Enthalpy Change Using Hess's Law?

Question

Calculate the enthalpy change for the reaction

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

I'm trying to follow the steps in my textbook to do this, but it's just not making sense. The answer is -1300 kJ, but I can't figure out how to get there. If someone could explain it to me I'd really appreciate it

Thank you

given the following enthalpies of reaction:

P4(s) +3O2(g) --> P4O6(s) delta H = -1640.1kJ

P4(s)+5O2(g) --> P4O10(s) delta H = -2940.1kJ

Answer 1

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This video will explain it to you, clearly: http://www.youtube.com/watch?v=u7aTBxA7sL8

Answer 2

Final equation: P4O6(s) +2 O2(g) → P4O10(s)

Equation 1: P4(s) +3 O2(g) → P4O6(s) -------------------- ΔH1 = –1640.1kJ

Equation 2: P4(s) + 5 O2(g) → P4O10(s) ------------------ ΔH2 = –2940.1kJ

Since “P4O6(s)” appears on the right side of Equation 1, and on the left side of the final Equation, lets reverse Equation 1 and change the sign of the ΔH1:

Equation 1: P4O6(s) → P4(s) + 3 O2(g) -------------------- ΔH1 = +1640.1kJ

Equation 2: P4(s) + 5 O2(g) → P4O10(s) ------------------ ΔH2 = –2940.1kJ

Look at both equations at the same time. If there are like terms on both sides of either equation, subtract them. Since “P4(s)” appears on the right side in Equation 1, and on the left side in Equation 2, subtract them. Likewise “3 O2(g)” appears on the right side of Equation 1, and “5 O2(g)” appears on the left side of Equation 2. When subtracting “3 O2” from both sides, it disappears from Equation 1, but leaves “2 O2(g)” on the left side in Equation 2. Adding all remaining terms in both equations, leaves us with the final equation:

Final equation: P4O6(s) + 2 O2(g) → P4O10(s) ----------- ΔH = –1300.0 kJ

Answer: The enthalpy change for the reaction P4O6(s) + 2 O2(g) → P4O10(s) is –1300.0 kJ.

Source(s): personal knowledge