What go it mean when there is a negative delta $S$ $(-\Delta S)$? One human being asked if this implied negative entropy, yet I don"t see exactly how this can be possible.

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Does this indicate a an unfavorable change in entropy? If the former, just how can an unfavorable entropy exist?


Negative delta S ($\Delta S ) is a decrease in entropy in regard come the system.

For physical processes the entropy the the cosmos still walk up however within the boundaries of the mechanism being studied entropy decreases.

One example is a freezer through a cup of liquid water in it. The freezer will use the electric energy coming in come pump heat from the water till it becomes a solid (ice). In ~ which allude the entropy the the mechanism (the materials of the freezer) decreases, however the electrical energy needed to be produced to power the freezer such together coal (burning a solid to a gas) and heat to be wasted through the freezer in the process both of which produce larger amounts of entropy 보다 was decreased in the system by the freezer.

For chemical processes entropy can be a an excellent driver of many reactions yet it is no absolute. A system"s favorability come release power (enthalpy) competes with entropy. Because that example, one electron that hydrogen might have higher entropy if the drifts indigenous the core proton yet the electrostatic pressures (and quantum mechanics) energetically keep it bound to the atom. Because that isobaric processes, girlfriend much identify the adjust in Gibbs totally free energy for the reaction to know which method it is driven. For isochoric processes, you must recognize the Helmholtz cost-free energy to recognize which method a reaction is driven.

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One example is the oxidation of stole in air. When the oxygen is in the gas state the has greater entropy yet the energy of bonding v iron is so an excellent that at common pressures oxygen goes native the gas phase and the iron rusts ("enthalpy wins") as delta G is negative.

Now us must consider statistical thermodynamics, this process is press dependent. At typical atmospheric pressure, the forward price of oxygen entering the gas step is the same as the reverse process. If the iron oxide were organized in a saturated high vacuum the reverse process would occur and also the stole oxide would reduce ago to iron like plenty of asteroids ("entropy wins"; note: thermodynamics is equilibrium after an limitless time). Accounting for press modifies the Gibbs free energy equation to: $$\Delta G = \Delta G^\circ -RT \ln(P) = \Delta H^\circ -T\Delta S^\circ -RT \ln(P)$$

One point to keep in mind is the for chemistry reactions the entropy and also enthalpy values room for a conventional temperature (such together $298\ \textK$). Because that a spontaneous mechanism with $\Delta S^\circ enthalpy have to be negative, this heat in truth is soaked up by the mechanism or the environment and also produces entropy follow to: $$ \int \mathrm dS \equiv \int \fracC_vT\,\mathrm dT \equiv \int \frac\mathrm dQT $$

This in and also of itself produces part entropy in the cosmos though might not net above zero as the bonding power is quiet the major driving force.