91Ó°ÊÓ

The most common source of energy in mammals is glucose, which reacts with oxygen to produce carbon dioxide and water. The reaction equation is:

${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6+6{\mathrm{O}}_2→6{\mathrm{CO}}_2+6{\mathrm{H}}_2\mathrm{O}$

The enthalpy of the reactants and products differs by $∆\mathrm{H}$:

$\mathrm{Δ\pm á}={\mathrm{Δ\pm á}}_{\text{react}}-{\mathrm{Δ\pm á}}_{\mathrm{prod}}$

$\mathrm{Δ\pm á}=6{\mathrm{Δ\pm á}}_{{\mathrm{H}}_2\mathrm{O}}+6{\mathrm{Δ\pm á}}_{{\mathrm{CO}}_2}-{\mathrm{Δ\pm á}}_{{\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6}$

- For the creation of one mole of carbon dioxide from elemental carbon (solid) and oxygen (gas), the enthalpy is:

${\mathrm{O}}_2\left(\text{gas}\right)+\mathrm{C}\left(\text{solid}\right)→{\mathrm{CO}}_2\text{(gas)}$

${\mathrm{Δ\pm á}}_{{\mathrm{O}}_2}+{\mathrm{Δ\pm á}}_{\mathrm{C}}→{\mathrm{Δ\pm á}}_{{\mathrm{CO}}_2}$

$\left(0\right)+\left(0\right)→-393.51$

So,

${\mathrm{Δ\pm á}}_{{\mathrm{CO}}_2}-{\mathrm{Δ\pm á}}_{\mathrm{C}}-{\mathrm{Δ\pm á}}_{{\mathrm{O}}_2}$

$=-393.51-0-0$

$=-393.51\mathrm{kJ}$

$∆{\mathrm{H}}_{{\mathrm{CO}}_2}\left(\mathrm{formation}\right)=-393.51\mathrm{kJ}$

- For the creation of two moles of liquid water from elemental oxygen (gas) and hydrogen (gas), the enthalpy is:

${\mathrm{H}}_2\left(\text{gas}\right)+\frac{1}{2}\mathrm{O}\left(\text{gas}\right)→{\mathrm{H}}_2\mathrm{O}\left(\mathrm{gas}\right)$

${\mathrm{Δ\pm á}}_{{\mathrm{H}}_2}+\frac{1}{2}{\mathrm{Δ\pm á}}_{\mathrm{O}}→{\mathrm{Δ\pm á}}_{{\mathrm{H}}_2\mathrm{O}}$

$\left(0\right)+\left(0\right)→(-285.83)$

So ,

${\mathrm{Δ\pm á}}_{{\mathrm{H}}_2\mathrm{O}}-{\mathrm{Δ\pm á}}_{\mathrm{O}}-{\mathrm{Δ\pm á}}_{{\mathrm{H}}_2}=(-285.83)-0-0$

$=-285.83\mathrm{kJ}$

$∆{\mathrm{H}}_{{\mathrm{H}}_2\mathrm{O}}(\mathrm{formation})$$=-285.83\mathrm{kJ}$

In book,

$\mathrm{Δ\pm á}=6{\mathrm{Δ\pm á}}_{{\mathrm{H}}_2\mathrm{O}}+6{\mathrm{Δ\pm á}}_{{\mathrm{CO}}_2}-{\mathrm{Δ\pm á}}_{{\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6}$

$\mathrm{Δ\pm á}=6(-285.83)+6(-393.51)-(-1268)$

$=2808.04\mathrm{kJ}$