Problema 3X08

Combustão completa do octano gasoso: $$\ce{C8H18 + \frac{25}{2}O2 -> 8CO2 + 9H2O}$$ $$\Delta H_C=7\Delta H_L(\ce{C-C})+18\Delta H_L(\ce{C-H})+\frac{25}{2}\Delta H_L(\ce{O2})-(16\Delta H_L(\ce{C=O})+18\Delta H_L(\ce{O-H}))$$ $$\Delta H_C=7\cdot\pu{348 kJ//mol}+18\cdot\pu{412 kJ//mol}+\frac{25}{2}\cdot\pu{496 kJ//mol}-16\cdot\pu{803 kJ//mol}-18\cdot\pu{463 kJ//mol}$$ $$\Delta H_C=\pu{-5130 kJ//mol}\implies\frac{\Delta H_C}{N(H)}=\pu{-285 kJ//mol}$$ Célula de combustível de hidrogênio e oxigênio: $$\ce{H2 + \frac{1}{2}O2 -> H2O}$$ $$\Delta H=\Delta H_L(\ce{H2})+\frac{1}{2}\Delta H_L(\ce{O2})-2\Delta H_L(\pu{O-H})$$ $$\Delta H=\pu{463 kJ//mol}+\pu{248 kJ//mol}-\pu{926 kJ//mol}$$ $$\Delta H=\pu{-215 kJ//mol}\implies\frac{\Delta H}{N(H)}=\pu{-107,5 kJ//mol}$$ Logo, a razão pedida é dada por: $$r=\frac{\pu{-285 kJ//mol}}{\pu{-107,5 kJ//mol}}\approx 2,65$$