8
$\begingroup$

My teacher says $H$ is a state function but $\Delta H$ is not. But I think both of them should be state function. Is he correct?

$\endgroup$
8
$\begingroup$

$\Delta H$ is a function of two states, the initial state and the final state.

For a given final state, there can be infinite $\Delta H$ values depending upon what the inital state was.

For a given inital state, there can be infinite $\Delta H$ values depending upon what the final state is.

Therefore, $\Delta H$ is not a state function.

$\endgroup$
2
  • $\begingroup$ Then is it a path function? $\endgroup$ – Ayush Pateria Feb 23 '15 at 17:07
  • 5
    $\begingroup$ It only depends on the inital state and the final state, regardless of the path between the two states $\endgroup$ – DavePhD Feb 23 '15 at 17:09
2
$\begingroup$

I generally agree with DavePhD's answer but I also want to give the opposite view (!). Everyone agrees $H$ is a state function, but energy (and thus enthalpy) has no built in zero-scale. So really even in defining $H$, there is some built-in standard state that you are comparing the state of interest to. For example, the standard state might be all electrons and all nuclei infinitely separated from each other (this standard is commonly used in quantum chemistry).

But we could also specify other standard states. We could say the standard state is one where all electrons and all protons and neutrons are infinitely separated...but that's possibly only useful for nuclear chemistry. We could also a standard state where all the atoms in a sample are separated into pure chemical elements. So to compute "the" enthalpy $H$ of say $\ce{CH4}$ using this standard state, we first separate its atoms to $\ce{C}$ and $\ce{2H2}$, our standard state, and then see how much the enthalpy changes.

Regardless of how we definite the standard state, enthalpy is a state function. But "the" enthalpy $H$ that we compute from the last standard state is the same as the enthalphy change of formation, $\Delta H_{form}$! So if those two things are equal, and one of them is a state function, then the other one must be a state function too! So $\Delta H$ is also a state function.

$\endgroup$
5
  • 1
    $\begingroup$ That is just a way to say that dX is actually conceptually same with X if we insist that dX always should be used only with a single standard state. While technically you can call it correct, I think it is not helpful: 1) the OP's question is referring to a misunderstanding that you do not correct or explain, only bury; 2) dH is more often than not used for comparing specific, non-standard states $\endgroup$ – Greg Feb 24 '15 at 4:18
  • $\begingroup$ Greg, I'm not sure what misunderstanding I have buried -- I do think whether $\Delta H$ is a state function is intimately related to standard states, and also that if $X$ is a state function then its differential $dX$ and by implication $\Delta X$ should also be a state function. If you have more to add (and by no means do I think my answer is the best or final word!), you should add your own. I would probably upvote it, just like I did DavePhD's. $\endgroup$ – Curt F. Feb 24 '15 at 4:23
  • $\begingroup$ As I said, dH is generally not referring a standard state. Also whether something a state-function or not is nothing to do whether it has a "built in zero". This distinction in thermodynamics (Q, W, dH are not state-function etc) actually has a meaning you fully ignore. $\endgroup$ – Greg Feb 24 '15 at 4:31
  • $\begingroup$ When I ignore something, I always try to ignore it "fully". $\endgroup$ – Curt F. Feb 24 '15 at 4:33
  • $\begingroup$ The "built in" scale for energy is $E=mc^2$. If we have a block of U238 at a certain T and P it has an energy and an enthalpy. We can heat it, cool it or take it down other paths, and each path will correspond to a delta H, that is not solely a function of the initial state. $\endgroup$ – DavePhD Feb 24 '15 at 10:05
2
$\begingroup$

A state function is a property of a single state of the system.

A change in a state function is associated with a process that involves two states of a system.

You may be thinking that the change in enthalpy will be a state function because you can take any path you like between the initial and final states and still get the same enthalpy change. But that's beside the point; the enthalpy change isn't a property of a single state of the system, so it isn't a state function.

$\endgroup$
2
  • $\begingroup$ Fred, can you say a few words about the (interesting) new picture in your avatar? $\endgroup$ – ron May 31 '15 at 13:50
  • $\begingroup$ @ron. Ha. It's just a recent picture of me--this is the avatar I use on my Moodle sites. Funny, after I switched to this one, I got 37% fewer instant messages from students--I wonder why? $\endgroup$ – Fred Senese Jun 2 '15 at 4:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.