Standard enthalpy of formation

Chemical Thermodynamics: Standard enthalpy of formation

The standard enthalpy of formation (ΔHf°) for a reaction is the enthalpy change that occurs when 1 mol of a substance is formed from its component elements in their standard states.

When we say “The standard enthalpy of formation of methanol, CH₃OH(l) is –238.7 kJ”, it means:

C(graphite) + 2 H₂(g) + ½ O₂(g) CH₃OH(l)

has a value of ΔH of –238.7 kJ.Likewise, for ethanol ∆H_f° [C2H5OH(l)] = – 279 kJ/mol means that when the reaction below is carried out, 279 kJ of energy is released. We may treat ΔH_f° values as though they were absolute enthalpies, in order to determine enthalpy changes for reactions. Note that standard enthalpy of formation for an element under standard conditions is 0.Standard conditions refer to a pressure of 1 atm and a specified temperature of 298K (25°C).The standard state of oxygen at 1 atm is O₂(g) and for nitrogen, N₂(g).The stoichiometry for formation reactions indicates the formation of 1 mole of the respective compound. Hence enthalpies of formation are always denoted with kJ/mol.

The process of calculation of standard enthalpy of formation is as follows:

∆H°_rxn = ∑n_p x ∆H_f°(products) –∑n_r x ∆H_f°(reactants)

Where the symbol ‘∑’ signifies the summation of several variables. The symbol ‘n’ signifies the stoichiometric coefficient used in front of a chemical symbol or formula.

In other words

1. Add all the values for ΔH_f° of the products.
2. Add all the values for ΔH_f° of the reactants.
3. Subtract #2 from #1

Example:

Calculate a value for the standard enthalpy of formation of propanone, CH₃COCH₃(l), where the following are standard enthalpy changes of combustion:

ΔHc/kJmol^–1for C(s) = –394kJmol^–1, for H₂(g)= –286kJmol^–1andCH₃COCH₃(l) = –1821kJmol^–1

For the enthalpy change, write an equation representing this change including its state symbols. For the formation of propanone, one mol of the propanone should form the main equation.

3C(s) + 3H₂(g) + 0.5O₂(g) → CH₃COCH₃(l)

The reactants in this equation C(s) and H₂(g)can be burned to produce CO2(g) and H2O(l).

Other equations related to this reaction are:

3C(s) + 3O₂(g) → 3CO₂(g) ;nΔH= 3(–394)

where 3 is the stochiometric coefficient of carbon in the main equation.

3H₂(g) + 1.5O2(g) → 3H₂O(l) ;nΔH= 3(–286)

where 3 is the stochiometric coefficient of H₂ in the main equation.

Also, 1 mole of propanone is formed in the main equation so the third equation is be reversed. The sign of ΔH is also changed since the equation is reversed.

3CO₂(g) + 3H₂O(l) → CH₃COCH₃(l) + 4O₂(g) ;nΔH=+1821

When we add these equations together and cancel down the moles of any substance that appears on both sides of the equation, it results in the main equation and the total of the enthalpy changes gives the enthalpy change for the main equation.