Exercise 4 | Thermochemistry

Given that:

2 H(g) + O(g) → 2 H2O (g)                  ∆H1 = - 483.6 kJ

2 H(g) + O(g) → 2 H2O (l)                   ∆H2 = - 571.6 kJ


What is the energy required to evaporate 1 mole of liquid water?

Evaporation of liquid water: H2O (l) → H2O (g)                ∆H3 = ?


   2 H(g) + O(g) → 2 H2O (g)         ∆H1

+ 2 H2O (l) → 2 H(g) + O(g)        - ∆H2

= 2 H2O (l) → 2 H2O (g)                      ∆H4


So, ∆H4 = ∆H-  ∆H2 = 2 ∆H3

⇒ ∆H3 = 12 (∆H-  ∆H2)

⇒ ∆H3 =  x (-483.6 + 571.6) = 44.00 kJ