physical chemistry - Why does water evaporate spontaneously at room temperature despite ΔG > 0?

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?

Answer

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!