Octane (C₈H₁₈) undergoes combustion according to the following thermochemical equation: 2C₈H₁₈(l) + 25O₂(g) $\rarr$ 16CO₂(g) + 18H₂O(l), $\Delta$H^$\circ$_rxn = -11,020 kJ/mol.Given that $\Delta$H^$\circ$_f[CO₂(g)] = -393.5 kJ/mol and $\Delta$H^$\circ$_f[H₂O(l)] = -285.8 kJ/mol, calculate the standard enthalpy of formation of octane. A) -210 kJ/mol B) -11,230 kJ/mol C) 22,040 kJ/mol D) -420 kJ/mol E) 420 kJ/mol

Question

Quizplus · Accepted Answer

The standard enthalpy of formation of octane can be calculated using the given thermochemical equation and the enthalpies of formation for CO2 and H2O. The equation for calculating the enthalpy of formation of octane is based on the principle that the enthalpy change of the reaction ($\Delta H_{rxn}^\circ$) is equal to the sum of the enthalpies of formation of the products minus the sum of the enthalpies of formation of the reactants. Given $\Delta H_{rxn}^\circ = -11,020$ kJ for the combustion of 2 moles of octane, and the enthalpies of formation for CO2 and H2O are -393.5 kJ/mol and -285.8 kJ/mol respectively, we can set up the equation as follows:$$-11,020\, \text{kJ} = [16(-393.5\, \text{kJ/mol}) + 18(-285.8\, \text{kJ/mol})] - [2 \times \Delta H_f^\circ \text{(octane)} + 25 \times 0\, \text{kJ/mol}]$$Solving for $\Delta H_f^\circ \text{(octane)}$ gives approximately -210 kJ/mol, indicating that option A is correct.

Octane (C8H18) Undergoes Combustion According to the Following Thermochemical Equation →\rarr→

Octane (C₈H₁₈) Undergoes Combustion According to the Following Thermochemical Equation $\rarr$