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Enthalpy of sublimation of iodine is "24...

Enthalpy of sublimation of iodine is `"24 cal g"^(-1)" at " 200^(@)C`. If specific heat of `l_(2)(s) and l_(2)` (vap) are 0.055 and 0.031 respectively, then enthalpy of sublimation of iodine at `250^(@)C` in `"cal g"^(-1)` is:

A

5 . 7

B

22.8

C

11.4

D

2.85

Text Solution

Verified by Experts

The correct Answer is:
B
`I_2 (s) rarr I_(2)(g)`
`Delta H_(2) = Delta H_(1) + Delta C_(p) (T_2 - T_1)`
`Delta C_(p)= C_(p)I_(2)(s) = 0.055 - 0.031 = 0.024 cal g^(-1) k^(-1)`
`Delta H_(2) = 24 - 0.024 xx 50 = 22.8 cal g^(-1) K^(-1)`.
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