Molar Enthalpy of Fusion

Question

At standard conditions (1 bar), the melting point of ethanol is at -114.1 C. When a pressure of 3700 bar is applied to solid ethanol, its melting temperature rises to -92.2 C. Determine the molar enthalpy of fusion of ethanol.

M (ethanol)= 46.1 g/mol

P (solid) = 0.816 g/mL

P(liquid) = 0.789 g/mL

Answer 1

Clausius Clapeyron equation:

ln(P2/P1) = -∆H/R (1/T2 - 1/T1)

P1 = 1 bar

P2 = 3700 bar

∆H = ?

R = 8.314 J/Kmol

T1 = -114.1 C + 273 = 158.9K

T2 = -92 C + 273 = 181K

Plug values in and solve for ∆H...

ln (3700/1) = -∆H / 8.314 (1/181 - 1/158.9)

ln 3700 = -∆H / 8.314 (0.00552 - 0.00629)

ln 3700 = -∆H / 8.314 (-7.7x10^-4)

8.22 = 7.7x10^-4 ∆H / 8.314

∆H = 88,755 J / mole = 88.8 kJ/mole

(be sure the check all of the math)