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delta H TPR question.

Discussion in 'MCAT Study Question Q&A' started by pfaction, Mar 29, 2012.

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  1. pfaction

    pfaction

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    SDN Members don't see this ad. (About Ads)
    So we all know deltaH = Hprod - Hreacts.
    TPR GChem passage 36, Q5:
    Based on the following data:
    S(r)+O2->SO2 dH= - 297.0
    S(m)+O2->SO2 dH= - 297.1

    What is the enthalpy change in the transformation of S(r) to S(m)?
    So I saw that we'd need to flip the second equation for this to work, right? My new equation was:
    S(r)+O2->SO2->S(m)+O2
    Therefore, products - reactants gave me
    dH = 298.1 - (-297) = 594.1

    Intuition kicked in and I chose +0.1 instead of it and I was right....
    but why isnt it 594.1?
  2. chiddler

    chiddler

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    according to the numbers you gave both are negative!

    -297 - (-297.1) = -297 + 297.1 = 0.1

    so if the numbers you gave are accurate, then you just made a small error.
  3. MedPR

    MedPR

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    Not sure what you're doing, but all you have to do is switch the sign on both of the given dH values, since reverse reactions have the same magnitude of enthalpy change, and then do S(m) - S(r) to get 0.1.
  4. pfaction

    pfaction

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    Yeah, but I don't understand.

    dH = hProd - hReact.
    We have to flip the 2nd equation because it's the reverse that forms the product.
    So it would be +297.1 - (-)297.0 right?
  5. chiddler

    chiddler

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    the numbers you gave are BOTH negative. redo the same calculations knowing this.

    your calculations show that one is positive, and one is negative.
  6. MedPR

    MedPR

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    The formation of S(r) and S(m) are both endothermic. If you could magically convert S(r) into S(m) you would do S(m) - S(r) with both numbers positive and get 0.1.
  7. pfaction

    pfaction

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    But I'm not flipping the first equation. I'm keeping it the same, since it's that SO2 that I'm using in the 2nd equation. THAT's the one I'm flipping, and since it's my end reaction, I have to substract the first thing from it. Am I seriously missing something or am I doing it all wrong?

    Example:
    A + B -> C, dH = -1
    D + E -> C, dH = -2

    So if I want the dH of A to D, I would do:
    A+B->C->D+E
    ^reactants----^products
    this is flipping the 2nd equation
    so it would be
    dH = +2 minus -1 = 2+1 = 3?
  8. MedPR

    MedPR

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    Based on the following data:
    S(r)+O2->SO2 dH= - 297.0
    S(m)+O2->SO2 dH= - 297.1

    What is the enthalpy change in the transformation of S(r) to S(m)?

    When you breakdown S(r) into SO2, you lose 297.0 kJ of energy to the surroundings. When you form S(m) from SO2, you take away 297.1 kJ of energy from the surroundings. You now have 0.1kJ of energy in the system. You have to understand why/how the equations work rather than blindly plugging in numbers.
  9. chiddler

    chiddler

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    oh damn that's right! i think you're right, then.
  10. chiddler

    chiddler

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    S(r)+O2->SO2 dH= - 297.0
    S(m)+O2->SO2 dH= - 297.1

    For S(r) --> S(m), you have to flip the bottom equation

    S(r)+O2->SO2 dH= - 297.0
    SO2 ->S(m)+O2 dH= + 297.1

    cancel out commons

    S(r) -> S(m)

    so product is 297.1
    reactant is -297

    product minus reactant is 297 - (-297.1)
  11. MedPR

    MedPR

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    I guess if you two agree then I must be wrong. Idk, I still think its 0.1
  12. chiddler

    chiddler

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    wait that's not the point! o_o

    why do you think it's 0.1? what's wrong with the calculations?
  13. pm1

    pm1

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    I'm still confused about the calculations since I would also try to flip one of the reaction and get 594.1kJ/mol.

    However, this is what TPR has to say:
    "If the combustion of monoclinic sulfur releases 0.1kJ/mol more energy than the combustion of rhombic sulfur, then S(monoclinic) must have 0.1kJ/mil more energy locked up inside than S(rhombic) has. So, converting S(rhombic) to S(monoclinic) would require adding an extra 0.1kJ/mol."
  14. MedPR

    MedPR

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    You consume 297.0 in the degredation of S(r), then you absorb 297.1 in the formation of S(m). You lose 297, then gain 297.1. You are up 0.1.
  15. pm1

    pm1

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    Now that I'm thinking.. a way that would work would be to set up the equations and make an addition.

    S(rhombic) + O2 --> SO2(g) dH = -297.0kJ/M
    SO2 --> O2 + S(monoclinic) dH = +297.1kJ/M (sign changes bc equation was flipped)
    ---------------------------------------------------------------------
    S(rhombic) -----> S(monoclinic) dH +0.1kJ/M

    so could it be that when it is a dH of reaction to solve it like this and then when it is a dH of formation dH = dh products - dh reactants

    ??:confused::confused:
  16. circulus vitios

    circulus vitios

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    Your post confused the hell out of me until I realized what you (and I) were trying to do. What you're mistakenly doing is trying to find delta H when you are already given delta H.

    Going back to the original problem:

    S(r) + O2 ==> SO2; dH= -297.0
    S(m) ==> O2 + SO2; dH= -297.1

    You want to find S(r) ==> S(m). Rearrange the equations. Whenever you do this, you flip the signs; you are correct in this.

    S(r) + O2 ==> SO2; dH= -297.0
    SO2 ==> S(m) + O2; dH = +297.1

    Again, notice how you are given dH values which means change in enthalpy. So you don't need to use the equation dH = H products - H reacants. To find the answer you simply add up the equations:

    S(r) + O2 ==> SO2; dH= -297.0
    SO2 ==> S(m) + O2; dH = +297.1
    _________________________________
    S(r) ==> S(m); dH = +0.1
  17. chiddler

    chiddler

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    why don't you need to use final-initial?
  18. circulus vitios

    circulus vitios

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    Because you're already given delta H. What do you think dH stands for?
  19. MedPR

    MedPR

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    I was confused by trying to use the equation too, that's why I abandoned it and just thought about what's happening conceptually.
  20. chiddler

    chiddler

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    That's the definition of hess's law.

    Specifically,

    ΔH (reaction) = ΣΔH(final/products) - ΣΔH(initial/reactants)

    on the contrary, this ΔH refers to heat of formation which i'm not sure the ΔH given above is of. We're given heat of reaction. Ok I think this is why we can't use it. However, mr circulus vitios, I don't think your reasoning is entirely correct because hess's law does use ΔH. The distinction must be made with what kind of ΔH we use.

    I'd also like to add that H by itself is such a difficult quantity to measure, it is almost never used.

    I will definitely keep this in mind! I would like to do more practice questions and do both. But i'm really upset because i've not had problems with this before.
  21. pfaction

    pfaction

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    You're absolutely right when I think of it this way. It makes perfect sense, it's overall an endothermic reaction, when I do it this way in my head it makes 100% sense. :thumbup:


    Not sure if I can post this legally:
    [​IMG]

    (sorry I didn't rotate)
    It seems what you guys, esp vit was saying was true. I somehow combined the regular deltaH with Hess's law? Still not sure of the difference tbqh.
  22. circulus vitios

    circulus vitios

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  23. pfaction

    pfaction

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    [​IMG]

    Thanks a billion, guys. I think I'll also do what MedPR said, conceptual before numerical.

    So on that post; the first would be the dH'f so it would be products minus reactants. So +4.
    Second one, you flip the first equation and then just add, right?
  24. gettheleadout

    gettheleadout custom title blank Moderator

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    Yeah, this problem is just an application of Hess' Law to partial reactions. As long as you have the partial reactions going in the right direction, you just combine the enthalpy change values for each. It's just -297 + 297.1 = 0.1

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