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 $I_2$ :

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

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.