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Some chemical reactions while producing the product molecules releases heat to the surroundings and such reactions are called “exothermic reactions”. Is it true that such systems must be showing energy loss in the process meaning “ΔE ” must be negative?

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    $\begingroup$ You are right, if you admit that the energy $E$ included in your $"\Delta E" $ correspond to what chemists call "internal energy" at constant volume and "enthalpy" at constant pressure. $\endgroup$ – Maurice Nov 23 '20 at 12:27
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No. If means that to maintain the temperature of the products equal to the temperature of the reactants, you need to remove heat, and, form the first law, if you do this, then $\Delta E$ is negative. However, if you don't remove the heat, the temperature of the products will be higher than those of the reactants, and $\Delta E$ of the system will then be zero (as required by the first law).

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  • $\begingroup$ Yes , you are right to say that become from 1st law.It also depends on the Macroscopic work done on or by the system. $\endgroup$ – srijan Sri Nov 23 '20 at 17:39
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    $\begingroup$ @Robert Patrick I assumed he was talking about a rigid container. I removed that from my first draft to keep the answer simple. $\endgroup$ – Chet Miller Nov 23 '20 at 17:49
  • $\begingroup$ Ok Thank you . . $\endgroup$ – srijan Sri Nov 23 '20 at 18:18
  • $\begingroup$ $\delta$ E = q+W . So in your case q which is your “Heat loss” is negative but if somehow there is some W done on the bod( By compressing a piston ;one way ).Then your E can be positive.So it depends. $\endgroup$ – srijan Sri Dec 17 '20 at 10:23
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    $\begingroup$ @user277768 At the OPs level of understanding, I made the determination that it would be confusing to him to include work. Actually, the "heat of reaction" is defined as $\Delta H$, assuming constant pressure and temperature, in which case some work is already included. $\endgroup$ – Chet Miller Dec 17 '20 at 13:02

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