Question

Consider the reactionI2(g) + Cl2(g)2ICl(g)Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 2.41 moles of I2(g) react at standard conditions.S°surroundings = J/K

Answer 1

We know,

`\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`

For given reaction,
`I_2(g)+Cl_2(g)\ -->\ 2ICl(g)`

`\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* (-26.878)\ KJ\\\\\Delta H_(rxn)=-64.78\ KJ`

We know :

`\Delta S = -(\Delta H_(rxn))/(T)\\\\\Delta S = -(-64.78)/(298)\ KJ/K\\\\\Delta S =-0.21738 \ KJ/K\\\\\Delta S=-217.38\ J/K`

Hence, this is the required solution.