Energy Changes in Chemical Reactions

CXC Chemistry Syllabus Reference: Section B7 - Energy Changes in Chemical Reactions

Introduction to Energy and Chemical Reactions

Chemical reactions involve breaking and forming bonds between atoms. These processes either release energy to the surroundings (exothermic) or absorb energy from the surroundings (endothermic). Understanding the energy changes in chemical reactions is fundamental to predicting reaction spontaneity and applications in industry and everyday life.

Key Concepts to Understand:

Energy Transfers in Chemical Reactions

In all chemical reactions, energy is either released or absorbed. This energy commonly takes the form of heat, but can also appear as light, sound, or electricity.

Exothermic reactions release energy to the surroundings, usually causing a temperature increase.

Endothermic reactions absorb energy from the surroundings, usually causing a temperature decrease.

Examples of Exothermic Reactions:

Examples of Endothermic Reactions:

Example: When sodium hydroxide (NaOH) dissolves in water, the temperature of the solution increases significantly. This is an exothermic process.

NaOH(s) → Na+(aq) + OH-(aq) + heat

Enthalpy Changes

Enthalpy (H) is a measure of the total energy in a system. The change in enthalpy (ΔH) represents the heat transferred during a reaction at constant pressure.

Enthalpy Change (ΔH) = Final Enthalpy - Initial Enthalpy

For exothermic reactions: ΔH < 0 (negative value)

For endothermic reactions: ΔH > 0 (positive value)

Standard Enthalpy Changes:

Exothermic Reaction Reactants Products Activation Energy ΔH Endothermic Reaction Reactants Products Activation Energy ΔH

Note: In energy level diagrams:

Activation Energy

Not all collisions between reactant particles lead to product formation. For a reaction to occur, colliding particles must have:

Activation Energy (Ea) is the minimum energy required for a reaction to occur.

Factors Affecting Reaction Rate:

Example: The decomposition of hydrogen peroxide is very slow at room temperature:

2H2O2(aq) → 2H2O(l) + O2(g)

However, when manganese(IV) oxide is added as a catalyst, the reaction occurs rapidly as the activation energy is lowered.

Bond Energy and Enthalpy Calculations

The energy required to break or form chemical bonds can be used to calculate the enthalpy change of a reaction.

Bond Energy (Bond Enthalpy) is the energy required to break one mole of a particular bond in the gas phase.

To calculate ΔH using bond energies:

ΔH = Σ(Energy required to break bonds) - Σ(Energy released when bonds form)

Or simply:

ΔH = Σ(Bond energies of bonds broken) - Σ(Bond energies of bonds formed)

Bond Average Bond Energy (kJ/mol)
H-H 436
C-H 413
C-C 347
C=C 612
C≡C 838
C-O 358
C=O 743
O-H 463
O=O 497
N≡N 945

Example calculation: Calculate the enthalpy change for the combustion of methane:

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

Bonds broken:

Bonds formed:

ΔH = 2646 - 3338 = -692 kJ/mol

The negative value indicates an exothermic reaction.

Hess's Law

Hess's Law states that the enthalpy change for a reaction is the same whether it occurs in one step or multiple steps. The total enthalpy change depends only on the initial and final states, not the pathway taken.

This law allows us to calculate enthalpy changes that may be difficult to measure directly, by using known enthalpy values of related reactions.

Initial State Final State Intermediate State ΔH direct ΔH1 ΔH2 ΔH direct = ΔH1 + ΔH2

Example application of Hess's Law:

Calculate the enthalpy change for the reaction:

C(s) + 2H2(g) → CH4(g)

Using the following information:

  1. C(s) + O2(g) → CO2(g) ΔH = -394 kJ/mol
  2. H2(g) + ½O2(g) → H2O(l) ΔH = -286 kJ/mol
  3. CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ΔH = -890 kJ/mol

To find the target reaction, we can rearrange the equations:

CO2(g) → C(s) + O2(g) ΔH = +394 kJ/mol

2H2O(l) → 2H2(g) + O2(g) ΔH = +572 kJ/mol

CO2(g) + 2H2O(l) → CH4(g) + 2O2(g) ΔH = +890 kJ/mol

Adding these equations:

C(s) + 2H2(g) → CH4(g) ΔH = +394 + 572 + 890 = -76 kJ/mol

Experimental Determination of Enthalpy Changes

Calorimetry

Calorimetry is the measurement of heat changes in chemical reactions. A simple calorimeter can be made using a polystyrene cup with a lid.

Heat change (q) = mass × specific heat capacity × temperature change

q = m × c × ΔT

where:

Measuring the Enthalpy of Combustion

The enthalpy of combustion can be measured using a bomb calorimeter or a simpler method using a spirit burner.

Example experiment: Determining the enthalpy of combustion of ethanol.

A spirit burner containing ethanol was weighed before and after heating 100g of water. The temperature of the water increased from 20°C to 45°C, and the mass of ethanol used was 0.54g.

Heat transferred to water = 100g × 4.18 J/g°C × 25°C = 10,450 J

Moles of ethanol used = 0.54g ÷ 46g/mol = 0.0117 mol

Enthalpy of combustion = 10,450 J ÷ 0.0117 mol = 892,308 J/mol = -892 kJ/mol

(Note: The negative sign indicates an exothermic reaction)

Applications of Enthalpy Changes

1. Fuels and Energy Production

Fuels are selected based on their enthalpy of combustion, availability, cost, and environmental impact.

Fuel Enthalpy of Combustion (kJ/g)
Methane -55.5
Propane -50.3
Octane -47.8
Ethanol -29.7
Hydrogen -141.8

2. Food and Metabolism

The energy content of food is measured in Calories (1 Cal = 4.18 kJ).

Nutrient Energy Content (kJ/g)
Carbohydrates 17
Proteins 17
Fats 38

3. Hand Warmers and Cold Packs

Chemical hand warmers use exothermic reactions (e.g., oxidation of iron) to release heat. Cold packs use endothermic processes (e.g., dissolving ammonium nitrate in water) to absorb heat.

4. Industrial Processes

Understanding enthalpy changes helps optimize industrial processes to minimize energy costs and environmental impact.

Self-Assessment Questions

Question 1: Define the terms 'exothermic reaction' and 'endothermic reaction' and give one example of each.

Answer:

An exothermic reaction releases energy to the surroundings, typically as heat. Example: Combustion of methane gas.

An endothermic reaction absorbs energy from the surroundings. Example: Photosynthesis in plants.

Question 2: The standard enthalpy of combustion of methane is -890 kJ/mol. What does this value tell us?

Answer:

The value tells us that when one mole of methane burns completely in excess oxygen under standard conditions, 890 kJ of heat energy is released to the surroundings. The negative sign indicates that it is an exothermic process.

Question 3: Using bond energies, calculate the enthalpy change for the following reaction:

H2(g) + Cl2(g) → 2HCl(g)

Bond energies: H-H = 436 kJ/mol, Cl-Cl = 242 kJ/mol, H-Cl = 431 kJ/mol

Answer:

Bonds broken: H-H (436 kJ/mol) + Cl-Cl (242 kJ/mol) = 678 kJ/mol

Bonds formed: 2 × H-Cl = 2 × 431 kJ/mol = 862 kJ/mol

ΔH = 678 - 862 = -184 kJ/mol

The reaction is exothermic with an enthalpy change of -184 kJ/mol.

Question 4: Explain how a catalyst affects the activation energy and enthalpy change of a reaction.

Answer:

A catalyst lowers the activation energy of a reaction by providing an alternative reaction pathway. This increases the rate of reaction as more particles will have sufficient energy to react. However, a catalyst does not affect the enthalpy change of the reaction - the energy difference between reactants and products remains the same.

Question 5: Use Hess's Law to calculate the enthalpy change for the reaction:

C(s) + H2O(g) → CO(g) + H2(g)

Given:

  1. C(s) + O2(g) → CO2(g) ΔH = -394 kJ/mol
  2. H2(g) + ½O2(g) → H2O(g) ΔH = -242 kJ/mol
  3. CO(g) + ½O2(g) → CO2(g) ΔH = -283 kJ/mol

Answer:

Step 1: Rearrange the equations to get the target equation:

C(s) + O2(g) → CO2(g) ΔH = -394 kJ/mol (keep as is)

H2O(g) → H2(g) + ½O2(g) ΔH = +242 kJ/mol (reverse equation 2)

CO2(g) → CO(g) + ½O