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

The correct answer is - 137 KJ

Explanation:

In order to solve the problem we have to use the inverse of reaction 1, because the product of this reaction (C₂H₄) is a reactant in the reaction we need. When we use inverse reactions, the enthaphy is multiplied by - 1. Then, we add to reaction 1 the reaction 2 and we sum the enthalphy values.

C₂H₄ (g) → 2C (s) + 2H₂ (g) ΔH° = (-1) x 52.3 kJ

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

As the term 2H₂ (g) is both in the right side and the left side of the equation, we can cancel it. The resultant equation is C₂H₄ (g) + H₂ (g) → C₂H₆ (g)

The total change of standard enthalphy is:

ΔHºtotal = - 52.3 kJ - 84.7 kJ = - 137 kJ