Using the following thermochemical data, calculate delta H° for the following reaction...

Hi! You are thinking correctly. You would have to divide both equations in half and add them together to get the desired chemical equation. As a result, the delta H values would also be cut in half. Also, since you have to flip the first equation around, the sign is also flipped, so it becomes +1929.4kJ/mol. After all the calculations, the final answer should be 215.7 kJ/mol, which is answer choice c.

That is quite tricky, but it's not different from the previous exercise.

You just have to reverse - multiply - divide the equations given, so that, the sum of the equations will give us the desired equation.

Lets do it step by step
The desired equation is:
SmF3 (s) + 3HCl(g) >>> SmCl3(s) + 3HF(g)

The reactants are: SmF3 (s) + 3HCl(g).
As you see, the substance SmF3 (s) is only found at the 1st equation: 2Sm(s) + 6HF(g) >> 2SmF3 (s) + 3H2(g)
And, since the 2SmF3 (s) must be on the reactants (and not the products where it currently is) we must reverse the reaction:
2SmF3 (s) + 3H2(g)  >>   2Sm(s) + 6HF(g)  ΔH° = +1929.4 kJ/mol

In addition, in the desired equation only have one SmF3 (s), so we also need to divide the 1st equation by 2:
SmF3 (s) + (3/2)H2(g)  >>   Sm(s) + 3HF(g) ΔH° =+1929.4/2= 964.7 kJ/mol



Now, lets go to the products. The products of the desired equation are: SmCl3(s) + 3HF(g). As you see, SmCl3(s) is only present at the 2nd equation:
2Sm(s) + 6HCl(g) >> 2SmCl3 (s) + 3H2(g)
This time, the SmCl3 (s) is already at the products, so we won't reverse the equation. However, we only want one SmCl3 so we will divide the 2nd equation with 2:
Sm(s) + 3HCl(g) >> SmCl3 (s) + (3/2)H2(g) ΔH = –1498.0/2= -749 kJ/mol


So, finally, we have these equations:
SmF3 (s) + (3/2)H2(g)  >>   Sm(s) + 3HF(g) ΔH° =+1929.4/2= 964.7 kJ/mol
Sm(s) + 3HCl(g) >> SmCl3 (s) + (3/2)H2(g) ΔH = –1498.0/2= -749 kJ/mol

We will just add them, and pray to get the desired equation:
SmF3 (s) + (3/2)H2(g) +  Sm(s) + 3HCl(g) >> Sm(s) + 3HF(g)  +  SmCl3 (s) + (3/2)H2(g)

So, we have
SmF3 (s) + 3HCl(g) >>> SmCl3(s) + 3HF(g) and we add the ΔH from the equations we made:  ΔH = 964.7 - 749 = +215.7 kJ/mol

Soo shout out to you two! kappariin thank you for the reply and reassurance!

And Alexx thank you once again, the break down is really helpful.

Could you clarify why 3HCl(g) and 3HF(g) does not effect the delta H in the final equation?

The 3HCl(g) and 3HF(g) are part of the 2nd and 1st equation respectively. I guess you meant to ask why we did the calculations only based to SmF3(s) and SmCl3(s), and we didn't mention anything about 3HCl(g) and 3HF(g) ?

Well, in order to get the desired equation, we needed to put the SmF3(s) and SmCl3(s) of the separate equations into the correct place, and with the correct factors (by reversing/ dividing).
By doing that, we expected that all the other members of the equations (like the 3HCl(g) and 3HF(g)) would be also modified correctly so that they would give the desired result when adding the equations.
Is it certain that HCl(g) and HF(g) (and all other members) will be modified correctly to give the desired result? No, but if adding the equations doesn't give the desired result after these modifications, it would mean that the exercise cannot be solved. That is because, we already did the needed modifications for SmF3(s) and SmCl3(s), so any further modification would alter these members and the final equation would still be incorrect.
So, as I said earlier, you just focus your attention to one member per equation (in our case SmF3 (s) for the 1st and SmCl3 (s) for the 2nd) and, when finishing the modifications, you just expect that the result will be the final equation.

So, in our case, we choose SmF3 (s) SmCl3 (s). Could we make the divisions/reverses based to another member instead? Yes, we could also use other members, as long as these members are not present at the other equation. Based on that, we could not use 2Sm(s) or 3H2(g) for our calculations since they are present to both the equations (even if it was possible, it would make the calculations very complicated). But we could use 3HCl(g) and  3HF(g) with the same way we used SmF3 (s) SmCl3 (s).

Ahh yes, exactly what i was asking!

that cleared some of the confusion,

Thank you