Sign in to save

Bookmark this page so you can find it later.

Sign in to save

Bookmark this page so you can find it later.

Heat of formation and Hess's law help students calculate energy changes for chemical reactions that are hard to measure directly. This cheat sheet covers standard enthalpy of formation, reaction enthalpy, and how to combine thermochemical equations. Students need these tools to predict whether reactions release or absorb heat and to connect balanced equations with energy changes. It is especially useful for stoichiometry, thermochemistry, and lab analysis problems.

Key Facts

  • Standard enthalpy of formation, ΔHf\Delta H_f^\circ, is the enthalpy change when 1 mol1\ \text{mol} of a compound forms from its elements in their standard states.
  • The standard enthalpy of formation of any pure element in its standard state is ΔHf=0 kJ mol1\Delta H_f^\circ = 0\ \text{kJ mol}^{-1}.
  • The reaction enthalpy from formation values is ΔHrxn=nΔHf(products)nΔHf(reactants)\Delta H_{\text{rxn}}^\circ = \sum n\Delta H_f^\circ(\text{products}) - \sum n\Delta H_f^\circ(\text{reactants}).
  • Hess's law states that ΔHoverall=ΔH1+ΔH2+ΔH3+\Delta H_{\text{overall}} = \Delta H_1 + \Delta H_2 + \Delta H_3 + \cdots for any set of steps that add to the target reaction.
  • If a thermochemical equation is reversed, the enthalpy sign changes, so ΔHreverse=ΔHforward\Delta H_{\text{reverse}} = -\Delta H_{\text{forward}}.
  • If every coefficient in a thermochemical equation is multiplied by kk, the enthalpy change is also multiplied, so ΔHnew=kΔHold\Delta H_{\text{new}} = k\Delta H_{\text{old}}.
  • An exothermic reaction has ΔH<0\Delta H < 0 because heat is released to the surroundings.
  • An endothermic reaction has ΔH>0\Delta H > 0 because heat is absorbed from the surroundings.

Vocabulary

Enthalpy
Enthalpy, HH, is the heat content of a system at constant pressure.
Standard enthalpy of formation
Standard enthalpy of formation, ΔHf\Delta H_f^\circ, is the enthalpy change for forming 1 mol1\ \text{mol} of a substance from its elements in standard states.
Hess's law
Hess's law says the total enthalpy change of a reaction depends only on the initial and final states, not the pathway.
Thermochemical equation
A thermochemical equation is a balanced chemical equation that includes the enthalpy change, such as ΔH=286 kJ\Delta H = -286\ \text{kJ}.
Standard state
A standard state is the most stable form of a substance at 1 bar1\ \text{bar} pressure, usually with solutions at 1 M1\ \text{M} concentration.
Reaction enthalpy
Reaction enthalpy, ΔHrxn\Delta H_{\text{rxn}}, is the heat change for a reaction as written with its balanced coefficients.

Common Mistakes to Avoid

  • Forgetting to multiply ΔHf\Delta H_f^\circ by coefficients is wrong because formation values are given per 1 mol1\ \text{mol}, while the balanced equation may use several moles.
  • Adding reactants minus products is wrong for formation calculations because the correct formula is ΔHrxn=nΔHf(products)nΔHf(reactants)\Delta H_{\text{rxn}}^\circ = \sum n\Delta H_f^\circ(\text{products}) - \sum n\Delta H_f^\circ(\text{reactants}).
  • Not changing the sign when reversing an equation is wrong because reversing a process changes heat released into heat absorbed, or absorbed into released.
  • Using nonzero ΔHf\Delta H_f^\circ values for elements in standard states is wrong because substances such as O2(g)\text{O}_2(g), H2(g)\text{H}_2(g), and C(s,graphite)\text{C}(s,\text{graphite}) have ΔHf=0 kJ mol1\Delta H_f^\circ = 0\ \text{kJ mol}^{-1}.
  • Ignoring the phrase 'as written' is wrong because ΔH\Delta H depends on the exact balanced coefficients in the thermochemical equation.

Practice Questions

  1. 1 Calculate ΔHrxn\Delta H_{\text{rxn}}^\circ for CH4(g)+2O2(g)CO2(g)+2H2O(l)\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l) using ΔHf[CH4(g)]=74.8 kJ mol1\Delta H_f^\circ[\text{CH}_4(g)] = -74.8\ \text{kJ mol}^{-1}, ΔHf[CO2(g)]=393.5 kJ mol1\Delta H_f^\circ[\text{CO}_2(g)] = -393.5\ \text{kJ mol}^{-1}, and ΔHf[H2O(l)]=285.8 kJ mol1\Delta H_f^\circ[\text{H}_2\text{O}(l)] = -285.8\ \text{kJ mol}^{-1}.
  2. 2 Given C(s)+O2(g)CO2(g)\text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g) with ΔH=393.5 kJ\Delta H = -393.5\ \text{kJ} and CO(g)+12O2(g)CO2(g)\text{CO}(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CO}_2(g) with ΔH=283.0 kJ\Delta H = -283.0\ \text{kJ}, use Hess's law to find ΔH\Delta H for C(s)+12O2(g)CO(g)\text{C}(s) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CO}(g).
  3. 3 If 2H2(g)+O2(g)2H2O(l)2\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(l) has ΔH=571.6 kJ\Delta H = -571.6\ \text{kJ}, what is ΔH\Delta H for H2O(l)H2(g)+12O2(g)\text{H}_2\text{O}(l) \rightarrow \text{H}_2(g) + \frac{1}{2}\text{O}_2(g)?
  4. 4 Explain why Hess's law can be used to calculate the enthalpy change for a reaction even when the reaction does not happen easily in one direct step.