Question

Consider the combustion of 1.26 mol of liquid ethanol (C₂H₅OH) to gaseous water and carbon dioxide. Calculate the enthalpy chagne for this reaction. The pertinent enthalpies of formation are

∆Hf H₂O=241.8

∆Hf CO₂=393.5

∆Hf C₂H₅OH=-277.7

Answer 1

The enthalpy change for the combustion of 1.26 moles of ethanol is -1723.68 kJ. This is calculated using the provided enthalpies of formation and the balanced chemical equation for the combustion of ethanol.

Step-by-step explanation:

To calculate the enthalpy change for the combustion of ethanol, we use the provided enthalpies of formation and apply the formula for the enthalpy of a reaction: ΔH = ∑ΔHf(products) - ∑ΔHf(reactants).

First, we need to write the balanced equation for the combustion of ethanol:

C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l)

Next we calculate the enthalpy change:

ΔHcombustion = [2(ΔHf CO₂) + 3(ΔHf H₂O(l))] - [1(ΔHf C₂H₅OH(l))]

ΔHcombustion = [2(-394 kJ/mol) + 3(-286 kJ/mol)] - [1(-278 kJ/mol)]

ΔHcombustion = (-788 kJ + -858 kJ) - (-278 kJ)

ΔHcombustion = -1646 kJ + 278 kJ

ΔHcombustion = -1368 kJ for 1 mole of ethanol

However, we need the enthalpy change for 1.26 moles of ethanol, so we multiply the enthalpy change for 1 mole by 1.26:

ΔHcombustion for 1.26 moles = -1368 kJ * 1.26

ΔHcombustion for 1.26 moles = -1723.68 kJ