The enthalpy of formation (in kcal/mol) of C 2 H 4 (g) is

2C + 2H 2 → C 2 H 4 . Δ f H = [2(-94.0) + 2(-68.3)] - (-337.2) = -188.0 - 136.6 + 337.2 = +12.6 kcal/mol. Ans (d).

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Passage / Comprehension

Study the following thermochemical equations:

(a) C₂H₆(g) → C₂H₄(g) + H₂(g); ΔH = +32.7 kcal

(b) C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(l); ΔH = -337.2 kcal

(d) C(graphite) + O₂(g) → CO₂(g); ΔH = -94.0 kcal

The enthalpy of formation (in kcal/mol) of C₂H₄(g) is

Answer:

Solution:

2C + 2H₂ → C₂H₄. Δ_f H = [2(-94.0) + 2(-68.3)] - (-337.2) = -188.0 - 136.6 + 337.2 = +12.6 kcal/mol. Ans (d).

Stream:JEESubject:ChemistryTopic:ThermochemistrySubtopic:Enthalpy of Formation

⏱ 2mℹ️ Source: QB

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