Home / IB DP / IB DP Chemistry 2026, 2027 & 2028 / IB DP Chemistry SL & HL Study Notes / Energy cycles in reactions: R1.2.4 Hess’s Law with combustion and formation data IB DP Chemistry Study Notes

Energy cycles in reactions: R1.2.4 Hess’s Law with combustion and formation data IB DP Chemistry Study Notes

Energy cycles in reactions: R1.2.4 Hess's Law with combustion and formation data IB DP Chemistry Study Notes - New Syllabus 2025

Energy cycles in reactions – IB DP Chemistry- Study Notes

IITian Academy excellent Introduction to the Particulate Nature of Matter – Study Notes and effective strategies will help you prepare for your IB DP Chemistry 2025 exam.

IB DP Chemistry Study Notes – All Topics

Reactivity 1.2.4 – Hess’s Law using enthalpy of formation and combustion data

Reactivity 1.2.4 – Hess’s Law using enthalpy of formation and combustion data

Hess’s Law is a key tool in thermochemistry that allows us to calculate the enthalpy change of a reaction using enthalpy data from related reactions. When direct measurement is not feasible, we use known values of either:

  • Standard enthalpies of formation (\( \Delta H_f^\circ \))
  • Standard enthalpies of combustion (\( \Delta H_c^\circ \))

What Is \( \Delta H_f^\circ \)?

The standard enthalpy of formation, denoted \( \Delta H_f^\circ \), is the enthalpy change when 1 mole of a compound is formed from its constituent elements, all in their standard states under standard conditions (298 K and 100 kPa).

For any element in its standard state, \( \Delta H_f^\circ = 0 \).

Applying Hess’s Law with \( \Delta H_f^\circ \)

Hess’s Law states that the total enthalpy change for a chemical reaction is independent of the path taken.

In a Hess cycle that uses \( \Delta H_f^\circ \) values:

  • The base of the cycle consists of the elements in their standard states.
  • Arrows in the diagram point upward — from the elements to the reactants and the products.
  • This method builds the enthalpy change by imagining that all reactants are broken down into their elements and then reassembled into products.

What Is \( \Delta H_c^\circ \)?

The standard enthalpy of combustion, symbolized as \( \Delta H_c^\circ \), is the energy released when 1 mole of a substance undergoes complete combustion in excess oxygen under standard conditions (298 K and 100 kPa), with all reactants and products in their standard states.

Combustion values are always exothermic, so \( \Delta H_c^\circ \) values are negative.

Applying Hess’s Law with \( \Delta H_c^\circ \)

Hess’s Law allows us to calculate the enthalpy change of a reaction indirectly by comparing the combustion enthalpies of the substances involved.

In a Hess cycle based on \( \Delta H_c^\circ \):

  • The common combustion products — usually \( \text{CO}_2(g) \) and \( \text{H}_2\text{O}(l) \) — are placed at the bottom of the cycle.
  • All arrows point downwards from the reactants and products to the combustion products.
  • This method constructs a Hess cycle where both reactants and products are burned completely in oxygen to form common products (typically CO₂ and H₂O).
  • Because all combustion paths lead to the same final products, the relative energy differences can be used to deduce \( \Delta H^\circ \).

Important Notes:

  • Always use balanced chemical equations with correct stoichiometry.
  • Only use data for substances in their standard states — gas, liquid, solid, or aqueous, as specified.
  • Values of \( \Delta H_f^\circ \) and \( \Delta H_c^\circ \) are provided in the IB Chemistry data booklet.
  • Keep units consistent (usually kJ/mol).

Calculating Enthalpy Changes Using Enthalpy of Formation or Combustion Data

1. Using Standard Enthalpies of Formation (\( \Delta H_f^\circ \))

To determine the enthalpy change of a reaction using formation data, apply the equation:

\( \Delta H^\circ = \sum \Delta H_f^\circ(\text{products}) – \sum \Delta H_f^\circ(\text{reactants}) \)

This method considers the hypothetical cycle in which all reactants are broken into their elements and reassembled into products.

Steps:

  1. Write a balanced chemical equation.
  2. Look up the \( \Delta H_f^\circ \) values of all reactants and products.
  3. Multiply each value by the number of moles in the equation.
  4. Use the equation: total of products minus total of reactants.

Example 

Calculate the enthalpy change for the reaction:

\( \text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g) \)

Given data:

  • \( \Delta H_f^\circ(\text{NH}_3(g)) = -46 \, \text{kJ/mol} \)
  • \( \Delta H_f^\circ(\text{N}_2(g)) = 0 \, \text{kJ/mol} \)
  • \( \Delta H_f^\circ(\text{H}_2(g)) = 0 \, \text{kJ/mol} \)
▶️Answer/Explanation

Step 1: Apply the formula:

\( \Delta H^\circ = [2 \times (-46)] – [1 \times 0 + 3 \times 0] = -92 \, \text{kJ} \)

Final Answer: \( \Delta H^\circ = -92 \, \text{kJ} \)

This means the reaction is exothermic, releasing 92 kJ of energy per 1 mol of \( \text{N}_2 \) reacted.

 

2. Using Standard Enthalpies of Combustion (\( \Delta H_c^\circ \))

To calculate the enthalpy change of a reaction using combustion data, use:

\( \Delta H^\circ = \sum \Delta H_c^\circ(\text{reactants}) – \sum \Delta H_c^\circ(\text{products}) \)

In this method, all substances are hypothetically combusted in oxygen, and their combustion enthalpies are compared via Hess’s Law.

Steps:

  1. Write and balance the chemical equation for the reaction.
  2. Locate the \( \Delta H_c^\circ \) values (standard enthalpies of combustion) for all relevant substances.
  3. Multiply each \( \Delta H_c^\circ \) value by the number of moles present in the balanced equation.
  4. Apply the Hess’s Law equation: total combustion enthalpy of reactants minus products:

Example 

Determine the enthalpy change for the reaction:

\( \text{C}_2\text{H}_4(g) + \text{H}_2(g) \rightarrow \text{C}_2\text{H}_6(g) \)

Given data:

  • \( \Delta H_c^\circ(\text{C}_2\text{H}_4(g)) = -1411 \, \text{kJ/mol} \)
  • \( \Delta H_c^\circ(\text{H}_2(g)) = -286 \, \text{kJ/mol} \)
  • \( \Delta H_c^\circ(\text{C}_2\text{H}_6(g)) = -1560 \, \text{kJ/mol} \)
▶️Answer/Explanation

Step 1: Apply the formula:

\( \Delta H^\circ = [(-1411) + (-286)] – [-1560] \)
\( = -1697 – (-1560) = -137 \, \text{kJ} \)

Final Answer: \( \Delta H^\circ = -137 \, \text{kJ} \)

This means the hydrogenation of ethene is exothermic and releases 137 kJ/mol.

Example 

Calculate the enthalpy change for the reaction:

\( \frac{1}{2}\text{N}_2(g) + \frac{3}{2}\text{H}_2(g) \rightarrow \text{NH}_3(g) \)

This is the formation of 1 mole of ammonia.

Given data (from the IB data booklet):

  • \( \Delta H_f^\circ(\text{NH}_3(g)) = -46 \, \text{kJ/mol} \)
  • \( \Delta H_f^\circ(\text{N}_2(g)) = 0 \, \text{kJ/mol} \)
  • \( \Delta H_f^\circ(\text{H}_2(g)) = 0 \, \text{kJ/mol} \)
▶️Answer/Explanation

This is already a formation reaction, so the enthalpy change is simply the enthalpy of formation of ammonia.

\( \Delta H^\circ = \Delta H_f^\circ(\text{NH}_3(g)) = -46 \, \text{kJ/mol} \)

Final Answer: \( \Delta H^\circ = -46 \, \text{kJ} \)

Even with fractional coefficients, the logic holds as long as the equation forms exactly 1 mole of product from elements in standard states.

 

   Key Tips for Calculation:

  • Use balanced chemical equations with correct stoichiometric coefficients.
  • Always check that data corresponds to the correct physical states.
  • Ensure all enthalpy values are in kJ/mol as given in the IB Data Booklet.
  • Understand which formula to apply: use products − reactants for \( \Delta H_f^\circ \), and reactants − products for \( \Delta H_c^\circ \).
Scroll to Top