Answer
-70.48 kJ/mol.
The results are close.
Work Step by Step
We find:
$\Delta G^{\circ}=\Sigma n_{p}\Delta G_{f}^{\circ}(products)-\Sigma n_{r}\Delta G_{f}^{\circ}(reactants)$
$=[2\Delta G_{f}^{\circ}(NO_{2},g)]-[2\Delta G_{f}^{\circ}(NO,g)+\Delta G_{f}^{\circ}(O_{2},g)]$
$=[2(51.31\,kJ/mol)]-[2(86.55\,kJ/mol)+(0)]$
$=-70.48\,kJ/mol$
In Example 19-5, we found $\Delta G^{\circ}$ to be $-70.4\,kJ/mol$. Therefore, our result is very near to that value and is more precise.