What does it mean once tbelow is an unfavorable delta \$S\$ \$(-Delta S)\$? One perchild asked if this implied negative entropy, but I don"t watch how this could be possible.

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Does this indicate an unfavorable readjust in entropy? If the former, just how have the right to negative entropy exist? Negative delta S (\$Delta S ) is a decrease in entropy in regard to the system.

For physical processes the entropy of the universe still goes up but within the confines of the system being studied entropy decreases.

One instance is a freezer via a cup of liquid water in it. The freezer will certainly use the electrical power coming in to pump warm from the water until it becomes a solid (ice). At which suggest the entropy of the mechanism (the contents of the freezer) decreases, yet the electric energy essential to be produced to power the freezer such as coal (burning a solid to a gas) and also heat was wasted by the freezer in the procedure both of which create larger quantities of entropy than was reduced in the system by the freezer.

For chemical processes entropy can be a good driver of many kind of reactions but it is not absolute. A system"s favorcapacity to release energy (enthalpy) competes via entropy. For instance, an electron of hydrogen might have greater entropy if it drifts from the core proton however the electrostatic pressures (and also quantum mechanics) energetically save it bound to the atom. For isobaric procedures, you much determine the change in Gibbs cost-free energy for the reaction to understand which method it is moved. For isochoric processes, you must identify the Helmholtz free power to know which means a reaction is moved.

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One instance is the oxidation of iron in air. When the oxygen is in the gas state it has actually higher entropy yet the energy of bonding with iron is so good that at normal pressures oxygen goes from the gas phase and the iron rusts ("enthalpy wins") as delta G is negative.

Now we need to think about statistical thermodynamics, this procedure is pressure dependent. At normal atmospheric press, the forward price of oxygen entering the gas phase is the same as the reverse process. If the iron oxide were held in a sufficiently high vacuum the reverse procedure would happen and also the iron oxide would certainly alleviate earlier to iron favor many kind of asteroids ("entropy wins"; note: thermodynamics is equilibrium after an boundless time). Accounting for pressure modifies the Gibbs totally free energy equation to: \$\$Delta G = Delta G^circ -RT ln(P) = Delta H^circ -TDelta S^circ -RT ln(P)\$\$

One thing to note is that for chemical reactions the entropy and also enthalpy values are for a traditional temperature (such as \$298 extK\$). For a spontaneous mechanism with \$Delta S^circ enthalpy should be negative, this heat in fact is absorbed by the device or the environment and produces entropy according to: \$\$ int mathrm dS equiv int fracC_vT,mathrm dT equiv int fracmathrm dQT \$\$

This in and of itself produces some entropy in the universe though may not net over zero as the bonding power is still the major driving force.