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NotTim

  • one year ago

What's the unit for enthalpy?

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  1. NotTim
    • one year ago
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    MY textbook is being mean, so I can't remember what the unit for enthalpy is (for a reaction at equilibrium, if that clarifies anything). I'm thinking between kJ/mol, kJ*mol and J/mol. The data I'm given is T, in degrees Celsius, and Keq (uM), from a chart.

  2. anonymous
    • one year ago
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    The unit for enthalpy is joules or J. Enthalpy is the transfer of heat.

  3. NotTim
    • one year ago
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    I should have specified, this is true when it is deltaH too?

  4. anonymous
    • one year ago
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    It can be written as H or delta H Delta H would mean enthalpy change as H would be enthalpy. \[\Delta H= H _{2}-H _{1}\] \[\Delta H= 32.76J\]

  5. NotTim
    • one year ago
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    Ah, so units are the same regardless? In what situations would I consider using kJ/mol, or J/mol? I've seen that in some instances in the practice problems in the later chapters, but I don't understand why its like that.

  6. anonymous
    • one year ago
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    It can be written as \[32.76 J/mol\] in a chemical reaction.

  7. anonymous
    • one year ago
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    I was about to state that. Writing it as J/mol would mean solving for the amount of heat the product transferred or released during the chemical reaction.

  8. anonymous
    • one year ago
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    Or you want to know the mass per 1 joule of enthalpy change.

  9. anonymous
    • one year ago
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    It is quite simple really. You do not need to understand the concept deeply.

  10. anonymous
    • one year ago
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    For example: http://openstudy.com/study#/updates/55f89467e4b0a91f39321a0f

  11. NotTim
    • one year ago
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    I'm just working through making a graph for the van't Hoff equation.

  12. NotTim
    • one year ago
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    So for -deltah/(RT), I was suppose to give the units of both delta H, and -deltah/(RT)

  13. anonymous
    • one year ago
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    Yes.

  14. NotTim
    • one year ago
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    sorry, afk. washroom

  15. anonymous
    • one year ago
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    Do you know the units for that?

  16. anonymous
    • one year ago
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    I got to go so, R is (Joules) (Kelvin)^-1 (mol-)^1 -Delta H is J/mol T is temperature in Kelvins (K) So the units are obtained by canceling \[\frac{ Joules }{ mol \times mol ^{-1}\times K ^{-1} \times Joules \times K }\] \

  17. anonymous
    • one year ago
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    =\[\frac{ Joules \times mol \times K }{ Joules \times mol \times K }=1\] So no units at all.

  18. anonymous
    • one year ago
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    Just thought I'd add a little more, when you do an experiment at constant pressure and there's heat released or absorbed, the amount of heat is dependent on how much stuff you have. Enthalpy is an extensive property, which means it depends on how much stuff you have. Extensive properties aren't generally very useful to think about. It's like if I weigh some gold, it depends on how much of it I have, I would never write down "5 pounds of gold" in a text book for students to learn because it depends on the gold chunk I felt like picking up that day. However the density of gold is what's called an intensive property. We've said that now the mass per amount you have is what doesn't change, which is actually useful. I can weigh my gold to find out how much volume it has or find out how much volume it displaces to figure out how much mass it has, which might actually be important to do something. Wait what does this have to do with enthalpy? Well like I was saying, on its own Enthalpy changes aren't that interesting since the heat depends on how much stuff you have. So if you divide out the number of particles you used, you'll get a kind of "density" like property, an intensive property called the molar enthalpy. These are the kinds of values you can look up in a book too.

  19. NotTim
    • one year ago
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    Thanks Woodward

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