InterviewSolution
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InterviewSolution
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The amount of energy when million atoms of iodine are completely converted into `I^(-)` ions in the vapour state according to the equation,`I_((g))+e^(-)`to`I^(-)_((g)) ` is `4.9X10^(-13)` J.What would be the electron gain enthalpy of iodine in terms of KJ `mol^(-1)` and eV per atom?A. `295,3.06`B. `-295-3.06`C. `439,5.09`D. `-356,-7.08` |
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Answer» Correct Answer - B The electron gain enthalpy of iodine is equal to the energy released when 1 mole of iodine atoms in vapour state is converted into `I^(-)` ions `Delta_(eg)H=(4.9)xx(10)^(-13)xx6.023xx10^(23)/(10)^(6)=29.5xx10^(4) J mol^(-1) 295 KJ mol^(-1)` Thus,electron gain enthalpy of iodine=`-295 KJ mol^(-1)` We know that ` eV per atom=96.49 KJ `mol^(-1)` `therefore`Electron gain enthalpy of iodine in eV per atom `=-(295)/(96.49)=3.06 eV per atom. |
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