Key Takeaways
- Hess's Law: Enthalpy change is independent of the pathway, provided the initial and final states remain identical.
- State Function: Because enthalpy is a state function, we can use thermochemical cycles to calculate values that are impossible to measure directly.
- Cycle Rules: Construct formation cycles with elements at the base, and combustion cycles with oxides (CO₂ and H₂O) at the base.
- Sign Conventions: When moving against the direction of an arrow in a cycle, you must reverse the sign of its enthalpy change.
Contents
- What is Hess's Law?
- Why is Hess's Law So Useful?
- Key Definitions for Energetics Calculations
- 1. Calculations Using Enthalpies of Formation
- 2. Calculations Using Enthalpies of Combustion
- Summary of Hess's Law Formulas
- A-Level and IB Specification Alignment
- Practice Exam Question
- Next Steps in Your Revision
In chemical energetics, we often need to determine the enthalpy change of a reaction. However, carrying out a direct laboratory measurement is not always possible. Hess's Law provides an elegant mathematical solution by using thermochemical cycles to calculate these values indirectly.
What is Hess's Law?
Hess's Law states that the total enthalpy change for a chemical reaction is independent of the route taken, provided the initial and final conditions are the same.
Definition: Hess's Law
The total enthalpy change for a reaction is determined solely by the energy states of the reactants and products, not by the pathway or number of intermediate steps between them.
This principle is a direct consequence of the conservation of energy. Because enthalpy is a state function, its value depends only on the current state of the chemical system (its temperature, pressure, physical states, and composition) and not on the history of how it reached that state.
Why is Hess's Law So Useful?
Some chemical reactions are impossible to measure directly in a calorimeter. For instance:
- The reaction might be extremely slow, making it difficult to prevent heat loss to the surroundings during measurement.
- The reaction might require extremely high temperatures, preventing standard laboratory calorimeter setups.
- Competing side reactions might occur, producing a mixture of different products rather than the single compound under investigation.
Hess's Law lets us bypass these physical limitations. By measuring the enthalpy changes of alternative, easily manageable reactions, we can link them together in a cycle to find the enthalpy change of our target reaction.
Key Definitions for Energetics Calculations
To successfully set up thermochemical cycles, you must know the standard enthalpy definitions. The term "standard" (indicated by the superscript symbol o) refers to measurements taken under standard conditions: 298 K (25 °C), 100 kPa pressure, and a concentration of 1 mol dm⁻³ for any aqueous solutions.
- Standard Enthalpy Change (ΔHo): The heat energy change measured under constant pressure and standard conditions. A negative value represents an exothermic reaction (heat released), while a positive value represents an endothermic process (heat absorbed).
- Standard Enthalpy of Formation (ΔHfo): The enthalpy change when one mole of a compound is formed from its constituent elements in their standard states under standard conditions. By definition, the standard enthalpy of formation for any pure element in its standard state is exactly 0 kJ mol⁻¹.
- Standard Enthalpy of Combustion (ΔHco): The enthalpy change when one mole of a substance is completely burned in oxygen under standard conditions.
1. Calculations Using Enthalpies of Formation
When an exam question provides standard enthalpies of formation (ΔHfo), you should construct a cycle where the elements in their standard states sit at the bottom.
The Cycle Pathway:
The cycle involves breaking down the reactants into their elements (the reverse of formation) and then assembling those elements into the products (formation):
Reactants → Elements → Products
Because the arrows for enthalpy of formation point from elements up to the compounds, we go against the reactant arrows (reversing their algebraic sign) and with the product arrows (keeping their sign).
General Formula:
ΔHreactiono = ΣΔHfo(products) - ΣΔHfo(reactants)
Worked Example 1: Calculation using Enthalpies of Formation
Calculate the standard enthalpy change for the combustion of methane (CH₄) using these standard enthalpies of formation:
- ΔHfo[CH₄(g)] = -74.8 kJ mol⁻¹
- ΔHfo[CO₂(g)] = -393.5 kJ mol⁻¹
- ΔHfo[H₂O(l)] = -285.8 kJ mol⁻¹
Reaction Equation: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
Step-by-Step Solution:
- Identify that oxygen (O₂) is an element in its standard state, so its ΔHfo is 0 kJ mol⁻¹.
- Sum the enthalpies of formation for the products:
ΣΔHfo(products) = ΔHfo[CO₂] + 2 × ΔHfo[H₂O]
ΣΔHfo(products) = -393.5 + 2 × (-285.8) = -393.5 - 571.6 = -965.1 kJ mol⁻¹ - Sum the enthalpies of formation for the reactants:
ΣΔHfo(reactants) = ΔHfo[CH₄] + 2 × ΔHfo[O₂]
ΣΔHfo(reactants) = -74.8 + 2 × (0) = -74.8 kJ mol⁻¹ - Apply the formula:
ΔHreactiono = ΣΔHfo(products) - ΣΔHfo(reactants)
ΔHreactiono = -965.1 - (-74.8) = -965.1 + 74.8 = -890.3 kJ mol⁻¹
Conclusion: The standard enthalpy of combustion for methane is -890.3 kJ mol⁻¹, indicating a highly exothermic reaction.
2. Calculations Using Enthalpies of Combustion
If the question gives standard enthalpies of combustion (ΔHco), the substances in the reaction are burned to form their common combustion products (typically carbon dioxide and water). These combustion products sit at the bottom of your cycle.
The Cycle Pathway:
We combust the reactants to form the combustion products, and we also combust the products to form those same combustion products:
Reactants + O₂ → Combustion Products ← Products + O₂
Since the arrows for combustion point from the organic compounds down to the combustion products, we go with the reactant arrows (keeping their sign) and against the product arrows (reversing their sign).
General Formula:
ΔHreactiono = ΣΔHco(reactants) - ΣΔHco(products)
Notice that this is reactants minus products: the exact opposite order compared to the formation formula. Pay close attention to this distinction, as it is a very common source of mistakes in exams.
Worked Example 2: Calculation using Enthalpies of Combustion
Calculate the standard enthalpy of formation of liquid ethanol (C₂H₅OH) using these standard enthalpies of combustion:
- ΔHco[C₂H₅OH(l)] = -1367 kJ mol⁻¹
- ΔHco[C(s)] = -393.5 kJ mol⁻¹
- ΔHco[H₂(g)] = -285.8 kJ mol⁻¹
Target Reaction (Formation of Ethanol): 2C(s) + 3H₂(g) + 1/2O₂(g) → C₂H₅OH(l)
Step-by-Step Solution:
- Identify the reactants and products of the target equation. The reactants are carbon and hydrogen (oxygen does not undergo combustion, as it is the oxidizing agent). The product is ethanol.
- Sum the enthalpies of combustion for the reactants (taking stoichiometry into account):
ΣΔHco(reactants) = 2 × ΔHco[C] + 3 × ΔHco[H₂]
ΣΔHco(reactants) = 2 × (-393.5) + 3 × (-285.8) = -787.0 - 857.4 = -1644.4 kJ mol⁻¹ - Sum the enthalpies of combustion for the products:
ΣΔHco(products) = ΔHco[C₂H₅OH] = -1367 kJ mol⁻¹ - Apply the formula:
ΔHreactiono = ΣΔHco(reactants) - ΣΔHco(products)
ΔHreactiono = -1644.4 - (-1367) = -1644.4 + 1367 = -277.4 kJ mol⁻¹
Conclusion: The standard enthalpy of formation of ethanol is -277.4 kJ mol⁻¹.
Summary of Hess's Law Formulas
Use this table to quickly review which formula and cycle structure to apply depending on your given data:
| Data Type | Bottom Species in Cycle | Formula to Use | Arrow Directions |
|---|---|---|---|
| Enthalpies of Formation (ΔHfo) | Constituent elements in standard states | Products - Reactants | Point UP from elements to reactants/products |
| Enthalpies of Combustion (ΔHco) | Combustion products (CO₂, H₂O, etc.) | Reactants - Products | Point DOWN from reactants/products to combustion products |
A-Level and IB Specification Alignment
Hess's Law is a core topic across all advanced secondary chemistry specifications. Here is what is expected for your specific course:
- AQA A-Level Chemistry (Section 3.1.4): You must be able to define standard enthalpies of formation and combustion, construct enthalpy cycles, and calculate enthalpy changes. Questions often involve multi-step cycles or thermodynamic definitions.
- OCR A-Level Chemistry (Section 5.1.1): The specification requires you to construct enthalpy cycles and calculate enthalpy changes from both formation and combustion data, as well as use algebraic methods to solve cycles.
- Edexcel A-Level Chemistry (Topic 8): Expect questions requiring you to apply Hess's Law to cycles involving both elements (formation) and combustion products, often linked with experimental calorimetry data.
- IB Chemistry (Topics 5 and 15): Candidates are required to design and evaluate cycles, use Hess's Law to determine unknown enthalpy changes, and link these concepts to bond enthalpies and entropy calculations.
Practice Exam Question
Apply these principles to this exam-style practice question. Try to draw the cycle out before checking the worked mark scheme below.
Exam Style Question: Enthalpy of Decomposition (6 Marks)
Directly measuring the enthalpy change for the thermal decomposition of calcium carbonate is difficult due to the high temperature required:
CaCO₃(s) → CaO(s) + CO₂(g)
Instead, a student reacts calcium carbonate and calcium oxide separately with an excess of dilute hydrochloric acid and determines these enthalpy changes:
- CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g) ΔH₁ = -15.9 kJ mol⁻¹
- CaO(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) ΔH₂ = -193.0 kJ mol⁻¹
Calculate the standard enthalpy change for the thermal decomposition of calcium carbonate, showing your working clearly.
Click to reveal the worked mark scheme solution
Worked Solution & Marking Points
Step 1: Set up the cycle or algebraic route. [1 Mark]
The target reaction is: CaCO₃(s) → CaO(s) + CO₂(g)
We can construct a pathway using the acid reactions. The reactants of our target (CaCO₃) and the products (CaO + CO₂) can both be reacted with HCl to form the common products: CaCl₂(aq) + H₂O(l) + CO₂(g).
Step 2: Align the equations to match the target. [2 Marks]
- The first equation has CaCO₃(s) on the reactant side, matching our target. We keep this reaction as it is:
CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g) ΔH₁ = -15.9 kJ mol⁻¹ [1 Mark] - The second equation has CaO(s) on the reactant side. However, in our target, CaO(s) is on the product side. Therefore, we must reverse this equation (which changes the sign of the enthalpy change):
CaCl₂(aq) + H₂O(l) → CaO(s) + 2HCl(aq) -ΔH₂ = +193.0 kJ mol⁻¹ [1 Mark]
Step 3: Combine equations and cancel intermediate species. [1 Mark]
Add the two equations together:
CaCO₃(s) + 2HCl(aq) + CaCl₂(aq) + H₂O(l) → CaCl₂(aq) + H₂O(l) + CO₂(g) + CaO(s) + 2HCl(aq)
Cancel out the 2HCl(aq), CaCl₂(aq), and H₂O(l) appearing on both sides. This yields the target decomposition equation: CaCO₃(s) → CaO(s) + CO₂(g).
Step 4: Calculate the final enthalpy change. [2 Marks]
- Apply Hess's Law: ΔHreaction = ΔH₁ + (-ΔH₂)
ΔHreaction = -15.9 + 193.0 [1 Mark] - ΔHreaction = +177.1 kJ mol⁻¹ [1 Mark for correct value, sign, and unit]
Note: The positive sign is required to indicate that the decomposition is an endothermic process.
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