CALCULATE the enthalpy of vapourisation per mole for ethanol given delta s =109.8J/KM AND THE BOILING POINT OF ETHANOL 78.5 CELSIUS

Open in App

Solution

We know that,
Entropy of vaporization,
Î”_vapS = Î”_vapH/ T_b where, Î”_vapH = enthalpy of vaporization and T_b= boiling point (temperature) in Kelvin
Here, Î”_vapS = 109.8J/Kmol, and boiling point T_b = 273 + 78.5 =351.5 K
So, enthalpy of vaporization of ethanol per mole is , Î”_vapH = Î”_vapS x T_b = 109.8 x 351.5 = 38594.7 J/mol

Suggest Corrections

Similar questions

Δ S = 109.8 J K^{- 1} {m o l}^{- 1}

{78.5}^{o} C

k J / m o l

109 J K^{- 1} m o l^{- 1}

78.5^{o} C .

k J / m o l

109 J K^{- 1} m o l^{- 1}

78.5^{o} C .

k J / m o l

109 J K^{- 1} m o l^{- 1}

78.5^{o} C .

{78.4}^{o} C

42.4 k J m o l^{- 1} .

Join BYJU'S Learning Program