Use the data at the back of this book to determine $\mathrm{\Delta H}$for the combustion of a mole of glucose,

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

This is the (net) reaction that provides most of the energy needs in our bodies.

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

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

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

$\mathrm{\Delta H}={\mathrm{\Delta H}}_{\text{react}}-{\mathrm{\Delta H}}_{\mathrm{prod}}$

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

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

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

${\mathrm{\Delta H}}_{{\mathrm{O}}_2}+{\mathrm{\Delta H}}_{\mathrm{C}}\to {\mathrm{\Delta H}}_{{\mathrm{CO}}_2}$

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

So,

${\mathrm{\Delta H}}_{{\mathrm{CO}}_2}-{\mathrm{\Delta H}}_{\mathrm{C}}-{\mathrm{\Delta H}}_{{\mathrm{O}}_2}$

$=-393.51-0-0$

$=-393.51\mathrm{kJ}$

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

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

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

${\mathrm{\Delta H}}_{{\mathrm{H}}_2}+\frac{1}{2}{\mathrm{\Delta H}}_{\mathrm{O}}\to {\mathrm{\Delta H}}_{{\mathrm{H}}_2\mathrm{O}}$

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

So ,

${\mathrm{\Delta H}}_{{\mathrm{H}}_2\mathrm{O}}-{\mathrm{\Delta H}}_{\mathrm{O}}-{\mathrm{\Delta H}}_{{\mathrm{H}}_2}=(-285.83)-0-0$

$=-285.83\mathrm{kJ}$

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

In book,

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

$\mathrm{\Delta H}=6(-285.83)+6(-393.51)-(-1268)$

$=2808.04\mathrm{kJ}$