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The specific heat $I_{2}$ in vapour and solid state are 0.031 and 0.055 cal/g respectively. The heat of sublimation of iodine at $200^{0} C$ is 6.096kcal ${mol}^{- 1}$ The heat of sublimation of iodine at $250^{\circ} C$ will be

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$3.8 kcal {mol}^{- 1}$

$4.8 kcal {mol}^{- 1}$

$5.8 kcal {mol}^{- 1}$

$2.28 kcal {mol}^{- 1}$

answer is D.

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Detailed Solution

$\frac{∆ H - ∆ H_{1}}{T_{2} - T_{1}} = ACP ∴ \frac{∆ H_{250} - 6.096}{250 - 200} = (0.31 - 0.055) 254 \times 10^{- 3} ∴ ∆ H_{250} = 5.791$

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