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

The correct answer is -137 KJ

Step-by-step 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