Question

Calculate the enthalpy of vaporization for 1gm of water from the following H2 (g) +1/2O2 (g)=H2O (g) ?H=-51kcal H2 (g)+1/2 O2 (g)=H2O (l) ?H=-68.3kcal

Answer 1

Enthalpy of vaporisation is conversion of liquid water to gas. H2O(l)  → H2O(g)H2 + 12 O2 →H2O(g) :     ∆H = 51 kCal   −−−−−−−−1H2 + 12 O2 →H2O(l) :     ∆H = 68.3 kCal −−−−−2Reverse the second equation.H2O(l) →H2 + 12 O2        ∆H = −68.3 kCal −−−−−3Now add 1 and 3 equationH2 + 12 O2 →H2O(g) H2O(l) →H2 + 12 O2−−−−−−−−−−−−−−−−  H2O(l)  → H2O(g)−−−−−−−−−−−−−−−∆H =  51 + (−68.3) = − 17.3 kCal