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I know that:

ds = dq/T for reversible processes.

Can I write change in entropy of a system in which a chemical reaction is happening using this equation. Basically, my question is about whether a chemical reaction happening in an open vessel(isobarically and isothermally) be treated as a reversible process?

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  • $\begingroup$ Google van’t Hopf equilibrium box. $\endgroup$ Dec 9, 2019 at 18:52
  • $\begingroup$ You can't treat it as reversible. Any spontaneous chemical reaction at constant $T$ and $P$ is necessarily irreversible. $\endgroup$
    – ratsalad
    Dec 11, 2019 at 20:30
  • $\begingroup$ @ChetMiller van't Hoff $\endgroup$
    – Karsten
    Jan 29, 2022 at 17:23

2 Answers 2

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Can I write change in entropy of a system in which a chemical reaction is happening using this equation. Basically, my question is about whether a chemical reaction happening in an open vessel(isobarically and isothermally) be treated as a reversible process?

If the Gibbs energy of reaction is non-zero and you don't capture it as work, you can't (whenever the total entropy increases, it is non-reversible).

$dS = dq/T$ is most useful for calculating the entropy change of a waterbath that is heated or cooled reversibly.

There are other ways you could measure the entropy, though. Entropy is a state function, so if you can find a reversible path (electrochemistry perhaps?), you could measure the entropy using the given relation.

In most cases, it is easier to measure the enthalpy and Gibbs energy change, and determine the entropy change from those measurements (see answer by Maurice).

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The simplest way is to calculate $\Delta$H and $\Delta$G = $\pu{-RTln K}$ (or $\pu{\Delta G = -zEF}$) of the reaction, then calculate the difference $\Delta$H - $\Delta$G. And you divide this difference by the temperature T. The result is $\Delta$S of the reaction.

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