Acetic anhydride, C4H6O3, has a normal boiling point of 139.8 C (ie, at 1.013 bar) and a standard enthalpy of vaporization of 51. 9 KJ mol^-1. What is the vapour pressure of acetic anhydride at room temperature (298 K) in units of bar?
Can someone help me solve this question, it'll be deeply appreciated.
Use the van't Hoff equation.
To calculate the vapor pressure of acetic anhydride at room temperature (298 K) in units of bar, we can use the Clausius-Clapeyron equation:
ln(P2/P1) = (ΔHvap/R) * (1/T1 - 1/T2)
Where:
P1 = Pressure at temperature T1 (known, in this case, P1 = 1.013 bar at the boiling point)
P2 = Pressure at temperature T2 (unknown)
ΔHvap = Standard enthalpy of vaporization (given as 51.9 kJ/mol)
R = Ideal gas constant (8.314 J/(mol·K))
T1 = Boiling point temperature (139.8 °C converted to K, T1 = 139.8 + 273 = 412 K)
T2 = Room temperature (298 K)
Rearranging the equation to solve for P2:
P2 = P1 * exp((ΔHvap/R) * (1/T1 - 1/T2))
Plugging in the known values:
P2 = 1.013 * exp((51.9 * 10^3)/(8.314) * (1/412 - 1/298))
Calculating the equation will give the vapor pressure at room temperature.