5.4 Standard Enthalpy of Formation

Standard Enthalpy of Formation:

The standard enthalpy of formation is recognized as a change in enthalpy when one mole of a substance in the standard state (1 atm of pressure and 298.15 K) is formed from its pure elements under the same conditions.

The equation for the standard enthalpy of formation id given below:

ΔH^oreaction  =  ∑ΔH^of(products)  −  ∑ΔH^of(Reactants)

Remember that the change in enthalpy is considering the enthalpy as a State Function.

Examples:

1. Consider the two molecules Cl₂(g) and Cl₂(l) at 298.15 K, which substance has a nonzero standard enthalpy of formation?

Cl₂(g) is most stable. Therefore Cl₂(g) will have the lower enthalpy and ΔH^o_f  = 0.

Cl₂(l) is least stable and therefore will have the highest enthalpy and ΔH^o_f has a nonzero value.

2. Calculate the ΔH^o_(reaction) for the reaction below:

2  SO₂      +     O₂    ⟶   2  SO₃

ΔH^oreaction  =  ∑n_pΔH^of(products)  −  ∑n_rΔH^of(Reactants)

n_p  =  number of moles of products

n_r = number of moles of the reactants

ΔH^oreaction  = { 2 x ΔH^of[SO₃]  –  [2 x ΔH^of[SO₂]  +  ΔH^of[O₂]

Using the table below of the standard enthalpy of the formation:

ΔH^of[SO₃] = – 395.2 kJ / mole

ΔH^of[SO₂] = -296.1 kJ / mole

ΔH^of[O₂] = 0 kJ / mole

ΔH^oreaction  = { 2 mole x ΔH^of[SO₃]  –  [2 mole x ΔH^of[SO₂]  +  ΔH^of[O₂]]

ΔH^oreaction ={ – 2 x 395.2 kJ/mole} – {[( – 2 x 296.1 kJ/mole) + 0]} = – 790.4 + 592.2 = – 198.2 kJ

ΔH^oreaction has negative value and hence the reaction is exothermic.

3. Calculate the ΔH^o_(reaction) for the reaction below:

2  NO      +     O₂    ⟶   2  NO₂

ΔH^oreaction  =  ∑n_pΔH^of(products)  −  ∑n_rΔH^of(Reactants)

n_p  =  number of moles of products

n_r = number of moles of the reactants

Using the table below of the standard enthalpy of the formation:

ΔH^of[NO₂] = 33.18 kJ / mole

ΔH^of[NO] =  90.25 kJ / mole

ΔH^of[O₂] = 0 kJ / mole

ΔH^oreaction  = { 2 mole x ΔH^of[NO₂]  –  [2 x ΔH^of[NO]  +  ΔH^of[O₂]]

ΔH^oreaction ={2 mole x 33.18  kJ/mole} – {[(  2 mole x 90.25 kJ/mole) + 0]} = 66.36 – 180.50 = – 114.14 kJ

ΔH^oreaction has negative value and hence the reaction is exothermic.

The videos below illustrate the concept of the standard enthalpy of formation as well as go over some example calculations:

Enthalpies of Formation – Chemistry Tutorial

Enthalpy of Formation Reaction & Heat of Combustion, Enthalpy Change Problems Chemistry

Standard Enthalpy of Formation for various Compounds.

Reference: http://sistemas.eel.usp.br/docentes/arquivos/5817712/TDQ%20I/R-standard_enthalpy_of_formation.pdf

Writing chemical Equations for ∆ H^o_f

Examples:

Write ∆ H^o_f for the reactions below:

1. CH3-O-CH₃(l)

2 C(solid, graphite)  +  ½ O₂(gas)    +   3 H₂(g)    ⟶   CH3-O-CH₃(l)

2. CH3-CO-CH₃(l)

3 C(solid, graphite)  +  ½ O₂(gas)    +   3 H₂(g)    ⟶   CH3-CO-CH₃(l)

Evaluating an Enthalpy of Formation

Hydrogen gas, H₂, reacts explosively with gaseous chlorine, Cl₂, to form hydrogen chloride, HCl(g). What is the enthalpy change for the reaction of 1 mole of H₂(g) with 1 mole of Cl₂(g) if both the reactants and products are at standard state conditions? The standard enthalpy of formation of HCl(g) is −92.3 kJ/mol.

H₂(g) + Cl₂(g) ⟶ 2 HCl(g)                      ΔH°_f   =   –   92.3 kJ /mole                        

1 mole of H₂(g) will produce 2 moles of HCl(g)

Also 1 mole of Cl₂(g) will produce 2 moles of HCl(g)

[1 mole H₂(g)]  x  2 mole HCl(g) / 1 mole H₂(g)   = 2 mole HCl(g)

[2 mole HCl(g)]  x   [ – 92.3 kJ / mole H₂(g)]  =     –  184.6 kJ