In most cases, the enthalpy changes of a reaction cannot be found by experiments using calorimetry.
- For example, it is impossible to burn carbon in oxygen and solely form carbon monoxide. Thus, the enthalpy change in the formation of carbon monoxide cannot be determined directly.
- So, we use a different approach to calculate the enthalpy change of a reaction.
Hess’s law states that, as long as the initial and final conditions are the same, the total enthalpy change in a chemical reaction is independent of the route by which the chemical reaction takes place. Hess’s law is based on the conservation of energy.
- Assuming that A and B react to give a product C. Product C can also be obtained by going through intermediate products D and E.
  - Using Hess’s law,
- The enthalpy cycle is a pictorial representation showing the alternative routes of reaction between reactants and products.
Enthalpy change of formation of butane
- Data of a standard enthalpy of combustion.
- Here, carbon and hydrogen can be combusted to form carbon dioxide and water. Butane also forms carbon dioxide and water when combusted.
- Drawing an enthalpy cycle and applying Hess’s law.
  - First, write the equation for the enthalpy change of formation at the top.
  - Then, write down the combustion products at the bottom and add oxygen on both sides of the equation to balance the combustion reactions.
  - Next, balance all the elements and compounds.
  - After that, draw arrows, making sure they point in the correct directions.
  - Now, applying Hess’s law, the heat of formation of butane
    - The negative sign here is due to the arrow pointing in the opposite direction.
  - Now,
  - Next, we calculate ΔH₂ which is the enthalpy of combustion of butane.
  - Using Hess’s law,
The standard enthalpy change of formation of an element in its standard state is always zero.
- Example of how to solve a problem:
  Enthalpy change of the reaction,
Bond enthalpy
- Bond enthalpy values represent the energy required to break one mole of covalent bond.
- Numerical problem:
  
  For this reaction, the bond enthalpies values are:
  E(N ≡ N) = 944 kJmol^-1
  E(H-H) = 436 kJmol^-1
  E(N-H) = 390 kJmol^-1
  - The energy cycle is:
    
    Applying Hess’s law.
    ∆H_r = ∆H₁– ∆H₂
    = (944 + 3* 436) – (2*3*390)
    = -88kJmol^-1

5.03 - Hess’s law

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