Question

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)

Answer 1

answer is -1300 on edge 2021

Step-by-step explanation:

Answer 2

- 1300 kJ.

Step-by-step explanation:

• We should modify the given two reactions to reach to the final reaction:

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.

• If we add the two reactions after modification, we get:

P₄O₆(s) → P₄O₁₀(s).

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