Q.25 For propene at 298 K, the molar enthalpy of hydrogenation is −124.27 kJ mol⁻¹ and the standard enthalpy of formation is 20.42 kJ mol⁻¹. For propane at 298 K, the standard enthalpy of formation in kJ mol⁻¹ is ______.

Correct Answer: The standard enthalpy of formation of propane at 298 K is
−103.85 kJ mol⁻¹ (≈ −104 kJ mol⁻¹).

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Question Restatement

For propene at 298 K, the following data are given:

• Molar enthalpy of hydrogenation, ΔH_hyd = −124.27 kJ mol⁻¹
• Standard enthalpy of formation of propene,
  ΔH_f^°(C₃H₆) = +20.42 kJ mol⁻¹

Find the standard enthalpy of formation of propane,
ΔH_f^°(C₃H₈) at 298 K.

This is a well-known thermodynamics problem frequently asked in GATE and Life Sciences examinations.

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Thermodynamic Concept Used

The hydrogenation reaction of propene is:

C₃H₆(g) + H₂(g) → C₃H₈(g)

The standard enthalpy change for this reaction is equal to the molar enthalpy of hydrogenation:

ΔH_rxn^° = ΔH_hyd = −124.27 kJ mol⁻¹

According to Hess’s law:

ΔH_rxn^° = ΣΔH_f^°(products) − ΣΔH_f^°(reactants)

For this reaction:

• Product: C₃H₈(g)
• Reactants: C₃H₆(g) and H₂(g)

Since hydrogen is an element in its standard state:

ΔH_f^°(H₂(g)) = 0

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Step-by-Step Calculation

Write the enthalpy expression:

ΔH_hyd = ΔH_f^°(C₃H₈)
− [ΔH_f^°(C₃H₆) + ΔH_f^°(H₂)]

Substitute the given values:

−124.27 = ΔH_f^°(C₃H₈) − (20.42 + 0)

Solving for ΔH_f^°(C₃H₈):

ΔH_f^°(C₃H₈)
= −124.27 + 20.42
= −103.85 kJ mol⁻¹

This agrees with standard tabulated values (≈ −104 kJ mol⁻¹).

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Explanation of Each Quantity

Molar Enthalpy of Hydrogenation (−124.27 kJ mol⁻¹)

This is the energy released when one mole of propene undergoes hydrogenation.
The negative sign indicates an exothermic process due to saturation of the C=C bond.

Standard Enthalpy of Formation of Propene (+20.42 kJ mol⁻¹)

This represents the enthalpy change when one mole of propene forms from graphite and hydrogen.
The positive value indicates that propene is less stable than its constituent elements.

Standard Enthalpy of Formation of Propane (−103.85 kJ mol⁻¹)

Corresponds to the reaction:

3C(graphite) + 4H₂(g) → C₃H₈(g)

The negative sign shows propane is thermodynamically more stable than its elements.

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Typical MCQ Options and Why Only One Is Correct

Option	Value (kJ mol−1)	Explanation
(A)	−124.27	This is the hydrogenation enthalpy, not an enthalpy of formation.
(B)	+20.42	This is the enthalpy of formation of propene, not propane.
(C)	−103.85	Correct value obtained using Hess’s law.
(D)	0	Only elements in their standard states have zero enthalpy of formation.

Final Answer:
ΔH_f^°(C₃H₈) = −103.85 kJ mol⁻¹