Diethyl ether is a volatile organic compound. the vapor pressure of diethyl ether is 401 mm Hg at 18 degrees celsius and the change in H vapor = 26.0 kJ/mol. Calculate the vappor pressure of diethyl ether at 25 degrees C.
Won't the Clausius-Clapeyron equation solve this problem for you?
To calculate the vapor pressure of diethyl ether at 25 degrees Celsius using the given information, we can make use of the Clausius-Clapeyron equation:
ln(P2/P1) = -(ΔHvap/R)((1/T2) - (1/T1))
Where:
P1 = vapor pressure at the first temperature (18 degrees Celsius)
P2 = vapor pressure at the second temperature (25 degrees Celsius)
ΔHvap = enthalpy of vaporization (26.0 kJ/mol)
R = gas constant (8.314 J/(mol·K))
T1 = temperature in Kelvin corresponding to P1 (18 degrees Celsius + 273.15)
T2 = temperature in Kelvin corresponding to P2 (25 degrees Celsius + 273.15)
Now, let's substitute the given values into the equation and solve for P2:
ln(P2/401 mm Hg) = -((26.0 kJ/mol)/(8.314 J/(mol·K)))*((1/(25 + 273.15)) - (1/(18 + 273.15)))
Simplifying the equation:
ln(P2/401 mm Hg) = -9.003*(0.00369 - 0.00392)
Calculating the right-hand side:
ln(P2/401 mm Hg) = -9.003*(-0.00023)
ln(P2/401 mm Hg) = 0.002073
To solve for P2, we need to exponentiate both sides:
P2/401 mm Hg = e^(0.002073)
P2 = 401 mm Hg * e^(0.002073)
Now, let's calculate P2:
P2 = 401 * e^(0.002073) ≈ 401 * 1.002075 ≈ 401.841 mm Hg
Therefore, the vapor pressure of diethyl ether at 25 degrees Celsius is approximately 401.841 mm Hg.