NotTim
  • NotTim
What's the unit for enthalpy?
Chemistry
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jamiebookeater
  • jamiebookeater
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NotTim
  • NotTim
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.
anonymous
  • anonymous
The unit for enthalpy is joules or J. Enthalpy is the transfer of heat.
NotTim
  • NotTim
I should have specified, this is true when it is deltaH too?

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anonymous
  • anonymous
It can be written as H or delta H Delta H would mean enthalpy change as H would be enthalpy. \[\Delta H= H _{2}-H _{1}\] \[\Delta H= 32.76J\]
NotTim
  • NotTim
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.
anonymous
  • anonymous
It can be written as \[32.76 J/mol\] in a chemical reaction.
anonymous
  • anonymous
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.
anonymous
  • anonymous
Or you want to know the mass per 1 joule of enthalpy change.
anonymous
  • anonymous
It is quite simple really. You do not need to understand the concept deeply.
NotTim
  • NotTim
I'm just working through making a graph for the van't Hoff equation.
NotTim
  • NotTim
So for -deltah/(RT), I was suppose to give the units of both delta H, and -deltah/(RT)
anonymous
  • anonymous
Yes.
NotTim
  • NotTim
sorry, afk. washroom
anonymous
  • anonymous
Do you know the units for that?
anonymous
  • anonymous
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 \[\frac{ Joules }{ mol \times mol ^{-1}\times K ^{-1} \times Joules \times K }\] \
anonymous
  • anonymous
=\[\frac{ Joules \times mol \times K }{ Joules \times mol \times K }=1\] So no units at all.
anonymous
  • anonymous
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.
NotTim
  • NotTim
Thanks Woodward

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