Question

Calculate the enthalpy of formation of HI in the given reaction:

H₂ + I₂ → 2HI ΔH = 20.40 kcal/mol

1. 20.40 kcal/mol
2. 10.20 kcal/mol
3. (-) 10.20 kcal/mol
4. (-) 20.40 kcal/mol

Answer 1

Correct Answer - Option 2 : 10.20 kcal/mol

Concept:

Enthalpy of formation (Heat of formation):

• It is defined as the enthalpy change when one mole of a compound is formed from its elements.
• It is designated as ΔH_f.

_​∆H_f = ΔHf (products) - ΔHf (reactants)  

Where, ΔHf (products) = Enthalpy of formation of the product, ΔHf (reactants) = Enthalpy of formation of reactant

Standard enthalpy of formation:

• The standard enthalpy change for the formation of one mole of a compound from its elements in their most stable states of aggregation is called Standard Molar Enthalpy of Formation.
• Its symbol is ΔH_f^o.
• The enthalpy of formation of all free elements in their standard states is regarded as zero.

i.e., ΔHfo = ΔH_f^o _(products) - ΔHfo  _(reactants) 

Where, ΔHfo = ​Standard enthalpy of a formation, ΔHfo (products) = Standard enthalpies of formation of all products, ΔHfo  (reactants) = Standard enthalpies of formation of all reactants 

Calculation:

Given: H₂ + I₂ → 2HI   ; ΔH = 20.40 kcal/mol

We know that, ∆Hf = ΔHf (products) - ΔHf (reactants) 

⇒ ∆H_f = [2×∆H_f (HI) - ∆H_f (H₂) - ∆H_f (I₂)]

The enthalpy of formation of all free elements in their standard states is regarded as zero.

Therefore, ∆H(H2) = 0 and ∆H(I2) = 0

⇒ 20.40 kcal = 2×∆H_f (HI)

⇒ ∆H_f (HI) = 20.40/2

⇒ ∆Hf (HI) = 10.2 kcal/mol

Hence, the enthalpy of formation of HI is 10.2 kcal/mol.