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Can I prove spontaneity using the entropy change? 
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#1
Mar1012, 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 


#2
Mar1012, 03:01 PM

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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:
ΔS_{uni} = ΔS_{sys} + ΔS_{surr} >= 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: ΔS_{surr} = q/T_{surr}. Therefore, the quantity ΔH/T_{surr} will tell you about the change in entropy of the surroundings. Given that ΔS_{surr} = ΔH/T_{surr}, we can plug this value into our expression for the entropy of the universe to give: ΔS_{uni} = ΔS_{sys} + ΔS_{surr} = ΔS_{sys}  ΔH/T_{surr} 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. 


#3
Mar1012, 03:19 PM

P: 783

I see, thank you but there are still a few things I don't get. ΔS_{surr} = ΔH/T_{surr}, 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 ΔS_{uni} = ΔS_{sys} + ΔS_{surr} = ΔS_{sys}  ΔH/T_{surr} 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 


#4
Mar1012, 03:31 PM

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Can I prove spontaneity using the entropy change?
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 (T_{surr} = T_{sys}), but I'll have to check my chemistry texts to make sure. 


#5
Mar1012, 03:35 PM

P: 783

BiP 


#6
Mar1012, 10:56 PM

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P: 1,378

dS_{sys} = dq_{rev}/T is always true, where dq_{rev} represents the heat transferred from the surroundings to the system reversibly.
dS_{surr} = dq/T_{surr} 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. 


#7
Mar1112, 10:44 AM

P: 783

BiP 


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