Enthalpy of Reaction

Enthalpy of Reaction: MDCAT Chemistry notes

Enthalpy of Reaction MDCAT notes: enthalpies of formation, combustion, neutralization, atomization and solution, their signs and food energy sums.

Unit: Thermochemistry and Energetics of Chemical Reactions · Updated

Enthalpy and enthalpy of reaction

Enthalpy is the heat content of a system. The heat exchanged at constant pressure equals the enthalpy change, $\Delta H = q_p$. The enthalpy of reaction is the enthalpy change when the moles of reactants shown in a balanced equation react completely.

  • Exothermic: $\Delta H_{\text{products}} \lt \Delta H_{\text{reactants}}$, so $\Delta H$ negative.
  • Endothermic: $\Delta H_{\text{products}} \gt \Delta H_{\text{reactants}}$, so $\Delta H$ positive.

Types of standard enthalpy change

TypeDefinition (standard conditions)Sign
Formation, $\Delta H_f^\circ$1 mole of a compound formed from its elements in their standard statesUsually negative; can be positive
Combustion, $\Delta H_c^\circ$1 mole of a substance burnt completely in oxygenAlways negative
Neutralization, $\Delta H_n^\circ$1 mole of water formed from acid and alkaliAlways negative
Atomization, $\Delta H_{at}^\circ$1 mole of gaseous atoms formed from the elementAlways positive
Solution, $\Delta H_{sol}^\circ$1 mole of solute dissolved in excess solventPositive or negative

Formation

Only equations that make exactly one mole of a compound from elements in standard states are formation equations, e.g. $\mathrm{H_2(g) + \tfrac12 O_2(g) \rightarrow H_2O(l)}$, $\Delta H_f^\circ = -285.8\ \mathrm{kJ\,mol^{-1}}$. Equations starting from a compound (e.g. CO) are not. $\Delta H_f^\circ$ of every element in its standard state is zero. $\Delta H_f^\circ$ of CO is about $-110.5\ \mathrm{kJ\,mol^{-1}}$; it cannot be measured directly because burning carbon also forms $\mathrm{CO_2}$.

Neutralization

For any strong acid with any strong base, $\mathrm{H^+(aq) + OH^-(aq) \rightarrow H_2O(l)}$, $\Delta H_n^\circ = -57.4\ \mathrm{kJ\,mol^{-1}}$. The value is the same because the net reaction is always the same. For weak acids or bases it is slightly less, as some energy is used to ionize them. A diprotic acid neutralized by a dibasic base, e.g. $\mathrm{H_2SO_4 + Mg(OH)_2}$, forms two moles of water, so about twice the heat is released.

Atomization

Atomization always requires breaking bonds, so it is always endothermic.

Energy from food

Approximate energy values: carbohydrate $4\ \mathrm{kcal\,g^{-1}}$, protein $4\ \mathrm{kcal\,g^{-1}}$, fat $9\ \mathrm{kcal\,g^{-1}}$.

Worked example: a snack with 30 g carbohydrate, 5 g protein and 10 g fat supplies $30\times4 + 5\times4 + 10\times9 = 120 + 20 + 90 = 230\ \mathrm{kcal}$.

Key formulas

  • $\Delta H = q_p$
  • $\Delta H_{\text{reaction}} = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants})$
  • $q = mc\Delta T$ for calorimetry

Common MDCAT traps

  • Enthalpy of neutralization is per mole of water, not per mole of acid.
  • Always exothermic: combustion and neutralization. Always endothermic: atomization, fusion, sublimation.
  • Enthalpy of formation is not always exothermic (some compounds have positive values).
  • A formation equation must produce one mole of product from elements only.
  • Heat at constant pressure is enthalpy; heat at constant volume is internal energy.

Quick revision

  • $\Delta H_n^\circ$ for strong acid and strong base is $-57.4$ kJ/mol.
  • $\Delta H_f^\circ$ of an element in its standard state is zero.
  • $\Delta H_f^\circ$ of CO is about $-110$ kJ/mol.
  • Fat gives 9 kcal/g; carbohydrate and protein give 4 kcal/g.
  • Atomization always has a positive enthalpy.

Test yourself

More in Thermochemistry and Energetics of Chemical Reactions