Heat of Sublimation

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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

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Last Modified on: 01-May-2024

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