Question

Define enthalpy of combustion.

Answer 1

Enthalpy of combustion:

• The change in enthalpy of a system when one mole of a substance is completely burnt in oxygen or air at a given temperature is known as the enthalpy of combustion.
• The standard enthalpy of combustion is $∆{\mathrm{H}}_{\mathrm{c}}^{\mathrm{o}}$.
• You usually calculate the enthalpy change of combustion from enthalpies of formation.

$\mathrm{\Delta }{\mathrm{H}}_{\mathrm{c}}^{\mathrm{o}}=\sum \mathrm{\Delta }{\mathrm{H}}_{\mathrm{f}}^0\left(\mathrm{p}\right)-\sum \mathrm{\Delta }{\mathrm{H}}_{\mathrm{f}}^{\mathrm{o}}\left(\mathrm{r}\right)$

• where p stands for "products" and r stands for "reactants".
• For each product, you multiply its $\mathrm{\Delta }{\mathrm{H}}_{\mathrm{f}}^0$ by its coefficient in the balanced equation and add them together.
• Do the same for the reactants. Subtract the reactant sum from the product sum.

Step 1:

• In this step we are going to write the equations of combustion of methane, graphite, and dihydrogen at 298K:

Combustion of Methane:

${\mathrm{CH}}_4\left(\mathrm{g}\right)+2{\mathrm{O}}_2\left(\mathrm{g}\right)\to {\mathrm{CO}}_2\left(\mathrm{g}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

$\mathrm{C}{\left(\mathrm{s}\right)}_{}+{\mathrm{O}}_2\left(\mathrm{g}\right)\to {\mathrm{CO}}_2\left(\mathrm{g}\right)$

$2{\mathrm{H}}_2\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\to 2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

The standard enthalpies of formation of methane, carbon dioxide, water are −74.85 kJmol^-1,−393.5 kJmol^-1 and −286 kJmol^-1.

Step 2:

• In this step we will write the desired equation for the formation of ${\mathrm{CH}}_4$ (methane) by using the above equations:

$\mathrm{C}\left(\mathrm{s}\right)+2{\mathrm{H}}_2\left(\mathrm{g}\right)\to {\mathrm{CH}}_4\left(\mathrm{g}\right)$

Step 3:

• In this step we will write the desired equation for the enthalpy formation of ${\mathrm{CH}}_4$ (methane) by using the above equation:

$\mathrm{\Delta }{\mathrm{H}}_{\mathrm{c}}^{\mathrm{o}}=\sum \mathrm{\Delta }{\mathrm{H}}_{\mathrm{f}}^0\left(\mathrm{p}\right)-\sum \mathrm{\Delta }{\mathrm{H}}_{\mathrm{f}}^{\mathrm{o}}\left(\mathrm{r}\right)$

$∆{\mathrm{H}}_{\mathrm{c}}^{\mathrm{o}}=\left\{\right(-393.5-2\times 286)-(-74.85\left)\right\}{\mathrm{kJmol}}^{-1}$

$∆{\mathrm{H}}_{\mathrm{c}}^{\mathrm{o}}=-890.7{\mathrm{KJmol}}^{-1}$

Hence, required enthalpy of formation of methane, $∆{\mathrm{H}}_{\mathrm{c}}^{\mathrm{o}}=-890.7{\mathrm{KJmol}}^{-1}$.