Equilibrium and Energetics help!

[Scienuk]

These questions are from the Spanish Chemistry HL 2016 paper one exam, I would appreciate it if you could help me understand them:

This question translates to: Which is correct for an isolated system in equilibrium? First column is Gibbs free energy and second is Entropy.

Maxima = Maximum

Minima = Minimum

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The issue I have with this problem is that I only know the relationship between Gibbs free energy and equilibrium, I wasn't taught equilibrium's relationship with entropy however.

This question translates to: Which are the signs of changes in entropy associated with reaction

The first column is the change in entropy of the surroundings / area and the second column is the change in entropy of the system

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The change in entropy of the system is negative because of the change in state from (g) to (l), but what about the change in entropy of the surroundings / area, I was never taught that / studied that.

Thanks!

 

[Msj Chem]

For question 25, you just need to know that at minimum Gibbs free energy (at equilibrium), entropy is at a maximum. No explanation is required. If you want to know more then google the principle of maximum entropy but be aware it is very complex and beyond the level of HL chemistry.

For question 19, the process is condensation (gas to liquid), which is exothermic. The entropy of the system will decease (be negative) as liquids have lower entropy than gases. The entropy of the surroundings will increase (be positive) as heat is released from the system to the surroundings, which will increase the entropy of the surroundings. 

[Scienuk]

20 minutes ago, Msj Chem said:

For question 25, you just need to know that at minimum Gibbs free energy (at equilibrium), entropy is at a maximum. No explanation is required. If you want to know more then google the principle of maximum entropy but be aware it is very complex and beyond the level of HL chemistry.

For question 19, the process is condensation (gas to liquid), which is exothermic. The entropy of the system will decease (be negative) as liquids have lower entropy than gases. The entropy of the surroundings will increase (be positive) as heat is released from the system to the surroundings, which will increase the entropy of the surroundings. 

Thank you so much! If you don't mind me asking another question:

Question three part (d) in May 2015 TZ2 Chemistry HL asks to calculate the absolute entropy and in the mark scheme the solution is to calculate the change in entropy, so is the absolute entropy the same as the change in entropy?

[Msj Chem]

The absolute entropy of a perfect crystal at  0 K is zero. Everything else has a positive entropy value when compared to this. There is a table of absolute entropy values in the data booklet.

The entropy change of a reaction is the sum of the absolute entropy values of the products - the sum of the absolute entropy values of the reactants. 

Absolute entropy values are used to calculate the entropy change for a reaction.

This video covers calculating the entropy change for a reaction.

 

Edited August 2, 2016 by Msj Chem

[mrwengchemistry]

Hi Scienuk,

That sounds right. It's the same with enthalpy. We are actually measuring changes and don't know what their complete energy is. Ie. Hidden atomic nuclear energy.

Mr Weng

[kw0573]

@Msj Chem For #25, the actual explanation may be more complex, but the result can be easily derived from known relations.
One way is to look at ΔG = ΔH - TΔS. Enthalpy change at equilibrium is 0 (no net input or output of energy) and T is always positive for a reaction. Then at equilibrium, ΔG = -TΔS, ΔS is thus maximized when ΔG is minimized. 

Another, simpler explanation is by the 2nd law of thermodynamics, which in the simple form states "all spontaneous processes has a non-negative entropy". Because reaching equilibrium is a spontaneous process then for each moment in time the non-negative entropy accumulates, resulting in a maximum at the equilibrium. 

Please verify if either explanation makes sense! Thanks.