Question

Estimate ΔG°rxn for the following reaction at 775 K. 2Hg(g) + O2(g) → 2HgO(s) ΔH°= -304.2 kJ; ΔS° = -414.2 J/K

Answer 1

`\Delta G^0_(rxn)=16.81kJ`

Step-by-step explanation:

Hello!

In this case, since the Gibbs free energy of any process is related with the enthalpy change, temperature and entropy change as shown below:

`\Delta G^0=\Delta H^0-T\Delta S^0`

For a chemical reaction it is simply modified to:

`\Delta G^0_(rxn)=\Delta H^0_(rxn)-T\Delta S^0_(rxn)`

Thus, since the enthalpy of reaction is given as -304.2 kJ and the entropy as -414.2 J/K (-0.4142 kJ/K), at 775 K the Gibbs free energy of reaction turns out:

`\Delta G^0_(rxn)=-304kJ-775K*(-0.4142(kJ)/(K) )\\\\\Delta G^0_(rxn)=16.81kJ`

Whose result means this is a nonspontaneous reaction.

Best regards!