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.

sorry i was late, but i don't know what assignment your talking about, i'm in flvs as well, im in earth space science. but u are behind pace bc every class should be in the 3's or 4'

the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be