Calculate the ΔG°rxn using the following information at 298K. 2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l) ΔG°rxn = ? ΔH°f (kJ/mol) -207.0 91.3 33.2 -285.8 S°(J/mol∙K) 146.0 210.8 240.1 70.0 Calculate the ΔG°rxn using the following information at 298K. 2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l) ΔG°rxn = ? ΔH°f (kJ/mol) -207.0 91.3 33.2 -285.8 S°(J/mol∙K) 146.0 210.8 240.1 70.0 -151 kJ -85.5 kJ +50.8 kJ +222 kJ +101.6 kJ

ΔG°rxn = +50.8 kJ/mol

Explanation:

It is possible to obtain ΔG°rxn of a reaction at certain temperature from ΔH°rxn and S°rxn, thus:

ΔG°rxn = ΔH°rxn - T×S°rxn (1)

In the reaction:

2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l)

ΔH°rxn = 3×ΔHfNO2 + ΔHfH2O - (2×ΔHfHNO3 + ΔHfNO)

ΔH°rxn = 3×33.2kJ/mol + (-285.8kJ/mol) - (2×-207.0kJ/mol + 91.3kJ/mol)}

ΔH°rxn = 136.5kJ/mol

And S°:

S°rxn = 3×S°NO2 + S°H2O - (2×S°HNO3 + S°NO)

ΔH°rxn = 3×0.2401kJ/molK + (0.0700kJ/molK) - (2×0.146kJ/molK + 0.2108kJ/molK)

ΔH°rxn = 0.2875kJ/molK

And replacing in (1) at 298K:

ΔG°rxn = 136.5kJ/mol - 298K×0.2875kJ/molK

ΔG°rxn = +50.8 kJ/mol

Explanation:

Hello,

In this case, given the reaction:

We compute the ΔH°rxn by using the given enthalpies of formation:

Similarly, we compute ΔS°rxn:

Finally we compute ΔG°rxn:

Best regards.