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 should reverse the first reaction and so multiply its ΔH by (- 1):

P₄O₆(s) → P₄(s) + 3O₂(g), ΔH₁' = 1640.1 kJ.

The second reaction also should be reversed and so multiply its ΔH by (- 1):

P₄(s) + 5O₂(g) → P₄O₁₀(s), ΔH₂' = - 2940.1 kJ.

P₄O₆(s) → P₄O₁₀(s).

∴ ΔH = ΔH₁' + ΔH₂' = 1640.1 kJ + (- 2940.1 kJ) = - 1300 kJ.