Gibbs free energy going to zero (equilibrium)

In summary: Therefore, the Gibbs free energy for the products is still negative. In summary, the boiling point of water is 373K because at this temperature, the vapor pressure of the liquid and the atmospheric pressure are equal. The volume of water vapor is different at different pressures, causing a difference in entropy change. When ΔG=0, it only matters that one mole of liquid water is converted to one mole of gaseous water. At 298K, there is still vaporized water present but in equilibrium with the liquid, resulting in a negative Gibbs free energy for the products.
  • #1
sgstudent
739
3
Hi I have a 4 part question about phase changes and I hope you guys can help me out here thanks :)

The chemical equation for vaporization for water is H2O(l)-->H2O(g)

1) When ΔG is 0 during a phase change we would use the formula ΔH=TΔS and put in values to get the boiling/melting point by rearranging it to T=ΔH/ΔS. However, after learning that every reaction at any temperature would slowly go into equilibrium where ΔG=0, why would the boiling point of water be 373K since equilibrium can be reached at even 298K?

2)I read that the ΔS used represents the vapor at 1 atm which would be different from the vapor at a lower atm. But I don't quite understand why they would be different. Their explanation was that water vapor at 1 atm occupies more space than at a lower atm so the ΔS is greater. But since the temperature are different shouldn't the volume be equal to each other too? So why would the entropy change be different when the vapor pressures are different?

3)They also mentioned that the ΔG would go to zero only when it reaches vapor pressure for that specific temperature. I agree with that but since when using ΔG°=-RTlnKeq is it assumed that the reaction is for 1 mole of liquid water to 1 mole of gaseous water at 1atm. So when ΔG=0 does it matter if more than 1 mole has or less than 1 mole had been vaporized? I would think only 1 mole can be reacted since for ΔH we would use the vaporization of 1 mole. So would the container of the gas be fixed to be such that 1 mole would be vaporized at the pressure defined (like 1 atm) and at the temperature defined (like 373K)? If so then would there be any difference if more or less than 1 mole of water vaporizes instead? It would not right since the moles double so would ΔH and ΔS so there isn't a change? Or is there another reason?

4)But even so at 298K, Gibbs free energy can also be represented with ΔG=G products-G reactants so for G products it would water vapor at 298K which I thought was impossible as well?

Thanks so much for the help :)
 
Chemistry news on Phys.org
  • #2
1) The boiling point of water is 373K because that is the temperature at which the vapor pressure of the liquid and the atmospheric pressure are equal. At 298K, the vapor pressure of the liquid is much lower than the atmospheric pressure and so the reaction will not proceed to equilibrium. 2) Water vapor at different pressures does occupy different volumes. At lower pressures, the volume of the water vapor is smaller as the molecules are more spread out. This difference in volume causes the entropy change to be different at different pressures. 3) When ΔG=0, it only matters that one mole of liquid water is converted to one mole of gaseous water. The pressure or the temperature at which this happens is irrelevant. However, the pressure and the temperature will affect the magnitude of the ΔH and ΔS values. 4) At 298K, there is still vaporized water present but it is in equilibrium with the liquid. This means that the amount of water vapor present is very small and so the vapor pressure is also very low.
 

1. What is Gibbs free energy?

Gibbs free energy is a thermodynamic quantity that measures the amount of energy available in a system to do work. It takes into account both the enthalpy (heat content) and entropy (disorder) of a system.

2. How does Gibbs free energy relate to equilibrium?

In a closed system, at constant temperature and pressure, the Gibbs free energy reaches a minimum at equilibrium. This means that the system has reached a state of balance where the energy available to do work is equal to the energy required to maintain the system in its current state.

3. Why does Gibbs free energy approach zero at equilibrium?

In a system at equilibrium, there is no net change in the system and the forward and reverse reactions occur at the same rate. This means that the system is in a state of minimum energy, and therefore the Gibbs free energy approaches zero.

4. How is Gibbs free energy used to predict the direction of a reaction?

If the Gibbs free energy change (ΔG) for a reaction is negative, it means that the reaction will proceed spontaneously in the forward direction. If ΔG is positive, the reaction will not occur spontaneously and will require an input of energy to proceed. If ΔG is zero, the reaction is at equilibrium.

5. Can Gibbs free energy be used to determine the amount of product formed at equilibrium?

Yes, the value of ΔG at equilibrium can be used to calculate the equilibrium constant (K) for a reaction, which is a measure of the ratio of products to reactants at equilibrium. This can then be used to determine the amount of product formed at equilibrium.

Similar threads

Replies
4
Views
1K
Replies
1
Views
2K
Replies
3
Views
2K
Replies
15
Views
10K
Replies
11
Views
2K
  • Chemistry
Replies
9
Views
1K
  • Biology and Chemistry Homework Help
Replies
4
Views
1K
Replies
4
Views
2K
Replies
5
Views
2K
Replies
8
Views
2K
Back
Top