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Assume that the ΔHo and ΔSo of vaporization do not change significantly with temperature. Calculate the vapor pressure of CH3OH at 42 oC (in atm).

CH3OH (l) ↔ CH3OH(g) . . . ΔHo = 38.0 kJ and ΔSo = 112.9 J/K

Respuesta :

Hagrid
We are given with
ΔHo = 38.0 kJ
ΔSo = 112.9 
for these standard conditions:
To = 25 C
Po = 1 atm

Using the Clapeyron Equation
ln (P/Po) = (ΔHo / R)(1/T1  - 1/T2)
Substitute the values and solve for P (Temperature should be in Kelvin)

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