Chapter 19 - Spontaenous Change: Entropy and Gibbs Energy - Example 19-5 - Calculating (symbol)G for a Reaction - Page 837: Practice Example B

-70.48 kJ/mol. The results are close.

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.