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My chemistry textbook says that:

Enthalpy (Internal Energy) = chemical potential energy + thermal energy

Now, I know that Enthalpy and Internal Energy are different so I really don't understand what my text book is saying.

Also, from searching I understood that $\mathrm{U = q + w}$ since, $\mathrm{w = -p \Delta V}$

And $\mathrm{H = U + \Delta(PV)}$

So for standard pressure :

$\mathrm{\Delta H = q}$ , my text book has stated that

At standard temperature, the difference in enthalpy is equal to the difference in chemical potential energy.

Also I'm not sure I fully understand what it means by potential chemical energy, in general I'm just "super confused". My text book even says that

By potential chemical energy we mean the intermoleculer forces and etc. that exist between atoms and molecules.

But from what I read online, these forces are because of the Internal energy of a system not the chemical potential energy. So if somebody could explain these thing and relate them to my text book somehow I would be grateful :

  1. What is Chemical Potential Energy?

  2. What is Internal Energy?

  3. What is enthalpy? And how do I relate these varying definitions for it to each other?

Note

$\mathrm{U, q, w, H}$ have usual meanings.

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    $\begingroup$ It would be helpful if you could provide the exact reference (title, author(s), edition, and publication date), and also provide a picture of the paragraph(s) containing the quotes. Sometimes textbooks are misquoted, and sometimes the quotes are taken out of context. $\endgroup$ – theorist Jan 7 at 6:13
  • $\begingroup$ It wouldn't be of any help, English is technically my third language and my text book are in another language. $\endgroup$ – amir Jan 7 at 6:15
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    $\begingroup$ Understood. It would be hard for me to understand what the textbook is trying to say without actually reading the relevant paragraphs. For instance, enthalpy and internal energy are not the same but, under certain constraints ($\Delta PV = 0$), $\Delta H$ can have the same value as $\Delta U$. So I don't know if your textbook is trying to say that U and H are the same (which they're not), or if it is trying to say that, under certain constraints, their changes can be equal (which they can). Is this a well-established textbook in your country? And what does your professor say about this? $\endgroup$ – theorist Jan 7 at 6:23
  • $\begingroup$ My text book hasn't mentioned any constraints and has boldly stated said sentence. This is a high school text book and it's the only official text book in the country, but it's famous for being quite terrible compared to western text books but since the college entrance exam is based solely on this book we have no choice but to memorize its nonsense but what I'm trying to do is understand Enthalpy on a deeper level but since it seem my text book is out right flawed in this sence I may have to just accept what it's saying. $\endgroup$ – amir Jan 7 at 6:30
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    $\begingroup$ Is chemical potential energy a part of Internal energy? If so why do we represent it as U = q + w. What part of this formula accounts for the intermoleculer forces? Also is thermal energy just Q? $\endgroup$ – amir Jan 7 at 6:49
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To understand internal energy*, I prefer to start with this:

Consider a thermodynamic system. There are only three ways we can increase its internal energy (which is the sum of its internal potential energy and internal kinetic energy): flow heat into it; do work on it; or add matter to it (and the converse for decreasing its internal energy). If the system is closed (no matter can flow in or out), we are left with only the first two, which naturally leads to the following equation: $$\Delta U= q + w$$

Now, suppose you are carrying out a reaction under the common condition of constant pressure. The amount of heat flow associated with this reaction is an important and measurable quantity (measurable using a calorimeter). So it would be very nice if there were a state function (i.e., one that is independent of path) that happens to give us the heat flow under those conditions. It turns out that we can construct such a function by adding $pV$ to $U$. We call that function the enthalpy, $H$:

$$H = U + pV => \Delta H = \Delta U + \Delta(pV) = \Delta U + p \Delta V+ V \Delta p $$

At constant pressure, $\Delta p = 0$, so:

$$\Delta H_p = \Delta U + p \Delta V = q_p + w_p + p\Delta V$$

But at constant $p$, $p_{sys} = p_{ext}$. Thus, if we only have $pV$ work: $$w_p = -p_{ext} \Delta V = - p \Delta V$$

Hence:

$$\Delta H_p = q_p- p \Delta V+ p\Delta V = q_p$$

Let's compare a reaction at constant pressure with one at constant volume. At constant $V$, with $pV$-work only, $q_V = \Delta U$ (because, with $pV$-work only, $w_V = 0$). Suppose the reaction is exothermic, and the volume of the products is greater than that of the reactants, i.e., $\Delta V >0$. You can see from the equations that $\Delta H$ will be less negative than $\Delta U$, i.e., less heat is evolved at constant pressure. We can understand this physically from the fact that, at constant pressure, some of the thermal energy the system evolves goes into the work the system does on the surroundings (because the system expands). Thus less thermal energy is left over to flow into the surroundings. I.e., $\Delta H$ is just $\Delta U$ with a "correction factor" to account for $pV$ work.

The other reason that constructing $H$ is useful is that it is an intermediate step in constructing $G$, which is an extremely important function in thermodynamics, as it allows us to determine the direction of spontaneous change (and the equilibrium state) at constant $T$ and $p$. [For more on this, see my answer here: What is wrong in this argument that dG must always be zero? ]

Finally, a quick (and somewhat simplified) answer on potential vs. kinetic energy (it really merits a separate question): Kinetic energy is the energy bodies have as a result of their motion. This includes translational, vibrational, and rotational motion. Potential energy is the energy bodies have as a result of their position in a potential field. For example: Two atoms in a strong bond have a lower potential energy than they would when they are separated. Conversely, if you have particles that repel each other, you increase their potential energy when you compress the system and thus force them to be closer.

*Copying from one of my other answers: The term "internal" refers to energy internal to the system. For example, the internal kinetic energy is the energy the system has as a consequence of the thermal motion of its atoms and molecules, as contrasted with the external kinetic energy the system might have as a result of the motion of the system as a whole through space.

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The internal energy is defined in (classical) thermodynamics for a finite change as $\Delta U=q+w$ where $U$ is the energy contained in the 'system' called the internal energy, and $q$ is the heat absorbed by the system and $w$ the work done on the system. Often the word 'system' means some ideal gas (often imagined to be in a cylinder with a frictionless piston) that cannot exchange matter with its surroundings but can exchange heat and do work, usually this work is expansion/contraction or $pV$ work although it could also be electrochemical or gravitational work, i.e. used to lift a weight but this is not usually relevant in chemistry. A system that cannot exchange matter but can exchange energy is called 'closed'.

Notice that we do not need to know what the nature the internal energy is to use thermodynamics, i.e. the nature of atoms and molecules is not relevant as thermodynamics was developed before this was understood, and this abstractness is the great strength of thermodynamics, but can make it more difficult to understand. Nevertheless, we do want to understand what the internal energy is.

First, the internal comprises the kinetic energy of the atoms or molecules, and so defines the thermal energy so an increase in kinetic energy is observed as an increase in temperature. Secondly molecules can rotate in three dimensions thus have rotational kinetic energy and its atoms are bound together by bonds that vibrate thus adding potential and kinetic energy to the internal energy.

The enthalpy is used, as it were, to make a 'level playing field' when comparing experiments because it accounts for the work done $H=U+pV$ by a reaction so the first statement you give is not correct. The work done is small, however, for liquids and solids as $\Delta V$ is small so that $H \approx U$. (The Gibbs energy $G=U+pV-TS=H-TS$ is similarly used to account for changes in entropy and work done).

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    $\begingroup$ (1) The internal energy chg. is not generally $\Delta U = q + w$. That only applies to a closed system. (2) You'll want to amend your language, changing "isolated" to "closed". What you're describing is a closed system (no exchange of matter with the surroundings) rather than an isolated system (isolated from the surroundings, and thus can exchange neither matter nor energy). I.e., q and w are both necessarily zero for an isolated system. Also, while ideal gases are often used pedagogically, in the actual practice of thermodynamics, the system is typically a real-world reacting system. $\endgroup$ – theorist Jan 7 at 15:24
  • $\begingroup$ Still isolated means no matter no energy exchanges $\endgroup$ – Alchimista Jan 8 at 8:16
  • $\begingroup$ Thank you for these comments I have corrected my answer. $\endgroup$ – porphyrin Jan 8 at 9:16
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The internal energy U is the sum of all energies stored in the chemical bonds of the examined sample of matter. It is a potential energy. The enthalpy H is the same, but with a correction due to the presence of the atmosphere. The difference between H and U is similar to the difference between weight and apparent weight. Let me go back to this measurement.

If you want to know the mass (or the weight) of an object, you put it an a balance and you read its weight. But this value is not the real mass (or weight) of the object, it is only the apparent weight. If you had done all this job in water, with a balance under water, you would also have obtained a lower value than in air. You would have obtained the apparent weight, namely the weight of the object minus the weight of the same volume of water. This correction is called Archimede's push. In air the correction is about 1000 times smaller than in water, but is is not zero. So usually we ignore this Archimede's correction in air, and we accept the weight in air as if it has been measured in the vacuum.

It is the same for U and H. U is the sum of all energies stored in the chemical bonds of the object, but only in the vacuum. As we are working surrounded by an atmosphere, the object needs to repell this atmosphere simply to exist and to put its atoms and molecules where they are. This work costs a little bit. It costs much more if the object is a gas. If you produce by a chemical reaction a couple of liters of gas, you have to repell air simply to allow your gas to be where it is, and this costs some work, not much, but this work is not zero. That is why H is U plus a small correction, equal to PV. And when you heat an object in our lab, the heat is equal to Delta H and not Delta U.

OK ?

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    $\begingroup$ The internal energy is not the potential energy. The internal energy is the sum of the internal potential and kinetic energy. For instance, we will increase the internal energy of an ideal gas by heating it, yet an ideal gas has no internal potential energy; it only has internal kinetic energy. I.e., for an ideal gas, all of its internal energy is kinetic energy. $\endgroup$ – theorist Jan 7 at 15:01
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    $\begingroup$ As theorist said. According to this answer a monoatomic ideal gas would have zero U instead of its kinetic energy. The latter is. $\endgroup$ – Alchimista Jan 8 at 8:18

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