Ethyl alcohol is produced by the fermentation of glucose,C₆H₁₂O₆. C₆H₁₂O₆(s) → 2 C₂H₅OH(l) + 2 CO₂(g) ΔH° = -69.1 kJ
Given that the enthalpy of formation is -277.7 kJ/mol for C₂H₅OH(l) and -393.5 kJ/mol for CO₂(g) ,find the enthalpy of formation for C₆H₁₂O₆.

Question

Quizplus · Accepted Answer

Using the given thermochemical equation, we can calculate the enthalpy change for the reaction:

ΔH = products - reactants
ΔH = [2(-393.5 kJ/mol) + 2(277.7 kJ/mol)] - (-69.1 kJ/mol)
ΔH = -1273.3 kJ/mol

Since the enthalpy change is for one mole of ethyl alcohol, we can say that the enthalpy of formation for ethyl alcohol is -1273.3 kJ/mol.

To find the enthalpy of formation for glucose, we need to know the enthalpy change for the combustion of glucose. However, this is not given in the problem, so we cannot calculate the enthalpy of formation for glucose.

Therefore, the correct answer is B.

Ethyl Alcohol Is Produced by the Fermentation of Glucose,C6H12O6

Ethyl Alcohol Is Produced by the Fermentation of Glucose,C₆H₁₂O₆