Reaction and Formation Enthalpies

Enthalpy (symbol: H) is a thermodynamic quantity representing the total heat content of a system at constant pressure. It is a state function, depending only on the system’s current state, not the path taken to reach it.

Enthalpy Change (ΔH)

The enthalpy change (ΔH) reflects the heat absorbed or released during a process at constant pressure.

Standard Enthalpy of Formation (ΔH°f)

The standard enthalpy of formation is defined as the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states under standard conditions (298 K, 1 atm).

Example: C (s, graphite) + \(\ce{O2}\) (g) → \(\ce{CO2}\) (g) ΔH°f = –393.5 kJ/mol

Calculating Reaction Enthalpy Using Formation Enthalpies

\(\Delta \text{H}^{\circ} _{\text{rxn}} = \Sigma \Delta \text{H}^{\circ} _{\text{f}} ({\text{products}}) \;\; – \;\; \Sigma \Delta \text{H}^{\circ} _{\text{f}} ({\text{reactants}})\)

Each \(\text{H}^{\circ} _{\text{f}}\) value must be multiplied by its stoichiometric coefficient from the balanced chemical equation.

Hess’s Law

Hess’s Law states that the total enthalpy change of a reaction is independent of the route taken. The overall ΔH is the sum of the ΔH values of individual steps leading to the overall reaction.

\(\Delta \text{H}_{\text{total}} = \Delta \text{H}_1 + \Delta \text{H}_2 + ... + \Delta \text{H}_n\)

Bond Enthalpy Method

\(\Delta \text{H} = \Sigma \text{D} (\text{bonds broken}) \;\; – \;\; \Sigma \text{D} (\text{bonds formed})\)

Where D represents bond dissociation enthalpies. Bond breaking requires energy (endothermic), while bond formation releases energy (exothermic).

Key Considerations

Conclusion

Reaction and formation enthalpies are essential for understanding the energy flow in chemical processes. Mastery of this topic provides insight into reaction feasibility, stability, and thermal behavior.

Question: Calculate the enthalpy change for the reaction:

\(\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)}\)

Given the following standard enthalpies of formation:

\(\Delta H_f^\circ(\ce{CH4(g)}) = -74.8 \, \text{kJ/mol}\)
\(\Delta H_f^\circ(\ce{CO2(g)}) = -393.5 \, \text{kJ/mol}\)
\(\Delta H_f^\circ(\ce{H2O(l)}) = -285.8 \, \text{kJ/mol}\)
\(\Delta H_f^\circ(\ce{O2(g)}) = 0 \, \text{kJ/mol}\) (element in standard state)

Solution:

\[ \Delta H^\circ_{\text{rxn}} = [(-393.5) + 2(-285.8)] - [(-74.8) + 0] \]

\[ = (-393.5 - 571.6) - (-74.8) = -965.1 + 74.8 = -890.3 \, \text{kJ/mol} \]

Answer: \(\Delta H^\circ_{\text{rxn}} = -890.3 \, \text{kJ/mol}\)

Question: Estimate the enthalpy change for the reaction:

\(\ce{H2(g) + Cl2(g) -> 2HCl(g)}\)

Given bond enthalpies:

H–H = 436 kJ/mol
Cl–Cl = 243 kJ/mol
H–Cl = 431 kJ/mol

Solution:

Bonds broken: 1 H–H + 1 Cl–Cl = 436 + 243 = 679 kJ/mol

Bonds formed: 2 H–Cl = 2 × 431 = 862 kJ/mol

\[ \Delta H = \text{Bonds broken} - \text{Bonds formed} = 679 - 862 = -183 \, \text{kJ/mol} \]

Answer: \(\Delta H = -183 \, \text{kJ/mol}\)