Question

A chemist fills a reaction vessel withmercurous chloridesolid,mercury (I)aqueous solution, andchlorideaqueous solution at a temperature of.Under these conditions, calculate the reaction free energyfor the following chemical reaction:Use the thermodynamic information in the ALEKS Data tab. Round your answer to the nearest kilojoule.

Answer 1

Answer:

Free energy change of the reaction is +98.83 kJ/mol.

Explanation:

The dissociation reaction of mercury(I) aqueous solution is as follows.

$Hg_{2}Cl_{2}(s)\rightarrow Hg_{2}^{2+}(aq)+2Cl^{-}(aq)$

The standard free energy of the reaction is as follows.

$\Delta G_{rxn}^{o}=m\Delta G^{o}(products)-n\Delta G^{o}_{reactants}$

$=[(1\times \Delta G^{o}_{f(Hg_{2}^{2+})})+(2\times \Delta G^{o}_{f(Cl^{-})})-(1\times \Delta G^{o}_{f(Hg_{2}Cl_{2})})]$

$=[(1\times 153.5+2\times -131.25)-(1\times -210.78)]kJ=+101.78kJ$

$Gas\,constant"R"=8.314\times 10^{-3}kJK^{-1}mol^{-1}$

$Temperature=25^{o}c=25+273=298K$

$Free\,energy\,change\,of\,the\,reaction=\Delta G=\Delta G^{o}+RT\,lnQ$

$\Delta G=\Delta G^{o}+RT\,ln([Hg_{2}^{2+}][Cl^{-}]^{2})$

$=(101.78kJ)+8.314\times 10^{-3}kJK^{-1}mol^{-1}\times298K ln((0.926)(0.573)^{2})$

$=(101.78kJ)+(-2.95kJ)=+98.83kJ/mol$

Therefore, Free energy change of the reaction is +98.83 kJ/mol.