Enthalpy

Defining a new quantity, enthalpy, makes it more convenient to calculate thermodynamic changes for constant-pressure processes.

[@markpwoodward]

Minor point. In your derivation of dH, you mentioned "reversible" as a step to make -P_ext*dV and P*dV cancel, but this confuses the issue (i.e. does dH = dq still hold for irreversible processes). Since pressure is constant, the derivation still holds, P=Pext. So maybe just skip the "rev." label and description.

[@PhysicalChemistry]

Yes, that's an excellent point. Constant P is enough for this to be true, and reversibility is not needed. Thanks very much for the comment.

[@adetunjidaniel5064]

@@PhysicalChemistry then does this make the equation dh=dq untrue for a reversible reaction, since p is never constant in such reaction because p=nrt/v(and the v changes)

[@PhysicalChemistry]

@@adetunjidaniel5064 You're right: dH ≠ đq for processes in which P is not constant (because there is nonzero PV work)

[@pratyushbhattacharjee5532]

Then how can we arrive at the dH=đq_p? If mechanical equilibrium between the system and the surroundings (or reversibility) is not assumed, how can P_ext be equal to P, which seems to be a necessary step in the derivation? And if reversibility is assumed, we seem to be deliberately breaking the condition of the constancy of pressure with which the derivation was begun in the first place. Could you please help me out?

[@esoto1386]

1:50 In the H definition, where does the initial extra PV comes from?

[@PhysicalChemistry]

That's just how we choose to define H. The reason we choose to define it that way is so that (as shown in the next few lines) it works out to be the same as the heat for a constant-pressure process.

[@esoto1386]

Ok, thank you!

[@joshg.6536]

Great work! But i do not get the step: converting atm and L to 8.3145 J and 0.0802 L*atm? Thanks :)

[@PhysicalChemistry]

You can use the conversion 1 L atm = 101.325 J and you will get the same answer. The conversion I used is just a little trick I use to keep from having to memorize too many constants. You likely already know that the gas constant is 8.3145 J / (mol K). And also that it is 0.08206 L atm / (mol K). So those two values are equal to each other. And we can use them as the conversion factor to convert units of J to L atm.

[@user-uv2sx6yo2s]

why we used 3/2 in H law

[@PhysicalChemistry]

The internal energy of an ideal gas is U = 3/2 nRT. Here's the video where this is obtained: kzbin.info/www/bejne/qXqZemaAj5mnhpY

[@AshokKumar-kn9ku]

At 2:20 with respect to what we are taking differential change

[@PhysicalChemistry]

Nothing! It's a differential, not a derivative. Differentials (like dx) are not the same as derivatives (dy/dx). For a derivative, you do need to specify what other variable you are taking the derivative with respect to: for example, it is the change in y that happens when you make a small change in x. But a differential is just a small change. dx is just a small change in x. In this case, dH is any change in H, for any reason. Since H depends on U and P and V, then the small change dH might be because of a change in U or a change in P or a change in V.

[@AshokKumar-kn9ku]

Thanks a lot

[@GulzarAhmad-sw1kh]

If H = U+PV, doesn't this mean that enthalpy is not heat but a mix of heat & work? Explain, plz.

[@PhysicalChemistry]

Remember the caveat that enthalpy is heat *at constant pressure*. Under other conditions, it does have contributions from both heat and work. It just happens that, at constant pressure, the PV term cancels the PV work that is hiding inside U.

[@GulzarAhmad-sw1kh]

@@PhysicalChemistry BUT H has been derived from 1st law as below. dq = du+pdv = du+d(pv)=d(u+pv)=dH While you have expanded the H (derived from 1st law) and then plugged in 1st law again?

[@PhysicalChemistry]

The step where you say P dV = d(PV) is only true when P is constant. (Thus, your equation dq = dH is only true at constant pressure.) I'm not quite sure what you were asking, but perhaps that helps.