Answer:
(D) - 16.6 kcal
Explanation:
Hello,
In this case, the Gibbs free energy for the given reaction is computed in terms of the Gibbs free energy of formation of each species involved in the chemical reaction:
[tex]\Delta _rG=2\Delta _fG_{NO_2}-2\Delta _fG_{NO}-\Delta _fG_{O_2}[/tex]
Thus, it is found for nitrogen monoxide, oxygen and nitrogen dioxide the following Gibbs free energies of formation: 87.6, 0 and 51.3 kJ/mol respectively, therefore we compute:
[tex]\Delta _rG=2(51.3kJ/mol)-2(87.6kJ/mol)=-72.6kJ*\frac{1kcal}{4.184kJ} \\\\\Delta _rG=-17.35kcal[/tex]
The closest result is (D) - 16.6 kcal, as such difference is noticed when different sources for thermochemical data are used, in this case, the NIST data were used.
Best regards.