Question

The enthalpy changes of these reactions can be measured:

a) C₂H₄ (g) + 3O₂ (g) → 2 CO₂ (g) + 2 H₂O (l) ΔH₁ = -1411.1 kJ
b) C₂H₅OH (l) + 3O₂ (g) + 2 CO₂ (g) + 3 H₂O (l) ΔH₂ = -1367.5 kJ
Use these values and Hess's Law to determine the enthalpy change for the reaction C₂H₄ (g) + H₂O (l) + C₂H₅OH (g).

Answer 1

Using Hess's Law and reversing the second provided reaction, we find that the enthalpy change for the reaction C₂H₄ (g) + H₂O (l) → C₂H₅OH (g) is -43.6 kJ.

Step-by-step explanation:

To find the enthalpy change for the reaction C₂H₄ (g) + H₂O (l) → C₂H₅OH (g), we use Hess's Law and the given reactions:

• 
  
• (a) C₂H₄ (g) + 3O₂ (g) → 2 CO₂ (g) + 2 H₂O (l), ΔH₁ = -1411.1 kJ
• 
  
• (b) C₂H₅OH (l) + 3O₂ (g) → 2 CO₂ (g) + 3 H₂O (l), ΔH₂ = -1367.5 kJ

To find the enthalpy for the target reaction, perform the following steps:

1. 
   
3. Reverse reaction (b) so C₂H₅OH is on the reactant side and multiply its enthalpy change by -1.
4. 
   
6. Subtract the enthalpy change of reaction (a) from the reversed reaction (b) to find the enthalpy change for the target reaction.

After reversing reaction (b), the enthalpy change becomes +1367.5 kJ. Subtracting ΔH₁ from this gives us the enthalpy change for the target reaction:
+1367.5 kJ - (-1411.1 kJ) = -43.6 kJ.

Therefore, the enthalpy change for the reaction C₂H₄ (g) + H₂O (l) → C₂H₅OH (g) is -43.6 kJ.