Q. 1.51 - An Introduction to Thermal Physics

Question

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.

Step-by-Step Solution

Answer

The value of $∆\mathrm{H}$ is $2808.04 \mathrm{kJ}$.

Step 1: Expression for ∆ H

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 $∆\mathrm{H}$:

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

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

Step 2: The formation of carbon dioxide

- 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,

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

$=-393.51-0-0$

$=-393.51 \mathrm{kJ}$

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

Step 3: Calculation for water formation

- 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}$

Step 4: Calculation for ∆ H

In book,

${\mathrm{\Delta H}}_{{\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6}=-1268 \mathrm{kJ}$

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

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

$=2808.04 \mathrm{kJ}$