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escapeartistq

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Entropy question
« on: May 28, 2005, 01:04:16 PM »
I don't understand the following statement in the context of Thermodynamics
"All non-spontaneous reactions involve an increase in the energy of the system."

The second law says that dS(universe) > 0 in a spontaneous reaction.

But if the reaction is not spontaneous it should be dS(universe) = 0 right?
The system evolves to a maximum of entropy in the equilibrium, so

dS(universe) = dS(sys) + dS(ext) = 0 => dS(ext) = -dS(sys)

but because dS(ext) = -dH(sys)/T then dS(sys) = dH(sys)/T and this ratio will have a maximum value at the equilibrium.

Suppose the process is endothermic and the pressure is constant.
Then dU = dH + W so it is possible that AH > 0 and W < 0 (work done by the system) so that W = -dH and dU = 0 which means that the energy of the system remains constant.

Or am I wrong? I would appreciate some help here.




Offline eugenedakin

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Re:Entropy question
« Reply #1 on: June 16, 2005, 11:32:54 PM »
Hello Escapeartistq,

Yes, the context of Thermodynamics and its laws can be daunting at first. I will explain my version of the answers to your questions (if it doesnt make sense, feel free to ask more questions).

First question:
"All non-spontaneous reactions involve an increase in the energy of the system"
This example is only for explanation purposes.  Example:  Pour 1 tablespoon of gasoline in a metal spoon.  At room temperature, no exothermic reaction happens.  In order for the exothermic reaction to occur, we need to add a spark (fire) in the presence of oxygen and gasoline for combustion to occur.  In order for the exothermic reaction to occur, we need to increase the energy of the system (by adding the heat/spark energy from a match).  If there is no energy given to the gasoline, combustion does not occur.

Second question:
"But if the reaction is not spontaneous it should be dS(universe) = 0 right?"
Unfortunately, no.  Lets use the same example as above.  At room temperature, the gasoline remains in the spoon and a match is NOT available.  Gasoline will go from an ordered state (liquid) to a more disordered state (gaseous)  by evaporation.  There is the arguement that nothing has a dS of zero, but is very close to zero. If we use an example that takes a signifacently longer period of time, lets say the decay of the Uranium isotope, its half life is approximately 4,470,000,000 years.  The decay reaction could be close to zero, but is not actually zero, probably closer to 0.000000001 (or some such small number).  Since the universe is so large, the number would probabaly be a small number, but would not be zero.

Third Scenario:
Suppose the process is endothermic and the pressure is constant.
Then dU = dH + W so it is possible that AH > 0 and W < 0 (work done by the system) so that W = -dH and dU = 0 which means that the energy of the system remains constant.

Lets provide some fictional numbers to explain this scenario.   Lets use the example where two liquids (A, B) are added to each other in a coffee cup and the combined liquids (C) are cooler than the mean average of A and B.
A + B => C

Lets say that A has a total energy of 5 units, and B has a total energy of 6 units.  When A and B combine to form C, additional heat energy (2.0 units) is required to form C (13 units), which is visible in the lowering of the temperature of C.  Energy is always lost in any system, no matter how efficient it is. Lets say the reaction is 99.9% efficient (This is an extremely high value).  This means that A + B will have 4.995 + 5.994 = 10.989 units of available energy.  Since 13 units of energy is required to form C, this relates to 2.011 units of energy must be supplied for the formation of C to occur.  Energy, in the form of heat, is removed from the liquid, air, and coffee cup to produce C.  

The energy efficiency of the reaction (this is the 100.0 % - 99.9% = 0.1%) is required since molecules vibrate, have concentration gradients, and need to physically be in contact with one another for the reaction to occur.  All of these actions require energy which lowers the efficiency of the reaction.

Although the cooled liquid (C) is in a more ordered state, this comes at the expense of a more disordered state around it.  In this case, the extra 0.011 units of energy required is the increased disorder of the entire system.

I hope this helps, and do not hesitate to follow-up with more questions.

Sincerely,

Eugene Dakin Ph.D., P.Chem.
There are 10 kinds of people in this world: Those who understand binary, and those that do not.

Offline Juan R.

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Re:Entropy question
« Reply #2 on: June 17, 2005, 04:00:24 AM »

For a closed system

dS >= dU/T (*)

that is

dU =< TdS

If we are working at constant entropy (basically in a mechanistic approach) then

dU =< 0

that is, energy is constant or decrease in any spontaneous process when entropy is approx. constant. In fact, this is the well known law of mechanics. The increase of energy is not permited in any spontaneous process at constant S because would imply a violation of (*)

Note: A dS = 0 is also spontaneous, simply the reaction is reversible.

dS < 0 is an impossible reaction.
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