a conical flask with 50 cm³of sodium thiosulfate solution.

a. Fill a measuring cylinder with 5 cm³of diluted hydrochloric acid.

b. Place a piece of paper with a cross drawn on it underneath the flask.

c. Start the stopwatch as soon as you add the acid to the flask.

d. Examine the cross from above (ensuring a minimum distance of 20 cm from the top of the flask), and when it is no longer visible, stop the stopwatch.

e. Repeat with varying sodium thiosulfate solution concentrations (dilute with water and mix varying volumes of sodium thiosulfate solution).

Result When the concentration of a solution increases, the rate of reaction also increases.

This is because there will be more reactants in a given volume, allowing more frequent and efficient collisions that speed up the reaction.

3. Experiment to investigate the effect of Surface Area on the rate of reaction Procedure

a. A conical flask should be filled with diluted hydrochloric acid.

b. This flask should be connected to a measuring cylinder upside down in a bucket of water via a delivery line (downward displacement).

c. Quickly replace the bung into the conical flask after adding magnesium ribbon.

d. Using the measuring cylinder, determine the volume of gas generated i predetermined amount of time.

e. Repeat using pieces of magnesium ribbon of varying sizes (the same mass of magnesium must be used).

Result Because smaller sections of magnesium ribbon enhance the solid’s surface area, the rate of reaction will also rise. This is because a larger particle surface area will be in contact with the other reactant, resulting in more frequent and effective collisions that speed up the reaction.

4. Experiment to investigate the effect of catalyst on the rate of reaction Procedure

a. Put hydrogen peroxide in a conical flask.

a. This flask should be connected to a measuring cylinder upside down in a tub of water via a delivery line (downward displacement).

b. Quickly insert the bung into the conical flask after adding the catalyst manganese (IV) oxide.

c. Using the measuring cylinder, determine the amount of gas generated i predetermined amount of time.

d. Comparing the findings, repeat the experiment without the manganese (IV) catalyst.

Result The rate of reaction will be increased by using a catalyst. In order to give more colliding particles, the necessary activation energy to react, the catalyst will offer a different pathway with a lower activation energy. The rate of reaction will increase as a result of more frequent and efficient contacts.

Application of the factors that affect rate of reaction in everyday life

1. Temperature Food Preservation: The rate of spoiling reactions is slowed down by cold temperatures. Perishables are therefore kept in a refrigerator to prevent bacteria and their enzymes from functioning as well as they could. Fruits and vegetables should be kept in the refrigerator to prolong their freshness.

2. Concentration Cleaning: Concentrated solutions are frequently found in household cleaning goods. Their efficacy and reaction rates are altered when diluted with water.

3. Catalysts Chemical Industry

Catalysts are necessary for effective manufacturing. They expedite responses, enabling producers to satisfy demand while upholding safety regulations.

4. Surface area Medicine and Pills Medication that has been crushed or powdered dissolves faster in the stomach. They are absorbed into the bloodstream more quickly due to their increased surface area.

5. Pressure Fertiliser production The manufacture of ammonia for fertilisers is done in gaseous form so using a high pressure will increase the rate of reaction.

Activity 2.3 Explaining factors that affect reaction rate

1. a. In groups, state the factors that affect rate of chemical reaction.

b. Explain why powdered sugar dissolves faster than sugar cubes in the volume of water?

c. In the production of nitric acid, ammonia is combusted according to the reaction equation:

4NH₃ (g) + 5O₂ (g) → 4NO(g) + 6H₂O(g).

i. Write the equations that relate:

The rate of consumption of the reactants and the rate of formation of the products.

Activity 2.4 Investigating factors that affect rate of reaction Spend fifteen (15 minutes) on each activity and move to the next. All the experiments must be performed before leaving the laboratory.

1. In groups, discuss and write down the procedure to investigate the effects of each of the following on reaction rate:

a. Changing concentrations

b. Changing surface area

c. Changing temperature

d. Introduction of catalyst

2. Perform the experiment following the procedure you outlined.

Activity 2.5 Analysing experimental data and graphs of reactions Materials needed: Handouts of experimental data and graphs of reaction rates.

Procedure

1. In groups, study and discuss the data and graphs in the handouts and answer the questions below:

a. What are the variables (dependent and independent) and their units in the experiment?

b. How do the variables relate to each other?

c. What do you observe about the steepness of the graph?

d. Which reaction occurred fastest and why?

e. How could you extend this into an investigation?

Activity 2.6 Discussing practical applications of rate of reaction.

a. Discuss the effect of the factors of rate of reaction in the following.

i. Surface area and safety of grain mill factories.

ii. Why a glowing splint rekindles when it is put in a bottle of oxygen gas,

iii. Why smoking is forbidden in areas where bottled oxygen is in use, burning charcoal

iv. The use of antioxidants as competitive inhibitors to preserve food.

Collision theory explains how and why chemical reactions happen. It says that for reactants to turn into products, their particles must collide with enough energy and in the right orientation.

Key points of collision theory

1. For particles to react, they need to collide with the correct orientation and have energy equal to or greater than the activation energy (the minimum energy needed for the reaction to occur).

2. The activation energy is the least amount of energy particles need to collide with tom cause a reaction.

3. Different reactions have different activation energy levels.

4. If a collision leads to the reactants turning into products, it is called a successful or effective collision.

Factors Affecting Reaction Rate According to

Collision Theory

1. Temperature When the temperature increases, the particles gain more energy, leading to more collisions with enough energy to cause a reaction. This increases the rate of reaction because a greater fraction of particles will have energy greater than the activation energy.

2. Surface Area and Reactant Nature

A larger surface area allows more particles to collide, increasing the number of successful collisions and increasing the rate of the reaction. More exposed particles mean more frequent collisions, which raises the reaction rate.

3. Concentration Increasing the concentration of reactants increases the number of particles in a given space, causing more frequent collisions. For gases, increasing the pressure has the same effect as more particles are forced into a smaller space, leading to more collisions and a faster reaction.

4. Catalyst A catalyst speeds up a reaction by providing an easier pathway with a lower activation energy. This allows more particles to have enough energy to collide effectively, resulting in more successful collisions and a faster reaction.

Maxwell-Boltzmann energy distribution curve The effect of temperature on the rate of reaction can be shown using a Maxwell- Boltzmann distribution curve. A typical curve looks like this below:

Figure 2.4: Maxwell-Boltzmann energy distribution curve The area under the curve represents all the molecules, and must remain constant for a given scenario. At a higher temperature (T₂), more particles have energy greater than the activation energy. When drawing the curve at a higher temperature, the curve becomes lower, but the total area under the curve remains the same. This is because no mass can be lost or gained during a chemical reaction.

When the temperature increases, particles gain more kinetic energy, causing more collisions per second. Also, a larger number of particles have energy above the activation energy, meaning more collisions will lead to a reaction. This increases the rate of the reaction because there are more effective collisions.

Activity 2.7

1. Form groups of not more than five. Each group should brainstorm and list all interactions you can think of that might happen in a chemical reaction (e.g., atoms molecules colliding, bonds forming or breaking, energy transfer). Include ideas!

2. Each group will then share their list with the class. Record all interactions on the or on a large piece of paper.

3. Have a whole-class discussion to decide on the most important interactions i chemical reaction.

4. Use the agreed-upon interactions to introduce the collision theory. Think about this theory explains what happens during a chemical reaction.

5. Write the main points of the collision theory on the board as a class.

6. Use examples from the brainstorming session to help explain each key point of collision theory.

7. Each group should make a concept map or diagram that shows the main ideas of collision theory using examples from the discussion.

Activity 2.8 Collision theory- simulating how particles interact and react

1. In the same groups, choose a scenario for each below:

a. Particles with different energies and orientations.

b. Particles with various surface areas.

c. Particles with different concentrations.

2. Act out the collision theory in your groups by pretending to be particles interacting reacting. Discuss how different factors (like temperature, surface area concentration) might affect these interactions.

3. Use your scenarios to think about how each factor could change the rate of a reaction. Draw diagrams and sketches to support your explanations.

4. Draw a sketch of the Maxwell-Boltzmann energy distribution curve and discuss how relates to the collision theory.

5. Your instructor will wrap up with a class discussion to summarize the main ideas key takeaways from the activities.

Activity 2.9 Exploring Reaction Rates Using Collision Theory and Maxwell-B Distribution

1. Understand the basics of collision theory and how it explains reaction rates.

2. Use the Maxwell-Boltzmann energy distribution curve to explain how affects reaction rates.

3. Compare and analyse the effects of concentration and temperature on reaction using collision theory and the Maxwell-Boltzmann distribution.

Materials needed

• Chart or projector displaying the Maxwell-Boltzmann energy distribution curve

• Reaction rate simulation tools (online tools or hands-on demonstrations reacting chemicals)

• Models or diagrams of particle collisions

• Worksheets for group discussions and note-taking

• Whiteboard and markers for visual comparisons Understand the basics of collision theory and how it explains reaction rates

a. Explain the idea that chemical reactions occur when particles collide with energy (activation energy) and proper orientation after watching the video below.

https://youtu.be/daihzyMQ_X8

b. What happens to the rate of reaction if particles collide more frequently?

c. You will participate in a demonstration where we shake containers filled with numbers of particles. This will show how more frequent collisions lead to reactions.

d. In groups, discuss and explore the factors that can influence the frequency and of collisions (such as concentration, temperature, or surface area).

e. Based on the discussion, write down how collision theory explains why increasing concentration of reactants generally increases the reaction rate.

Understanding the Maxwell-Boltzmann Distribution Curve

Aim: Use the Maxwell-Boltzmann energy distribution curve to explain why an increase in temperature increases the rate of a chemical reaction.

a. Watch the video below attentively and carry out the activities that https://th.bing.com/th?&id= OVP.HyQXE9N3yjrqKjuvCs4l- HAHgFo&w= 323&h= 181&c= 7&pid= 2.1&rs= 1

b. Look at the Maxwell-Boltzmann energy distribution curve displayed. This curve how the kinetic energies of particles in a gas sample are distributed. Only with energy above the activation energy will react.

c. Compare two curves: one representing a lower temperature and one representing a higher temperature. Observe how the area under the curve changes shape, particularly the number of particles with energy above the activation energy.

d. How does increasing the temperature affect the number of particles with energy to react?

e. Each group will present their findings to the class, explaining how the shift in the at higher temperatures leads to more successful collisions and, therefore, a reaction rate.

Comparing the Effects of Concentration and Temperature

Compare and analyse the effects of concentration and temperature on reaction rates using both collision theory and the Maxwell-Boltzmann energy distribution curve.

a. Working in two teams, you will compare and contrast the effects of increasing concentration and increasing temperature on the rate of a chemical reaction using both collision theory and the Maxwell-Boltzmann energy distribution curve.

b. Team 1 (Concentration): Use collision theory to explain how increasing the number of particles (i.e., higher concentration) leads to more frequent collisions and a faster reaction rate.

c. Team 2 (Temperature): Use the Maxwell-Boltzmann distribution curve to show that at higher temperatures, more particles have energy greater than the activation energy, leading to a faster reaction rate.

d. After both teams present, engage in a class-wide discussion to compare and contrast the effects of temperature and concentration on reaction rates. Focus on the similarities and differences.

e. Which factor, temperature or concentration, do you think has a greater effect on reaction rates in typical conditions?

Finally, complete the worksheet provided, which asks you to compare and contrast the effects of concentration and temperature using both collision theory and the Maxwell-Boltzmann energy distribution curve.

1. Rate Equation

The rate equation or rate law for a chemical reaction is an expression that provides a relationship between the rate of the reaction and the concentration of the reactants participating in it.

Consider the hypothetical chemical reaction given by: aA + bB → cC + dD Where a, b, c and d are the stoichiometric coefficients of the reactants and products, then the rate equation for the reaction is given by:

Rate α [A]ˣ[B]ʸ Rate = k[A]ˣ[B]ʸ Where; [A] and [B] denotes the concentrations of reactants A and B.

‘x’ and ‘y’ denote the orders with respect to reactants A and B (which may or may not be equal to their stoichiometric coefficients a and b The proportionality constant ‘k’ is the rate constant of the reaction.

2. Order of reaction Order of a reaction with respect to each reactant is define as the exponent to which the concentration term of that reactant in the rate law is raised.

Consider the general reaction: aA + bB → cC + dD suppose the rate expression for this reaction is: rate = k[A]ˣ[B]ʸ, then the order with respect to A and B are x and y respectively.

a. Overall order of a reaction The overall order of a reaction is determined by adding up the powers (or exponents) of the concentration of all the reactants in the rate equation. To find the overall order, simply add up the individual orders of each reactant involved in the reaction. It is therefore defined as, the sum of the exponents to which the concentrations of all the reactants in the rate law are raised.

Considering the reaction, rate = k[A]ˣ[B]ʸas an example, the ove rall orde r of the re action = x + y

b. Zero order reaction For a chemical reaction in the form aA → product, if the order with respect to A is a zero order, the rate law for the reaction is: rate = k[A]⁰rate = k(1) = k.

This means that for a zero-order reaction, the rate is constant. It does not change with concentration. The rate is independent of the concentration of the reactants, meaning an increase in concentration of the reactant will not translate to an increase in rate of the reaction. It is worth noting that a zero-order reaction will obviously change rate when all of the reactant is exhausted (the reaction cannot continue at the constant rate past completion).

c. First-order reaction For a chemical reaction in the form aA → product, if the order with respect to A is a first order, then rate = −∆[A]_ ∆ t rate = k[A] For a first order reaction, doubling the concentration of a first order reaction causes the rate to double.

d. Second order reaction For a chemical reaction of the form aA → product, if the order with respect to A is a second order, rate = − k ∆[A]²_ ∆ t For a second order reaction, doubling the concentration of a second order reaction causes the rate to quadruple.

e. Half-life of a reaction The time required for a reactant to reduce to half its initial concentration is called the half-life of a reactant and is designated by the symbol t_(½).

For first order reaction, t₁__ 2 = 0.693_ k For a second order reaction, t₁__ 2 = 1_ k [A]ₒ

3. Rate determining step Think of a chemical reaction like an assembly line in a factory.

A whole reaction might be finished in one go, or it might take a series of small steps (each small step is called an 'elementary reaction'). This entire series of steps is known as the 'reaction mechanism'.

The speed of the entire assembly line (the overall rate of the reaction) is always limited by the slowest single worker or machine on that line.

In chemistry, the slowest step in the entire sequence is the most important one.

We call this the rate-determining step.

Simply put: We look at the reaction mechanism (the full sequence of steps), and the step that takes the longest is the one that sets the pace for the whole process.

That's the rate-determining step.

Graphical Determination of Reaction Order

In chemical kinetics, the order of a reaction describes how the rate of the reaction depends on the concentration of reactants. Graphical analysis allows us to determine the order of a reaction with respect to a specific reactant. Here is how it is done:

1. Experimental Data Collection: Conduct an investigation where the concentration of the reactant is monitored time. Record data points of concentration at different time intervals.

2. Plotting the Concentration vs. Time Graph: Create a graph with concentration (on the y-axis) against time (on the x-axis).

The resulting curve represents the change in concentration as the reaction progresses.

3. Drawing Tangents: Draw tangents (straight lines) to the curve at various points.

These tangents represent the instantaneous rates of reaction at those concentrations.

4. Calculating Tangent Gradients: Calculate the gradient (slope) of each tangent line.

The gradient at a specific concentration corresponds to the rate of reaction at concentration.

5. Interpreting the Graph Shape: The shape of the rate against concentration graph provides clues about the order.

a. Zero Order: A straight horizontal line indicates zero order with respect to the reactant.

rate Concentration rate = k[A]⁰rate = k.

b. First Order: A linear (straight-line) graph suggests first order.

rate Concentration rate = − ∆[A]_ ∆ t rate = k[A]

c. Second Order: A curved graph (concave upward) indicates second order.

rate Concentration rate = − k ∆[A]²_ ∆ t Calculations involving the rate law expression

Example 1

Consider data for the reaction: A + B → products Experiment [A] / moldm⁻³[B] / moldm⁻³Rate / moldm⁻³s⁻¹1 0.002 0.001 5 × 10⁻⁴ 2 0.002 0.002 10 × 10⁻⁴ 3 0.004 0.001 20 × 10⁻⁴ Determine,

i. the rate law

ii. the overall order of the reaction

iii. the specific rate constant k Answer

i. Determine the exponents separately Substituting data from experiment 1 and 2 into the general rate equation, Rate = k[A]ˣ[B]ʸ, 5 × 10⁻⁴= k(0.002)ˣ(0.001)ʸ… … … … … … . equation 1.

10 × 10⁻⁴= k(0.002)ˣ(0.002)ʸ… … … … … … . equation 2.

Divide equation 2 by1 10 × 10⁻⁴_______ 5 × 10⁻⁴= k(0.002)ˣ(0.002)ʸ____________ k(0.002)ˣ(0.001)ʸ2 = 2ʸy = 1 Also, substituting data from experiment 1 and 3 into the rate law expression, 5 × 10⁻⁴= k(0.002)ˣ(0.001)ʸ… … … … … … . . equation 3 20 × 10⁻⁴= k(0.004)ˣ(0.001)ʸ… … … … … … … equation 4 Divide equation 4 by 3 20 × 10⁻⁴_______ 5× 10⁻⁴= k(0.004)ˣ(0.001)ʸ____________ k(0.002)ˣ(0.001)ʸ4 = 2ˣ2²= 2ˣx = 2 Substituting the exponents into a rate law Rate = k[A]ˣ[B]ʸRate = k[A]²[B]¹ii. The overall order of the reaction is = y + x = 1 + 2 = 3 Therefore, the reaction is in 3ʳᵈorder.

iii. Pick any complete set of experimental data, substitute it into the rate equation and calculate the value of the constant k Substituting data from experiment three into the rate equation:

Rate = k[A]²[B] 5 × 10⁻⁴= k(0.004)²(0.001) k = 20 x 10⁻⁴_____________ ( 0.004)²(0.001) k = 1.25 × 10⁵Units of k = moldm⁻³s⁻¹_______________________ (moldm⁻³) (moldm⁻³× moldm⁻³) k = mol⁻²dm⁻⁶s⁻¹Therefore, k = 1.25 × 10⁵mol⁻²dm⁻⁶s⁻¹Activity 2.10 Visualising Activation Energy, Collision Theory, and Reaction Rate Influence

1. Reflect on an analogy, such as climbing a hill, to illustrate the concept of energy to initiate a reaction. Write a brief explanation on how this analogy represents concept of activation energy in chemical reactions.

2. Create a table listing various factors (e.g., concentration, surface area, catalysts) that influence reaction rates. Use visual aids, such as particle collision diagrams, to represent how each factor impacts collisions and reaction rates.

3. If possible, use interactive software to manipulate temperature and observe changes in the Maxwell-Boltzmann distribution curve in real time. Record observations on how temperature influences the curve shape and discuss the implications for reaction rates.

4. Practice interpreting concentration vs. time graphs by analysing slopes and trends. Use these graphs to deduce the reaction order and write an explanation of how the graph’s shape correlates to different reaction orders.

Activity 2.11 Exploration and Reflection on Rate of Chemical Reactions

1. Use your own experiences to brainstorm definitions for key terms, such as rate constant (k), order of reaction, and rate-determining step (RDS).

Write out your ideas and consider how each term applies to chemical reactions you have learned about.

2. After completing the above activities, summarise the key points and insights gained from each. Reflect on any areas where you would like further clarification and note down questions to discuss with your teacher or peers.

Activity 2.12 Experimental Data Analysis on Rate of Chemical Reactions Given the following experimental data for the reaction A→B, where the concentration of A is measured at different times:

Time (s) [A] mol/dm³0 0.50 10 0.40 20 0.32 30 0.25 40 0.20

a. Use the data provided to deduce the rate expression for the reaction by the reaction order with respect to A.

b. Show your calculations and reasoning, and explain the process used to arrive at rate expression.

c. Sketch the expected concentration vs. time graphs for zero-order, first- order, and second-order reactions. Describe the shape of each graph and how the slope or curve provides insight into the reaction order.

d. Given the half-life data for a first-order reaction with an initial concentration of 0.40 mol/dm³and a half-life of 15 seconds, calculate the rate constant and use it to confirm the reaction order.

e. Discuss how this half-life concept is applied in real-life scenarios, such as radioactive decay, and note any insights or questions that arise.

Review Questions 2.1

1. Define the term rate of reaction.

2. Compare and contrast the use of changes in volume of gas evolved verses formation of precipitate to measure rate of reaction.

3. Design an experiment to measure the rate of reaction using colour change.

4. State at least three factors that affects rate of chemical reactions

5. Discuss the rationale behind using powdered versus solid reactants.

6. Consider the spoilage of milk, a perishable item that undergoes a chemical reaction involving the enzyme lactase. Explain how refrigeration slows down the spoilage process, using concepts related to rate of reaction, activation energy. Include a graphical representation of the effect of temperature on the rate of reaction.

7. Consider the following reaction in aqueous solution:

5 Br−(aq) + Br O₃ −(aq) + 6 H+ (aq) → 3 Br₂(aq) + 3 H₂ O(l).

If the rate of disappearance of Br– (aq) at a particular moment during the reaction is 3.5x10⁻⁴mol dm⁻³s⁻¹, what is the rate of appearance of Br₂(aq) at that moment?

Review Questions 2.2

1. What is the collision theory, and how does it explain the rate of a chemical reaction?

1. Compare and contrast the effects of increasing concentration and increasing temperature on the rate of a chemical reaction, using collision theory and the Maxwell Boltzmann energy distribution curve to support your answer.

2. Design an experiment to investigate how changing the surface area of reactants affects the rate of reaction and explain how the collision theory supports your hypothesis.

3. Why does increasing the concentration of a solution increase the rate reaction?

4. Why do catalysts increase a rate of reaction?

5. Describe the Maxwell-Boltzmann distribution curve and explain how you can estimate the fraction of particles with sufficient energy to react.