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)

 

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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