Calculate the heat of formation of gaseous ethyl alcohol,
2C(graphite) + 3H2(g) +12 O2(g) → C2H5OH (g) given that enthalpy of sublimation of graphite is 714 kJ/mol and bond
enthalpies of H – H, O = O, C – C, C – H, C – O and O – H are respectively
435 kJ/mol–1, 498 kJ/mol–1, 347 kJ/mol–1, 415 kJ/mol–1, 355 kJ/mol–1 and
462 kJ/mol–1 respectively
Answers
heat of formation of ethyl alcohol is 257 KJ/mol
we have to find heat of formation of ethyl alcohol (C2H5OH)
heat of formation is the amount of energy released when one mole of compound is formed with help of its elements.
i.e., 2C(g) + 3H2(g) + /1/2 O2 (g) ⇒C2H5OH (g)
∆H_(formation) = ∆H_(product) - ∆H_(reactants)
= 1 × (C - C) + 5 × (C - H) + 1 × (C - O) + 1 × (OH) - [2 × enthalpy of sublimation of C + 3 × (H - H) + 1/2 × (O = O) ]
= 1 × 347 + 5 × 415 + 1 × 355 + 1 × 462 - [2 × 714 + 3 × 435 + 1/2 × 498 ]
= 257 KJ/mol
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