# Need Help understanding energy and enthalpy please

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1. Oct 21, 2015

I was studying about Ionization enthalpy and I found that in many books it is described in terms of enthalpy change while in others it is described in terms of energy, although in all of them the basic definition is almost same. It is the enthalpy change when ..... or it is the energy required to .... I know both enthalpy and energy, in most of the cases, mean the same thing but which is the more appropriate way to describe Ionization Enthalpy, in terms of enthalpy change or in terms of energy ?
Any help will be greatly appreciated !

2. Oct 22, 2015

### DrDu

Ionization enthalpy is clearly an enthalpy. However, the difference between ionization energy and ionization enthalpy usually makes no difference as long as you consistently use one or the other. I am referring here to the Born Haber process. There, the amount of atoms + ions remains the same, while the amount of electrons drops out. Hence there is no overall change in volume and the difference between enthalpy and energy vanishes (at least for constant pressure).

3. Oct 25, 2015

### CrazyNinja

Enthalpy is heat transfer at constant pressure. As most of the experiments are done at constant pressure, ionisation enthalpy is a right term. But the correct definition is as given by DrDu.

4. Nov 2, 2015

### akashram

Enthalpy combines internal energy and energy due to the volume of the system. Most systems don't exist in a vacuum. They have a creation energy, PV, associated with them that describes the energy needed to push the atmosphere out of the way so the can occupy volume V. It is common for systems to change volume. If they are not in a vacuum then enthalpy accounts for the energy associated with volume changes and internal energy changes. Enthalpy isn't a required quantity but it makes the equations neater. The concept of an isenthalpic process is important as well since it closely matches many real gas flow systems such as a Joule Thompson valve.

5. Nov 2, 2015

### James Pelezo

I'm confused, what do you mean by "while the amount of electrons drops out... Hence there is no overall change in volume and ∆U(internal energy) = ∆H(enthalpy)." How do "electrons dropping out" define a closed system? The Born-Haber Cycle is a closed system because the volume gained by 'Basic Elements in Standard States' in transition into the gas phase to be ionized is lost upon ionic bond formation giving the ionic compound in solid state. That is, Volume Gained on atomization = Volume Lost on bond formation of the ionic compound. ∆U = ∆H - P∆V, if ∆V =Vatomization = Vbonding = 0 and P∆V = 0 and ∆U(internal energy) = ∆H(enthalpy). In this condition ∆U(internal energy) = ∆H(enthalpy) and the Energy vs Enthalpy are interchangeable. These are Enthalpy of Atomization (metal) and (nonmetal), Ionization enthalpy of Metals (oxidation), Electron Affinity (an exothermic enthalpy on gain of electron during reduction and reduction) and finally Enthalpy of Bond Formation,

6. Nov 3, 2015

### DrDu

I meant that in a full Born Haber cycle, you consider a step, say, Na(g) + Cl(g) -> Na$^+$(g)+Cl$^-$(g). As the number of particles does not change, there is also no change in molar volume $\Delta V_m$ and hence $\Delta U=\Delta H$ for this step. This does not hold for the single ionization steps Na(g)-> Na$^+$(g) + e$^-$ and Cl(g)+e$^-$-> Cl$^-$(g), but in forming the complete cycle, it won't matter whether to use the energy or enthalphy of ionization (as long as they are consistently used for both reactions).

7. Nov 3, 2015

### James Pelezo

OK, I'll buy that. The electrons 'dropping out' thing threw me. Thanks