For water ∆h°vap = 40.7 kj/mol at 100.°c, its boiling point. calculate ∆s° for the vaporization of 1.00 mol water at 100.°c and 1.00 atm pressure.

Answer : The value of change in entropy for vaporization of water is

Explanation :

Formula used :

where,

= change in entropy  of vaporization = ?

= change in enthalpy of vaporization = 40.7 kJ/mol

= boiling point temperature of water =

Now put all the given values in the above formula, we get:

Therefore, the value of change in entropy for vaporization of water is

"among the transition metal elements, manganese and zinc have a slight lowering of electronegativities. the reason behind this is that the nuclear charges of these elements are slightly weaker than the other transition metal elements due to the way their valence electrons are filled up. both manganese and zinc have their valence electrons filled up to the 4th shell with 1 pair of electrons (4s2) completely occupying it. although the electrons in the 3rd shell (3d) are yet to be paired since the outer most shell is filled up, the ability of the nucleus to attract electrons is lessened. hence,   the electronegativity (or the ability to attract electrons) is lessened."