The enthalpy of combustion of ethyl alcohol is −1380 kJ mole-1. If e

1.	The enthalpy of combustion of ethyl alcohol is −1380 kJ mole-1. If enthalpies of formation of CO2 and H2O are −394.5 and −286.5 kJ mole-1 respectively. Calculate the enthalpy of formation of ethyl alcohol.
Answer» 2C(s) + 3H₂ (g) + \(\frac{1}{2}\)O2 (g) → C₂H₅OH(l)∆H_f⁰=? C(s) + 2H₂ (g) + 2H₂ (g) → CO₂ (g) + 2H₂O (l) ∆H = [2X(−286) + (−394.5)] = −572 − 394.5 = −966.5 kJ/mol 2C(s) + 3H₂ (g) + \(\frac{1}{2}\)O₂ (g) → C₂H₂OH(l) ∆H = −966.5 − (−1380) = −966.5 + 1380 = 413.5 kJ/mole

The enthalpy of combustion of ethyl alcohol is −1380 kJ mole-1. If enthalpies of formation of CO2 and H2O are −394.5 and −286.5 kJ mole-1 respectively. Calculate the enthalpy of formation of ethyl alcohol.

Answer»

2C(s) + 3H₂ (g) + \(\frac{1}{2}\)O2 (g) → C₂H₅OH(l)∆H_f⁰=?

C(s) + 2H₂ (g) + 2H₂ (g) → CO₂ (g) + 2H₂O (l)

∆H = [2X(−286) + (−394.5)]

= −572 − 394.5

= −966.5 kJ/mol

2C(s) + 3H₂ (g) + \(\frac{1}{2}\)O₂ (g) → C₂H₂OH(l)

∆H = −966.5 − (−1380)

= −966.5 + 1380

= 413.5 kJ/mole

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The enthalpy of combustion of ethyl alcohol is −1380 kJ mole-1. If enthalpies of formation of CO2 and H2O are −394.5 and −286.5 kJ mole-1 respectively. Calculate the enthalpy of formation of ethyl alcohol.

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