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? A)-687.6 kJ/mol B)-277.6 kJ/mol C)687.6 kJ/mol D)1,367 kJ/mol E)3,010 kJ/mol

Question

Quizplus · Accepted Answer

The standard enthalpy of formation of ethanol can be calculated using Hess's Law: $\Delta H°_{combustion} = \Delta H°_{f}[CO_2] + \Delta H°_{f}[H_2O] - \Delta H°_{f}[C_2H_5OH]$. Solving for $\Delta H°_{f}[C_2H_5OH]$, we get -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 $\Delta$