A reaction has Δ H < 0 and Δ S < 0: it releases heat and reduces disorder. When does it run?

At low temperature only, since TΔ S grows with T

At any temperature, since releasing heat is enough on its own

At high temperature only, since reactions generally go faster when hot

Never, because a reaction cannot reduce disorder

Physics I / Integrated chemical reasoning

Practice question · Multiple choice

A reaction has $\Delta H < 0$ and $\Delta S < 0$ : it releases heat and reduces disorder. When does it run?

Hints

Write $\Delta G = \Delta H - T\Delta S$ and put in the signs.
With $\Delta S$ negative, $-T\Delta S$ is positive and grows. What happens to $\Delta G$ as $T$ rises?

Show the answer

D. At low temperature only, since $T\Delta S$ grows with $T$

Why

Both terms matter and temperature is the referee. $-T\Delta S$ is positive here and grows linearly, so $\Delta G$ starts negative and crosses zero at $T = \Delta H/\Delta S$ . Exothermic is not sufficient, and disorder can fall locally provided the surroundings gain more, which is what the heat released does. This is why freezing is spontaneous below 0 °C and not above.

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A reaction has ΔH<0\Delta H < 0ΔH<0 and ΔS<0\Delta S < 0ΔS<0: it releases heat and reduces disorder. When does it run?

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A reaction has $\Delta H < 0$ and $\Delta S < 0$ : it releases heat and reduces disorder. When does it run?