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I don't understand why enthalpy change won't change at an equilibrium?

The reason I ask this is because of this question: This question

At first I thought the answer is C, but I realised that the amount depends on where the position of equilibrium lies. The correct answer is D, but doesn't enthalpy change (delta H) also depend on where the position of equilibrium lies?

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  • $\begingroup$ (C) can be incorrect as amount might refer to mass formed, which won't be the same at all. $\endgroup$ – Tamoghna Chowdhury Dec 31 '15 at 8:28
  • $\begingroup$ Trick questions. (D) is correct since the equilibrium position is reversed as the reaction itself is reversed, so the enthalpy change is also reversed (by your own logic). $\endgroup$ – Tamoghna Chowdhury Dec 31 '15 at 8:30
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The answer should be D. Because of the principle of microscopic reversibility, the forward enthalpy is opposite to the reverse enthalpy. This also follows from the conservation of energy if you will.

In this StackExchange question, there is a nice picture given of a potential energy surface. There you can readily see that the forward reaction enthalpy is opposite to the backward reaction enthalpy.

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The answer to this question is related to the definition of $\Delta H^0$. It is precisely defined as the enthalpy change between the following two thermodynamic equilibrium states:

State 1: Pure reactants (say, in separate containers) in stochiometric proportions at 1 bar and 25C

State 2: Pure products (say, in separate containers) in corresponding stochiometric proportions at 1 bar and 25 C

From this, you can see that the enthalpy change from State 2 to State 1 is minus the enthalpy change from State 1 to State 2

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