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

Option 1)
$2.85$
Option 2)
$22.8$
Option 3)
$5.7$
Option 4)
$11.4$

Answers (1)

best_answer

Enthalpy of Sublimation -

Amount of enthalpy change to sublimise 1 mole solid into 1 mole vapour at a temperature below its melting point

- wherein

$H_{2}O_{(s)}\rightarrow H_{2}O_{(g)}$

$\Delta H_{sublimation}= 46.6\, kj/mol$

$I_{2}\left ( s \right )\rightarrow I_{2}\left ( g \right ),\bigtriangleup H_{200^{\circ}}=24cal/g$

Specific heat of $I_{2}\left ( s \right )$ & $I_{2}\left ( vap \right )$ are respectively, $0.055calg^{-1}k^{-1}$ & $0.031calg^{-1}k^{-1}$

$\therefore \bigtriangleup S=$ Heat capacity of Product - Reactant

$\therefore \bigtriangleup S=0.031-0.055$

$=-0.024$

Applying Kirchoff's equation, we get:

$\bigtriangleup H_{2}=\bigtriangleup H_{1}+ \bigtriangleup S\left ( T_{2}-T_{1} \right )$

$\bigtriangleup H_{2}=24+\left ( -0.024 \right )\times 50$

$=24-1.2=22.8$

$\therefore$ Enthalpy of sublimation of iodine at

$250^{\circ}C=22.8cal/g$

Option 1)

$2.85$

Option 2)
$22.8$
Option 3)

$5.7$

Option 4)

$11.4$

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