Ethanol undergoes combustion in oxygen to produce carbon dioxide gas and liquid water.The standard heat of combustion of ethanol, C₂H₅OH(l), is -1366.8 kJ/mol. Given that $\Delta$H°_f[CO₂(g)] = -393.5 kJ/mol and $\Delta$H°_f[H₂O(l)] = -285.8 kJ/mol, what is the standard enthalpy of formation of ethanol?

Question

Quizplus · Accepted Answer

The standard enthalpy of formation of ethanol can be calculated using the equation: $\Delta H°_{combustion} = \Delta H°_{f}[products] - \Delta H°_{f}[reactants]$. Rearranging gives $\Delta H°_{f}[C_2H_5OH(l)] = \Delta H°_{f}[CO_2(g)] + 3\Delta H°_{f}[H_2O(l)] - \Delta H°_{combustion}$. Substituting the given values: $-393.5 + 3(-285.8) - (-1366.8) = -277.6$ kJ/mol.

Ethanol Undergoes Combustion in Oxygen to Produce Carbon Dioxide Gas Δ\DeltaΔ

Ethanol Undergoes Combustion in Oxygen to Produce Carbon Dioxide Gas $\Delta$