Question

The enthalpy of combustion of ethyl alcohol is -1380 kJ/mol. If enthalpies of formation of CO and H₂O are -394.5 and -286.5 kJ/mol, respectively, calculate the enthalpy of formation of ethyl alcohol.

a. -270.5 kJ/mol
b. -398.5 kJ/mol
c. -812.5 kJ/mol
d. -1086.5 kJ/mol

Answer 1

To find the enthalpy of formation of ethyl alcohol, we use Hess's Law and the given enthalpies of combustion and formation for CO2 and H2O to solve for the unknown enthalpy of formation of ethanol. The calculated value is -268.5 kJ/mol, therefore the closest provided answer choice is -270.5 kJ/mol.

Step-by-step explanation:

The student is asking for the calculation of the enthalpy of formation of ethyl alcohol (ethanol) given its enthalpy of combustion and the enthalpies of formation for CO and H₂O. According to Hess's Law, the enthalpy change for a reaction is the same no matter the route taken. In the context of the provided values, we can express the combustion of ethanol as:

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

Now we use the enthalpy values to write an equation that relates the enthalpy of formation of ethanol (ΔHf, ethanol) to the known enthalpies:

ΔHc = 2ΔHf [CO₂(g)] + 3ΔHf [H₂O(l)] - ΔHf [C₂H₅OH(l)]

-1380 kJ/mol = (2 x -394.5 kJ/mol) + (3 x -286.5 kJ/mol) - ΔHf [C₂H₅OH(l)]

-1380 kJ/mol = -789 kJ/mol - 859.5 kJ/mol - ΔHf [C₂H₅OH(l)]

-1380 kJ/mol = -1648.5 kJ/mol - ΔHf [C₂H₅OH(l)]

ΔHf [C₂H₅OH(l)] = -1648.5 kJ/mol + 1380 kJ/mol

ΔHf [C₂H₅OH(l)] = -268.5 kJ/mol

Therefore, the closest answer choice to -268.5 kJ/mol is -270.5 kJ/mol, which is option (a).