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I was asked this question at class,

What is the thermal state of a reaction at equilibrium, the forward reaction of which, is exothermic?

I conjectured that since the forward (exothermic) reaction and the backward (endothermic) reaction occur at equal rates, the reaction at equilibrium would be athermic since there doesn't seem to be any net heat evolution. However, my teacher said this was not true.

Could someone tell me where I'm going wrong with my reasoning?

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    $\begingroup$ Define "thermal state"... $\endgroup$ – Zhe Nov 15 '17 at 18:31
  • $\begingroup$ Well what we usually understand by thermal state at my school is the sign of enthalpy change, if it is negative we say it is exothermic, if positive we say it is endothermic, and if nil we say it is athermic. $\endgroup$ – George daou Nov 15 '17 at 20:50
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    $\begingroup$ If the sign of the enthalpy change is all you want, then even at equilibrium, the enthalpy change of the forward reaction will remain the same, and that of the backward reaction will also remain the same. Now, you can't add forward plus backward and say that it is athermic; enthalpy is defined only when we consider a certain direction of the reaction. Also, see this related question: chemistry.stackexchange.com/questions/43039/… $\endgroup$ – Abhigyan Chattopadhyay Nov 16 '17 at 1:58
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You're right, as far as I can see. If we look at a particular transformation $A \leftrightharpoons B$ there is a forward reaction and a backwards reaction. They are associated with enthalpy change of different signs and the same magnitude. At equilibrium the rates are the same, so the forwards and backwards reactions release and absorb the same amount of heat and it appears as if nothing is happening.

There is no separate "reaction at equilibrium" with its own enthalpy change. It's not really clear to me what "thermal state of a reaction" means, as systems have states and reactions are pathways between systems.

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