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

Chemistry

Jovani Branch

5 December, 02:22

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.

+3

Answers (1)

Know the Answer?

Harrison Mcconnell · Answer 1

-1273.3

Explanation:

Enthalpy of formation of a compound is the amount of heat absorbed or evolved when one mole of the compound is formed from other compounds.

enthalpy of formation Of CO2 = 2 X - 393.5 = - 787

enthalpy of formation Of C2H5OH = 2 X - 277.7 = - 555.4

enthalpy of formation Of C6H12O6 = 69.1 (reverse sign) + (-787 + - 555.4) = - 1273.3 Joules