Thermochemistry

10 min
Video + Practice
SC-18

Target Objective

Calculate enthalpy changes using Hess's law

Thermochemistry

Every chemical reaction involves energy changes. When you burn wood for warmth, energy is released. When you use a cold pack for an injury, energy is absorbed. Thermochemistry studies the heat energy associated with chemical reactions.

Enthalpy (H)

Enthalpy is the total heat content of a system at constant pressure. We cannot measure absolute enthalpy, but we can measure the change in enthalpy (delta H).

delta H = H(products) - H(reactants)

Exothermic and Endothermic Reactions

Exothermic reactions release heat to the surroundings. delta H is negative.

  • Example: Combustion of methane: CH₄ + 2O₂ --> CO₂ + 2H₂O, delta H = -890 kJ/mol
  • The surroundings get warmer

Endothermic reactions absorb heat from the surroundings. delta H is positive.

  • Example: Thermal decomposition of CaCO₃: CaCO₃ --> CaO + CO₂, delta H = +178 kJ/mol
  • The surroundings get cooler

Hess's Law

The total enthalpy change for a reaction is the same regardless of the pathway taken, as long as the initial and final conditions are the same.

delta H (overall) = delta H₁ + delta H₂ + delta H₃ + ...

This is a consequence of enthalpy being a state function (depends only on initial and final states, not the path).

Worked Example (Hess's Law)

Given:

  • C + O₂ --> CO₂, delta H₁ = -393 kJ
  • CO + 1/2 O₂ --> CO₂, delta H₂ = -283 kJ

Find delta H for: C + 1/2 O₂ --> CO

Solution: Using Hess's law: delta H₁ = delta H(target) + delta H₂

delta H(target) = delta H₁ - delta H₂ = -393 - (-283) = -110 kJ

Bond Energy

Bond energy is the energy required to break one mole of a particular bond in the gaseous state.

delta H(reaction) = Sum of bond energies broken - Sum of bond energies formed

Breaking bonds requires energy (endothermic). Forming bonds releases energy (exothermic).

Example: For H₂ + Cl₂ --> 2HCl

  • Bonds broken: H-H (436 kJ) + Cl-Cl (242 kJ) = 678 kJ
  • Bonds formed: 2 x H-Cl (2 x 431) = 862 kJ
  • delta H = 678 - 862 = -184 kJ (exothermic)

Standard Enthalpy of Formation

The enthalpy change when 1 mole of a compound is formed from its elements in their standard states (25 C, 1 atm).

delta H(reaction) = Sum of delta Hf(products) - Sum of delta Hf(reactants)

By definition, delta Hf of elements in their standard state = 0.

Nepal Connection

Biogas plants in rural Nepal use the exothermic decomposition of organic waste to produce methane (CH₄), which is then burned for cooking. Understanding thermochemistry helps optimize these energy systems.

Key Takeaways

  • Exothermic: delta H < 0 (releases heat); Endothermic: delta H > 0 (absorbs heat)
  • Hess's law: total enthalpy change is path-independent
  • Bond energy method: delta H = bonds broken - bonds formed
  • Standard enthalpy of formation is the basis for many calculations

Quick Quiz

1. An exothermic reaction has:

2. Hess's law is based on the fact that enthalpy is a:

3. If breaking bonds requires 500 kJ and forming bonds releases 700 kJ, the reaction is: