Enthalpy of sublimation of iodine is 24 cal g^-1 at 200°C. If specific heat of I_2 (s ) and I_2 ( vap ) are 0.055 and 0.031 calg^-1K^-1 respectively, then enthalpy of sublimation of iodine at 250°C i…

Question

Enthalpy of sublimation of iodine is 24 cal g^-1 at 200°C. If specific heat of I_2 (s ) and I_2 ( vap ) are 0.055 and 0.031 calg^-1K^-1 respectively, then enthalpy of sublimation of iodine at 250°C in cal g^-1 is:

Accepted Answer

Correct answer: C. (c) I_2 (s ) ⟶I_2 (g ) Heat of reaction depend upon temperature i.e., it varies with temperature, as given by Kirchoff's equation, H_T_2= H_T_1+ 0.05em C_p dT Where C_p=C_p of product -C_p of reactant ∴ C_p=0.031-0.055=-0.024 cal/g Now H_T_2- H_T_1= C_p(T_2-T_1) H_ (250 )- H_ (200 )=-0.024 (523-473 ) H_ (250 )=24-50 0.024=22.8 cal/g

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

Solution

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