Given thath 2 (g)+ f 2 (g)⟶2hf(g) δ h ∘ rxn =−546.6 kj2 h 2 (g)+ o 2 (g)⟶2 h 2 o(l) δ h ∘ rxn =−571.6 kjcalculate the value of δhrxn∘ for2 f 2 (g)+2 h 2 o(l)⟶4hf(g)+ o 2 (g)δhrxn°= kj

-521.6 kJ

Explanation:

So we have:

H2 + F2 --> 2HF where ΔH = -546.6 kJ

2H2 + O2 --> 2H20 where ΔH= -571.6 kJ

Then what is:

2F2 + 2H20 --> 4HF + O2 : ΔH = ?

So to do this we have to rearrange the above equations and then combine them using Hess's Law

So first we need to look at a few things:

In the equation we need and from the one we have we have HF on the correct side but it only has a coefficient of 2. We need the HF in H2 + F2 --> 2HF to have a coefficient of 4. So we need to multiply this whole reaction by 2.

Giving us:

2H2 + 2F2 --> 4HF

Remember this also applies to the ΔH so now we have:

2H2 + 2F2 --> 4HF where ΔH = -546.6 kJ x 2 (we will evaluate this later)

So now we are set with the top equation.

Looking at the second one we see that our 2H20 is on the wrong side, we need it to be a reactant rather than a product. To do this we flip the equation and then multiply our ΔH by -1. So:

2H20 --> 2H2 + O2 where ΔH = -571.6 kJ x (-1)

So now lets take another look at what we have:

2H2 + 2F2 --> 4HF where ΔH = -546.6 kJ x 2

2H20 --> 2H2 + O2 where ΔH = -571.6 kJ x (-1)

Now we can see that if we added these two equations we would get what we wanted! This is because our 2H2 would cancel and it is a product on one side and a reactant on the other.

So we now have:

2H20 + 2F2 --> 4HF + O2

We can now evaluate and add together our ΔH's

So we have ΔH   = (-546.6 x 2) + (-571.6 x -1)

This gives: -1093.2 + 571.6

Which leads to: -521.6 kJ

So our ΔH for this reaction is -521.6 kJ

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