Answer:
-1,300 kJ
Explanation:
P₄ + 3 O₂ ⇒ P₄O₆ ΔH = -1,640.1 kJ (EQ 1)
P₄O₁₀ ⇒ P₄ + 5 O₂ ΔH = 2,940.1 kJ (EQ 2)
These are the equations you are given. You need to make these equations into the equation below.
P₄O₆ + 2 O₂ ⇒ P₄O₁₀ (end equation)
Look at the product side of the end equation. You need to produce P₄O₁₀. In the EQ 2, P₄O₁₀ is on the reactant side. Flip the equation. Since you flipped the equation, the enthalpy will have the opposite sign.
P₄ + 5 O₂ ⇒ P₄O₁₀ ΔH = -2,940.1 kJ
On the reactant side of the end equation, you need P₄O₆ and 2 O₂. First, rearrange the equation so that P₄O₆ is on the right side. In EQ 1, P₄O₆ is on the product side. Flip the equation. Like the last one, the sign will change.
Now, cancel out all possible values. P₄ will cancel out since there is one on each side of the equation. Since there is 5 O₂ on one side and 3 O₂ on the other, subtract the two and put the remainder on the side of the larger value.
P₄ + 5 O₂ ⇒ P₄O₁₀ ΔH = -2,940.1 kJ
<u>P₄O₆ ⇒ P₄ + 3 O₂ ΔH = 1,640.1 kJ </u>
P₄O₆ + 2 O₂ ⇒ P₄O₁₀
This should be the resulting equation. Now, add the two enthalpies together to find the overall enthalpy.
-2,940.1 kJ + 1,640.1 kJ = -1,300 kJ
The overall enthalpy is -1,300 kJ.
