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

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Answer 1 · 2018-08-08T17:30:04

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

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 $+40,680J or +40.68KJ$