1 Answer

Stefan Rusek · Answer 1 · 2021-11-09T15:59:21+0000

Answer: The value of
$\Delta G^o$ of the reaction is 28.38 kJ/mol

Step-by-step explanation:

For the given chemical reaction:

$SO_2(g)+Cl_2(g)\rightarrow SO_2Cl_2(g)$

$\Delta H^o_(rxn)=\sum [n* \Delta H^o_f_((product))]-\sum [n* \Delta H^o_f_((reactant))]$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_(rxn)=[(1* \Delta H^o_f_((SO_2Cl_2(g))))]-[(1* \Delta H^o_f_((SO_2(g))))+(1* \Delta H^o_f_((Cl_2(g))))]$

We are given:

$\Delta H^o_f_((SO_2Cl_2(g)))=-364kJ/mol\\\Delta H^o_f_((SO_2(g)))=-296.8kJ/mol\\\Delta H^o_f_((Cl_2(g)))=0kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_(rxn)=[(1* (-364))]-[(1* (-296.8))+(1* 0)]=-67.2kJ/mol=-67200J/mol$

$\Delta S^o_(rxn)=\sum [n* \Delta S^o_f_((product))]-\sum [n* \Delta S^o_f_((reactant))]$

The equation for the entropy change of the above reaction is:

$\Delta S^o_(rxn)=[(1* \Delta S^o_((SO_2Cl_2(g))))]-[(1* \Delta S^o_((SO_2(g))))+(1* \Delta S^o_((Cl_2(g))))]$

We are given:

$\Delta S^o_((SO_2Cl_2(g)))=311.9J/Kmol\\\Delta S^o_((SO_2(g)))=248.2J/Kmol\\\Delta S^o_((Cl_2(g)))=223.0J/Kmol$

Putting values in above equation, we get:

$\Delta S^o_(rxn)=[(1* 311.9)]-[(1* 248.2)+(1* 223.0)]=-159.3J/Kmol$

To calculate the standard Gibbs's free energy of the reaction, we use the equation:

$\Delta G^o_(rxn)=\Delta H^o_(rxn)-T\Delta S^o_(rxn)$

where,

$\Delta H^o_(rxn)$ = standard enthalpy change of the reaction =-67200 J/mol

$\Delta S^o_(rxn)$ = standard entropy change of the reaction =-159.3 J/Kmol

Temperature of the reaction = 600 K

Putting values in above equation, we get:

$\Delta G^o_(rxn)=-67200-(600* (-159.3))\\\\\Delta G^o_(rxn)=28380J/mol=28.38kJ/mol$

Hence, the value of
$\Delta G^o$ of the reaction is 28.38 kJ/mol

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