Answer to Question #87470 in General Chemistry for jay

Question

(1) 4C(s) + 5H2(g)C4H10(g)...... ΔH° = -125.6 kJ

(2) C2H4(g)2C(s) + 2H2(g)......ΔH° = -52.3 kJ

what is the standard enthalpy change for the reaction:

(3) 2C2H4(g) + H2(g) --------> C4H10(g) ΔH° = ?

_______ kJ

Answer 1

The standard enthalpy of formation of elementary substance is zero.

(1)         4C(s) + 5H₂(g) = C₄H₁₀(g)...... Δ_rH° = –125.6 kJ

Δ_rH° = Δ_fH°(C₄H₁₀) – 4 Δ_fH°(C) – 5 Δ_fH°(H₂) = Δ_fH°(C₄H₁₀) = –125.6 kJ.

Δ_fH°(C₄H₁₀) = –125.6 kJ.

(2)        C₂H₄(g) = 2C(s) + 2H₂(g)......ΔH° = –52.3 kJ

Δ_rH° = 2Δ_fH°(C) + 2 Δ_fH°(H₂) – Δ_fH°(C₂H₄) = – Δ_fH°(C₂H₄)  = –52.3 kJ

Δ_fH°(C₂H₄)  = 52.3 kJ.

(3)        2C₂H₄(g) + H₂(g) ----> C₄H₁₀(g)

Δ_rH° = Δ_fH°(C₄H₁₀) – 2Δ_fH°(C₂H₄) – Δ_fH°(H₂) = Δ_fH°(C₄H₁₀) – 2Δ_fH°(C₂H₄) = –125.6 kJ – 2×52.3 kJ = –230.2 kJ.

Δ_rH° = –230.2 kJ.