Answer:
-1300. kJ
Explanation:
We have two equations:
1. P₄(s) +3O₂(g) ⟶ P₄O₆(s); ΔH₁ = -1640.1 kJ
2. P₄O₁₀(s) ⟶ P₄(s) + 5O₂(g); ΔH₂ = 2940.1 kJ
From these, we must devise the target equation:
3. P₄O₆(s) + 2O₂(g) ⟶ P₄O₁₀(s); ΔH = ?
The target equation has P₄O₆(s) on the left, so you reverse Equation 1.
When you reverse an equation, you reverse the sign of its ΔH.
4. P₄O₆(s) ⟶ P₄(s) +3O₂(g); ΔH₁ = 1640.1 kJ
Equation 4 has P₄ on the right. That is not in the target equation.
You need an equation with P₄ on the left, so you reverse Equation 2.
5. P₄(s) + 5O₂(g) ⟶ P₄O₁₀(s); ΔH₂ = -2940.1 kJ
Now, you add equations 4 and 5, cancelling species that appear on opposite sides of the reaction arrows.
When you add equations, you add their ΔH values.
You get the target equation 3:
4. P₄O₆(s) ⟶ P₄(s) + 3O₂(g); ΔH₁ = 1640.1 kJ
5. P₄(s) + 2(5)O₂(g) ⟶ P₄O₁₀(s); ΔH₂ = -2940.1 kJ
3. P₄O₆(s) + 2O₂(g) ⟶ P₄O₁₀(s); ΔH = -1300. kJ
ΔH for the reaction is -1300. kJ
