Heat of Sublimation

Calculate the enthalpy of sublimation of Iodine from the following data:

H₂(g) + I₂(s) → 2 HI(g)      ΔH_R = 51.9 kJ

H₂(g) + I₂(g) → 2 HI(g)      ΔH_R = -9.2 kJ

H₂(g) + I₂(s) → 2 HI(g) -- I      ΔH_R = 51.9 kJ

H₂(g) + I₂(g) → 2 HI(g) -- II      ΔH_R = -9.2 kJ

From Hess's law of constant heat of summation,

Subtracting equation II from I gives,

I₂(s) - I₂(g) = 0      ΔH_R = 51.9 - (-9.2) = 61.1 kJ

i.e.,

I₂(s) → I₂(g)      ΔH_R = 61.1 kJ

That is enthalpy of sublimation (transformation of iodine from solid phase to directly vapor phase) of iodine is 61.1 kJ