The specific heats of iodine vapours and solid are 0.031 and 0.055 cal/g, respectively. If the enthalpy of sublimation of iodine is 24 cal/g at 200°C, then the enthalpy of sublimation of iodine at 250

Question

The specific heats of iodine vapours and solid are 0.031 and 0.055 cal/g, respectively. If the enthalpy of sublimation of iodine is 24 cal/g at 200°C, then the enthalpy of sublimation of iodine at 250°C should be

Prashant Kotian · Accepted Answer

Correct answer: B. The process is sublimation of iodine: (s) → (g). The change in specific heat is ΔCp = (g) - (s) = 0.031 - 0.055 = -0.024 cal/(g °C). Using Kirchhoff's equation: ΔH(T2) = ΔH(T1) + ΔCp × (T2 - T1). Given T1 = 200°C, T2 = 250°C, ΔH(200°C) = 24 cal/g: ΔH(250°C) = 24 + (-0.024) × (250 - 200) = 24 - 0.024 × 50 = 24 - 1.2 = 22.8 cal/g.

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