Given the standard enthalpy changes for the following two reactions

Question

Given the standard enthalpy changes for the following two reactions:

(1) 2C(s) + 2H₂(g)C₂H₄(g)...... ΔH° = 52.3 kJ

(2) 2C(s) + 3H₂(g)C₂H₆(g)......ΔH° = -84.7 kJ

what is the standard enthalpy change for the reaction:

(3) C₂H₄(g) + H₂(g)C₂H₆(g)......ΔH° = ?

 kJ

Answer 1

Use Hess's Law of Summation to determine ΔH° for reaction (3).

In reaction 3, C₂H₄ has to be on the reactant side, but in reaction 1, C₂H₄ is on the product side. Therefore, reaction 1 needs to be reversed. Since the forward reaction is endothermic (positive ΔH°), the reverse reaction will be exothermic (negative ΔH°).

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

In reaction 3, C₂H₆ is on the product side, therefore, reaction 2 will remain as-is. Now add both reactions together.

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

2C (s) + 3H₂ (g) → C₂H₆ (g) ΔH° = -84.7 kJ

2C will cancel since they are on opposite sides of the equation. Cancelling 2H₂ will leave H₂ on the reactant side. Net equation is:

C₂H₄ (g) + H₂ (g) → C₂H₆ (g) ΔH° = -52.3 kJ + (-84.7 kJ) = -137.0 kJ