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JEE Main & Advanced JEE Main Paper (Held on 12-4-2019 Morning)

  • question_answer
    Enthalpy of sublimation of iodine is \[24\,ca\operatorname{l}\,{{g}^{-1}}\] at \[200{}^\circ C.\]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 \[250{}^\circ C\] in \[cal\,{{g}^{-1}}\]is : [JEE Main Held on 12-4-2019 Morning]

    A) 2.85                

    B) 11.4

    C) 5.7                              

    D) 22.8

    Correct Answer: D

    Solution :

    \[{{I}_{2(s)}}\to {{I}_{2(g)}}:\Delta {{H}_{1}}=24cal/g\,\text{at}\,\text{20}{{\text{0}}^{o}}C\]           \[\Delta {{H}_{2}}=\Delta {{H}_{1}}+\Delta {{C}_{{{P}_{rxn}}}}({{T}_{2}}-{{T}_{1}})\] \[=24+\left( 0.0310.055 \right)\times 50\] \[=241.2\] \[=22.8\text{ }Cal/g\]


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