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 2025 SL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Chemistry 2025 SL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
- IB DP Chemistry 2025 HL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Chemistry 2025 HL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
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:
- Write a balanced chemical equation.
- Look up the \( \Delta H_f^\circ \) values of all reactants and products.
- Multiply each value by the number of moles in the equation.
- 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:
- Write and balance the chemical equation for the reaction.
- Locate the \( \Delta H_c^\circ \) values (standard enthalpies of combustion) for all relevant substances.
- Multiply each \( \Delta H_c^\circ \) value by the number of moles present in the balanced equation.
- 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 \( \Delta H_c^\circ \).