I know that a closed adiabatic container acts as an isolated system, so I expect the internal energy of the system to be a constant as in an isolated system. But since it is an exothermic reaction, temperature will increase, and since internal energy is a function of temperature it should increase. This is a direct contradiction. Can anyone please help?
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$\begingroup$ Have you heard about energy conversion ? $\endgroup$– PoutnikCommented Apr 25, 2021 at 11:09
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Temperature will increase. If system is adiabatic, it doesn't neccesarily mean that temperature of the system can't change because even though system can't exchange energy with surroundings, processes inside can generate heat. In your example, exothermic reaction. Note that, internal energy in your case stays the same even though temperature increased. This is because internal energy is the sum of kinetic energy of all molecules and atoms and all potential enegy of interaction between atoms chemically bonded and between all molecules (intermolecular bonds). When exothermic reaction happens, potential energy of interaction between atoms is partially converted into heat because if reaction is exothermic, chemical bonds formed (of products)are stronger than of reactants. Potential energy is only converted to heat which is kinetic energy of atomic and molecular motion (energy conservation - 1st law of thermodynamics). Since internal energy is sum of potential and kinetic energy of atoms and molecules it stays the same, only potential enegy is smaller and kinetic is bigger after exothermic reaction. If you have any misunderstandings, fire them to me.
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$\begingroup$ If a reaction happens in an isolated system will it reach equilibrium? We know that entropy must be maximized. So what is the equilibrium state that maximizes entropy when a reaction takes place inside an isolated system? $\endgroup$– AntonCommented Jun 29, 2021 at 14:12
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$\begingroup$ Yes. You would need to solve for what extent of reaction entropy reaches its maximum, dS/dX = 0 (X is extent of reaction) in the same way as you needed to solve dG/dX = 0 for closed systems at constant p and T, to find equilibrium X. $\endgroup$ Commented Jul 3, 2021 at 14:24