$\Delta H_{rxn}=2\Delta H_{ICl(g)}-\Delta H_{I_2(g)}-\Delta H_{Cl_2(g)}\\\\\Delta H_{rxn}=2(17.78)-0-62.438\ KJ/mol\\\\\Delta H_{rxn}=-26.878\ KJ/mol$

For , 2.41 moles of $I_2$ :

$\Delta H_{rxn}=2.41\times (-26.878)\ KJ\\\\\Delta H_{rxn}=-64.78\ KJ$

We know :

$\Delta S = -\dfrac{\Delta H_{rxn}}{T}\\\\\Delta S = -\dfrac{-64.78}{298}\ KJ/K\\\\\Delta S =-0.21738 \ KJ/K\\\\\Delta S=-217.38\ J/K$

Hence, this is the required solution.

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Explanation:

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