Ch 2 · Hess's Law · AS & A Level Chemistry 9701
Topic 2 of 2 in Chemical energetics — AS & A Level Chemistry 9701 — 35 lessons.
Introduction
Many enthalpy changes cannot be measured directly, for example the enthalpy of formation of a hydrocarbon, because carbon and hydrogen do not combine cleanly in a calorimeter. Hess's Law provides a way to calculate such values indirectly, by building an energy cycle from enthalpy changes that can be measured, most often standard enthalpies of formation or combustion.
Key ideas — Hess's Law and conservation of energy
Hess's Law states that the enthalpy change for a reaction is independent of the route taken, provided the initial and final conditions are the same. This follows directly from the principle of conservation of energy: enthalpy is a state function, so the total energy of a closed system depends only on its current state, not on the path used to reach it. If two different routes between the same reactants and the same products gave different overall enthalpy changes, energy could be created or destroyed simply by choosing one route over the other, which would violate the conservation of energy.
🧑🏫 Teacher's explanation
Imagine you are hiking from a valley to a mountain peak. You can take a steep, direct path, or a long, winding trail that circles the mountain. Even though the paths are different, the change in your altitude (the vertical distance climbed) is exactly the same because you start at the same point and end at the same point. In chemistry, enthalpy is like that altitude.
📖 Definitions
🔍 Worked example — Easy
Calculate the enthalpy change for the reaction: C(s) + 1/2O_2(g) arrow CO(g) given that Δ H_f^° [CO(g)] = -110 kJ mol^-1 and Δ H_f^° [C(s)] = 0 kJ mol^-1 and Δ H_f^° [O_2(g)] = 0 kJ mol^-1.
🔍 Worked example — Standard
Using Hess's Law, calculate Δ H for 2C(s) + O_2(g) arrow 2CO(g) given: \n(i) C(s) + O_2(g) arrow CO(g), Δ H = -110 kJ mol^-1 \n(ii) CO(g) + 1/2O_2(g) arrow CO_2(g), Δ H = -394 kJ mol^-1 \n(iii) C(s) + O_2(g) arrow CO_2(g), Δ H = -394 kJ mol^-1
🔍 Worked example — Exam-style
Calculate Δ H for the reaction CH_4(g) + 2O_2(g) arrow CO_2(g) + 2H_2O(l) using the following data: \nΔ H_f^° [CH_4(g)] = -71.6 kJ mol^-1 \nΔ H_f^° [CO_2(g)] = -393.5 kJ mol^-1 \nΔ H_f^° [H_2O(l)] = -285.8 kJ mol^-1
⚠️ Common errors
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🎯 Exam question
Key ideas — Constructing an energy cycle from enthalpies of formation
To find an enthalpy change of reaction, Δ H_r, from standard enthalpies of formation, Δ H_f^ominus, draw the reactants and products of the target equation on a horizontal line, with an arrow for Δ H_r between them. Below this, draw a line for the constituent elements in their standard states, since every substance can be formed from its elements. Arrows point down from elements to reactants (using Δ H_f^ominus of the reactants) and down from elements to products (using Δ H_f^ominus of the products). Hess's Law then gives:
🧑🏫 Teacher's explanation
Imagine you are traveling from City A to City B. You can take a direct flight (the direct route) or you can fly from City A to City C, then from City C to City B (the indirect route). Even though the paths are different, the total distance traveled remains exactly the same. In chemistry, enthalpy is like the 'altitude' or 'energy level' of a substance. The total energy change depends only on where you start and where you end, not how you get there.
📖 Definitions
⚠️ Common errors
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🎯 Exam question
Key ideas — Constructing an energy cycle from enthalpies of combustion
Where enthalpies of combustion, Δ H_c^ominus, are more readily available, for example for hydrocarbons and alcohols, the elements' row is replaced by a row of the same combustion products (usually CO_2(g) and H_2O(l)) below the reactants and products, since every reactant and product can, in principle, be burned completely in oxygen. Both reactants and products are connected downward to this common set of combustion products, and Hess's Law gives:
Key ideas — Combining Hess cycles with bond energy data
An energy cycle can equally be built from average bond energies, with a row of gaseous atoms below the reactants and products, since breaking every bond in the reactants (and separately in the products) produces the same set of free atoms. This is the origin of the bond-energy expression Δ H = Σ(bonds broken) - Σ(bonds formed) met in the previous lesson: it is simply a Hess cycle in which the common intermediate is the set of gaseous atoms rather than elements or combustion products. Whichever intermediate is used, the two routes around the cycle — direct, and via the common intermediate — must give the same total enthalpy change.
🧑🏫 Teacher's explanation
Imagine you are traveling from London to Paris. You can take a direct flight, or you can fly from London to Dublin and then from Dublin to Paris. Even though the routes are different, the total distance traveled is the same because the starting point and the destination haven't changed. In chemistry, the 'distance' is the enthalpy change (Δ H).
📖 Definitions
🔍 Worked example — Easy
Calculate the enthalpy change for the reaction H_2(g) + Cl_2(g) arrow 2HCl(g) using bond enthalpies: H-H = 436 kJ/mol, Cl-Cl = 242 kJ/mol, and H-Cl = 431 kJ/mol.
🔍 Worked example — Standard
In a reaction, the enthalpy change is -250 kJ/mol. If the energy required to break all reactant bonds is 1200 kJ/mol, calculate the energy released when the product bonds are formed.
🔍 Worked example — Exam-style
Using bond enthalpies, calculate Δ H for the combustion of methane: CH_4(g) + 2O_2(g) arrow CO_2(g) + 2H_2O(g). Given: C-H = 413, O=O = 498, C=O = 799, O-H = 464 (all in kJ/mol).
⚠️ Common errors
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🎯 Exam question
Worked example
Question. Use the standard enthalpies of formation below to calculate the standard enthalpy change of reaction for C_2H_4(g) + H_2(g) arrow C_2H_6(g). Δ H_f^ominus[C_2H_4(g)] = +52.3\ mathrmkJ\ mol^-1; Δ H_f^ominus[C_2H_6(g)] = -84.7\ mathrmkJ\ mol^-1; Δ H_f^ominus[H_2(g)] = 0\ mathrmkJ\ mol^-1 (element in its standard state). [3]
🔍 Worked example — Standard
Using the following data, calculate the enthalpy change for C(s) + O_2(g) arrow CO_2(g): Δ H_f^ominus[CO_2(g)] = -393.5 kJ mol^-1 and Δ H_f^ominus[C(s)] = 0 kJ mol^-1.
🔍 Worked example — Exam-style
Calculate the standard enthalpy change for the reaction C_2H_4(g) + H_2(g) arrow C_2H_6(g) given: Δ H_f^ominus[C_2H_4(g)] = +52.3 kJ mol^-1, Δ H_f^ominus[C_2H_6(g)] = -84.7 kJ mol^-1, and Δ H_f^ominus[H_2(g)] = 0 kJ mol^-1.
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