Given the standard enthalpy changes for the following two reactions:
(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° = ?
(3) 2C2H4(g) + H2(g) ----> C4H10(g)
ΔrH° = ΔfH°(C4H10) – 2ΔfH°(C2H4) – ΔfH°(H2) = ΔfH°(C4H10) – 2ΔfH°(C2H4) = –125.6 kJ – 2×52.3 kJ = –230.2 kJ.
ΔrH° = –230.2 kJ.
Comments
Leave a comment