Calculate the enthalpy of sublimation of Iodine from the following data:
H2(g) + I2(s) → 2 HI(g) ΔHR = 51.9 kJ
H2(g) + I2(g) → 2 HI(g) ΔHR = -9.2 kJ
Calculations:
H2(g) + I2(s) → 2 HI(g) -- I ΔHR = 51.9 kJ
H2(g) + I2(g) → 2 HI(g) -- II ΔHR = -9.2 kJ
From Hess's law of constant heat of summation,
Subtracting equation II from I gives,
I2(s) - I2(g) = 0 ΔHR = 51.9 - (-9.2) = 61.1 kJ
i.e.,
I2(s) → I2(g) ΔHR = 61.1 kJ
That is enthalpy of sublimation (transformation of iodine from solid phase to directly vapor phase) of iodine is 61.1 kJ
Last Modified on: 04-Feb-2022
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