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Heat of Sublimation

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

H_{2}(g) + I_{2}(s) →
2 HI(g) ΔH_{R} = 51.9 kJ

H_{2}(g) + I_{2}(g) →
2 HI(g) ΔH_{R} = -9.2 kJ

Calculations:

H_{2}(g) + I_{2}(s) →
2 HI(g) -- I ΔH_{R} = 51.9 kJ

H_{2}(g) + I_{2}(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,

I_{2}(s) - I_{2}(g) = 0 ΔH_{R} = 51.9 - (-9.2) = 61.1 kJ

i.e.,

I_{2}(s) →
I_{2}(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**

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Last Modified on: 04-Feb-2022

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