Enthalpy change during a reaction is the same regardless of the number of steps.
Example:
Single step: A → D, ΔH = 24 kJ.
Multiple steps: A → B → C → D, where ΔH = ΔH1 + ΔH2 + ΔH3 (12 kJ + 8 kJ + 4 kJ = 24 kJ).
Second Statement:
Total enthalpy change in a cyclic process is zero.
Mathematically, ∑ΔH = 0.
Important Example: Formation of Carbon Dioxide
One-Step Reaction:
C (s) + O2 (g) → CO2 (g)
ΔH = -393.7 kJ/mole (exothermic).
Two-Step Reaction:
Step 1: C (s) + 1/2 O2 (g) → CO (g), ΔH1 = -111.052 kJ/mole.
Step 2: CO (g) + 1/2 O2 (g) → CO2 (g), ΔH2 = -283 kJ/mole.
Combined: ΔH = ΔH1 + ΔH2 = -393.052 kJ/mole.
Importance of Studying Hess's Law
Used for reactions where enthalpy change cannot be measured directly with a calorimeter (e.g., formation of CCl4).
Fast Method for Numerical Problems in Hess's Law
Standard Enthalpy of Formation of Methanol (CH3OH)
Components: C (s), H2 (g), O2 (g).
Balanced Reaction: C + 2H2 + 1/2 O2 → CH3OH.
Reverse reactions as needed, changing enthalpy signs.
Combine equations, canceling spectator molecules, and calculate ΔH.
Result: ΔH = -239 kJ.
Difficult Numerical Problem
Required reaction: Fe2O3 + 3CO → 2Fe + 3CO2.
Use enthalpy changes from given reactions, modifying as needed (e.g., reversing, multiplying).
Result: ΔH = +6 kJ.
Conclusion
Hess's Law provides a systematic approach for calculating enthalpy changes for reactions through various pathways, reinforcing the principle of energy conservation within chemical processes.