Question

Calculate the enthalpy of vapourization of 1 gm of water from the following...

H2(g) +1/2O2(g)----> H2O(g)
ΔH= -51 kcal


H2(g) +1/2O2(g)----> H2O(l)
ΔH= -68.3 kcal

Answer 1

enthalpy of vaporization :- the change in enthalpy for the formation of one mole of vapour from liquid at constant temperature is known as enthalpy of vaporization .
for water :-
H2O → H2O ( g)

H2 + 1/2O2 → H2O ( g) ------(1) ∆H1 =-51 kcal

H2 +1/2 O2 → H2O ( l ) ∆H2 =-68.3 kcal
so,

H2O (l ) → H2 + 1/2O2 ∆H2" = 68.3 kcal
-----(2)

now add equations (1) and (2)

H2O( l ) → H2O ( g)

∆vH = ∆H1 + ∆H2"
= -51 kcal +68.3 kcal = 17.3 kcal

the enthalpy of vaporisation of one mole of water = 17.3 kcal

so, enthalpy of vaporisation of 1gm of water = 17.3 × 1/18 kcal
= 0.9611 kcal

Answer 2

Answer:

0.9611 kcal..... steps in attachment.