Question

Isooctane burns in air to give water and carbon dioxide (∆rH
0 = -10 992kJ/mol-rxn). What is the
enthalpy change if you burn 1.00 L of isooctane?

Answer 1

The balanced reaction equation of isooctane combustion is:

2C₈H₁₈+27O₂$\rightarrow$ 16CO₂+18H₂O.

Therefore, the enthalpy in kJ/mol-rxn refers to 2 molecules of iso-octane and can be written as: -10 992 kJ/(2 mol C₈H₁₈).

The density $d$ of isooctane is 690 g/L. To find the mass of isooctane, the following formulae can be used:

$m = d·V = 690\text{ g/L}·1.00\text{ L} = 690$ g.

In order to convert the mass to the number of the moles, we must divide the mass by the molar mass of iso-octane: 114.23 g/mol.

$n = \frac{m}{M} =\frac{690 \text{ g}}{ 114.23\text{ g/mol}} = 6.04$ mol.

Finally, the enthalpy change when 6.04 mol of isooctane is:

$∆H = -10992\text{ kJ}·\frac{6.04\text{ mol}}{2\text{ mol }C_8H_{18}} = -33198$ kJ.

Answer: if you burn 1.00 L of isooctane, the enthalpy change is -33198 kJ.