If we take an isolated system at constant volume in which chemical reaction happens than it is said that heat of reaction changes internal energy of the system. As far as I know internal energy takes into account all microscopic energy of the system such as kinetic energy of atomic and molecular motion and potential energy of interaction of all bonds in the system (intermolecular and chemical). If we have an reaction which releases heat (exothermic) than potential energy of chemical bonds goes down and microscopic kinetic energy goes up (temperature increases). Change in micro kinetic energy is the same as change in potential energy due to energy conservation. Since internal energy takes into account both types of these energies, its sum must remain the same due to energy conservation. If so, than internal energy doesn't change. What is a problem here?
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$\begingroup$ Who say / where it is said that if we take a rigid isolated system in which a chemical reaction happens then the heat of reaction changes the internal energy of the system? Your entire discussion is correct, it is basically the first principle. It is just the premise, which I don't know from where is coming from, that is wrong. $\endgroup$– AlchimistaCommented Apr 29, 2021 at 12:04
3 Answers
When they say that the heat of reaction changes the internal energy, what they mean is that the internal energy (or the enthalpy) changes at constant temperature (i.e., if the initial and final temperatures are held the same). If the change in internal energy is forced to be zero in an adiabatic enclosure, then the temperature must change to offset the heat of reaction. For an exothermic reaction, this means that, for the internal energy to remain constant, the temperature must rise to offset the negative heat of reaction.
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$\begingroup$ Hi Chet. As you know, I am not a chemist like you. But can I think of the heat of combustion as the potential for heating due to temperature increase (the potential for energy transfer by heat) even if that potential cannot be realized in the case of combustion in an adiabatic enclosure? The phrase "heat of reaction" always bugs me since the reaction need not result in heating. Some others (chemistry minded) on this site keep insisting that "heat content" means containing heat. And that heat can be considered a property of the system as in, for example, the case of an isochoric process. $\endgroup$– Bob DCommented Apr 29, 2021 at 12:19
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$\begingroup$ @BobD in adiabatic enclosure that heating certainly causes the T to rise. Why it can't be “realised“? $\endgroup$ Commented Apr 29, 2021 at 12:58
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$\begingroup$ @Alchimista. how you are defining "heating"? $\endgroup$– Bob DCommented Apr 29, 2021 at 13:12
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$\begingroup$ @BobD Actually, I'm a chemical engineer. I would not think of the heat of combustion in that way. I would think of the heat of combustion as the amount of heat that has to be removed to hold the system at constant temperature (and pressure). I'm not bugged by the phrase heat of reaction if I regard it as the change in enthalpy between the products and the reactants, which is equal to the amount of heat that must be added to the reaction mixture in a constant pressure process to hold the mixture at constant temperature. $\endgroup$ Commented Apr 29, 2021 at 13:17
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$\begingroup$ So the first part of this refers to the change in a state property (enthalpy) between well-specified end states of a system, and the second part refers to how it is would be measured. $\endgroup$ Commented Apr 29, 2021 at 13:17
Heat is also a form of Energy. It is defined as the energy that flows into or out of a system because of difference in temperature between the thermodynamic system and its surrounding. It occurs at the boundary of the system.
Now, as you have taken the system to be Adiabatic or isolated there is no heat released into the surrounding, it's still contained in the system as the heat is the exchange of energy between the system and the surrounding but that's not possible.
So, basically Q=0 and the energy that is released in breaking of bonds completely converts to Kinetic Energy making change in internal energy=0.
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$\begingroup$ This what I said in my question. After talking about this question with others, answer to this question seems to be in how we define internal energy or what is our reference state. In thermodynamics, energy stored in chemical bonds is usually not taken into account as internal energy because we usually define reference state as ideal gas. In ideal gas there are no intermolecular interactions. Such reference state is good because we can track/calculate temperature changes in system via changes in internal energy. $\endgroup$ Commented Apr 29, 2021 at 10:40
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$\begingroup$ So, we can pretend like heat of reaction is like heat exchanged with surroundings regardless of system being isolated since it has the same effect. If we pick reference state like I did than we can't calculate temperature changes via internal energy changes since according to my definiton internal energy doesn't change. $\endgroup$ Commented Apr 29, 2021 at 10:44
Since internal energy takes into account both types of these energies, its sum must remain the same due to energy conservation. If so, than internal energy doesn't change. What is a problem here?
That is correct and there is no problem. Heating is energy transfer due solely to temperature difference. Whether or not heat is actually released in a reaction depends on how you define the system and where exactly heat transfer, if any, occurs.
FIG 1 below is a sketch of an Oxygen Bomb Calorimeter. It is a device that measures the heat of chemical reactions or physical changes.
The combustion process takes place in the steel bomb. The steel bomb is immersed in water. Both are placed in a rigid insulated outer container. We define the contents of the outer container as our system. Since the outer container is both insulated and rigid, our system is isolated, i.e., it can not exchange heat or work with the environment outside the outer container.
Combustion takes place. The internal chemical potential energy of the fuel is converted to internal kinetic energy and the temperature of the contents of the steel bomb suddenly increases. The temperature of the steel bomb is now greater than the water it is immersed in so there is heat transfer from the bomb to the water. All of the heat transfer occurs within the isolated system. Heat lost by the bomb equals heat gained by the water and internal energy is unchanged.
Now consider FIG 2. It is the same steel bomb but not immersed in water and now with thermal insulation around the bomb. Again we have an isolated system. Combustion takes place as before converting potential to kinetic energy and raising the temperature of the bomb. There is no change in internal energy of the bomb. However there is no heat transfer involved as occurred in FIG 1. The fact that there is no heat transfer involved does not change the heating potential or heating value of the combustion process. In fact, that potential is determined by the Oxygen Bomb Calorimeter of FIG 1.
Hope this helps.
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$\begingroup$ Thank you for detailed answer, I actually figured it out later. It all depends on how we define internal energy or what do we include within it. In thermodynamics, energy stored in chemical bonds is not usually taken into account as internal energy. We usually take ideal gas as a reference in which there are no intermolecular interactions. So, in example of exothermic reaction we can pretend that heat of reaction is like heat which came from surroundings regardless of system being isolated since in both scenarios it has the same effect. $\endgroup$ Commented Apr 29, 2021 at 18:04
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$\begingroup$ Picking reference state as ideal gas is actually much better than how I defined it because we can calculate temperature change via change of internal energy which can't be done if you pick my reference state since internal energy doesn't change according to my reference. In order to calculate it, we would need to know energy of all chemical bonds in our reaction and calculate changes in average kinetic energy of atoms and molecules which is much more difficult and less practical especially for engineers like myself. $\endgroup$ Commented Apr 29, 2021 at 18:04
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1$\begingroup$ @DarioMirić There are a number of statements here that I take issue with, but by response would be too long to be appropriate in comments. Would you be interested in discussing this in Chat? $\endgroup$– Bob DCommented Apr 29, 2021 at 20:13
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$\begingroup$ @DarioMirić take O2. Even treating it as ideal, the fact that there'are not intermolecular interactions (which by the way must be corrected depending on conditions, as usual) isn't relevant concerning the potential energy stored within the molecule. We do not pretend energy comes from the surrounding. This is indeed molecular (within the molecule). $\endgroup$ Commented Apr 30, 2021 at 9:09
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$\begingroup$ @Bob D, no problem although I don't know how to start one. $\endgroup$ Commented Apr 30, 2021 at 10:30