How to Calculate Enthalpy Change
Learn how to calculate enthalpy change (ΔH) using standard enthalpies of formation, Hess's Law, and calorimetry. Includes worked examples for exothermic and endothermic reactions.
What Is Enthalpy?
Enthalpy (H) is a thermodynamic quantity representing the total heat content of a system at constant pressure. The enthalpy change of a reaction (ΔH) is the heat absorbed or released when the reaction proceeds at constant pressure. A negative ΔH indicates an exothermic reaction (heat released to surroundings), while a positive ΔH indicates an endothermic reaction (heat absorbed from surroundings). Enthalpy is measured in joules (J) or kilojoules (kJ).
Using Standard Enthalpies of Formation
The standard enthalpy of formation (ΔHf°) is the enthalpy change when one mole of a compound is formed from its elements in their standard states at 298 K and 1 bar. The standard enthalpy of any element in its most stable form is defined as zero. The standard enthalpy of reaction is calculated as: ΔH°rxn = Σ ΔHf°(products) - Σ ΔHf°(reactants), where each term is multiplied by the stoichiometric coefficient from the balanced equation.
Step-by-Step Example: Combustion of Methane
For CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l), the standard enthalpies of formation are: CH4(g) = -74.8 kJ/mol, O2(g) = 0 kJ/mol, CO2(g) = -393.5 kJ/mol, H2O(l) = -285.8 kJ/mol. ΔH°rxn = [1(-393.5) + 2(-285.8)] - [1(-74.8) + 2(0)] = [-393.5 - 571.6] - [-74.8] = -965.1 - (-74.8) = -890.3 kJ/mol. The reaction releases 890.3 kJ per mole of methane burned.
Hess's Law
Hess's Law states that the total enthalpy change for a reaction is the same regardless of whether the reaction occurs in one step or multiple steps, because enthalpy is a state function. To apply Hess's Law, arrange known reactions (and their ΔH values) algebraically so they sum to the target reaction. If a reaction is reversed, the sign of ΔH changes; if a reaction is multiplied by a factor, ΔH is multiplied by the same factor.
Hess's Law Example
To find ΔH for C(s) + 2 H2(g) → CH4(g), use: (1) C(s) + O2(g) → CO2(g), ΔH1 = -393.5 kJ; (2) H2(g) + 1/2 O2(g) → H2O(l), ΔH2 = -285.8 kJ; (3) CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l), ΔH3 = -890.3 kJ. Take (1) + 2x(2) - (3): C + O2 + 2 H2 + O2 - CH4 - 2 O2 → CO2 + 2 H2O - CO2 - 2 H2O, simplifying to C + 2 H2 → CH4. ΔH = -393.5 + 2(-285.8) - (-890.3) = -74.8 kJ/mol.
Calorimetry: Measuring Enthalpy Experimentally
In a coffee-cup calorimeter (constant pressure), the heat of reaction is measured using q = mcΔT, where m is the mass of the solution, c is the specific heat capacity (4.184 J/g·°C for dilute aqueous solutions), and ΔT is the temperature change. The enthalpy change per mole is then ΔH = -q / n, where n is moles of the limiting reagent and the negative sign reflects the sign convention (heat released by reaction = heat absorbed by solution). Bomb calorimeters measure heat at constant volume (ΔU) rather than constant pressure (ΔH).
Bond Enthalpy Method
Enthalpy change can be estimated using average bond enthalpies: ΔH ≈ Σ (bonds broken) - Σ (bonds formed). Energy is required to break bonds (endothermic) and released when bonds form (exothermic). For H2 + Cl2 → 2 HCl: bonds broken = H-H (436 kJ/mol) + Cl-Cl (243 kJ/mol) = 679 kJ. Bonds formed = 2 x H-Cl (432 kJ/mol) = 864 kJ. ΔH ≈ 679 - 864 = -185 kJ. This method gives an estimate; actual values from formation enthalpies are more accurate.
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