Consider the chemical equations shown here. P4(s) + 302(g) - P406(S) AH1 = -1,640.1 kJ
P4010(s) - P4(s) + 5O2(g) AH2 = 2,940.1 kJ
What is the overall enthalpy of reaction for the equation shown below?
Round the answer to the nearest whole number.
P406(S) + 2O2(g) → P4010(s)
DONE

-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

P₄O₆  ⇒  P₄  +  3 O₂     ΔH = 1,640.1 kJ    

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.

they are all solids.