Yh Hoo said:
When impurities was added to the water, it tends to increase the boiling point of the water to 102 degree celsius and lower the melting point of the water to -2 degree celsius! Why this happen? Is it because the impurities tends to absorb the heat supplied to boil the water causing it to take in more heat energy in oder to vapourize?
I think its useful to consider the problem in terms of entropy. The boiling point of a liquid is defined as the temperature at which the liquid and gas phases are at equilibrium. Mathematically, this means that the change in free energy from going to liquid to gas is zero (ΔG = 0).
Now, recall that ΔG is composed of two components: enthalpy (ΔH) and entropy (ΔS). They are related by the equation ΔG = ΔH - TΔS. Since ΔG = 0 at the boiling point, we can solve for the boiling point: T = ΔH/ΔS.
What happens to this value when we add an impurity into the liquid phase? The change in enthalpy (ΔH) describes the amount of heat required to break the interactions between liquid molecules in order to become a gas. This value does not change much when you add an impurity into the water. The
change in entropy (ΔS) describes the entropy gained when going from the much more ordered liquid phase to the much less ordered gas phase. Adding a (non-volatile) impurity to the liquid phase increases the entropy of the liquid without affecting the entropy of the gas; the end result is that ΔS smaller for an impure liquid than a pure liquid.
Adding an impurity to the liquid phase causes ΔS to decrease without changing ΔH. As you can see from the equation above, this situation must cause the boiling point (T) to increase. Therefore, adding an impurity to water, will cause it to boil at a higher temperature.
In essence, vaporization is a process that is driven by the increase in entropy associated with going from the liquid phase to the gas phase. By making the liquid phase more disordered, this gain in entropy becomes smaller, and vaporization becomes slightly less favorable.