Enthalpy of sublimation of iodine is $24 \mathrm{cal} \mathrm{g}^{-1}$ at $200^{\circ} \mathrm{C}$. If specific heat of $\mathrm{I}_{2}(\mathrm{~s})$ and $\mathrm{I}_{2}$ (vap) are 0.055 and $0.031 \mathrm{cal} \mathrm{g}^{-1} \mathrm{~K}^{-1}$ respectively, then enthalpy of sublimation of iodine at $250^{\circ} \mathrm{C}$ in $\mathrm{cal} \mathrm{g}^{-1}$ is :

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Chemical Thermodynamics

Enthalpy of sublimation of iodine is

24 cal g^{- 1}

20 0^{\circ} C

. If specific heat of

I_{2} (s)

and

I_{2}

(vap) are 0.055 and

0.031 cal g^{- 1} K^{- 1}

respectively, then enthalpy of sublimation of iodine at

25 0^{\circ} C

cal g^{- 1}

is :

Answer: (D)

Use Kirchhoff's law for how an enthalpy change varies with temperature,

Δ H_{2} = Δ H_{1} + Δ C_{P} Δ T

. For sublimation

Δ C_{P} = C_{P} (vapour) - C_{P} (solid) = 0.031 - 0.055 = - 0.024 cal g^{- 1} K^{- 1}

, and the temperature rise is

Δ T = 250 - 200 = 50 K

. Therefore

Δ H_{2} = 24 + (- 0.024) (50) = 24 - 1.2 = 22.8 cal g^{- 1}

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Single correct option