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Standard Gibbs free energy of formation of liquid water at $\pu{298 K}$ is $\pu{−237.17 kJ mol-1}$ and that of water vapour is $\pu{−228.57 kJ mol-1}$ therefore,

$$\ce{H2O (l) -> H2O (g)}\qquad\Delta G =\pu{8.43{kJ mol-1}}$$

Since $\Delta G>0$, it should not be a spontaneous process but from common observation, water does turn into vapour from liquid over time without any apparent interference. Why does this happen?

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  • $\begingroup$ possible duplicate of Why does water evaporate at room temperature? $\endgroup$ – M.A.R. Apr 8 '15 at 16:33
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    $\begingroup$ @MARamezani The marked question does not discuss the process in terms of the second/third law of thermodynamics. (i wish to know about holes in my understanding of free energy) $\endgroup$ – Apoorv Apr 8 '15 at 16:37
  • $\begingroup$ Also, take a look at physics.stackexchange.com/questions/106754/… also. AFAIK, this thing has nothing to do with $\Delta G$ $\endgroup$ – M.A.R. Apr 8 '15 at 16:50
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    $\begingroup$ I retracted my close vote. The other answers didn't mention how the Delta G issue can be explained. This question does require a different approach, anyway. $\endgroup$ – M.A.R. Apr 8 '15 at 17:12
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    $\begingroup$ Maybe an appropriate example $\endgroup$ – ChemExchange Apr 10 '15 at 17:16
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Standard Gibbs free energy of formation of liquid water at 298 K is −237.17 kJ/mol and that of water vapour is −228.57 kJ/mol. Therefore, $$\ce{H2O(l)->H2O(g)}~~\Delta G=8.43~\mathrm{kJ/mol}$$

Since $\Delta G>0$, it should not be a spontaneous process but from common observation, water does turn into vapour from liquid over time without any apparent interference.

Your math is correct but you left out a very important symbol from your equations. There is a big difference between $\Delta G$ and $\Delta G^\circ$. Only $\Delta G^\circ$ means the Gibbs energy change under standard conditions, and as you noted in the question, the free energy values you quoted are the standard gibbs free energy of water and water vapor.

Whether or not something is spontaneous under standard conditions is determined by $\Delta G^\circ$. Whether something is spontaneous under other conditions is determined by $\Delta G$. To find $\Delta G$ for real conditions, we need to know how they differ from standard conditions.

Usually "standard" conditions for gases correspond to one bar of partial pressure for that gas. But the partial pressure of water in our atmosphere is usually much lower than this. Assuming water vapor is an ideal gas, then the free energy change as a function of partial pressure is given by $G = G^\circ + RT \ln{\frac{p}{p^\circ}}$. If the atmosphere were perfectly 100% dry, then the water vapor partial pressure would be 0, so $\ln{\frac{p}{p^\circ}}$ would be negative infinity. That would translate to an infinitely negative -- i.e. highly spontaneous -- $\Delta G$ for the water evaporation reaction.

Small but not-quite zero dryness in the atmosphere would still lead to the $\Delta G$ of water vapor that is more negative than liquid water. So water evaporation is still spontaneous.

Extra credit: given the standard formation energies you found, and assuming water is an ideal gas, you could calculate the partial pressure of water vapor at which $\Delta G = 0$ for water evaporation. And the answer had better be the vapor pressure of water, or else there is a thermodynamic inconsistency in your data set!

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There are two considerations about why water evaporates but only one of them has to do with the Gibbs free energy of vaporisation.

As the first answer correctly states, there is an equilibrium (heavily biased in favour of the liquid not the vapour) at room temperature. In equilibrium (which is what the $\Delta G$ measures) there is enough thermal energy to put a small amount of water into the vapour phase but most water will remain liquid. This is the correct analysis in a closed vessel. The water doesn't continue to evaporate once the equilibrium is reached.

But, mostly, you don't observe water in closed vessels. This explains why the water continues to evaporate. The point is that, when the vapour can be removed from the vessel containing the liquid, the normal equilibrium is not established and the water can slowly evaporate (at least as long as the surrounding environment has a lower humidity than the equilibrium vapour pressure).

In open systems the simple thermodynamic numbers don't correctly describe the long term behaviour because equilibrium isn't what happens.

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    $\begingroup$ I think this example might be appropriate: Water in an open vessel evaporates (open system) while the one in ones bottle doesn't (closed system). Very good answer! $\endgroup$ – ChemExchange Apr 10 '15 at 17:12
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$\Delta G = -RT \ln K$

Which means there for any given process an equilibrium constant can be deduced. Given the products are 'gaseous' they will escape the system, and further drive the equilibrium.

In this case,

$K = 0.05$

Applied pragmatically, we define a system as 'near the puddle'. When the water evaporates, it can either condense again, or move 'far from the puddle' and hence leaves the system. Over time the 'moving away' portion of this process leads to what we observe humanisticly as the puddle disappearing.

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  • $\begingroup$ So basically, the vapour leaves the system and is hence out of consideration? $\endgroup$ – Apoorv Apr 8 '15 at 18:59
  • $\begingroup$ Roughly yes. Even though a small amount is vapor, that vapor is constantly leaving at some rate, and the equilibrium shifts to more vapor to try to balance. $\endgroup$ – Lighthart Apr 8 '15 at 19:40
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The Gibbs energy is defined for systems at a specified temperature and pressure. Think of a container closed to any transfer of material, immersed in a heat bath at a specified temperature, but with adjustable volume, like a cylinder with a movable piston. Put nothing but water in it, set the heat bath to 25 C, and adjust the volume so the pressure is 1 atmosphere. Since the vapor pressure of water at 25 C is 0.0313 atmospheres, there will be no vapor, the piston will be in contact with the liquid water, pressing down with one atmosphere pressure. Your equation and conclusions are correct.

If you lower the pressure to 0.0313 atm by increasing the volume , then you can set the volume to anything between the volume of the water and the volume of the vapor when the last bit of water just evaporated and still maintain 0.0313 atm. If you increase the volume more, you are setting the pressure below the vapor pressure, and all the water will evaporate.

When you note that water will evaporate at 25 C, you are specifying a huge volume with air at 1 atm, and a little water, so you are effectively setting the partial pressure of the vapor to something very small, and of course, it will evaporate.

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