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Thermodynamics question

2019 · 12 Apr · Shift 1 · Q17
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Thermodynamics question

2019 · 12 Apr · Shift 1 · Q17

JEE MainChemistryThermodynamicsMCQ+4 / −1
Enthalpy of sublimation of iodine is 24 cal g–1 at 200 oC. If specific heat of I2I_2I2​(s) and l2 (vap) are 0.055 and 0.031 cal g–1K –1 respectively, then enthalpy of sublimation of iodine at 250 oC in cal g–1 is :
  1. A
    2.85
  2. B
    22.8
  3. C
    11.4
  4. D
    5.7
View written solutionFree

Correct answer: B

  1. Use Kirchhoff’s relation for temperature dependence of enthalpy change

For sublimation, I2(s)→I2(g)I_2(s) \rightarrow I_2(g)I2​(s)→I2​(g)

The change of enthalpy of sublimation with temperature is: ΔHsub(T2)=ΔHsub(T1)+∫T1T2ΔCp dT\Delta H_{sub}(T_2)=\Delta H_{sub}(T_1)+\int_{T_1}^{T_2} \Delta C_p\, dTΔHsub​(T2​)=ΔHsub​(T1​)+∫T1​T2​​ΔCp​dT where ΔCp=Cp(vap)−Cp(solid)\Delta C_p = C_p(\text{vap})-C_p(\text{solid})ΔCp​=Cp​(vap)−Cp​(solid)

  1. Given data

At 200∘C200^\circ C200∘C, ΔHsub(200∘C)=24 cal g−1\Delta H_{sub}(200^\circ C)=24\ \text{cal g}^{-1}ΔHsub​(200∘C)=24 cal g−1

Specific heats: Cp(I2,s)=0.055 cal g−1K−1C_p(I_2,s)=0.055\ \text{cal g}^{-1}\text{K}^{-1}Cp​(I2​,s)=0.055 cal g−1K−1 Cp(I2,vap)=0.031 cal g−1K−1C_p(I_2,vap)=0.031\ \text{cal g}^{-1}\text{K}^{-1}Cp​(I2​,vap)=0.031 cal g−1K−1

Required: ΔHsub\Delta H_{sub}ΔHsub​ at 250∘C250^\circ C250∘C

Temperature change: ΔT=250−200=50 K\Delta T = 250-200 = 50\ \text{K}ΔT=250−200=50 K

  1. Calculate ΔCp\Delta C_pΔCp​

ΔCp=0.031−0.055=−0.024 cal g−1K−1\Delta C_p = 0.031-0.055 = -0.024\ \text{cal g}^{-1}\text{K}^{-1}ΔCp​=0.031−0.055=−0.024 cal g−1K−1

  1. Apply Kirchhoff’s equation

ΔHsub(250∘C)=24+(−0.024)(50)\Delta H_{sub}(250^\circ C)=24+(-0.024)(50)ΔHsub​(250∘C)=24+(−0.024)(50)

=24−1.2=24-1.2=24−1.2

=22.8 cal g−1=22.8\ \text{cal g}^{-1}=22.8 cal g−1

  1. Match with options

22.8 cal g−122.8\ \text{cal g}^{-1}22.8 cal g−1 corresponds to Option B.

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