Consider the chemical equations shown here.

P4 (s) + 3O2 (g) - -->P4O6 (s) ΔH1 = - 1,640.1 kJ

P4O10 (s) → P4 (s) + 5O2 (g) ΔH2 = 2,940.1 kJ

What is the overall enthalpy of reaction for the equation shown below?

Round the answer to the nearest whole number.

P4O6 (s) + 2O2 (g) - -->P4O10 (s)

-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