Answer:
The correct option is;
By first reversing Reaction 2 and then adding the enthalpy of Reaction 1 and the enthalpy of reaction 2
Explanation:
The given reactions are;
1) C₂H₂ (g) + 2H₂ (g) → C₂H₆ (g) [tex]{}[/tex] ΔH = -75 kcal
2) C₂H₄ (g) + H₂ (g) → C₂H₆ (g) [tex]{}[/tex] ΔH = -33 kcal
3) C₂H₂ (g) + H₂ (g) → C₂H₄ (g) [tex]{}[/tex] ΔH = ?
By reversing reaction (2) and subtracting the reactants from the products we then have followed by adding the enthalpy of reaction 1 to the value of the enthalpy of the (reversed) reaction 2, we have;
C₂H₂ (g) + 2H₂ (g) → C₂H₆ (g) [tex]{}[/tex] ΔH = -75 kcal
C₂H₆ (g) → C₂H₄ (g) + H₂ (g) [tex]{}[/tex] ΔH = 33kcal
Adding, and cancelling out the italicized terms which are repeated in the product and reactants, we get
C₂H₂ (g) + H₂ (g) → C₂H₄ (g) [tex]{}[/tex] ΔH = -75 kcal + 33 kcal = -42 kcal
Therefore, the correct option is "by first reversing Reaction 2 and then adding the enthalpy of Reaction 1 and the enthalpy of reaction 2"
