Hess's Law in Combustion Reactions
HSC Chemistry Syllabus
- Apply Hess’s Law to simple energy cycles and solve problems to quantify enthalpy changes within reactions, including but not limited to:
– Heat of combustion
Hess's Law in Photosynthesis & Respiration
This video explains how to use Hess's Law to determine overall enthalpy change of two complex processes: photosynthesis and respiration.
Hess's Law Reviewed
Hess's Law states that the total enthalpy change of a reaction is independent of the pathway taken, provided the initial and final conditions remain the same. This principle is based on the law of conservation of energy, which ensures that energy cannot be created or destroyed but can only be transferred or transformed.
Mathematically, Hess’s Law can be written as:
$$\Delta H_{\text{total}} = \Delta H_1 + \Delta H_2 + ...$$
This means that if a chemical reaction can be expressed as the sum of multiple reactions, the total enthalpy change for the reaction is the sum of the enthalpy changes of those individual reactions.
Hess’s Law is particularly useful in determining enthalpy changes for reactions that are difficult to measure directly, such as photosynthesis and respiration.
Enthalpy Change of Combustion
A combustion reaction is a type of exothermic chemical reaction in which a substance (typically a fuel) reacts with oxygen to produce energy in the form of heat and light.
There are two types of combustion:
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Complete Combustion:
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Occurs when there is sufficient oxygen.
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Produces carbon dioxide (CO₂) and water (H₂O) as products.
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Releases the maximum amount of energy.
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Incomplete Combustion:
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Occurs when oxygen supply is limited.
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Produces carbon monoxide (CO) or even solid carbon (soot) instead of CO₂.
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Releases less energy compared to complete combustion because the enthalpy of formation of CO and soot are less than that of carbon dioxide.
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The enthalpy change of incomplete combustion for a given fuel is always less than its complete combustion. This is because the enthalpy of formation of carbon dioxide releases more energy than that of carbon monoxide.
The enthalpy of formation of carbon monoxide (CO) is -110 kJ/mol. This value represents the enthalpy change when 1 mole of carbon monoxide is formed from its elements in their standard states:
$$C(s) + \frac{1}{2} O_2(g) \rightarrow CO(g) \hspace{2cm} \Delta H_f = -110 \text{ kJ/mol}$$
The enthalpy of formation of carbon dioxide is -394 kJ/mol.
Enthalpy Change of Complete Combustion of Methane
Methane is a common hydrocarbon fuel. The complete combustion of methane occurs as follows:
$$CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$$
This exothermic reaction releases a significant amount of energy as heat. However, the enthalpy change for this reaction can be determined indirectly using Hess’s Law by summing the enthalpy changes of known reactions.
Step 1: Write the formation reactions of the products and reactants from their constituent elements in standard states:
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Formation of carbon dioxide:
$$C(s) + O_2(g) \rightarrow CO_2(g) \hspace{2cm} \Delta H_f = -394 \text{ kJ/mol}$$
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Formation of water:
$$H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(l) \hspace{2cm} \Delta H_f = -286 \text{ kJ/mol}$$
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Formation of methane:
$$C(s) + 2H_2(g) \rightarrow CH_4(g) \hspace{2cm} \Delta H_f = -75 \text{ kJ/mol}$$
The constituent elements in standard states can react to form the reactants first, which then react to form the products. Alternatively, they can react to form the products directly, These two pathways are shown in the following diagram.
Step 2: Apply Hess’s Law
By Hess's law, the enthalpy change is the same regardless of the pathway taken, as long as the initial and final conditions of the pathways are identical. In the diagram above, the overall enthalpy change of the two reactions represented by the red and orange arrows is equal to the enthalpy change of the blue arrow. This means the enthalpy change of combustion, i.e. heat of combustion of methane, can be determined by finding the difference between the enthalpy change of formation of the products (blue arrow) and reactants (red arrow).
Using the enthalpy of formation values, we can calculate the enthalpy change for the combustion of methane:
$$\Delta H_{\text{combustion}} = \sum \Delta H_f \text{(products)} - \sum \Delta H_f \text{(reactants)}$$
Thus, the enthalpy change for the complete combustion of methane is –891 kJ/mol, indicating that the reaction releases 891 kJ of energy per mole of methane combusted.