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Can I prove spontaneity using the entropy change?

by Bipolarity
Tags: entropy, prove, spontaneity
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Bipolarity
#1
Mar10-12, 02:07 PM
P: 783
According to the second law of thermodynamics, for any spontaneous process,
[itex] \Delta S_{uni}> 0 [/itex] .

And for any reversible process,
[itex] \Delta S_{uni}= 0 [/itex] .

This means that no process can be reversible and spontaneous at the same time.

However, what I don't understand is the connection to [itex]\Delta G [/itex]. Suppose I know that [itex] \Delta S_{uni}> 0 [/itex], i.e. the process is spontaneous. Can I use this equation to prove that for the system in question, [itex]\Delta G < 0[/itex]. ? Using only the definition of the terms and knowing that the process is spontaneous?

Thanks!

BiP
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Ygggdrasil
#2
Mar10-12, 03:01 PM
Other Sci
Sci Advisor
P: 1,378
The relationship ΔG <= 0 (i.e. that reactions seek to minimize their free energy) is derived from the second law of thermodynamics (specifically for the case of a reaction occurring at constant pressure). The second law states that the entropy of the universe is always increasing. There are two components to the entropy of the universe which we must consider, the entropy of the system and the entropy of the surroundings. By splitting the entropy of the universe into these two components, we can write the second law as follows:

ΔSuni = ΔSsys + ΔSsurr >= 0

From this equation, we can see that there will be trade offs between the entropy of the system and the entropy of the surroundings. In order for the system to lose entropy and become more ordered, the surroundings must gain entropy to compensate and vice versa.

The formula for the Gibbs free energy reflects these trade offs between the entropy of the system and the entropy of the surroundings. Recall that for a reaction occurring at constant pressure, the change in enthalpy is equal to the amount of heat released/absorbed by the system, ΔH = q. Further, remember that the change in enthalpy of the surroundings is given by the following formula: ΔSsurr = q/Tsurr. Therefore, the quantity ΔH/Tsurr will tell you about the change in entropy of the surroundings.

Given that ΔSsurr = -ΔH/Tsurr, we can plug this value into our expression for the entropy of the universe to give:

ΔSuni = ΔSsys + ΔSsurr = ΔSsys - ΔH/Tsurr

Applying the second law, we obtain the inequality:

ΔS - ΔH/T >= 0

Or, equivalently:

ΔH - TΔS <= 0

Which is exactly the equation that explains why spontaneous reactions tend to minimize their free energy.
Bipolarity
#3
Mar10-12, 03:19 PM
P: 783
I see, thank you but there are still a few things I don't get. ΔSsurr = -ΔH/Tsurr, isn't this only true for reversible processes? Why are we allowed to use it here? I thought we were dealing with a process that we assumed to be irreversible (i.e. universe's entropy must rise).

Also, I can't seem to fllow ΔSuni = ΔSsys + ΔSsurr = ΔSsys - ΔH/Tsurr to ΔS - ΔH/T >= 0. Yes it's just plugging everything in, but doesn't that mean the Gibbs Free Energy is defined in terms of the temperature of the surroundings? When you write ΔS - ΔH/T >= 0, are you specifically referring to the temperature of surrounding, or temperature of system? Or are they the same thing?

BiP

Ygggdrasil
#4
Mar10-12, 03:31 PM
Other Sci
Sci Advisor
P: 1,378
Can I prove spontaneity using the entropy change?

Quote Quote by Bipolarity View Post
I see, thank you but there are still a few things I don't get. ΔSsurr = -ΔH/Tsurr, isn't this only true for reversible processes? Why are we allowed to use it here? I thought we were dealing with a process that we assumed to be irreversible (i.e. universe's entropy must rise).

Also, I can't seem to fllow ΔSuni = ΔSsys + ΔSsurr = ΔSsys - ΔH/Tsurr to ΔS - ΔH/T >= 0. Yes it's just plugging everything in, but doesn't that mean the Gibbs Free Energy is defined in terms of the temperature of the surroundings? When you write ΔS - ΔH/T >= 0, are you specifically referring to the temperature of surrounding, or temperature of system? Or are they the same thing?

BiP
ΔSsurr = -q/Tsurr should be true for all processes. ΔSsys = q/Tsurr is true only for reversible processes.

Yes, you are correct that, in my derivation, the T in ΔS - ΔH/T >= 0 refers to the temperature of the surroundings whereas the ΔS and ΔH refer to the entropy and enthalpy of the system. Perhaps another assumption required for the ΔG < 0 proof is that the system is in thermal equilibrium with the surroundings (Tsurr = Tsys), but I'll have to check my chemistry texts to make sure.
Bipolarity
#5
Mar10-12, 03:35 PM
P: 783
Quote Quote by Ygggdrasil View Post
ΔSsurr = -q/Tsurr should be true for all processes. ΔSsys = -q/Tsurr is true only for reversible processes.

Yes, you are correct that, in my derivation, the T in ΔS - ΔH/T >= 0 refers to the temperature of the surroundings whereas the ΔS and ΔH refer to the entropy and enthalpy of the system. Perhaps another assumption required for the ΔG < 0 proof is that the system is in thermal equilibrium with the surroundings (Tsurr = Tsys), but I'll have to check my chemistry texts to make sure.
Ah I see! Thank you so much! But why should ΔSsurr = -q/Tsurr should be true for all processes? I'm sorry but I think I some basic mental gap in my understanding of entropy. Exactly when do you know whether you can use the Q/T definition and when you can not? Sorry if it's dumb question!

BiP
Ygggdrasil
#6
Mar10-12, 10:56 PM
Other Sci
Sci Advisor
P: 1,378
dSsys = dqrev/T is always true, where dqrev represents the heat transferred from the surroundings to the system reversibly.

dSsurr = -dq/Tsurr is always true, even if the system undergoes an irreversible change. This happens because we assume the surroundings are an infinitely large heat sink, so any heat transferred to/from the surrounds will not change the properties of the surroundings. Thus, heat absorption/release from the surroundings will always be reversible.
Bipolarity
#7
Mar11-12, 10:44 AM
P: 783
Quote Quote by Ygggdrasil View Post
dSsys = dqrev/T is always true, where dqrev represents the heat transferred from the surroundings to the system reversibly.

dSsurr = -dq/Tsurr is always true, even if the system undergoes an irreversible change. This happens because we assume the surroundings are an infinitely large heat sink, so any heat transferred to/from the surrounds will not change the properties of the surroundings. Thus, heat absorption/release from the surroundings will always be reversible.
I see! Thanks!~

BiP


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