Gibbs Free Energy Equation: why is the entropy change of the system not q/T ?

In summary, the entropy change of the system in the Gibbs Free Energy Equation is given by ΔS(syst) = ΔH(syst)/T for systems involving a phase change, but for systems involving a chemical reaction, G changes as a result of changes in the amounts of the various species present and the equation S=q/T only holds true for reversible reactions in equilibrium.
  • #1
SPG
1
0
Homework Statement
The way we stipulate the conditions for the Gibbs Free Energy Equation (GFEE) - it seems to me that ΔS(system) must always equal negative ΔS(surroundings) ? Obviously this is incorrect but I can't see why. The GFEE says the magnitude of the entropy change of the surroundings is given by

ΔS(surr) = ΔH(syst)/T

Since the only exchange between the system-surroundings is the heat of ΔH(syst), then the change of heat in the system is -tautologically- also the same ΔH that the surroundings experience. So then surely the magnitude of the entropy change of the system is also the same quantity as the above expression for that of the surroundings. Because both system and surroundings have the same magnitude of heat change at the same temperature with no PV work or matter exchange i.e. it erroneously appears to me that

ΔS(syst) = ΔH(syst)/T = ΔS(surr)

But of course if this were true it would render the GFEE useless since ΔG would always be zero.
So why does the normal entropy equation ΔS=q/T not work for the entropy change of the SYSTEM in the GFEE?
What instead is the expression (or way to consider) the entropy change of the system in the GFEE ?
I have never seen this discussed in my Thermodynamics text books or anywhere online.
Relevant Equations
ΔS(univ) = ΔS(surr) + ΔS(syst)
ΔS = q/T
q=ΔH
ΔS(surr) = ΔH(syst)/T
leading towards Gibbs:
ΔS(univ) = ΔH(syst)/T + ΔS(syst)
multiply by T to give Gibbs Free Energy Equation
T.ΔS(univ) = ΔG(univ) = ΔH(syst) - T.ΔS(syst)
What is the entropy change of the system in the Gibbs Free Energy Equation?
The general expression for entropy change is ΔS=q/T
The only exchange between the system and the surroundings is ΔH done reversibly, with no PV work and no matter transfer, therefore
q(syst) = ΔH(syst)
therefore surely the entropy change of the system is given by
ΔS(syst) = q(syst)/T
therefore
ΔS(syst) = ΔH(syst)/T
but I know this isn't correct, i just can't see why this expression for entropy change of the system is incorrect
 
Physics news on Phys.org
  • #2
SPG said:
Problem Statement: The way we stipulate the conditions for the Gibbs Free Energy Equation (GFEE) - it seems to me that ΔS(system) must always equal negative ΔS(surroundings) ? Obviously this is incorrect but I can't see why. The GFEE says the magnitude of the entropy change of the surroundings is given by

ΔS(surr) = ΔH(syst)/T

Since the only exchange between the system-surroundings is the heat of ΔH(syst), then the change of heat in the system is -tautologically- also the same ΔH that the surroundings experience. So then surely the magnitude of the entropy change of the system is also the same quantity as the above expression for that of the surroundings. Because both system and surroundings have the same magnitude of heat change at the same temperature with no PV work or matter exchange i.e. it erroneously appears to me that

ΔS(syst) = ΔH(syst)/T = ΔS(surr)

But of course if this were true it would render the GFEE useless since ΔG would always be zero.
So why does the normal entropy equation ΔS=q/T not work for the entropy change of the SYSTEM in the GFEE?
What instead is the expression (or way to consider) the entropy change of the system in the GFEE ?
I have never seen this discussed in my Thermodynamics textbooks or anywhere online.
Relevant Equations: ΔS(univ) = ΔS(surr) + ΔS(syst)
ΔS = q/T
q=ΔH
ΔS(surr) = ΔH(syst)/T
leading towards Gibbs:
ΔS(univ) = ΔH(syst)/T + ΔS(syst)
multiply by T to give Gibbs Free Energy Equation
T.ΔS(univ) = ΔG(univ) = ΔH(syst) - T.ΔS(syst)

What is the entropy change of the system in the Gibbs Free Energy Equation?
The general expression for entropy change is ΔS=q/T
The only exchange between the system and the surroundings is ΔH done reversibly, with no PV work and no matter transfer, therefore
q(syst) = ΔH(syst)
therefore surely the entropy change of the system is given by
ΔS(syst) = q(syst)/T
therefore
ΔS(syst) = ΔH(syst)/T
but I know this isn't correct, i just can't see why this expression for entropy change of the system is incorrect
If you are talking about a system where a phase change is involved, then it is correct. If you are talking about a system involving a chemical reaction, then G changes as a result of changes in the amounts of the various species present. Which are you referring to?

See the following thread: https://www.physicsforums.com/threads/thermochemistry-challenge-problem-chets-paradox.913567/
 
  • #3
The equation S=q/T holds true only for reversible reactions, i.e. for a reaction which moves through a sequence of equilibrium states. And in deed in equilibrium ##\Delta G=0##. When you see some ##\Delta G## values, they typically refer to some standard state, where all reactants are at a concentration (more precisely activity) of 1 mole/L, which is but rarely an equilibrium state.
 

Related to Gibbs Free Energy Equation: why is the entropy change of the system not q/T ?

1. Why is the entropy change of the system not q/T?

The entropy change of the system, denoted as ΔS, is not equal to q/T because q/T only represents the change in entropy of the system at constant temperature. However, the change in entropy of a system also depends on the amount of energy transferred as heat, the temperature, and the reversible work done on or by the system.

2. How is the Gibbs free energy equation related to entropy change?

The Gibbs free energy equation, ΔG = ΔH - TΔS, is directly related to entropy change. This equation shows that the change in Gibbs free energy (ΔG) is equal to the change in enthalpy (ΔH) minus the change in entropy (ΔS) multiplied by the temperature (T). This relationship highlights the importance of entropy in determining the spontaneity of a process.

3. What does a negative value of ΔS indicate in the Gibbs free energy equation?

A negative value of ΔS in the Gibbs free energy equation indicates that the total entropy of the system and its surroundings is decreasing. This means that the process is not spontaneous and will not occur without an external input of energy. This could be achieved by increasing the temperature or performing work on the system.

4. How does temperature affect the entropy change in the Gibbs free energy equation?

Temperature plays a crucial role in determining the entropy change in the Gibbs free energy equation. A higher temperature results in a larger value for the q/T term, meaning that the change in entropy of the system is more dependent on the amount of heat transferred. This also means that at higher temperatures, the system is more likely to have a positive value for ΔS, making the process more spontaneous.

5. Can the Gibbs free energy equation be used to predict the direction of a reaction?

Yes, the Gibbs free energy equation can be used to predict the direction of a reaction. If the value of ΔG is negative, the reaction is spontaneous and will proceed in the forward direction. If ΔG is positive, the reaction is not spontaneous and will not occur without an external input of energy. If ΔG is zero, the system is at equilibrium and the reaction is reversible.

Similar threads

  • Biology and Chemistry Homework Help
Replies
10
Views
397
  • Biology and Chemistry Homework Help
Replies
13
Views
346
  • Biology and Chemistry Homework Help
Replies
2
Views
219
  • Biology and Chemistry Homework Help
Replies
1
Views
2K
  • Biology and Chemistry Homework Help
Replies
3
Views
2K
Replies
11
Views
623
  • Introductory Physics Homework Help
Replies
2
Views
774
  • Biology and Chemistry Homework Help
Replies
6
Views
2K
Replies
4
Views
2K
  • Biology and Chemistry Homework Help
Replies
9
Views
2K
Back
Top