Given thath2(g) + f2(g) -> 2hf(g) => ∆h = -546.6 kj . mol-12h2(g) + o2(g) -> 2h20(l) => ∆h = -571.6 kj. mol-1calculate the value of ∆h for: 2f2(g) + 2h20(l) -> 4hf(g) + o2(g)

The correct answer is -521.6 KJ/mol

Explanation:

In order to solve the problem, we have to rearrange and to add the reactions with the aim to obtain the requested reaction.

We have:

1) H₂(g) + F₂ (g) → 2HF(g) => ∆H₁ = -546.6 kJ/mol

2) 2H₂(g) + O₂(g) → 2H₂0(l) => ∆H₂ = -571.6 kJ/mol

If we multiply reaction 1 by 2 and add the inverse reaction of 2, we obtain:

2H₂(g) + 2F₂ (g) → 4HF(g) => 4 x ∆H₁ = 4 x (-546.6 kJ/mol)

2H₂0(l) → 2H₂(g) + O₂(g) => (-1) x ∆H₂ = (-1) x (571.6 kJ/mol)

2H₂(g) + 2F₂ (g)+  2H₂0(l) → 4HF(g) + 2H₂(g) + O₂(g)

We eliminate 2H₂(g) is repeated in both members of the reaction, and we obtain: 2F₂(g) + 2H₂0(l) -> 4HF(g) + O₂(g)

Finally, the ΔH of the reaction is calculated as follows:

ΔHtotal= (2 x ΔH₁) + ((-1)ΔH₂

ΔHtotal= (2 x (-546.6 KJ/mol)) + ((-1) x (-571.6 KJ/mol)

ΔHtotal= -521.6 KJ/mol

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.