Question

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

Answer 1

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`