We know,
[tex]\Delta H_{I_2(g)}=62.438\ KJ/mol\\\\\Delta H_{Cl_2(g)}= 0.0\ KJ/mol\\\\\Delta H_{ICl(g)}=17.78\ KJ/mol[/tex]
For given reaction, [tex]I_2(g)+Cl_2(g)\ -->\ 2ICl(g)[/tex]
[tex]\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[/tex]
For , 2.41 moles of [tex]I_2[/tex] :
[tex]\Delta H_{rxn}=2.41\times (-26.878)\ KJ\\\\\Delta H_{rxn}=-64.78\ KJ[/tex]
We know :
[tex]\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[/tex]
Hence, this is the required solution.
