[brinraeven]

I've done several of these problems before and done fine, but I can't figure out this one, at least if the practice test answer sheet is correct:

12. Given the following reactions with enthalpy of reaction, solve for the missing value of enthalpy:

6 H2O(l) + 6 H2(g) + 3 O2(g)                   ΔH = ?

CO2(g) --> C(s) + O2(g)                                   ΔH = 221.3 kJ
C2H6(g) --> 2 C(s) + 3 H2(g)                           ΔH = 47.6 kJ
2 C2H6(g) + 7 O2(g) --> 4 CO2(g) + 6 H2O(l)           ΔH = −1755.6 kJ

To get my answer, I multiplied 221.3 by 3 (663.9) to get 3O2. I also multiplied 47.6 (95.2) by 2 to get 6H2. Lastly, I reversed the third formula to get +1755.6.

663.9+95.2+1755.6 = 2514.7

My answer is 2514.7, but the answer sheet says 956.6. Would you help me figure out what I'm doing wrong?

Thanks,

Steph

[Enthalpy]

Your logic is hard to follow, so getting wrong results isn't so surprising. For instance, 47.6kJ is the heat for a reaction, not for H₂, so doubling it won't suffice to inject in the unknown line.

More fundamentally, there is no ΔH associated with a species or a group of species as you try to write in the text. ΔH is associated with a transformation, here a chemical reaction, as in the title. So much that, when tables give "enthalpies of formation", they tell explicitly what the reference is - most often, the transformation is the formation of the compound, starting from the elements in their normal state (molecules of solid, liquid, gas), and often the heat is given for elements and compounds at 298K.

----------

  -47.6kJ is the heat of formation of ethane from the elements
-1755.6kJ is its heat of combustion

From both, you can extract the heat of formation of the combustion products together.

-221.3kJ is the heat of formation of CO₂, so you also deduce the heat of formation of H₂O.

Reversing and multiplying the heat of formation of H₂O, you get the heat of the unknown transformation.

[brinraeven]

Wow, do you always insult people who are genuinely trying to learn? I believe you are talking about Hess's law in reference to determining enthalpies of reaction  from standard enthalpies of formation, while I was tasked with solving the enthalpy of the reaction by summing the series of steps (i.e., the sum of the heats of the reactions).

So, my TA got back to me, and I was actually right to multiply the second formula by 2, but I needed to multiply the first formula by -4 to cancel 4CO2.

Thanks for your input.

[Borek]

I don't see insults in Enthalpy's post, but I see where the problem is. You say

I multiplied 221.3 by 3 (663.9) to get 3O2

when you mean "I multiplied the first equation by three to get 3O₂ on the right side". Subtle difference, but I see why it can be confusing.

[cseil]

I've done several of these problems before and done fine, but I can't figure out this one, at least if the practice test answer sheet is correct:

12. Given the following reactions with enthalpy of reaction, solve for the missing value of enthalpy:

6 H2O(l) + 6 H2(g) + 3 O2(g)                   ΔH = ?

CO2(g) --> C(s) + O2(g)                                   ΔH = 221.3 kJ
C2H6(g) --> 2 C(s) + 3 H2(g)                           ΔH = 47.6 kJ
2 C2H6(g) + 7 O2(g) --> 4 CO2(g) + 6 H2O(l)           ΔH = −1755.6 kJ

To get my answer, I multiplied 221.3 by 3 (663.9) to get 3O2. I also multiplied 47.6 (95.2) by 2 to get 6H2. Lastly, I reversed the third formula to get +1755.6.

663.9+95.2+1755.6 = 2514.7

My answer is 2514.7, but the answer sheet says 956.6. Would you help me figure out what I'm doing wrong?

Thanks,

Steph

You should apply a more accurate method to solve the exercise.
Anyway, let's analyse your reactions.

You know the ΔH of combustion of C₂H₆.

ΔcH=6ΔfH(H₂O)+4ΔfH(CO₂)-2ΔfH(C₂H₆)

You want to calculate the 6ΔfH of water (actually the opposite, so you're looking for -6ΔfH).
So.. do you know the other terms? Yes, you do.

Because ΔfH(C₂H₆) is -47.6kJ and ΔfH(CO₂) is -221.3kJ.

Now you can calculate 6ΔfH(H₂O) for the previous equation. You'll get -965.6. The correct answer (of course) is 956.6 because you're looking for -6ΔfH.