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$$\Delta G = \Delta H_\text{system}-T\Delta S_\text{system}$$

$$ \begin{array}{ccc} \hline \text{Sign of}~ΔH & \text{Sign of}~ΔS & \text{Spontaneity} \\ \hline + & + & \text{The reaction is spontaneous at high temperature} \\ + & - & \text{The reaction is never spontaneous} \\ - & - & \text{The reaction is spontaneous at low temperature} \\ - & + & \text{The reaction is always spontaneous} \\ \hline \end{array} $$

When enthalpy change is negative, the reaction is exothermic, which means it releases energy into the surroundings. If the system is losing energy, shouldn’t the entropy of the system always decrease? I understand mathematically that $\mathrm{d}S = \mathrm{d}Q/T$, and if there is heat exchange, then entropy change can be positive.

But intuitively, if energy is taken away from the system, shouldn’t the entropy of that same system decrease? For some reason, I always thought that more energy means more entropy.

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For example, suppose you have a solid block of TNT.

It explodes and releases much energy. $\Delta H$ is negative.

Gaseous products like nitrogen, carbon dioxide and water vapor are formed. The system has become more disordered, so entropy has increased.

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  • $\begingroup$ "The system has become more disordered, so entropy has increased" I understand this statement intuitively, but how to actually express this mathematically? Does this mean that reversible conversion of TNT to gases requires $\triangle H$ to be positive? $\endgroup$
    – Eisenstein
    Commented Mar 28 at 4:34
  • $\begingroup$ @Eisenstein Mathematically, $\Delta H$ is negative and $\Delta S$ is positive and $\Delta G$ is negative. Reversible would be much different than an explosion. Reversible means the change is slow. $\Delta H$ , $\Delta S$ and $\Delta G$ would all be the same, but more work on surroundings would be obtainable if it was a reversible process. $\endgroup$
    – DavePhD
    Commented Mar 28 at 12:15

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