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Thermodynamics
Entropy and Spontaneity
2 questions, each with its answer.
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In which case change in entropy is negative?Evaporation of water: $H_2O(l)\rightarrow H_2O(v)$, $\Delta S>0$ Expansion of a gas at constant temperature: $\Delta S>0$ Sublimation of solid to gas: $\Delta S>0$ $2H(g)\rightarrow H_2(g)$: $\Delta S<0$ (since $\Delta n_g<0$)
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Considering entropy (S) as a thermodynamic parameter, the criterion for the spontaneity of any process is :-According to the second law of thermodynamics, a process is spontaneous if the total entropy of the universe increases.
$\Delta S_{total} = \Delta S_{universe} = \Delta S_{system} + \Delta S_{surroundings}$
For a spontaneous process: $\Delta S_{system} + \Delta S_{surroundings} \gt 0$. At equilibrium $\Delta S_{total} = 0$.
Entropy of the system alone or of the surroundings alone is not enough; for example, freezing of water below $0^\circ C$ is spontaneous although $\Delta S_{system}$ is negative, because $\Delta S_{surroundings}$ is larger and positive.
So the criterion is $\Delta S_{system} + \Delta S_{surroundings} \gt 0$.
That is every question in Entropy and Spontaneity.
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