JEE Main · 2019 · Shift-IhardTHERMO-125

Enthalpy of sublimation of iodine is 24\,cal g-1 at 200C. If specific heat of I2(s) and I2 (vap) are 0.055 and…

Thermodynamics · Class 11 · JEE Main Previous Year Question

Question

Enthalpy of sublimation of iodine is 24 cal g−124\,\text{cal g}^{-1} at 200∘C200^\circ\text{C}. If specific heat of I2(s)\text{I}_2(s) and I2\text{I}_2 (vap) are 0.055 and 0.031 cal g−1K−10.031\,\text{cal g}^{-1}\text{K}^{-1} respectively, then enthalpy of sublimation of iodine at 250∘C250^\circ\text{C} in cal g−1\text{cal g}^{-1} is:

Options
  1. a

    11.4

  2. b

    2.85

  3. c

    22.8

    ✓
  4. d

    5.7

Correct Answerc

22.8

Detailed Solution

🧠 Kirchhoff's law: ΔHT2=ΔHT1+ΔCP(T2−T1)\Delta H_{T_2} = \Delta H_{T_1} + \Delta C_P(T_2 - T_1) The enthalpy of sublimation changes with temperature at a rate equal to the difference in heat capacities between the vapor and solid phases.

🗺️ Calculation ΔCP=CP(vapor)−CP(solid)=0.031−0.055=−0.024 cal g−1K−1\Delta C_P = C_P(\text{vapor}) - C_P(\text{solid}) = 0.031 - 0.055 = -0.024\,\text{cal g}^{-1}\text{K}^{-1}

ΔH250=24+(−0.024)(250−200)=24−1.2=22.8 cal g−1\Delta H_{250} = 24 + (-0.024)(250 - 200) = 24 - 1.2 = 22.8\,\text{cal g}^{-1}

Answer: (c) 22.8 cal g−1\boxed{\text{Answer: (c) 22.8 cal g}^{-1}}

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