Question

07.0 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

Answer 1

Answer:

$\Delta G\°_{rxn}=50.8\frac{kJ}{mol}$

Explanation:

Hello,

In this case, given the reaction:

$2 HNO_3(aq) + NO(g) \rightarrow 3 NO_2(g) + H_2O(l)$

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

$\Delta H\°_{rxn}=3*\Delta _fH_{NO_2}+\Delta _fH_{H_2O}-2\Delta _fH_{HNO_3}-\Delta _fH_{NO}\\\\\Delta H\°_{rxn}=3*33.2+(-285.8)-2*(-207.0)-91.3=136.5kJ/mol$

Similarly, we compute ΔS°rxn:

$\Delta S\°_{rxn}=3*S_{NO_2}+S_{H_2O}-2S_{HNO_3}-S_{NO}\\\\\Delta S\°_{rxn}=3*240.1+70.0-2*146.0-210.8 =287.5J/mol*K$

Finally we compute ΔG°rxn:

$\Delta G\°_{rxn}=\Delta H\°_{rxn}-T\Delta S\°_{rxn}=136.5\frac{kJ}{mol}-298K*287.5\frac{J}{mol*K}*\frac{1kJ}{1000J}\\ \\\Delta G\°_{rxn}=50.8\frac{kJ}{mol}$

Best regards.

Answer 2

Answer:

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