What is the boiling point in K of Zn, if the entropy change for evaporation of .8564 mole at its boiling ponit is 95.067K? The heat of vaporization of Zn is 131jk/mole.
To find the boiling point in Kelvin (K) of Zinc (Zn), we can use the equation:
∆S = ∆H_vap / T
Where:
∆S = entropy change
∆H_vap = heat of vaporization
T = boiling point in Kelvin
Given:
∆S = 95.067 K
∆H_vap = 131 J/mol
Rearranging the equation, we have:
T = ∆H_vap / ∆S
Substituting the given values into the equation, we get:
T = 131 J/mol / 95.067 K
Now, we simply need to perform the calculation to find the boiling point.