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SPM WikiChemistryChapter 3: Thermochemistry

Chapter 3: Thermochemistry

Understand heat changes in chemical reactions, exothermic and endothermic processes, enthalpy calculations, and practical applications of thermochemistry in everyday life and industrial processes.

Chapter 3: Thermochemistry

Overview

Thermochemistry is the study of heat changes that occur during chemical reactions. This fundamental branch of chemistry helps us understand energy transfers in reactions, classify reactions as exothermic or endothermic, and calculate energy changes quantitatively. From cooking food to industrial manufacturing, thermochemical principles are essential for understanding how energy is stored, released, and utilized in chemical processes. This chapter explores the concepts of enthalpy, different types of heats of reaction, and their practical applications in our daily lives.

Learning Objectives

After studying this chapter, you should be able to:

  • Distinguish between exothermic and endothermic reactions
  • Understand the concept of enthalpy and enthalpy change (ΔH)
  • Calculate heat changes using calorimetry and thermometric data
  • Identify and calculate different types of heats of reaction
  • Apply thermochemical principles to real-world problems
  • Understand practical applications of exothermic and endothermic processes

3.1 Heat Changes in Reactions

What is Thermochemistry?

Thermochemistry is the study of heat changes that accompany chemical reactions and physical changes. It helps us understand:

  • How much heat is absorbed or released
  • Why some reactions release heat while others absorb heat
  • How to control reaction conditions for optimal energy use

Thermochemistry Overview:

Exothermic Reactions

Definition: Reactions that release heat to the surroundings

Characteristics:

  • Temperature of surroundings increases
  • Reaction vessel becomes hot
  • Chemical energy → Heat energy
  • Products have lower energy than reactants
  • Enthalpy change (ΔH) is negative (ΔH < 0)

Energy diagram:

Exothermic Reaction Process:

Examples:

  • Combustion reactions: CH4(g)+2O2(g)CO2(g)+2H2O(l)ΔH=890kJ/mol\text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(l)} \quad \Delta H = -890 \text{kJ/mol}
  • Neutralization reactions: HCl(aq)+NaOH(aq)NaCl(aq)+H2O(l)ΔH=57.1kJ/mol\text{HCl(aq)} + \text{NaOH(aq)} \rightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)} \quad \Delta H = -57.1 \text{kJ/mol}
  • Addition of water to concentrated acids: H2SO4(l)+H2O(l)H2SO4(aq)ΔH=94.3kJ/mol\text{H}_2\text{SO}_4\text{(l)} + \text{H}_2\text{O(l)} \rightarrow \text{H}_2\text{SO}_4\text{(aq)} \quad \Delta H = -94.3 \text{kJ/mol}
  • Reactive metals with acids: Zn(s)+2HCl(aq)ZnCl2(aq)+H2(g)ΔH=153kJ/mol\text{Zn(s)} + 2\text{HCl(aq)} \rightarrow \text{ZnCl}_2\text{(aq)} + \text{H}_2\text{(g)} \quad \Delta H = -153 \text{kJ/mol}

Exothermic Applications:

Endothermic Reactions

Definition: Reactions that absorb heat from the surroundings

Characteristics:

  • Temperature of surroundings decreases
  • Reaction vessel becomes cold
  • Heat energy → Chemical energy
  • Products have higher energy than reactants
  • Enthalpy change (ΔH) is positive (ΔH > 0)

Energy diagram:

Endothermic Reaction Process:

Examples:

  • Thermal decomposition: CaCO3(s)CaO(s)+CO2(g)ΔH=+178kJ/mol\text{CaCO}_3\text{(s)} \rightarrow \text{CaO(s)} + \text{CO}_2\text{(g)} \quad \Delta H = +178 \text{kJ/mol}
  • Dissolving ammonium salts: NH4NO3(s)+H2O(l)NH4+(aq)+NO3(aq)ΔH=+25.7kJ/mol\text{NH}_4\text{NO}_3\text{(s)} + \text{H}_2\text{O(l)} \rightarrow \text{NH}_4^+\text{(aq)} + \text{NO}_3^-\text{(aq)} \quad \Delta H = +25.7 \text{kJ/mol}
  • Photosynthesis: 6CO2(g)+6H2O(l)lightC6H12O6(aq)+6O2(g)ΔH=+2803kJ/mol6\text{CO}_2\text{(g)} + 6\text{H}_2\text{O(l)} \xrightarrow{\text{light}} \text{C}_6\text{H}_{12}\text{O}_6\text{(aq)} + 6\text{O}_2\text{(g)} \quad \Delta H = +2803 \text{kJ/mol}
  • Evaporation: H2O(l)H2O(g)ΔH=+44.0kJ/mol\text{H}_2\text{O(l)} \rightarrow \text{H}_2\text{O(g)} \quad \Delta H = +44.0 \text{kJ/mol}

Endothermic Applications:

Comparison Diagram:

Key Terms

  • Enthalpy (H): Total energy content of a system at constant pressure
  • Enthalpy Change (ΔH): Heat absorbed or released during a reaction at constant pressure
  • Activation Energy (Ea): Minimum energy required for a reaction to occur

Key Terms Diagram:


Did You Know?

The human body constantly undergoes exothermic reactions to maintain body temperature. The average adult human generates about 100 watts of heat energy at rest, equivalent to a bright light bulb. This is why we feel warm when we exercise!

3.2 Heat of Reaction

What is Heat of Reaction?

Heat of reaction (or enthalpy change) is the heat change that occurs when a reaction is carried out under standard conditions. It can be measured experimentally using calorimetry.

Heat of Reaction Formula:

ΔH=qn\Delta H = \frac{q}{n}

Experimental Measurement Using Calorimetry

Step-by-step calculation:

  1. Calculate heat change (q): q=mcθq = mc\theta

    • mm = mass of solution (g) ≈ volume of solution (cm3m^3) for water
    • cc = specific heat capacity of water = 4.2 J g⁻¹ °C⁻¹
    • θ\theta = temperature change (°C)
  2. Calculate number of moles (n): Based on the limiting reactant

  3. Calculate enthalpy change (ΔH): ΔH=qn\Delta H = \frac{q}{n}

    • Units: J/mol or kJ/mol
    • Negative sign (-) for exothermic reactions
    • Positive sign (+) for endothermic reactions

Calorimetry Process Diagram:

Heat Calculation Formula:

q=mcΔTq = mc\Delta T

Where:

  • qq = heat energy (J or kJ)
  • mm = mass (g)
  • cc = specific heat capacity (J g⁻¹ °C⁻¹)
  • ΔT\Delta T = temperature change (°C)

Example Calculation

Problem: When 50 cm3m^3 of 2.0 mol dm⁻³ HCl is added to 50 cm3m^3 of 2.0 mol dm⁻³ NaOH, the temperature rises from 25.0°C to 31.5°C. Calculate the heat of neutralization.

Solution:

  1. Calculate heat change:

    m=50+50=100 g (assuming density ≈ 1 g/cm3)m = 50 + 50 = 100 \text{ g (assuming density ≈ 1 g/c$m^3$)} θ=31.525.0=6.5°C\theta = 31.5 - 25.0 = 6.5°\text{C} q=mcθ=100×4.2×6.5=2730 J=2.73 kJq = mc\theta = 100 \times 4.2 \times 6.5 = 2730 \text{ J} = 2.73 \text{ kJ}
  2. Calculate moles of acid and base:

    moles HCl=0.050 dm3×2.0 mol dm⁻³=0.100 mol\text{moles HCl} = 0.050 \text{ d$m^3$} \times 2.0 \text{ mol dm⁻³} = 0.100 \text{ mol} moles NaOH=0.050 dm3×2.0 mol dm⁻³=0.100 mol\text{moles NaOH} = 0.050 \text{ d$m^3$} \times 2.0 \text{ mol dm⁻³} = 0.100 \text{ mol}
  3. Calculate ΔH:

    ΔH=qn=2.73 kJ0.100 mol=27.3 kJ/mol\Delta H = \frac{q}{n} = \frac{-2.73 \text{ kJ}}{0.100 \text{ mol}} = -27.3 \text{ kJ/mol}

    (Negative because it's exothermic)

Step-by-Step Calculation Process:

Sign Convention:

  • Negative ΔH: Heat released (exothermic)
  • Positive ΔH: Heat absorbed (endothermic)

Types of Heats of Reaction

1. Heat of Precipitation (ΔHₚᵣₑc)

Definition: Heat change when 1 mole of precipitate forms from its ions in aqueous solution

Example: Ag+(aq)+Cl(aq)AgCl(s)ΔH=65.5kJ/mol\text{Ag}^+\text{(aq)} + \text{Cl}^-\text{(aq)} \rightarrow \text{AgCl(s)} \quad \Delta H = -65.5 \text{kJ/mol}

Precipitation Process:

Ag+(aq)+Cl(aq)AgCl(s)+heat\text{Ag}^+\text{(aq)} + \text{Cl}^-\text{(aq)} \rightarrow \text{AgCl(s)} + \text{heat}

2. Heat of Displacement (ΔHₜₛₚ)

Definition: Heat change when 1 mole of metal is displaced from its salt solution by a more electropositive metal

Example: Zn(s)+CuSO4(aq)ZnSO4(aq)+Cu(s)ΔH=217kJ/mol\text{Zn(s)} + \text{CuSO}_4\text{(aq)} \rightarrow \text{ZnSO}_4\text{(aq)} + \text{Cu(s)} \quad \Delta H = -217 \text{kJ/mol}

Displacement Process:

Zn(s)+Cu2+(aq)Zn2+(aq)+Cu(s)+heat\text{Zn(s)} + \text{Cu}^{2+}\text{(aq)} \rightarrow \text{Zn}^{2+}\text{(aq)} + \text{Cu(s)} + \text{heat}

3. Heat of Neutralization (ΔHₙₑᵤₜ)

Definition: Heat change when 1 mole of water is formed from the reaction between acid and base

Strong acid + Strong alkali: ΔH ≈ -57 kJ/mol (constant because net reaction is H++OHH2O\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}) Weak acid or weak alkali: ΔH < -57 kJ/mol (additional heat absorbed to ionize the weak acid/alkali)

Example: HCl(aq)+NaOH(aq)NaCl(aq)+H2O(l)ΔH=57.1kJ/mol\text{HCl(aq)} + \text{NaOH(aq)} \rightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)} \quad \Delta H = -57.1 \text{kJ/mol}

Neutralization Process:

H+(aq)+OH(aq)H2O(l)ΔH=57.1kJ/mol\text{H}^+\text{(aq)} + \text{OH}^-\text{(aq)} \rightarrow \text{H}_2\text{O(l)} \quad \Delta H = -57.1 \text{kJ/mol}

4. Heat of Combustion (ΔH꜀ₒₘb)

Definition: Heat released when 1 mole of substance burns completely in excess oxygen

Examples:

  • CH4(g)+2O2(g)CO2(g)+2H2O(l)ΔH=890kJ/mol\text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(l)} \quad \Delta H = -890 \text{kJ/mol}
  • C3H8(g)+5O2(g)3CO2(g)+4H2O(l)ΔH=2220kJ/mol\text{C}_3\text{H}_8\text{(g)} + 5\text{O}_2\text{(g)} \rightarrow 3\text{CO}_2\text{(g)} + 4\text{H}_2\text{O(l)} \quad \Delta H = -2220 \text{kJ/mol}

Combustion Process:

Fuel+OxygenCO2+H2O+heat\text{Fuel} + \text{Oxygen} \rightarrow \text{CO}_2 + \text{H}_2\text{O} + \text{heat}

5. Heat of Solution (ΔHₛₒₗ)

Definition: Heat change when 1 mole of solute dissolves in a solvent

Example: NaOH(s)Na+(aq)+OH(aq)ΔH=44.5kJ/mol\text{NaOH(s)} \rightarrow \text{Na}^+\text{(aq)} + \text{OH}^-\text{(aq)} \quad \Delta H = -44.5 \text{kJ/mol} (exothermic)

Solution Process:

Solute+SolventDissolved Species±heat\text{Solute} + \text{Solvent} \rightarrow \text{Dissolved Species} \pm \text{heat}

Types of Heats Summary:


SPM Exam Tips

  • Always include the sign (+/-) when reporting ΔH values
  • For neutralization reactions, remember that strong acid + strong alkali ≈ -57 kJ/mol
  • ΔH is negative for exothermic reactions (heat released)
  • ΔH is positive for endothermic reactions (heat absorbed)
  • When calculating ΔH, make sure to use the limiting reactant to find moles

3.3 Applications of Exothermic and Endothermic Reactions

Exothermic Applications

1. Instant Hot Packs

Composition: Contains calcium chloride or magnesium sulfate Process: When water is added, dissolution is exothermic Chemistry: CaCl2(s)+H2O(l)Ca2+(aq)+2Cl(aq)+heatΔH=81.3kJ/mol\text{CaCl}_2\text{(s)} + \text{H}_2\text{O(l)} \rightarrow \text{Ca}^{2+}\text{(aq)} + 2\text{Cl}^-\text{(aq)} + \text{heat} \quad \Delta H = -81.3 \text{kJ/mol} Use: Muscle relief, pain relief, hand warmers

Hot Pack Process:

2. Combustion of Fuels

Characteristics: High energy density, readily available fuels Examples:

  • Petrol: (C8H18)+252O28CO2+9H2O+heatΔH=5470kJ/mol(\text{C}_8\text{H}_{18}) + \frac{25}{2}\text{O}_2 \rightarrow 8\text{CO}_2 + 9\text{H}_2\text{O} + \text{heat} \quad \Delta H = -5470 \text{kJ/mol}
  • Natural gas (CH4H_4): CH4+2O2CO2+2H2O+heatΔH=890kJ/mol\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} + \text{heat} \quad \Delta H = -890 \text{kJ/mol}
  • Coal: Various hydrocarbons + O2CO2+H2O+heat\text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} + \text{heat}

Applications: Transportation, electricity generation, heating

Combustion Applications:

3. Self-Heating Food Containers

Process: Chemical reaction generates heat to cook food Example: CaO(s)+H2O(l)Ca(OH)2(aq)+heatΔH=65.2kJ/mol\text{CaO(s)} + \text{H}_2\text{O(l)} \rightarrow \text{Ca(OH)}_2\text{(aq)} + \text{heat} \quad \Delta H = -65.2 \text{kJ/mol}

Endothermic Applications

1. Instant Cold Packs

Composition: Contains ammonium nitrate or urea Process: When water is added, dissolution is endothermic Chemistry: NH4NO3(s)+H2O(l)NH4+(aq)+NO3(aq)heatΔH=+25.7kJ/mol\text{NH}_4\text{NO}_3\text{(s)} + \text{H}_2\text{O(l)} \rightarrow \text{NH}_4^+\text{(aq)} + \text{NO}_3^-\text{(aq)} - \text{heat} \quad \Delta H = +25.7 \text{kJ/mol} Use: Sports injuries, first aid, cooling

Cold Pack Process:

2. Refrigeration and Air Conditioning

Principle: Endothermic process absorbs heat from surroundings Example: NH3(g)+H2O(l)NH4+(aq)+OH(aq)heat\text{NH}_3\text{(g)} + \text{H}_2\text{O(l)} \rightarrow \text{NH}_4^+\text{(aq)} + \text{OH}^-\text{(aq)} - \text{heat}

Refrigeration Cycle:

3. Cooking

Process: Heat energy absorbed by food for chemical reactions Examples:

  • Baking (endothermic decomposition of baking soda)
  • Grilling (heat transfer to food)

Fuel Values

Definition: Amount of heat energy released when 1 gram of fuel burns completely Unit: kJ/g

Fuel Value Formula:

Fuel Value=Heat Released (kJ)Mass of Fuel (g)\text{Fuel Value} = \frac{\text{Heat Released (kJ)}}{\text{Mass of Fuel (g)}}

Comparative fuel values:

  • Hydrogen: 142 kJ/g (highest, but difficult to store)
  • Methane: 55.5 kJ/g
  • Ethanol: 29.7 kJ/g (biofuel)
  • Wood: 15-20 kJ/g
  • Coal: 25-35 kJ/g

Fuel Comparison Chart:

Biofuels: Renewable energy sources from biological materials

  • Ethanol: From fermentation of sugars
  • Biodiesel: From transesterification of vegetable oils

Biofuel Production Process:

Applications Summary:


Safety Reminder

When working with thermochemical experiments:

  • Use proper eye protection and lab coats
  • Be careful with hot equipment and hot solutions
  • Handle concentrated acids and bases with care
  • Use appropriate calorimeters (polystyrene cups for simple calorimetry)
  • Never mix unknown chemicals
  • Follow proper waste disposal procedures

3.4 Calorimetry Experiments

Simple Calorimetry Using Polystyrene Cup

Materials:

  • Polystyrene cup (good insulator)
  • Thermometer (0-100°C)
  • Measuring cylinders
  • Stopwatch

Calorimetry Setup:

Procedure for Neutralization:

  1. Measure equal volumes of acid and alkali (e.g., 50 cm3m^3 each)
  2. Measure initial temperature of both solutions
  3. Mix solutions quickly and start stopwatch
  4. Record temperature every 10 seconds for 5-10 minutes
  5. Plot temperature vs. time graph
  6. Calculate ΔT from maximum temperature

Key considerations:

  • Assume density of solution = 1 g/cm3m^3
  • Use specific heat capacity of water = 4.2 J g⁻¹ °C⁻¹
  • Heat loss to surroundings must be minimized
  • Stirring ensures even temperature distribution

Temperature vs Time Graph:

Example: Heat of Combustion Experiment

Aim: To determine the heat of combustion of ethanol

Reaction: C2H5OH(l)+3O2(g)2CO2(g)+3H2O(l)ΔH=1367kJ/mol\text{C}_2\text{H}_5\text{OH(l)} + 3\text{O}_2\text{(g)} \rightarrow 2\text{CO}_2\text{(g)} + 3\text{H}_2\text{O(l)} \quad \Delta H = -1367 \text{kJ/mol}

Materials:

  • Ethanol burner
  • Copper calorimeter
  • Thermometer
  • Water
  • Electronic balance

Combustion Calorimetry Setup:

Procedure:

  1. Weigh empty calorimeter with water (initial mass)
  2. Measure initial temperature
  3. Burn ethanol for fixed time
  4. Record final temperature
  5. Weigh calorimeter again (final mass)
  6. Calculate mass of ethanol burned

Calculations:

q=mcΔT (heat absorbed by water)q = mc\Delta T \text{ (heat absorbed by water)} methanol=initial massfinal massm_{\text{ethanol}} = \text{initial mass} - \text{final mass} ΔH=qmolesethanol\Delta H = -\frac{q}{\text{moles}_{\text{ethanol}}}

Combustion Calculation Steps:

Error Sources and Minimization:


Summary

Key Concepts

  1. Thermochemistry studies heat changes in chemical reactions
  2. Exothermic reactions: Release heat (ΔH < 0)
  3. Endothermic reactions: Absorb heat (ΔH > 0)
  4. Enthalpy change (ΔH): Heat absorbed or released at constant pressure
  5. Types of heats of reaction:
    • Heat of precipitation
    • Heat of displacement
    • Heat of neutralization
    • Heat of combustion
    • Heat of solution

Concept Summary Diagram:

Experimental Skills

  • Perform calorimetry experiments using simple apparatus
  • Calculate heat changes using q=mcΔTq = mc\Delta T
  • Determine enthalpy changes from experimental data
  • Identify and control sources of error in calorimetry

Experimental Skills Flowchart:

Practical Applications

  • Hot packs: Exothermic dissolution reactions
  • Cold packs: Endothermic dissolution reactions
  • Fuel combustion: Energy release for heating and power
  • Cooking: Heat transfer for food preparation
  • Refrigeration: Endothermic processes for cooling

Applications Matrix:

Problem-Solving Strategy

  1. Identify the type of reaction (exothermic/endothermic)
  2. Calculate heat change using q=mcΔTq = mc\Delta T
  3. Determine moles of limiting reactant
  4. Calculate ΔH using ΔH=qn\Delta H = \frac{q}{n}
  5. Include appropriate sign (+/-) and units

Problem-Solving Steps:

Practice Questions

  1. Explain the difference between exothermic and endothermic reactions with energy diagrams.

  2. When 25 cm3m^3 of 1.0 mol dm⁻³ HCl is added to 25 cm3m^3 of 1.0 mol dm⁻³ NaOH, the temperature rises from 24.0°C to 28.2°C. Calculate the heat of neutralization.

  3. Describe an experiment to determine the heat of combustion of a candle wax.

Final Summary:


Related Topics: