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In Atkin's Chemical Principles: The Quest for Insight, the Clausius inequality is said to be

$$\Delta S \geq \frac qT$$ where equality applies to a reversible process.

Why would the change in entropy be different for a reversible process compared to an irreversible process if entropy is a state function, and the initial and final properties of the system remain the same?

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The variation of the entropy for a reversible process is indeed $\Delta S_{rev} = \frac{\Delta Q_{rev}}{T}$, which is a trivial thermodynamic fact. During the process, even if the initial and final elements are conserved, the number of microstates of the system can change. If the initial state has $\Omega_1$ microstates and the final state has $\Omega_2$ microstates, the variation in entropy will be $\Delta S_{irr} = k_b \log \frac{\Delta_2}{\Delta_1}$. The latter is crucial, if there is no microstate created $\Omega_2 = \Omega_1$, the process is reversible so that $\Delta S = \Delta S_{rev}$. However, when microstates are created, $\Delta S > \Delta S_{rev} = \frac{\Delta Q_{rev}}{T}$, the process is irreversible, because $\Delta S_{irr} > 0$ cannot be destroyed because it contributes to the free energy.

At $T=0\text{ K}$, the increase in entropy has no noticeable effect $\Delta S_{irr}$, there is no need to create new microstates, some processes are reversible but at a finite temperature $T>0$ the increase in entropy reduces the free energy more or less considerably with $-T\Delta S$, a disordered system is more favorable and the processes are less reversible.

An example will be identical polarized dipoles (either electric or magnetic) at $T=0\text{ K}$, some will flip to form antiparallel configuration at $T>0 \text{ K}$, creating more information and reducing the free energy.

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You're right and the textbook is also right because you're considering two different things. You're considering a process that's a complete thermodynamic cycle where the state of the system (which is the substance itself in chemistry texts) is the same before as after the process. On the other hand, the textbook is considering only the part of a thermodynamic cycle when the substance's entropy is changing due to the flow of heat into or out of the substance.

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    $\begingroup$ I don't think OP was referring to a cycle in which the end state is the same as beginning state, but rather comparing reversible and irreversible processes that convert the system from the same starting state to the same ending state, which comes down to a difference in w vs q for the processes. $\endgroup$
    – Andrew
    Jul 10, 2023 at 15:32
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    $\begingroup$ The entropy of a given state is constant. However, returning to a given states usually requires that the entropy of the universe increases. Imagine the last slice of bread on your plate; it has an entropy. Replace it tomorrow with an identical slice; it has the same entropy. What happened to the entropy of the universe? In everyday life this is reflected in the efficiency of energy use in an operation. Greater efficiency less entropy gain. $\endgroup$
    – jimchmst
    Jul 11, 2023 at 8:49

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