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Energetics / Thermochemistry - Hess’s Law

Grade 11IBChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Hess’s Law states that the total enthalpy change (ΔH\Delta H) for a chemical reaction is independent of the route by which the chemical reaction occurs, provided the initial and final states are the same.

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Enthalpy is a state function, meaning the value of ΔH\Delta H depends only on the identity of the reactants and products, not on the intermediate steps.

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The law is a specific application of the Law of Conservation of Energy (the First Law of Thermodynamics).

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Standard Enthalpy of Formation (ΔHf⊖\Delta H_f^\ominus): The enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions (100 kPa100\text{ kPa} and 298 K298\text{ K}). By definition, ΔHf⊖\Delta H_f^\ominus of any element in its standard state is 00.

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Standard Enthalpy of Combustion (ΔHc⊖\Delta H_c^\ominus): The enthalpy change when one mole of a substance is burned completely in excess oxygen under standard conditions.

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Algebraic manipulation: If a reaction is reversed, the sign of ΔH\Delta H is flipped. If the coefficients of a reaction are multiplied by a factor nn, ΔH\Delta H must also be multiplied by nn.

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Cycle construction: When using formation data, the arrows in the cycle point from the elements to the reactants and products. When using combustion data, the arrows point from the reactants and products to the combustion products (e.g., CO2\text{CO}_2 and H2O\text{H}_2\text{O}).

📐Formulae

ΔHreaction⊖=∑nΔHf⊖(products)−∑mΔHf⊖(reactants)\Delta H_{\text{reaction}}^\ominus = \sum n\Delta H_f^\ominus(\text{products}) - \sum m\Delta H_f^\ominus(\text{reactants})

ΔHreaction⊖=∑nΔHc⊖(reactants)−∑mΔHc⊖(products)\Delta H_{\text{reaction}}^\ominus = \sum n\Delta H_c^\ominus(\text{reactants}) - \sum m\Delta H_c^\ominus(\text{products})

ΔHcycle=ΔH1+ΔH2+⋯+ΔHn=0\Delta H_{\text{cycle}} = \Delta H_1 + \Delta H_2 + \dots + \Delta H_n = 0

💡Examples

Problem 1:

Calculate the standard enthalpy of formation of methane (CH4\text{CH}_4) given the following standard enthalpies of combustion: ΔHc⊖(C, graphite)=−394 kJ mol−1\Delta H_c^\ominus(\text{C, graphite}) = -394\text{ kJ mol}^{-1}, ΔHc⊖(H2(g))=−286 kJ mol−1\Delta H_c^\ominus(\text{H}_2\text{(g)}) = -286\text{ kJ mol}^{-1}, and ΔHc⊖(CH4(g))=−891 kJ mol−1\Delta H_c^\ominus(\text{CH}_4\text{(g)}) = -891\text{ kJ mol}^{-1}.

Solution:

The equation for the formation of methane is: C(s)+2H2(g)→CH4(g)\text{C(s)} + 2\text{H}_2\text{(g)} \rightarrow \text{CH}_4\text{(g)}. Using combustion data: ΔHf⊖=∑ΔHc⊖(reactants)−∑ΔHc⊖(products)\Delta H_f^\ominus = \sum \Delta H_c^\ominus(\text{reactants}) - \sum \Delta H_c^\ominus(\text{products}). ΔHf⊖(CH4)=[1×(−394)+2×(−286)]−[1×(−891)]\Delta H_f^\ominus(\text{CH}_4) = [1 \times (-394) + 2 \times (-286)] - [1 \times (-891)]. ΔHf⊖(CH4)=[−394−572]+891=−966+891=−75 kJ mol−1\Delta H_f^\ominus(\text{CH}_4) = [-394 - 572] + 891 = -966 + 891 = -75\text{ kJ mol}^{-1}.

Explanation:

To find the enthalpy of formation, we sum the combustion enthalpies of the constituent elements (reactants) and subtract the combustion enthalpy of the target compound (product). Note that the coefficient for H2\text{H}_2 is 22, so its combustion value must be doubled.

Problem 2:

Calculate the enthalpy change for the reaction 2CO(g)+O2(g)→2CO2(g)2\text{CO(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{CO}_2\text{(g)} given ΔHf⊖(CO)=−110.5 kJ mol−1\Delta H_f^\ominus(\text{CO}) = -110.5\text{ kJ mol}^{-1} and ΔHf⊖(CO2)=−393.5 kJ mol−1\Delta H_f^\ominus(\text{CO}_2) = -393.5\text{ kJ mol}^{-1}.

Solution:

ΔHrxn⊖=∑ΔHf⊖(products)−∑ΔHf⊖(reactants)\Delta H_{rxn}^\ominus = \sum \Delta H_f^\ominus(\text{products}) - \sum \Delta H_f^\ominus(\text{reactants}) ΔHrxn⊖=[2×(−393.5)]−[2×(−110.5)+0]\Delta H_{rxn}^\ominus = [2 \times (-393.5)] - [2 \times (-110.5) + 0] ΔHrxn⊖=−787.0−(−221.0)=−566.0 kJ\Delta H_{rxn}^\ominus = -787.0 - (-221.0) = -566.0\text{ kJ}

Explanation:

We use the formation data formula. The value for O2\text{O}_2 is 00 because it is an element in its standard state. The stoichiometric coefficients (22) must be multiplied by the respective ΔHf⊖\Delta H_f^\ominus values.

Hess’s Law - Revision Notes & Key Formulas | IB Grade 11 Chemistry