$$\ce{N2(g) + O2(g) → 2NO(g)}$$

My thinking: since gases are combined together, so there randomness or disorder gets decreased. Hence the sign of entropy change of the system should be negative.

Book Solution: the sign of entropy change is impossible to predict.

Why can't we predict the sign of entropy change for the given system?

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    $\begingroup$ Why would randomness get decreased? What's "together" about it? We had two gas molecules, and we still have two gas molecules. $\endgroup$ – Ivan Neretin Mar 28 at 5:58
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    $\begingroup$ If the reaction reaches equilibrium (and temperature constant) then entropy will have increased as there are three types of molecules present. If it goes to completion so that only NO is present then entropy decreases as there is only one type of molecule left and not two. If the reaction were A+B$\to $ C+D then the entropy would not change. $\endgroup$ – porphyrin Mar 28 at 8:42
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    $\begingroup$ It is impossible to predict based on the simple basis covered by the book at that point, chapter or paragraph (entropy as disorder, indistinguishable particles). $\endgroup$ – Alchimista Mar 28 at 9:03
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    $\begingroup$ The entropy of mixing is not the only factor here. There is also the entropy change as a result of breaking and making new chemical bonds (and energetic effects associated with this). $\endgroup$ – Chet Miller Mar 28 at 12:04
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    $\begingroup$ I would think that the entropy should increase in this example to first order. The NO molecule is less symmetric, so by rough hand-waving there are more microstates in the NO system. $\endgroup$ – Zhe Mar 31 at 16:03

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