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) -2

Question

3 years ago

5

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) -2

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

Chemistry

2 answers:

siniylev [52] · Answer 1 · 2022-02-03T12:17:30+03:00

siniylev [52]3 years ago

8 0

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:

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