Enthalpy of Reactions Hess's Law

Question

A. Given the standard enthalpy changes for the following two reactions:

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

(2) C₂H₂(g)2C(s) + H₂(g)......ΔH° = -226.7 kJ

what is the standard enthalpy change for the reaction:

(3) 2C₂H₂(g) + 3H₂(g)C₄H₁₀(g)......ΔH° = kj?

Answer 1

2C₂H₂(g) + 3H₂(g) ==> C₄H₁₀(g) ... TARGET EQUATION

Given:

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

(2) C₂H₂(g) ==> 2C(s) + H₂(g) ..... ΔH° = -226.7 kJ

Procedure:

copy (2)x2: 2C₂H₂(g) ==>4C(s) + 2H₂(g) ..... ΔH° = -453.4 kJ

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

Add them together...

2C₂H₂(g) + 4C(s) + 5H₂(g) ==> 4C(s) + 2H₂(g) + C₄H₁₀(g)

Cancel and/or combine like items to get ...

2C₂H₂(g) + 3H₂(g) ==> C₄H₁₀(g) ... TARGET EQUATION

∆H = sum the ∆H values...

-453.4 kJ + -125.6 kJ = -579 kJ = ∆H