How do you calculate enthalpy of vaporization for 1 mole of water?

1 Answer
Aug 8, 2018

Answer:

The enthalpy of vaporization is #+40,680J or +40.68KJ#

Explanation:

Enthalpy or Heat of vaporization is given by;

#Q = mH_v#

Where;

#H_v = 2260 Jg^-1 ("water at" 100^o C)#

#Q = "Enthalpy or Heat of vaporization"#

#m = "mass"#

Now we are given #1# mole of water..

Recall;

#"No of moles" = "mass"/"molar mass"#

#n = m/(Mm)#

#:. m = n xx Mm#

#n = 1"moles"#

#Mm color(white)x of color(white)x H_2O = (1 xx 2) + 16 = 2 + 16 = 16gmol^-1#

Hence;

#m = 1cancel(mol) xx 18gcancel(mol^-1)#

#m = 18g#

Now plugging it into the formula;

#Q = 18g xx 2260 Jg^-1#

#Q = 18cancelg xx 2260 Jcancel(g^-1)#

#Q = 40,680J or 40.68KJ#

Therefore the enthalpy of vaporization is #+40,680J or +40.68KJ#