Given: Hsub [C(graphite)] = 710 kJ mol–1 C–HH = 414 kJ mol–1 H–HH = 436 kJ mol–1 C=CH = 611kJ mol–1 The Hf for CH2=CH2 is ____________ kJ mol–1 (nearest integer value)

Question

Prashant Kotian · Accepted Answer

Correct answer: .. # Solution: Finding ΔHf for Ethylene (CH₂=CH₂) **Step 1: Write the formation equation** C(graphite) + H₂(g) → CH₂=CH₂(g) **Step 2: Apply Hess's Law using bond energies** ΔHf = Energy required to break bonds - Energy released when forming bonds ΔHf = ΔHsub[C] + ΔH dissociation[H₂] - (bonds formed in C₂H₄) **Step 3: Identify bonds formed in ethylene** CH₂=CH₂ contains: - 1 C=C double bond (611 kJ/mol) - 4 C–H single bonds (4 × 414 = 1,656 kJ/mol) - Total bonds formed = 611 + 1,656 = 2,267 kJ/mol **Step 4: Identify bonds broken** - 1 C atom sublimated: 710 kJ/mol - 1 H–H bond broken: 436 kJ/mol - Total bonds broken = 710 + 436 = 1,146 kJ/mol **Step 5: Calculate ΔHf** ΔHf = 1,146 - 2,267 = -1,121 kJ/mol **Step 6: Round to nearest integer** ΔHf ≈ **-1,121 kJ/mol** (or **-1,120 kJ/mol** depending on rounding) Therefore, the answer is **-1,121** kJ mol⁻¹ (or **-1,120** kJ mol⁻¹)

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